Solution_Manual for Stoichiometry- Bhatt and Thakore



Dimensions and Units EXERCISE 1.1 (a) Wave length 5500 Å = 5500 ´ 10–8 cm ´

1 m 10 +9 nm 10 +2 cm 1 m

= 550 nm

Ans.

(b) Moisture content of humid air 175 grain = 1 lb dry air =

175 grain 1 lb ´ 1 lb dry air 7000 grains

=

1 lb moisture 40 lb dry air

or

kg moisture kg dry air

1 kg moisture 1000 g ´ 1 kg 40 kg dry air = 25 g moisture/kg dry air

Ans.

Vacuum = 475 Torr Absolute pressure = 760 – 475 = 285 Torr º 285 ´ 101.325/760 º 38 kPa or 0.38 bar 1 Pa = 1.450 377 ´ 10–4 psi Pressure = 38 000 ´ 1.450 377 ´ 10–4 = 5.511 psia

Ans.

=

EXERCISE 1.2

EXERCISE 1.3 F (5)2 4 = 19.635 cm2 Pressure = Force/Area = 192.6 ´ 104/(19.635 ´ 106) = 0.0981 MPa º 0.981 bar

Cross sectional area of piston =

2 Solutions Manual—Stoichiometry

In FPS units, pressure = 0.0981 ´ 106 ´ (1.450 377 ´ 10–4) = 14.228 psi

Ans.

EXERCISE 1.4 Weight = 500 lb º 500 ´ 0.4536 º 226.8 kg Volume = 29.25 L = 0.029 25 m3 Density of Fe, r = 226.8/0.029 25 = 7753.85 kg/m3 º 7.754 t/m3

Ans.

EXERCISE 1.5 Diameter = 5 ft = 1.524 m Height of tank = 6 ft 6 inch = 1.981 m Volume of tank = (p/4)(Dia)2(Height) = (3.1416/4)(1.524)2(1.981) = 3.614 m3 Volume of CCl4 liquid in the tank = 3.614 ´ 0.75 = 2.71 m3 Density, r = 1600 kg/m3 Mass of CCl4 in the tank = 2.71 ´ 1600 = 4336 kg

Ans.

EXERCISE 1.6 Pressure drop = 0.05 ´

lbf × min (Pa. in 2 ) ´ 6894.759 (in 2 × US gal) lbf

m3 1 US gal 1 h ´ ´ 10 3 –3 h m 60 min 3.785 412 ´ 10

1 kPa 1000 Pa = 15.178 kPa

´

Ans.

EXERCISE 1.7 Total exposed area = 2 ´ 7.595 ´ 1.276 + 2 ´ 7.595 ´ 0.1535 + 2 ´ 1.276 ´ 0.1535 = 22.106 cm2 Weight loss = 14.9412 – 14.6254 = 0.3158 g Density = 7753.85 kg/m3 0.3158 1 10 6 365 1 ´ ´ ´ ´ 1000 7753.85 50 22.106 1 = 0.0134 cm/a º 0.0053 in/year º 5.3 mpy

Corrosion rate =

Ans.

Dimensions and Units 3

EXERCISE 1.8 Let p¢ be the vapour pressure in kPa and T be the temperature in K 101 325kPa = 760 Torr or 1 kPa = 7.500 62 Torr p = 7.500 62 p¢ T = t + 273.15 or t = T – 273.15 Substitute the values in the equation, 1211.0 log10 (7.500 62 p¢) = 6.9057 – (T - 27315 . + 220.8) log10 p¢ + 0.8751 = 6.9057 –

1211.0 (T - 52.35)

log10 p¢ = 6.0306 –

1211.0 (T - 52.35)

Ans.

EXERCISE 1.9 Let Cmo p¢ be the heat capacity of n-butane in SI units. 1 Btu/(lb mole × °R) = 1 kcal/(kmol × K) = 4.1868 kJ/(kmol × K) Cmo p¢ /4.1868 = Cmo p T = 1.8 T ¢ Substituting the values in the equation, ( Cmo p¢ /4.1868) = 4.429 + 40.159 ´ 10–3 ´ 1.8 T ¢ – 68.562 ´ 10–7 (1.8 T¢)2 Simplifying, Cmo p¢ = 18.5433 + 302.6479 ´ 10–3 T ¢ – 93.006 ´ 10–6 T ¢2 Ans.

EXERCISE 1.10 Let Dp¢, v¢ and L¢ be pressure drop in cm WC, gas velocity is m/s and liquid flow rate in m3/m3 gas flow. Dp¢ = Dp ´ 2.54 v = v¢ ´ 3.281 L¢ = (L/264.172) 35.314 67 = 0.133 68 L Substitute above values in the Calvert equation.

Dp ¢ L¢ = 5 ´ 10–5 ´ (v¢ ´ 3.281)2 ´ 0.133 68 2.54 Dp¢ = 1.0227 ´ 10–2 v¢2 L¢

Ans.

4 Solutions Manual—Stoichiometry

EXERCISE 1.11 Let h¢ = heat transfer coefficient, kW/(m2 × K) G¢ = mass velocity of fluid, kg/(m2 × s) Cp¢ = specific heat, kJ/(kg × K) k¢ = thermal conductivity of fluid, kW/(m × K) D¢ = diameter of tube, m m¢ = viscosity of fluid, kg/(m × s) h=

3412.142 ´ 10-3 h¢ = 0.1761 h¢ 10.7639 ´ 1.8

G=

2.204 62 ´ 3600 G¢ = 737.28 G¢ 10.7639

Cp =

9.478 172 ´ 10 -4 = 2.2885 ´ 10–4 Cp¢ 2.204 62 ´ 1.8

k=

3412.142 ´ 10-3 k¢ = 0.5778 k¢ 3.2808 ´ 1.8

D = 3.2808 D¢ m¢ =

2.204 62 ´ 3600 m¢ = 2419.11 m¢ 3.2808

Substituting the values, 0.1761 h¢ =

0.023( 737.28G ¢ )0.8 (0.5778k ¢ )0.67 (2.3885 ´ 10 -4 C p¢ )0.33 (3.2808 D¢ )0.2 (2419.11m ¢ )0.47

or h¢ = 0.023 G¢0.8 k¢0.67 Cl¢0.33 / (D¢0.2 m¢0.47) Thus the equation does not change when consistent SI units are used. This is because the equation is a simplified form of Sieder-Tate equation which is madeup of three dimensionless numbers. Ans.

Basic Chemical Calculations EXERCISE 2.1 Molar mass of oxygen = 2 ´ 16 = 32 g/mol Amount of oxygen = 500/32 = 15.625 mol

Ans.

EXERCISE 2.2 1 mol C is present in 1 mol CO2. 12 g C º 44 g CO2 Carbon content in 264 g CO2 = (12 ´ 264/44) = 72 g

Ans.

EXERCISE 2.3 Molar mass of KMnO4 = 39 + 55 + 4 ´ 16 = 158

Ans.

EXERCISE 2.4 Molar mass of HNO3 = 63 Molar mass of H2SO4 = 98 100 g HNO3 = 100/63 = 1.5873 mol 100 g H2SO4 = 100/98 = 1.0204 mol Excess = 1.5873 – 1.0204 = 0.5669 mol HNO3 Excess atoms = 0.5669 ´ 6.022 142 ´ 1023 = 3.414 ´ 1023 atoms of HNO3

EXERCISE 2.5 Molar mass of CS2 = 12 + 2 ´ 32 = 76 76 g CS2 º 12 g C

Ans.

6 Solutions Manual—Stoichiometry

1 mol CS2 = 1 mol C 3.5 kmol C = 3.5 kmol CS2 º 3.5 ´ 76 º 266 kg CS2

Ans.

EXERCISE 2.6 Molar mass of Al2(SO4)3 = 2 ´ 27 + 3(32 + 4 ´ 16) = 342 Valence of Al2(SO4)3 = 6 Equivalent mass of Al2(SO4)3 = 342/6 = 57

Ans.

EXERCISE 2.7 Molar mass of KMnO4 = 158 Valence of KMnO4 = 5 (based on oxidation number) Equivalent mass of KMnO4 = 158/5 = 31.6 500 g KMnO4 = 500/31.6 = 15.82 g eq

Ans.

EXERCISE 2.8 Basis: 100 kg magnesite ore Compound

Molar mass

kg

kmol

mole %

MgCO3 SiO2 H2O

84.3 60 18

81 14 5

0.961 0.233 0.278

65.28 15.83 18.89

Total

—

100

1.472

100.00

Ans.

EXERCISE 2.9 Basis: 100 kg glass Compound

Molar mass

Na2O MgO ZnO Al2O3 B2O3 SiO2

62 40.3 81.4 102.0 69.6 60.0

Total

—

kg

kmol

mole %

7.8 7.0 9.7 2.0 8.5 65.0

0.1258 0.1737 0.1192 0.0196 0.1221 1.0833

7.65 10.57 7.25 1.19 7.43 65.91

1.6437

100.00

100

Ans.

Basic Chemical Calculations 7

EXERCISE 2.10 Concentration of solids = 35 000 ppm or mg /L º (35 000 ´ 100)/106 º 3.5% For watery solutions, 10 000 ppm = 1% by mass

Ans.

EXERCISE 2.11 Basis: 100 kg limestone 1 kmol CaO = 1 kmol CaCO3 Molar mass of CaCO 3 ´ 54.5 CaCO3 in limestone = Molar mass of CaO = 100 ´ 54.5/56 = 97.32% (by mass)

Ans.

EXERCISE 2.12 (a) In one mole of ammonium sulphate, two atoms (or one mole) nitrogen are present. Nitrogen content of commercial ammonium sulphate = Molar mass of N 2 ´ 96 28 ´ 96 = Molar mass of (NH 4 ) 2 SO 4 132 = 20.36% Ans. (b) One mole of sodium nitrate contains half mole of nitrogen. Nitrogen content of pure sodium nitrate = 0.5 ´ Molar mass of N 2 ´ 100 0.5 ´ 28 ´ 100 = Molar mass of NaNO 3 85

= 16.47%

Ans.

EXERCISE 2.13 Chemical equation: 2 NaOH ¾ ¾¾® Na2O + H2O NaOH content of flakes =

2 ´ Molar mass of NaOH ´ 74.6 Molar mass of Na 2 O

2 ´ 40 ´ 74.6 62 = 96.26%

=

Ans.

8 Solutions Manual—Stoichiometry

EXERCISE 2.14 Basis: 1 kmol azeotropic mixture Compound

kmol

Molar mass

kg

% by mass

H2O HNO3

0.622 0.378

18 63

11.196 23.814

31.98 68.02

Total

1.000

—

35.010

100.00

Ans.

EXERCISE 2.15 Basis: 100 kg saline solution Compound

kg

Molar mass

kmol

mole %

NaCl H2O

25 75

58.5 18

0.4274 4.1667

9.3 90.7

Total

100

—

4.5941

100.0

Ans.

EXERCISE 2.16 Basis: 100 g water Component

mass, g

HCl KCl H2O

4.00 19.61 100.0

Total

123.61

mass %

Molar mass

mole

mole %

3.24 15.86 80.90

36.4609 74.5513 18.0153

0.1097 0.2630 5.5508

1.85 4.44 93.71

100.00

—

5.9235

100.00

Ans.

EXERCISE 2.17 Basis: 100 kg solution Compound

kg

Nitrogen content kg/kmol comp.

Molar mass

Nitrogen Content kg

NH3 NH4NO3 NH2CONH2

19 65.6 6

14 28 28

17 80 60

15.647 22.960 2.800

Total

90.6

—

—

41.407

Nitrogen content of the solution = 41.41%

Ans.

Basic Chemical Calculations 9

EXERCISE 2.18 In one mole of ethanol, 2 moles of carbon are present. TOC of the solution = 2 ´ 12 ´ 1000/46 = 522 mg/L Oxidation (combustion) reaction: C2H5OH + 3 O2 ¾ ¾¾® 2 CO2 + 3 H2O 46

3 ´ 32

2 ´ 44

3 ´ 18

ThOD of the solution = 3 ´ 32 ´ 1000/46 = 2087 mg/L

Ans.

EXERCISE 2.19 Basis: 100 kg Phosphoric acid of 35% P2O5 strength 2 H3PO4 ¾ ¾¾® P2O5 + 3 H2O 2 ´ 98

142

3 ´ 18

H3PO4 content of the acid = 2 ´ 98 ´ 35/142 = 48.31%

Ans.

EXERCISE 2.20 Basis: 100 kg spent acid Free acid (H2SO4) content = 20 kg NH4HSO4 content = 45 kg +

H ¾® H SO + NH + NH4HSO4 ¾ ¾¾® NH4+ + HSO–4 ¾ ¾¾ 2 4 4 115

18

97

98

Chemically bound acid = 98 ´ 45/115 = 38.35 kg Total acid content = 20 + 38.35 = 58.35 kg or 58.35%

18

Ans.

EXERCISE 2.21 Basis: 1 m3 of aqueous TEA solution Mass of TEA = 0.47 ´ 1.00 ´ 1125 = 528.75 kg Mass of water = 0.53 ´ 1.00 ´ 1000 = 530 kg mass % TEA = 528.75 ´ 100/(528.75 + 530) = 49.94% Ans. Note: Volumes are strictly not additive but for an ideal solution, they can be considered additive.

10 Solutions Manual—Stoichiometry

EXERCISE 2.22 Basis: 100 L wine Mass of alcohol = 20 ´ 0.79 = 15.80 kg Mass of alcohol-free liquid = (100 – 20) ´ 1.0 = 80 kg mass % alcohol =

15.80 ´ 100 = 16.49 (15.80 + 80)

Ans.

EXERCISE 2.23 Component CaCO3 Na2CO3 MgSO4 (a) Concentration of Na2CO3 = = (b) Concentration of MgSO4 = =

Equivalent mass 100/2 = 50 106/2 = 53 120.3/2 = 60.15 50 ´ 800/53 754.7 ppm or mg/L as CaCO3 50 ´ 85/60.15 70.7 ppm or mg/L as CaCO3

Ans.

EXERCISE 2.24 (a) Equivalent mass of H2SO4 = 98/2 = 49 Concn. of H2SO4 in solution = 294/49 = 6 mol/L º6N

Ans.

(b) Equivalent mass of CaCl2 = 112/2 = 55.5 Concn. of CaCl2 in solution = 4.8 ´ 1000/(1000 ´ 55.5) = 0.0865 mol/L º 0.0865 N (c) Equivalent mass of H3PO4 = 98/3 = 32.67 Concn. of H3PO4 in solution = 5 N º 5 ´ 32.67 º 163.35 g/L (d) Equivalent mass of HCl = 36.5 Concn. of HCl in solution = 54.75/36.5 = 1.5 g/L º 1.5 N º 1.5 M (e) Molar mass of K2SO4 = 174 Concn. of K2SO4 in solution = 174 ´ 3 = 522 g/L

Ans.

Ans.

Ans.

Basic Chemical Calculations 11

EXERCISE 2.25 Basis: 100 kg aqueous acetic acid solution Molar mass of acetic acid = 60 Acetic acid content = 35/60 = 0.5833 kmol Volume of 100 kg solution = 100/1.04 = 96.15 L Molarity = 0.5833 ´ 1000/96.15 = 6.066 M Since acetic acid is monovalent. Normality of the solution = 6.066 N Molality = 0.5833 ´ 1000/65 = 8.974

Ans.

EXERCISEE 2.26 Basis: 1 litre solution CO2 dissolved (in gas form) = 40 L at NTP = 40/22.414 = 1.785 mol MEA content = 1 ´ 1.011 ´ 0.2 = 0.2022 kg º 202.2 g Molar mass of MEA = 61 Moles of MEA in solution = 202.2/61 = 3.315 mol CO2 concentration = 1.785/3.315 = 0.5385 mol/mol MEA

Ans.

EXERCISE 2.27 Basis: 100 L of 60 volume H2O2 solution O2 liberated in total = 60 ´ 100 = 6000 L Sp. volume of ideal gas at STP = 23.69 L/mol O2 liberated = 6000/23.69 = 253.27 mol H2O2 ¾ ¾¾® H2O + 1/2 O2 H2O2 decomposed = 2 ´ 253.27 = 506.54 mol Mass of H2O2 decomposed = 506.54 ´ 34 = 17 222.36 g Mass of H2O2 solution = 100 ´ 1.075 = 107.5 kg mass % H2O2 in solution = (17 222.36 ´ 100)/(107.5 ´ 1000) = 16.02

Ans.

EXERCISE 2.28 Boiling point of 40% solution = 402.2 K(129.2°C). (Ref. Fig. 2.3) Boiling point of pure water at atm. pressure = 100°C (373.15 K) Boiling point elevation = 402.2 – 373.2 = 29 K or 29°C

Ans.

12 Solutions Manual—Stoichiometry

EXERCISE 2.29 Since solution is an ideal solution, it follows Raoult's law. Vapour pressure of solution, pvs = 0.5 ´ 37.2 + 0.5 ´ 12.3 = 24.75 kPa

Ans.

EXERCISE 2.30 Basis: 100 kg urea solution containing 25 % urea Urea present in solution =

25 = 0.4167 kmol 60

75 = 4.1667 kmol 18 Quantum of solution = 0.4167 + 4.1667 = 4.5834 kmol

Water present in solution =

4.1667 4.5834 = 0.9091 Since vapour pressure of urea at 60°C (333.15 K) is negligible, Vapour pressure of solution = 0.9091 ´ 19.92 = 18.11 kPa

Mole fraction of water in solution =

Ans.

EXERCISE 2.31 1 bar = 10.1972 m H2O 1000 m H2O = 98.066 bar According to Henry's law, mole fraction of solute =

Partial pressure of solvent Henry's constant

Assuming partial pressure of solvent (water) to be vapour pressure (as its mole fraction is near unity), mole fraction of nitrogen in water at 1000 m depth =

98.066 = 1.119 ´ 10–3 87 650

mole fraction of helium in water at 1000 m depth =

98.066 = 0.775 ´ 10–3 126 600

Since helium is much less dissolved in water, it is preferred as a mixture with oxygen. Ans.

Basic Chemical Calculations 13

EXERCISE 2.32 Mole fraction of NH3 in solution = 0.02 Roult's law: yi p = xi pi yi = =

[Ref. Eq. (2.18)]

xi pi p

11.58 ´ 0.02 1.01 325

= 0.2286

Henry's law:

pi = = = = yi =

Hi xi 0.861 ´ 0.02 0.01722 bar 1.722 kPa

[Ref. Eq. (2.44)]

pi 0.01722 = p 1.013 25

= 0.017

Ans.

EXERCISE 2.33 Basis: 100 kmol gas Gas

Formula

Molar mass

kmol

kg

mass %

Ethylene Benzene Oxygen Methane Ethane Nitrogen

C2H4 C6H6 O2 CH4 C2H6 N2

28 78 32 16 30 28

30.6 24.5 1.3 15.5 25.0 3.1

856.8 1911.0 41.6 248.0 750.0 86.8

22.00 49.07 1.07 6.37 19.26 2.23

Total

—

—

100.00

3894.2

100.00

Average molar mass = 3894.2/100 = 38.942 Ideal gas occupies 22.414 m3/kmol at NTP. Density of gas mixture = 38.942/22.414 = 1.737 kg/m3

Ans.

14 Solutions Manual—Stoichiometry

EXERCISE 2.34 Basis: 100 kmol sewage gas Gas

Formula

Molar mass

kmol

kg

mass %

Methane Carbon dioxide Ammonia

CH4 CO2

16 44

68 30

1088 1320

44.55 54.05

NH3

17

2

34

1.40

Total

—

—

100

2442

100.00

Average molar mass = 2442/100 = 24.42 Density of sewage gas = 24.42/22.414 = 1.09 kg/m3

Ans.

EXERCISE 2.35 Basis: 1.10 kg CO2 Moles of CO2 = 1.10/44 = 0.025 kmol Volume occupied = 33 L = 0.033 m3 V = 0.033/0.025 = 1.32 m3/kmol T = 300 K van der Waals equation:

F p + a I (V – b) = RT H VK F p + 3.6 I (1.32 – 0.043) = 0.008 314 ´ 300 ´ 1000 H 1.32 K 2

2

Solving the equation, p = 1951.104 kPa º 19.51 bar

Ans.

EXERCISE 2.36 For chlorine gas: (a) Ideal gas law: p = 15.2 MPa a, T = 503.15 K (230°C) V = RT/p = 0.008 314 ´ 503.15/15.2 = 0.2752 m3/kmol 71 Molar mass = = 258.0 kg/m3 0.2752 V (b) van der Waals equation: For chlorine, pc = 79.77 bar a and Tc = 416.90 K a = 27 R2 T2/64 pc

Density =

Ans.

Basic Chemical Calculations 15

= 27(0.083 14)2(416.90)2/(64 ´ 79.77) = 6.353 72 (m3 )2× bar/(kmol)2 b = RTc/8pc = 0.083 14 ´ 416.9/(8 ´ 79.77) = 0.054 314 m3/kmol Substituting in the van der waals equation 6.353 72 152 + (V – 0.054 314) = 0.083 14 ´ 503.15 = 41.831 89 V2 Simplifying, 152 V3 – 5.0075 V2 + 0.6354 V – 0.03451 = 0 Solving the equation by Newton-Raphson method. V = 0.152 m3/kmol Density = 71/0.152 = 467.1 kg/m3 Mathcad Solution:

FG H

I K



Density = 71/0.1536 = 462.24 kg/m3

Ans.

EXERCISE 2.37 Equation:

Pressure p = 73 bar, Temperature T = 423.15 K p = [RT(1 – e)/ v2] [V + b] – A/v2 a A = A0 1 = (5.88) [1 – (0.058 61/V)] V b = 0.094 [1 – (0.019 15/V)] B = B0 1 V

F H F H

I K I K

90 ´ 104 V (423.15)9 0.9 0.011 89 = = 75.687V V

e = c/VT3 =

Substituting the values.

é æ 0.011 89 ö 2 ù /V ú 73 = ê0.08314 ´ 423.15 ç1 è ø÷ V ë û

[V + 0.094 – (0.0018/V)] – [5.88 – (0.344 63/V)]/V2

16 Solutions Manual—Stoichiometry

Simplifying and solving the equation by Newton-Raphson method, V = 0.4 m3/kmol Molar mass of ethane = 30.0704 Density of ethane = 30.0704/0.4 = 75.176 kg/m3 Mathcad Solution:

Density = 30.0704/0.40045 = 75.092 kg/m3

Ans.

Ans.

EXERCISE 2.38 For dimethyl ether (DME): p = 15 bar a T = 353.15 K pc = 53.4 bar a Tc = 400.2 K Acentric factor w = 0.192 Tr = f ° = 0.1445 –

T 353.15 = = 0.8824 Tc 400.2

0.607 ´ 10 –3 0.1385 0.0121 0.330 – – – (0.8824)8 0.8824 (0.8824) 2 (0.8824)3

= – 0.4262 f 1 = 0.0637 +

0.8 ´ 10 –2 0.331 0.423 – – (0.8824)8 (0.8824) 2 (0.8824)3

= – 0.148 B=

0.08314 ´ 400.2 [– 0.4262 + 0.192 (– 0.148)] 53.4

= – 0.283 26

Basic Chemical Calculations 17

Z=1–

( - 0.283 26)15 0.08314 ´ 353.15

= 1.1447 V=

ZRT 1.1447 ´ 0.08314 ´ 353.15 = p 15

= 2.241 m3/kmol

Ans.

EXERCISE 2.39 (a) Ideal gas law: p = 7.09 + 1.013 25 = 8.103 25 T = 923.15 K (650 °C) V=

0.08314 ´ 923.15 RT = 8.103 25 p

= 9.472 m3/kmol Density = 27.587/9.472 = 2.912 kg/m3 (b) van der Waals equation: a=

27 R 2Tc2 64 pc

and b = RTc / 8pc

where pc and Tc are pseudo critical properties. Gas

Mole fraction ni

Molar mass Mi

Mass kg n × Mi

N2 O2 H2O NH3

0.705 0.188 0.012 0.095

28 32 18 17

19.740 6.016 0.216 1.615

Total

1.000

—

27.587

Critical temp Tc, K

Critical pressure pc, bar

Tc

ni × Tc

pc

88.893 29.061 7.718 38.523

33.94 50.42 221.2 113.50

126.09 154.58 643.30 405.50

164.195

ni × pc 23.93 9.48 2.65 10.78 46.84

Pseudo critical temperature, Tc = 164.195 K Pseudo critical pressure, pc = 46.84 bar a = [27(0.083 14)2(164.195)2]/(64 ´ 46.84) = 1.678 (m3)2× bar/(kmol)2 b = (0.083 14 ´ 164.195)/(8 ´ 46.84) = 0.036 42 m3/kmol

18 Solutions Manual—Stoichiometry

Substituting the values in the equation, 1.679 (V – 0.036 43) = 0.083 14 ´ 923.15 = 75.7507 8.1 + V2 Mathcad Solution:

F H

I K

Average molar mass = 27.587 Density = 27.587/9.3626 = 2.9465 kg/m3 Specific gravity of the gas mixture = 27.587/28.96 = 0.952

Ans.

EXERCISE 2.40 Concentration of water vapours = 0.216 kg/kmol º (0.216/9.3626)106 º 23 071 mg/m3 23.071 ´ 27.587 2.9465 ´ 18 = 12000 ppm

In terms of ppm concentration =

Ans.

EXERCISE 2.41 Basis: 9.082 g furfural-n-butane mixture Partial pressure of water = 12.5 Torr at 295.2 K Partial content of n-butane = 763.2 – 12.5 = 750.7 torr n-butane content of vapours = pV/RT 750.7 ´ 105.7 = 760 ´ 1000 ´ 0.082 06 ´ 295.2 = 0.004 314 mol Mass of n-butane = 0.004 314 ´ 58 = 0.2501 g Mass of furfural = 9.082 – 0.2501 = 8.8319 g mass % n-butane = 0.2501 ´ 100/9.082 = 2.75% Moles of furfural = 8.8319/96 = 0.092 mol Total moles = 0.092 + 0.004 314 = 0.096 314 mol mole % n-butane = 0.004 314 ´ 100/0.096 314 = 4.48 Ans.

Basic Chemical Calculations 19

EXERCISE 2.42

Basis: 100 kmol flue gases Gas CO2 O2 N2 SO2

Molar mass 44 32 28 64

Total

—

Avg. molar mass = Partial pressure of water = = Absolute humidity = =

EXERCISE 2.43

kmol 10.00 7.96 82.00 0.04

kg 440.00 254.72 2296.00 2.56

100.00

2993.28

29.933 (of dry gas) Vapour pressure of water at DP 10.612 kPa (ref. Chapter 6) [10.612 ´ 18]/[(100 – 10.612) 29.933] 0.071 39 kg/kg º 71.39 g/kg Ans.

Concentration of SO2 in flue gases = (0.04/100) ´ 106 = 400 ppm (v/v) = 2.56/2993.28 = 0.8552 ´ 10–3 kg/kg º 855.2 mg/kg or ppm (by mass) Volume of gas = 0.082 06 ´ 463.15 ´ 760/750 = 38.513 m3/kmol mg kmol 2.56 kg 1 ´ 106 ´ kg 100 kmol 38.513 m 3 3 = 664.7 mg/m Ans.

Concentration of SO2 =

EXERCISE 2.44

Basis: 733 kg mixture Component

kg

Molar mass

CH3CH Steam Oxygen (from air) Nitrogen (from air)

200 133

32 18

Component O2 N2 CH3OH H2O Total

kmol 6.25 7.39 0.21 ´ 400/29 = 2.897 0.79 ´ 400/29 = 10.897

Total pressure = 68.6 kPa g Absolute Pressure = 68.6 + 101.325 = 169.925 kPa kmol mole % Partial pressures, kPa 2.897 10.56 17.94 10.897 39.72 67.49 6.250 22.78 38.71 7.390 26.94 45.78 27.434 100.00 169.92 Ans.

20 Solutions Manual—Stoichiometry

EXERCISE 2.45 Basis: 1 kmol LPG mixture at 313.15 K (40 °C). Gas

kmol

Molar mass

kg

C3H8 n-C4H10 i-C4H10

0.30 0.45 0.25

44.0956 58.1222 58.1222

13.229 26.155 14.531

Total

1.00

53.915

Average molar mass = 53.915 Specific gravity = Gas

kmol

C3H8 n-C4H10 i-C4H10

0.30 0.45 0.25

Total

1.00

Ans. (a)

53.915 = 1.861 28.97 Vapour pressure bar 13.975 3.773 5.290

Pressure in LPG cylinder = 7.2129 bar

Ans. (b) Partial pressure bar 4.1925 1.6979 1.3225 7.2129

Ans. (c)

EXERCISE 2.46 Basis: 5 m3 solution Total pressure = 6.77 MPa g = 6.77 + 0.1013 = 6.8713 MPa a Solubility of N2 at 0.1013 MPa and 305.75 K = 1.35 ´ 5/100 = 0.0675 Nm3/5 m3 soln. Partial pressure of N2 at 6.77 MPa g total pressure = (6.8713 ´ 0.206)/(1 – 0.022) = 1.4473 MPa (on ammonia-free basis) Solubility of N2 at 6.77 MPa g and 305.75 K = 0.0675 ´ 1.4473/0.1013 = 0.9644 Nm3/5 m3 solution In a similar manner, solubility can be calculated for all the gases as shown in the following, table,

Basic Chemical Calculations 21

Gas

Solubility at 0.1013 MPa a and 305.75 K/5 m3 soln.

N2 H2 Ar CH4 Total

Partial pr. at 6.77 MPa g

Solubility at 305.75 K and 6.77 MPa g/5 m3 solution

0.0675 0.0800 0.1375 0.1400

1.4473 4.3560 0.2881 0.7800

0.964 3.440 0.391 1.078

0.4250 m3

6.8714 MPa a

5.873 Nm3

Ans.

EXERCISE 2.47 Refer Example 2.25 pi = 7 bar g = 8.013 bar a , pf = 1.013 bar a T f = Ti – m (pi – pf) = 313.15 – 0.21 (8.013 – 1.013) = 311.68 K or 38.53°C

Ans.

EXERCISE 2.48 p1 = 101 atm a p2 = 1 atm a Ti = 308.15 K T f = 308.15 – 0.169 (101 – 1) = 291.25 K or 18.1°C

Ans.

EXERCISE 2.49 Assume ideal gas law. Basis: Receiver of 2 m3 capacity Volume of air in the receiver at 7.5 bar g and 313.15 K, 8.513 25 ´ 2 ´ 27315 . v1 = 31315 . ´ 1.013 25 = 14.657 Nm3 Volume of air in the receiver at 1 bar g and 313.15 K, v2 =

2.013 25 ´ 2 ´ 27315 . 31315 . ´ 1.013 25

= 3.466 Nm3 Volume of air pressurized = 14.657 – 3.466 = 11.191 Nm3 in 4 min 11191 . ´ 60 4 = 167.9 Nm3/h

Capacity of air compressor =

Ans.

22 Solutions Manual—Stoichiometry

EXERCISE 2.50 qL =

(1 ´ 10-3 - 2 ´ 10-4 )25 7.5 ´ 60

= 4.444 ´ 10–5 (mbar × L)/s

Ans.

3

Material Balances without Chemical Reaction EXERCISE 3.1 Basis: 4 kg feed water Product water quantity = 3 kg Material balance of dissolved solids (DS): Feed water quantity ´ concentration of DS in feed water = product water quantity ´ concentration of DS in product water + concentrated stream quantity ´ concentration of DS in concentrated stream Let concentration of DS in concentrated stream = x 4 ´ 500 = 3 ´ 600 + 1 ´ x x = 18 200 ppm Ans.

EXERCISE 3.2 Basis: Feed rate of 20 000 kg/h solution containing 91% ethanol by mass Non-permeate from pervaporation unit contains all ethanol of the feed. Ehanol in non-permeate = 20 000 ´ 0.91 = 18 200 kg/h Purity of non-permeate stream = 99.7% 18 200 Flow rate of non-permeate stream = 0.997 = 18 254.8 kg/h Let flow rate of vapour entering pervaporation unit = x kg/h and flow rate of permeate = y kg/h x = y + 18 254.8 (1) Ethanol balance: x ´ 0.78 = y ´ 0.029 + 18 200 (2) Solving both equations, y = 5274.7 kg/h Ans.

24 Solutions Manual—Stoichiometry

EXERCISE 3.3 Basis: 1000 kg solid caustic soda to be processed/d. When the evaporator system concentrates 4% solution to 25% concentration, evaporation = 100 – (4/0.25) = 84 kg/100 kg original solution However, the capacity of the evaporation system is 4 kg solid caustic soda per 100 kg original solution. Consider another case in which 5% solution is concentrated to 50% concentration the evaporation = 100 – (5/0.5) = 90 kg/100 kg original solution The plant capacity is thus 5 kg caustic soda. It may be noted that the plant capacity in terms of evaporation remains constant, i.e. the plant has actual capacity of 84 kg evaporation per 100 kg solution. In second case, for 84 kg evaporation, Caustic soda processed = 5´ 84/90 = 4.667 kg. Plant Capacity is terms of solid caustic soda = 4.667 ´ 1000/4 = 1166.7 kg/d Ans.

EXERCISE 3.4 Basis: 100 kg coal Carbon content of coal = 67.2 kg Ash, present in coal = 22.3 kg Ash does not take part in combustion. Ash content of refuse = 92.9% Quantity of refuse = 22.3/0.929 = 24.0 kg Carbon content of refuse = 24.0 – 22.3 = 1.7 kg % carbon, unburnt = 1.7 ´ 100/67.2 = 2.53

Ans.

EXERCISE 3.5 Basis: 100 kg flaked soybean seeds Solids (unextractable, i.e. inerts) = 69 kg Final deoiled cake after extraction (meal) = 69/0.877 = 78.68 kg Oil retained in DOC = 78.68 ´ 0.008 = 0.63 kg Oil recovery = (18.6 – 0.63)100/18.6 = 96.6%

EXERCISE 3.6 Basis: 100 kmol feed gas (vent stream) H2 in vent stream = 66 kmol

Ans.

Material Balances without Chemical Reaction 25

Recovery of hydrogen = 66 ´ 0.85 = 56.1 kmol Purity of hydrogen stream = 98% Quantity of hydrogen stream = 56.1/0.98 = 57.24 kmol CH4 in hydrogen stream = 57.24 – 56.1 = 1.14 kmol Composition of Reject Stream Component H2 CH4 Other components

kmol 66 – 56.1 = 9.9 33 – 1.14 = 31.86 1.00

Total

42.76

mole % 23.15 74.51 2.34 100.00

Ans.

EXERCISE 3.7 Basis: 100 kg deodorizer distillate Fatty acids content = 19.8 + 56.1 = 75.9 kg Let O be kmol of distillate from SPU and R be kmol of residue from SPU. O + R = 100

(1)

0.95O + 0.1R = 75.9

(2)

Solving two simultaneous equations, O = 77.53 kg and R = 22.47 kg Ratio O : R = 77.53 : 22.47 = 3.45 : 1 by mass

Ans.

EXERCISE 3.8 Basis: 1000 kg saturated solution of magnesium chloride in ethanol at 298.15 K (25 ºC) Let x = mass of crystals of MgCl2 6H2O, kg y = mass of ethanol, kg x + y = 1000 (1) Crystal/solvent = 190/100 = 1.9 = x/y or x = 1.9y (2) solving the equations, x = 655.5 kg and y = 344.5 kg MgCl2 present in solution = 95.32 ´ 655.5/203.32 = 307.3 kg Water present in solution = 655.5 – 307.3 = 348.2 kg

26 Solutions Manual—Stoichiometry

Compound

Formula

molar mass

mass, kg

mass %

kmol

mole %

Magnesium chloride Water Ethanol

MgCl2 H2O C2H5OH

95.32 18 46

307.3 348.2 344.5

30.73 34.82 34.45

3.224 19.344 7.490

10.72 64.36 24.92

Total

—

—

1000.0

100.00

30.057

100.00

Ans.

EXERCISE 3.9 Basis: 1 kmol initial solution at 298.15 K (25 ºC) Compound

Formula

Chloroacetic acid (CAA) Ether (E)

ClCH2COOH (CH3CH2)2O

Total

—

molar mass

kmol

mass

mass %

94.5 74.0

0.2 0.8

18.9 59.2

24.2 75.8

—

1.0

78.1

100.0

New Basis: 290 kg final solution, saturated at 298.15 K (25 ºC) Compound

mass %

kmol

mole %

CAA E

Mass kg 190 100

65.52 34.48

2.01 1.35

59.82 40.18

Total

290

100.00

3.36

100.00

Let quantity of spent solution = x kg and quantity of make-up CAA = y kg x + y = 500 Balance of CAA: 0.242 x + y = 0.6552 ´ 500 = 327.6 Solving the equations, x = 227.44 kg y = 272.56 kg

(1) (2) Ans.

EXERCISE 3.10 Basis: 100 kg mixture Let x = mass of CuSO4 × 5H2O in the mixture, kg y = mass of FeSO4 × 7H2O in the mixture, kg x + y = 100 After dehydration, the mixture will contain CuSO4 and FeSO4 alone. Molar Mass of CuSO 4 Molar Mass of FeSO 4 + = 59.78 Molar mass of CuSO 4 × 5H 2 O Molar mass of FeSO 4 × 7H 2 O

(1)

Material Balances without Chemical Reaction 27

(159.5/249.5) ´ x + (152/278) ´ y = 59.78 0.6393 x + 0.5468 y = 59.78

(2)

Solving the equation, x = 55.15 kg y = 44.85 kg Ratio, CuSO4/FeSO4 = x/y = 1.23 : 1 by mass

Ans.

EXERCISE 3.11 Basis: 100 kmol flue gases Let x = amount of N2 in gas mixture, kmol and y = average molar mass of nitrogen-free flue gas Amount of nitrogen-free gases = 100 – x kmol If correct molar mass of nitrogen is considered, 28 ´ x (100 - x ) y = 30.08 + 100 100 If incorrect molar mass (14) of nitrogen is considered, 14 ´ x (100 - x ) y = 18.74 + 100 100 Solving two equations, x = 81 kmol or 81% nitrogen Substitute the value of x in Eq. (1), (100 - 81) y = 30.08 – (28 ´ 81)/100 100 = 7.4 y = 7.4/0.19 = 38.95 Let m and n be the amounts of CO2 and O2, respectively. m + n = 100 – 81 = 19 44 m + 32 n = 38.95 ´ 19 = 740.05 Solving Eqs. (3) and (4), m = 11 kmol or 11% CO2 n = 8 kmol or 8% O2

(1)

(2) Ans. (a)

(3) (4) Ans. (b)

EXERCISE 3.12 Basis: 100 kg vapour mixture, entering overhead condenser Let top layer = x kg bottom layer = y kg x + y = 100 Benzene balance: 0.81 x + 0.1 y = 74 Solving two equations, x = 90.14 kg

(1) (2)

28 Solutions Manual—Stoichiometry

y = 9.86 kg Ratio, x/y = 9.142:1 (by mass)

Ans.

EXERCISE 3.13 Basis: 100 kg complex of acetic acid and oil Let

x = quantity of upper layer, kg y = quantity of lower layer, kg x + y = 100

(1)

0.0962 x + 0.9303 y = 0.634 ´ 100 = 63.4

(2)

Acetic acid balance: Solving the equation,

x = 35.5 kg and y = 64.5 kg Ratio x/y = 35.5/64.5 = 0.5504 (by mass)

Ans. (a)

The new complex with solvent (chloroform) contains 57.8% acetic acid. Quantity of new complex = 63.4/0.578 = 109.7 kg Mass of chloroform added = 109.7 – 100 = 9.7 kg Let

Ans. (c)

m = quantity of upper layer in new complex, kg n = quantity of lower layer in new complex, kg m + n = 109.7

(3)

Balance of acetic acid: 0.145 m + 0.875 n = 63.4 Solving the equations,

(4)

m = 44.64 kg and n = 65.06 kg

Ratio,

m/n = 44.64/65.06 = 0.686 (by mass)

Ans. (b)

For solving the example with the help of coordinate plots, following values are tabulated. Eq. (1)

Eq. (2)

Eq. (3)

x

30

32

34

36

38

40

y

70

68

66

64

62

60

x

30

32

34

36

38

40

y

65.05

64.84

64.63

64.43

64.22

64.01

m

40

42

44

46

48

50

n

69.7

67.7

65.7

63.7

61.7

59.7

Material Balances without Chemical Reaction 29

Eq. (4)

m

40

42

44

46

48

50

n

65.83

65.5

65.17

64.83

64.50

64.17

Plot of first two equations are given in Fig. E3.1. Similar plot for Eq. (3) and Eq. (4) can be drawn. x = 35.52, y = 84.48, m = 44.64 and n = 65.06, all in kg Ans. 700

68

66 (35.52, 64.48) 64

62

60 30

32

34

36

Fig. E3.1 Solution of Exercise 3.13

EXERCISE 3.14 Basis: 100 kg spent lye Glycerol content of lye = Loss of glycerol by entertainment = Glycerol content of final solution = = Quantity of final solution =

9.6 kg 9.6 ´ 0.045 = 0.432 kg 9.6 – 0.432 9.168 kg 9.168/0.8 = 11.46 kg

38

40

30 Solutions Manual—Stoichiometry

Salt content of final solution = 11.46 ´ 0.06 = 0.69 kg Salt crystallized in salt box = 10.30 – 0.69 = 9.61 kg Evaporation = 100 – 11.46 – 9.61 = 78.93 kg For the feed rate of 5000 kg/h, Rate of crystallization = 5000 ´ 0.0961 = 480.5 kg/h Rate of evaporation = 5000 ´ 0.7893 = 3946.5 kg/h

Ans. (b) Ans. (a)

EXERCISE 3.15 Basis: 100 kg mixed fertiliser of grade 10 : 26 : 26 Nitrogen content of fertilizer = 10 kg as N Phosphorous content of fertilizer = 26 kg as P2O5 Potassium content of fertilizer = 26 kg as K2O Element/compound Atomic mass/molar mass N 14 142 P2O5 94.2 K2O 98 H3PO4 HCl 74.6 17 NH3 Urea 60 (a) Anhydrous NH3 requirement = 17 ´ 10/14 = 12.14 kg 2 kmol H3 PO4 º 1 kmol P2O5 Anhydrous H3PO4 requirement = 98 ´ 2 ´ 26/142 = 35.89 kg 2 kmol KCl º 1 kmol K2O 100% pure KCl requirement = 74.6 ´ 2 ´ 26/94.2 = 41.18 kg Total of three is 89.21 kg/h. Balance is called 'filler'. (b) Urea requirement = (60/28) ´ 10 = 21.43 kg (c) 98% KCl requirement = 41.18/0.98 = 42.02 kg Aqueous H3PO4 requirement = 100 – 12.14 – 42.02 = 45.84 kg Concentration of H3PO4 in aqueous acid = 35.89 ´ 100/45.84 = 78.09%

EXERCISE 3.16 Basis: 100 kg mixed acid Quantity of 68.3% HNO3 required Sulphuric acid requirement Water in 68% HNO3 Water content of mixed acid

= 39/0.683 = 57.1 kg = 100 – 57.1 = 42.9 kg = 57.1 (1.000 – 0.683) = 18.1 kg = 19 kg

Ans. Ans.

Ans.

Material Balances without Chemical Reaction 31

Water from H2SO4 = 19 – 18.1 = 0.9 kg H2SO4 content of the acid = 42.9 – 0.9 = 42.0 kg Strength of the acid required = 42.0 ´ 100/42.9 = 97.9% Ans. (a) Mass ratio of the acids = 57.1/42.9 = 1.33 : 1 Ans. (b) This problem can be conveniently solved on a triangular plot. Refer Fig. E3.2. Point A represents 68.3% HNO3. Point B represents the desired mixed acid containing 39% HNO3 and 42% H2SO4. Join AB which intersects the axis at point C, representing 97.9% H2SO4. 100% H2O 90

10

80

20

30

H

2O

70 60

40

50

60

40 A

70

SO 4 H2

30

50

80

20

90

10

C 100% HNO3

90

80

70

60 HNO3

50

40

30

20

10

100% H2SO4

Fig. E3.2 Preparation of Mixed Acid Mass ratio = BC/AC = 7.9 units/6.0 units = 1.32:1

EXERCISE 3.17 Basis: 100 kg oleum Oleum contains 10% SO3. This means 110 kg oleum contains 100 kg 100% H2SO4 and 10 kg SO3. H2SO4 content of 100 kg oleum = 90.91 kg SO3 content of 100 kg oleum = 100 – 90.91 = 9.09 kg SO3 will combine with H2O to form H2SO4 when mixed with spent acid and aqueous HNO3.

32 Solutions Manual—Stoichiometry

Let

m = quantity of oleum used, kg n = quantity of spent acid, kg p = quantity of aqueous 90% HNO3, kg m + n + p = 1000 (1) Total H2SO4 available after blending = 0.0091 m + [0.09 + 98/(80 ´ 100)] m = 1.0205 m H2SO4 balance: 1.0205 m + 0.444 n = 600 (2) HNO3 balance 0.113 n + 0.9 p = 320 (3) Mathcad Solution: 1 1 O LM1 M := 1.0205 0.444 0 P MM0 P 0.113 0.9 PQ N LM1000OP v := 600 MM 320 PP N Q LMmOP MM np PP := M × v NQ LMmOP LM532 OP 128.589 all in kg MM np PP := MM339 P N Q N .41 PQ –1

m = 532.0 kg, n = 128.6 kg and p = 339.4 kg

Ans.

This problem cannot be solved directly on a triangular chart. This is because total H2SO4 content on equivalent basis is 102.05% and the property of the triangular chart is that each component can be represented upto 100% on each axis. However, it is possible to express H2SO4 concentration in equivalent SO3 concentration and then the triangular chart can be used. Since conversion of H2SO4 concn. to SO3 concn. is required for use of the chart, it is recommended that the algebraic method be preferred.

EXERCISE 3.18 Refer the triangular chart Fig. E 3.3. Points A, B and C represent 99% H2SO4, pure water and 95% HNO3 respectively. Point F1 represents original spent acid. Point F4 represents the final fortified mixed acid (containing 50% H2SO4, 40% HNO3 and 10% H2O). Join AC. Join F1 and F4 and extend F1 F4 to meet at F3 on AC.

Material Balances without Chemical Reaction 33

Fig. E3.3 Fortifying of Spent Acid Basis: 1000 kg original spent acid

Blend F3 5.35 units = Blend F1 1.25 units Blend F3 = (5.35/1.25)1000 = 4280 kg

99% H 2SO 4 10.25 units = 95% HNO3 9.1 units Thus (10.25 + 9.1 =) 19.35 kg blend F3 will contain 10.25 kg 99% H2SO4 and 9.1 kg 95% HNO3. Hence 4280 kg blend F3 will contain 10.25 ´ 4280/19.35 = 2267 kg 99% H2SO4 and 4280 – 2267 = 2013 kg 95% HNO3. Final acid after evaporation 10.75 units = original spent acid 17.75 units

Ans. (a)

Therefore final fortified acid = 10.75 ´ 1000/17.75 = 605.6 kg Water can be evaporated = 1000 – 605.6 = 393.4 kg

Ans. (b)

34 Solutions Manual—Stoichiometry

EXERCISE 3.19 Basis: 100 kg feed containing 38.2 mass % benzene Component

kg

Molar mass

kmol

mole %

Benzene

38.2

78

0.4897

40.0

Cyclohexane

61.8

84

0.7357

60.0

1.2254

100.0

Total

100.0

Refer Fig. E3.4 which is a triangular chart. Points A, B and C represent 100% cyclohexane, benzene and acetone each. Point D represents the feed mixture containing 40 mole % benzene while point E represents the azeotropic mixture containing 74.6 mole % acetone. Join CD which intersects BE at M which represents mixture after adding acetone to the feed. CYCLOHEXANE A 90

10

80

20

30

70 D 60

40

50

50

40 30

60

A

70 E

M

20

80 90

10 B BENZENE

90

80

70

60

50

40

30

20

C 10 ACETONE

Note : All Percentages Are Mole %.

Fig. E3.4 Mixing of Benzene-Acetone-Cyclohexane

Acetone to be added 9 units = Feed (Point D) 5 units Acetone to be added = (9/5)1.2254 = 2.2057 kmol Molar mass of acetone = 58

Material Balances without Chemical Reaction 35

Mass of acetone = 127.9 kg

Ans.

Note: It may be noted that the triangular chart is plotted with mole % compositions.

EXERCISE 3.20 Basis: 1000 kg dry building boards Mass of moist boards = 1000/0.84 = 1190.5 kg Final mass of boards after drying = 1000/0.995 = 1005 kg Evaporation = 1190.5 – 1005 = 185.5 kg Humidity difference in outgoing and incoming air = 0.09 – 0.02 = 0.07 kg/kg dry air Air required for drying = 185.5/0.07 = 2650 kg/h (dry air) Specific volume of air at 301 K (28 °C) and 0.02 kg/kg dry air = 0.882 m3/kg dry air (ref. Fig. 6.14) Volumetric flow rate of incoming air = 2650.0 ´ 0.882 = 2337.3 m3/h

Ans.

EXERCISE 3.21 Basis: 100 kg lard Molar mass of all the components are listed below. Component

Molar mass (rounded off values)

Palmitodistearin Stearodipalmitin Oleodistearin Oleopalmitostearin Palmirodiolein Palmitic acid (C15H31COOH) Stearic acid (C17H35COOH) Oleic acid (C17H33COOH)

862 834 888 860 858 256 284 282

Mass, kg 3 2 2 11 82 ? ? ?

Consider palmitodistearin (a tri-glyceride): 862 kg palmitodistearin liberates 256 kg plamitic acid. 3 kg palmitodistearin will therefore liberate 256 ´ 3/862 = 0.891 kg palmitic acid. Similarly, 862 kg palmitodistearin liberates (284 ´ 2) = 568 kg stearic acid. 3 kg palmitodistearin will therefore liberate 568 ´ 3/862 = 1.977 kg stearic acid. In this manner, the fatty acids present in each tri-glyceride can be calculated which are summerised follows.

36 Solutions Manual—Stoichiometry

Tri-glyceride

Fatty acid content, kg palmitic

stearic

oleic

Total

Palimitodistearin Steorodipalmitin Oleodistearin Oleopalmitostearin Palmitodiolein

0.891 1.228 — 3.274 24.466

1.977 0.681 1.279 3.633 —

— — 0.635 3.607 53.902

2.867 1.909 1.914 10.514 78.368

Total

29.859

7.570

58.144

95.573

% palmitic acid = 29.859 ´ 100/95.573 = 31.24 % stearic acid = 7.570 ´ 100/95.573 = 7.92 % cleic acid = 100 – 31.24 – 7.92 = 60.84

Ans.

EXERCISE 3.22 Basis: 1 litre water NaCl in water = 58.5 ´ 775.6/35.5 = 1278.1 mg/L Na2SO4 in water = 142 ´ 230.4/96 = 340.8 mg/L Na2CO3 in water = 53 ´ 80.5/50 = 85.3 mg/L NaHCO3 in water = 84 ´ 208.1/50 = 349.6 mg/L Mg (HCO3) in water = 73.14 ´ 194/50 = 283.8 mg/L Calcium hardness in water = 284 – 194 = 90 mg/L as CaCO3 Ca(HCO3)2 in water = 81 ´ 90/50 = 145.8 mg/L Dissolved solids in water = 1278.1 + 340.8 + 85.3 + 349.6 + 283.8 + 145.8 = 2483.4 mg/L Na in NaCl = 1278.1 – 775.6 = 502.5 mg/L Na in Na2SO4 = 340.8 – 230.4 = 110.4 mg/L Na in Na2CO3 = 46 ´ 85.30/106 = 37.0 mg/L Na in NaHCO3 = 23 ´ 349.6/84 = 95.7 mg/L Total sodium (Na) content = 110.4 + 502.5 + 37.0 + 95.7 = 745.6 mg/L º 32.41 meq/L Equivalent of calcium (Ca) = 90/50 = 1.8 meq/L Equivalent of magnesium (Mg) = 194/50 = 3.88 meq/L Total cations = 32.41 + 1.8 + 3.88 = 38.09 meq/L % sodium (Na) = 32.41 ´ 100/38.09 = 85.09 Ans.

EXERCISE 3.23 Basis: 1 litre dyehouse effluent Na in effluents = 245.7/23 = 10.68 meq/L

Material Balances without Chemical Reaction 37

Ca in effluents = 37.5 ´ 2/40 = 1.88 meq/L % Na in effluents = 10.68 ´ 100/(10.68 + 1.88) = 85.04 If sodium content is to be brought down to 60%, total cations should equal to 10.68/0.6 = 17.80 meq/L Additional Ca to be added = 17.80 – 10.68 – 1.88 = 5.24 meq/L º 104.8 mg/L as Ca CaSO4 to be dissolved = 136 ´ 104.8/40 = 356.3 mg/L Ans.

EXERCISE 3.24

A graph (/Fig. E3.5) is plotted of time (in s on x-axis) vs oxygen concentration (in volume % on y-axis).

Oxygen Concentration, vol. %

25

Air

Air in N2

N2

21

20

Oxygen consumed from air

15

Effluent oxygen concentration

10 Oxygen consumed from 5 % mixture

Start 100 % air feed

5

Stop air feed

Start 5 % O2 in feed 0

0

600 1200 Decoking Time, s

1800

2400

3000

Fig. E3.5 Decoking of catalyst

Area under the diagonal lines are as follows: x-axis coverage

Area, square units

0 – 1800 s 0.6619 Considering 5% concentration of O2 in the feed stream, area upto nitrogen introduction = 1044 square units

38 Solutions Manual—Stoichiometry

Area for air introduction (i.e. from 1800 to 3600 s) = 2.099 square units Basis: Nitrogen supply @ 0.025 L/s at NTP. Molar flow of nitrogen =

0.025 ´ 10-3 22.414

= 0.1115 ´ 10–5 kmol/s Air with 21% O2 is mixed with N2. Let the flowrate of air be x kmol/s. x ´ 0.21 = (x + 0.1115 ´ 10–5) 0.05 x = 0.3484 ´ 10–6 kmol/s In 1800 s, O2 supply = 0.3484 ´ 10–6 ´ 0.21 ´ 1800 = 0.132 ´ 10–3 kmol º 0.7496 square units 0.6619 ´ 0.132 ´ 10-3 0.7496 = 0.117 ´ 10–3 kmol º 0.117 mol

O2 consumed =

After stoppage of N2 supply, air flow = 0.025 L/s at NTP = 0.1115 ´ 10–5 kmol/s O2 supply during the period 1800 s to 3600 s, = 0.1115 ´ 10–5 ´ 1800 ´ 0.21 = 0.42 ´ 10–3 kmol º 2.099 square units O2 consumed =

0.5495 ´ 0.42 ´ 10-3 4384.8

= 0.11 ´ 10–5 kmol º 0.011 mol Total oxygen consumed = 0.117 + 0.011 = 0.128 mol º 4.096 g or 4.096 ´ 10–3 kg

EXERCISE 3.25 Basis: 1000 kg/h feed rate Material balance across mixer: Let recycle and mixed feed be R and M kg/h, respectively.

Ans.

Material Balances without Chemical Reaction 39

Overall balance: M = 1000 + R

(1)

Balance of solids: 0.15 M = 300 + 0.03 R M = 2000 + 0.2 R

(2) Equating Eq. (1) and (2); 1000 + R = 2000 + 0.2 R 0.8 R = 1000 or R = 1250 kg/h and M = 2250 kg/h Balance across calciner: Let the evaporation in calciner be W kg/h and product from the divider by P kg/h. M=W+P+R But P = 700/0.97 = 721.65 kg/h W = M – P – R = 2250 – 721.65 – 1250 = 278.35 kg/h Total dry product from the calciner = P + R = 1250 + 721.65 = 1971.65 kg/h Recycle fraction = 1250 ´ 100/1971.65 = 63.4% Ans.

EXERCISE 3.26 Basis: Cloth speed = 1 m/s Production of dry cloth = 1 ´ 3600 ´ 90 ´ 90/(100 ´ 1000) = 291.6 kg/h Evaporation in the stenter = 291.6(0.80 – 0.08) = 210 kg/h Ans. (a) Fresh air required = 210/(0.1 – 0.015) = 2470.6 kg dry air/h Ans. (b) Specific volume of air = 8.314 ´ 303/(100 ´ 29) = 0.8687 m3/kg dry air Volumetric flow rate of dry air = 2470.6 ´ 0.8687 = 2146.2 m3/h Ans. (c) Material balance of moisture across points 1 and 3: R ´ 0.1 + 0.015 ´ 2470.6 = (R + 2470.6) 0.095 R = 39 530 kg dry air/h Ans.(d)

40 Solutions Manual—Stoichiometry

EXERCISE 3.27 Basis: 10 m3/L product water, nett of regeneration System-I: Ion Exchange based demineralization Effluent generation = 10 ´ 0.2 = 2 m3/h Feed (raw) water requirement = 10 + 2 = 12 m3/h Ans. (a) System-II: RO + Ion Exchange Effluent generation from ion-exchange system = 10 ´ 0.1 = 1 m3/h Feed to ion-exchange system = 10 + 1 = 11 m3/h In RO module, permeate recovery is 80% and DS rejection is 95%. Thus for 10 m3 input raw water with 2500 mg/h DS, 8 m3 permeate is produced. 8 ´ xDS = 10 ´ 2500 (1 – 0.95) = 1250 xDS = 156.25 mg/L in permeate In reject water, DS concentration can be calculated. 2 ´ zDS = 10 ´ 2500 ´ 0.95 = 23 750 zDS = 11 875 mg/L Ans. (d) Let P1 be flow of permeate water from RO module which is mixed with bypass flow to achieve 11 m3/h feed rate to ion-exchange column. P1 ´ 156.25 + (11 – P1) 2500 = 11 ´ 500 P 1 = 9.387 m3/h Bypass flow = 11 – 9.387 = 1.613 m3/h Ratio of permeate/bypass = 9.387/1.613 = 5.82 m3/m3 Ans. (c) Raw water input to RO module = 9.387/0.80 = 11.734 m3/h Total raw water feed rate = 11.734 + 1.613 = 13.347 m3/h Ans. (b)

EXERCISE 3.28 Refer Example 3.18. Calculations upto R2 = 1.25 m3/h are common. Recovery from RO Module I = 60%:

P1 = F + R2 F= R2/F = R1 =

0.6 or

6.25 = 0.6 F + 1.25

9.17 m3/h 1.25/9.17 = 0.136 m3/m3 F – P2 = 9.17 – 5 = 4.17 m3/h

Material Balances without Chemical Reaction 41

Fx F = R1 xR + P2 xP 1 2 9.17 ´ 4200 = 4.17 ´ xR + 5 ´ 5 1 xR = 9230 mg /L or g/m3 1

5´5 = 267 g/m3 or mg/L 0.015 ´ 6.25 = P2 xP + R2 xR

DS in P1, xP = 1

P1 xP

1

2

2

6.25 ´ 267 = 5 ´ 5 + 1.25 ´ xR

2

xR = 1315 g/m3 or mg/L 2

(unchanged)

DS in mixed feed = F xF + R2 xR 2 = 9.17 ´ 4200 + 1.25 ´ 1315 = 40 158 g Total mixed feed (MF) to RO module I = 9.17 + 1.25 = 10.42 m3/h Concentration DS in MF =

40 158 10.42

= 3854 g/m3 or mg/L DS in R1 = 4.17 ´ 9230 = 38 489 g Rejection RO Module I =

38 489 ´ 100 40 158

= 95.84% Rejection from Module I = 70%

Ans. (a)

6.25 = 0.7 F + 1.25 F = 7.68 m3/h R2/F = 1.25/7.68 = 0.163 m3/m3 R 1 = F – P2 = 7.68 – 5 = 2.68 m3/h F xF = R1 xR + P2 xP 1 2 7.68 ´ 4200 = 2.68 ´ xR + 5 ´ 5 1 xR = 12 026 mg/L or g/m3 1 DS is mixed feed = F xF + R2 xR 2 = 7.68 ´ 4200 + 1.25 ´ 1315 = 33 900 g Concentration of DS in MF =

33 900 (7.68 + 1.25)

42 Solutions Manual—Stoichiometry

= 3796 g/m3 or mg/L DS in R1 = R1 xR = 2.68 ´ 12 026 1 = 32 230 g Rejection in RO Module I =

32 230 ´ 100 33 900

= 95.07% Ans. Note: Recovery from RO Module I can be varied by varying feed pressure (PI1) and selecting right RO memberane. It can be seen that this does not affect % DS rejection significantly. However, it makes significant changes in other parameters (F and R1).

EXERCISE 3.29 Basis: 100 kg final 24% NaOH solution Let mass of solution in the dissolution tank = W3 kg Balance of NaOH: W3 ´ 0.5 + W2 ´ 0 = 100 ´ 0.24 W3 = 48 kg of 60% solution Overall material balance across the dilution tank: W2 + W3 = 100 or W2 = 100 – 48 = 52 kg water Balance of NaOH across the dissolution tank: W1 = W3 ´ 0.5 = 48 ´ 0.5 = 24 kg water W3 /W2 = 24/52 = 0.462:1

EXERCISE 3.30 Basis: 3000 Nm3/h fresh feed Overall material balance: . Let enriched CH4 stream from M-2 = n1 Nm3/h . and reject stream from M-3 = n2 Nm3/h . . n1 + n2 = 3000 CH4 balance: . . 0.985 n1 + 0.04 n2 = 3000 ´ 0.88 . Solving two equations, n1 = 2666.67 Nm3/h . n2 = 333.33 Nm3/h Module-2: Permeate from module = 12 % Non-permeate from module = 100 – 12 = 88 % Non-permeate from M-2 is the enriched CH4 stream.

(1) (2)

Material Balances without Chemical Reaction 43

. 2666.67 Feed rate to M-2, n3 = = 3030.31 Nm3/h 0.88 . . R 1 = n3 – n1 = 3030.31 – 2666.67 = 363.64 Nm3/h

Module-3: Permeate from module-1 = 44 % Permeate from M-3 is the rejedct stream.

. 333.33 n4 = = 757.568 Nm3/h 0.44 . . R 2 = n4 – n2 = 757.568 – 333.33 = 424.238 Nm3/h Mixed feed (MF) to M-1 = F + R1 + R2 = 3000 + 363.64 + 424.238 = 3787.878 Nm3/h . . CO2 contents of n4 and n3 streams are 60.2 % and 4.25 % by mole, respectively CO2 content of MF = x1 (mole fraction) x1 ´ 3787.878 = 757.568 ´ 0.602 + 3030.31 ´ 0.0425 x1 = 0.1544 Mole fraction of CH4 in MF = 1 – 0.1544 = 0.8456 . CH4 in n3 = 3030.31 (1 – 0.0425) = 2901.522 Nm3/h CH4 in R1 = 2901.522 – 2666.67 ´ 0.985 = 274.852 Nm3/h Feed rate to M-3,

CH4 concentration of R1 =

274.852 ´ 100 = 75.58 % 363.64

. CH4 in n4 = 757.568 (1 – 0.602) = 301.512 Nm3/h CH4 in R2 = 301.512 – 333.33 ´ 0.04 = 288.179 Nm3/h CH4 content of R2 =

288.179 ´ 100 = 67.93 % 424.238

Ans.

EXERCISE 3.31 Basis: 20 L/s weak liquor Mass flow rate wate = 20 ´ 3600 = 72 000 kg/h (SG = 1.0) SO2 present in the weak liquor = 72 000 ´ 0.005 = 360 kg/h SO2 present in the rich liquor = 72 000 ´ 0.01 = 720 kg/h SO2 absorbed in the absorber = 720 – 360 = 360 kg/h Now in the incoming 100 kmol gas, 17 kmol SO2 are present. Out of 17 kmol, 75% SO2 is absorbed.

44 Solutions Manual—Stoichiometry

SO2 absorbed = 17 ´ 0.75 = 12.75 kmol º 816 kg Outgoing gas mixture = (100 – 17) + (17 – 12.75) = 87.25 kmol SO2 content of outgoing gas mixture = 4.25 ´ 100/87.25 = 4.87% Gas flow rate = 100 ´ 360/816 = 44.12 kmol/h º 12.26 ´ 10–3 kmol/s Ans. (a) Specific volume of gas mixture, V = 8.314 ´ 308.15/(101.3 + 50) = 16.925 m3/kmol Volumetric flow rate = 44.12 ´ 16.925 = 746.73 m3/h º 207.43 dm3/s Ans. (b)

EXERCISE 3.32 Basis: 1 litre effluent sample Let x = content of methanol, mg/L y = content of formaldehyde, mg/L Carbon content of methanol = 12/32 = 0.375 mg/mg Carbon content of formaldehyde = 12/30 = 0.4 mg/mg TOC of the sample = 0.375 x + 0.4 y = 258.3 Oxidation reactions: CH3OH + 1.5 O2 ¾ ¾¾® CO2 + H2O 32

HCHO 30

+

1.5 ´ 32

O2

32

44

¾ ¾¾®

CO2 +

44

(1)

18

H2O 18

Oxygen demand of methanol = 48/32 = 1.5 mg/mg Oxygen demand of formaldehyde = 32/30 = 1.068 mg/mg ThOD of sample = 1.5 x + 1.068y = 956.3 Solving the equations, x = 535 mg/L y = 144 mg/L

EXERCISE 3.33 Basis: 16.5 Nm3/s of moist gas flow rate Molar flow rate = 16.5/22.414 = 0.736 kmol/s º 2650 kmol/h Water vapours, present in moist gas = 4.2 Nm3/s = 0.1874 kmol/s º 674.6 kmol/h Solids, present in the gas mixture = (16.5 – 4.2) ´ 9000/106 = 0.1107 kg/s º 398.5 kg/h Chlorides in gas mixture = 1.5 g/s = 5.4 kg/h Dry gas flow rate = 2650 – 674.6 = 1975.4 kg/h

(2) Ans.

Material Balances without Chemical Reaction 45

Solids in outgoing gas mixture = 120 ´ (16.5 – 4.2)/106 = 1.476 ´ 10–3 kg/s º 5.314 kg/h Solids scrubbed = 398.5 – 5.314 = 393.186 kg/h Vapour pressure of water at 348 K (75 ºC) = 38.549 kPa (ref. Chapter 6) Humidity = 38.549/(101.325 – 38.549) = 0.614 kmol/kmol dry gas Moisture in outgoing gas mixture = 0.614 ´ 1975.4 = 1213 kmol/h Moisture added in gas mixture = 1213 – 674.6 = 538.4 kmol/h º 9691.2 kg/h Let x be purge rate (bleed) in kg/h. It will purge solids, equal to the amount scrubbed (i.e. 393.186 kg/h). Chlororides in purge = 200 ppm = 200 mg/kg as Cl Balance of chlorides: Chlorides in purge = chlorides from incoming gas mixture + chlorides from make-up water – chlorides loss in evapn. x ´ 200/106 = 5.4 + (x + 9691.2) 50/106 – 0 x = 39 230 kg/h º 10.9 kg/s Solids concn. in circulating solution = 393.186 ´ 100/39 230 = 1.00% SO3 in incoming gas mixture = 1600 ´ (16.5 – 4.2) ´ 3600/106 = 70.85 kg/h SO3 scrubbed = 70.85 ´ 0.8 = 56.68 kg/h H2SO4 produced = 98 ´ 56.68/80 = 69.43 kg/h which is purged out in the bleed. H2SO4 concentration =

69.43 ´ 100 = 0.18% 39 230

Make-up water = 9691 + 39 230 = 48 921 º 13.59 kg/s

Ans.

EXERCISE 3.34 Let the baterial suspended in total combined feed Bacterial suspended solids in total combined feed For minimum recycle ratio, xR Bacterial suspended solids in fresh feed Concentration of solids in clarifier overflow

= xR mg/L = x0 mg/L = 5x0 =0 =0

(1)

46 Solutions Manual—Stoichiometry

Concentration of solids (bacterial and non-bacterial) in the total combined feed = 2% = 20 000 mg/L Balance of total solids at the entrance of aeration tank: Total solids in fresh feed + total solids in recycle stream = total solids in combined feed 485 QF + QR xR = (QF + QR) 20 000 Dividing by QF and substituting the recycle ratio, r = QR/QF 485 + r xR = (1 + r)20 000 xR =

20 000 (1 + r ) - 485 r

(2)

Balance of bacterial suspended solids at the entrance of aeration tank: Total solids in combined feed – fresh feed solids (non-bacterial) = total bacterial suspended solids 20 000(QR + QF) – 485 QF = (QF + QR) x0 20 000(1 + r) – 485 = (1 + r) x0 x0 = 20 000 – [485/(1 + r)]

(3)

Substitute Eqs. (2) and (3) in Eq. (1). 20 000 (1 + r ) + 485 = 5[20 000 – (485/(1 + r)}] r

Simplifying, 80 000 r2 + 58 060 r – 19 515 = 0 r = 1/4

or

–0.975 75

Discarding negative solution, minimum recycle ratio = 1/4

Ans.

EXERCISE 3.35 At constant temperature T: (a) Let initial quantum of the component in vessel = x kmol Total quantum of gas in vessel = n1 kmol Concentration of component at p1 pressure, c1 = x/n1 kmol/kmol At an absolute pressure p1, n1 = p1 × V/RT where V = vessel volume When the vessel is pressurised to absolute pressure p2 with inert gas, quantum of gas in the vessel will be n2 = p2 × V/RT Concentration of component at p2 pressure, c2 = x/n2 kmol/kmol During depressurisation of the vessel there will be no change in concentration c2.

Material Balances without Chemical Reaction 47

Ratio, c2/c1 = n1/n2 = p1 × RT/(p2 × RT ) = p1/p2 Thus final concentration of the component will be (p1/p2) times the initial concentration. This is for one cycle of pressurisation/depressurisation. If this cyclic process is carried out k times, concentration of the component after k cycles (ck) will be (p1/p2)k times the original concentration (c1). Quantum of inert gas required = n2 – n1 kmol for one pressurization Ratio of inert gas required to initial gas quantum = [(n2 – n1)/n1]V  = [(p2 – p1)/p1] V per cycle

Ans.

(b) At sub-atmospheric absolute pressure p (in kPa), quantum of gas in vessel, n1 = pV/RT kmol. Note that R will have consistent units. Concentration of component at p pressure, c1 = x/n1 kmol/kmol During evacuation, there will be no change in concentration c1. Now the vessel is pressurised with inert gas to atmospheric pressure. At atmospheric pressure, total gas = n2 = 101.3 × V/RT kmol Concentration of component at atmospheric pressure, c2 = x/n2 kmol/ kmol Ratio, c2/c1 = n1/n2 = p/101.3 per cycle of evacuation/pressurisation

If k cycles are carried out, final concentration will be (p/101.3)k times c1. Ratio of inert gas required to bring the vessel to atmospheric pressure = [(n2 – n1)/n1]V = [(101.3 – p)/101.3]V per cycle

Ans.

If pressure p is expressed in atmospheres, c2/c1 = p and inert gas requirement = (1 – p)V per cycle Note: In industrial practice, pressure cycle method is more common in use as compared to vacuum cycle method or the atmospheric method (i.e. Example 3.20 in the text). Why?

EXERCISE 3.36 Empty height of isotank, h1 = 2.5 ´ 0.15 = 0.375 m 2.5 = 1.25 m 2 h1 0.375 = = 0.3 1.25 R From Chemical Engineers' Handbook, Ed. R.H. Perry, 5th Ed., p, 1-22, 1973,

Radius of isotank, R =

48 Solutions Manual—Stoichiometry

cross -sectional area of empty segment, A = 0.2955 R2 A = 0.2955 (1.25)2 = 0.461 72 m2

Volume of air before purging = 0.461 72 ´ 6 = 2.7703 m3 Based on Exercise 3.35 (a),

p1 = 1 atm p2 = 2 bar g = 1.973 85 atm g = 2.973 85 atm a

Before pressure purging, c1 = 20.946 vol.% (O2 in air) After first cycle of purging, c2 =

20.946 2.973 85

= 7.043 vol.% 7.043 2.973 85 = 2.368 vol.% 2.368 After third cycle of purging, c3 = 2.973 85 = 0.796 vol.% (i.e. < 1%) Thus 3 cycles of purging are required.

After second cycle of purging, c3 =

No. of empty vessel volumes, required for purging = (p2 – p1)/p1 = 1.973 85 times per cycle Total nitrogen required for purging = 1.973 85 ´ 3 ´ 2.7703 = 16.404 m3 at 1 atm and 308.15 K (35°C) At NTP, nitrogen requirement = 16.404 ´ 273.15/308.15 = 14.541 Nm3

EXERCISE 3.37 Basis: 100 kmol dry gas mixture at time of shutdown. Steam present = 100 ´ 1.2 = 120 kmol

Ans.

Material Balances without Chemical Reaction 49

Component

kmol

mole % on wet basis

H2 CO CO2 N2 Ar H2O

56.0 15.0 7.0 21.7 0.3 120.0

25.45 6.82 3.18 9.86 0.14 54.55

Total

220.0

100.00

Initially when HT Shift Converter is pressure reduced to 0.25 bar g, there will be no change in concentrations. Since presence of H 2O is objectional, its concentration will be considered for calculations. c1 = 0.5455 mole fraction p1 = 0.25 bar g = 0.246 73 atm g = 1.246 73 atm a p2 = 5 bar g = 4.934 62 atm g = 5.934 62 atm a 1.246 73 = 0.210 08 p1/p2 = 5.934 62 c2 = (p1/p2) c1 = 0.210 08 ´ 0.5455 = 0.1146 mole fraction c3 = 0.210 08 ´ 0.1146 = 0.024 07 c4 = 0.210 08 ´ 0.024 07 = 0.005 06 (i.e. < 0.01) Thus 3 cycles of purging are required. Final concentration of water = c4 = 0.5% (by volume) Final concentration of carbon monoxide = (0.210 08)3 0.0682 = 0.0006 mole fraction or 0.06% by volume Number of (empty) void volumes required for purging = (p2 – p1)/p1 = (5.934 62 – 1.246 73)/1.246 73 = 3.76 times Nitrogen requirement = 3.76 ´ 3 ´ 40 = 451.2 m3 at 1.246 73 atm a and 308.15 K (35 °C) 451.2 ´ 1.246 73 ´ 27315 . At NTP, nitrogen requirement = 308.15 = 498.63 Nm3 Ans.

EXERCISE 3.38

Alternate 1: Methane addition Basis: 200 kmol mixture H2 present = 100 kmol or 50% Air present = 100 kmol

50 Solutions Manual—Stoichiometry

Since H2 is 50% in the mixture, it is within the flammability envelope. Let x kmol CH4 are to be added to raise the upper limit of flammability.

LM 100 OP 100 LM x OP 100 N 200 + x Q + N 200 + x Q = 1 75

1400

14

LM 100 OP + 7500LM x OP = 14 ´ 75 = 1050 N 200 + x Q N 200 + x Q

14 ´ 100 + 75 ´ x = 10.50 (200 + x) 1400 + 75 x = 2100 + 10.5 x x = 10.85 kmol

Ans. (b)

Alternate 2: Oxygen present = 100 ´ 0.2095 = 20.95 kmol Let y be kmol of the total mixture. 0.05 y = 20.95 y = 419 kmol N2 to be added = 419 – 200 = 219 kmol

Ans. (a)

EXERCISE 3.39 At any time G (h) the concentration of oxygen in the chamber is y mol/mol of mixture. Total air present initially = pV/RT p = (53.3 + 101.3) = 154.6 kPa Air present = 154.6 ´ 4500/(8.314 ´ 293 ´ 1000) = 0.2856 kmol Astronaut inhales 123 kg air and exhales 112 kg air. The difference between two is the oxygen consumption by the astronaut. Oxygen consumed from the chamber = (123 – 112)/24 = 0.4583 kg/h º 0.0143 kmol/h CO2 coming to the chamber = 8.8/(44 ´ 24) = 0.0083 kmol/h At time G, total moles of dry gas, present in the chamber (N) = 0.2856 – 0.0143 G + 0.0083 G = 0.2856 – 0.006 G kmol Oxygen balance at time G: Consumption of oxygen = input oxygen – output oxygen

Material Balances without Chemical Reaction 51

Since mole fraction of oxygen is y in the chamber at time G, average molar mass of gas mixture in the chamber, excluding nitrogen = 32 y + (1 – y)44 = 44 – 12y Air output (inhalation) from the chamber = 123 kg/d = 5.125 kg/h Equivalent oxygen output = 5.125y/(44 – 12y) kmol/h Air input (exhalation) to the chamber = 112 kg/d = 4.667 kg/h Equivalent oxygen input = 4.667y/(44 – 12y) kmol/h Input – output = – 0.458y/(44 – 12y) kmol/h dy But consumption of oxygen = N dG 0.458 y dy (0.2856 – 0.006 q) =– dG ( 44 - 12 y) dG -dy( 44 - 12 y) = (0.2856 - 0.006 G ) 0.45 y

Rearranging,

When G = 0, i.e. start of the mission, y = 1.0, i.e. pure oxygen. At the end of mission, i.e. G = G, y = 0.2. G

dG ò (0.2856 - 0.006G ) = 0

z

0.2

1

( 44 - 12 y)dy 0.458 y

– (1/0.006) ln (1 – 0.021 G ) = 133.655 ln (1 – 0.021 G ) = – 133.655 ´ 0.006 = – 0.801 93 1 – 0.021 G = 0.4485 0.021 G = 0.5515 G = 26.264 h º 26 h 16 min Ans. (a) This means the mission will last for 1 day 2 hours and 16 minutes before O2 concentration will reduce to 20% (v/v). At the end of 26.264 h, CO2 accumulated = 26.264 ´ 0.0083 = 0.218 kmol Oxygen present = 20% Total gas mixture = 0.218/0.8 = 0.2725 kmol nRT 0.2725 ´ 8.314 ´ 293 ´ 100 = V 4500 = 147.5 kPa Ans. (b) Partial pressure of O2 = 147.5 ´ 0.2 = 29.5 kPa Partial pressure of CO2 = 147.5 – 29.5 = 118 kPa Ans. (c)

Pressure in the system =

EXERCISE 3.40 Basis: 1.65 kg/s 10% NaOH solution

52 Solutions Manual—Stoichiometry

The solution contains 0.165 kg/s NaOH. This solution is prepared by mixing (0.165/0.5 =) 0.33 kg/s 50% lye and (1.65 – 0.33 =) 1.32 kg/s water. Hold-up in the tank = 1900 ´ 1.1 = 2090 kg NaOH balance of unsteady-state process when the flow of 50% lye will stop: Let c = concentration of NaOH by mass in the tank at any time G (in seconds) and dG = time interval in seconds 1.32 ´ 0 – c ´ dG = 2090 dc

z

z

c2

G

dG = (2090/0.165)

0

dc/c = 12 667 ln (c1/c2)

c1

G = 12 667 ln (0.1/0.08) = 2827 s

Ans.

EXERCISE 3.41 Basis: 150 Sm3/h air at 101.325 kPa and 288.7 K (15 °C) Specific volume, V = RT/p = 8.314 ´ 288.7/101.325 = 23.689 m3/kmol Molar flow rate of moist air = 150/23.689 = 6.332 kmol/h Inlet pressure = 710 kPa g = 811.325 kPa a Vapour pressure of water at 318 K (45 °C) = 9.582 kPa (ref. Chapter 6) Humidity of inlet air = 9.582/(811.325 – 9.582) = 0.011 95 kmol/kmol dry air º 0.0118 kmol/kmol moist air Total moisture in the inlet air = 6.332 ´ 0.0118 = 0.0747 kmol/h º 1.345 kg/h Dry air flow rate = 6.332 – 0.075 = 6.257 kmol/h º 181.45 kg/h Step-I: Increase of moisture content of silica gel from 2.5 to 3.75 kg per 100 kg desiccant Mixture picked-up by silica gel = (3.75 – 2.5) 220/100 = 2.75 kg Average moisture content of outlet air = (5 + 10)/2 = 7.5 mg/Sm3 Moisture, removed from air in G1 h = [1.345 – (7.5 ´ 150)/106] G1 = 1.3439 G1 kg 1.3439 G1 = 2.75 G1 = 2.046 h

Material Balances without Chemical Reaction 53

Similar iterations can be performed. Step II: Increase in moisture content of silica gel l from 3.75 kg to 5.625 kg per 100 kg desiccant G2 = 3.071 h G3 = 1.537 h G4 = 1.281 h Total duration = å G i = 2.046 + 3.071 + 1.537 + 1.281 = 7.935 h Ans.

EXERCISE 3.42 Basis: Cooling water (CW) hold-up in system = 3000 m3 Consider a small time interval DG = 1 h G1 = 1 h Initial total suspended solids in cooling water = 3000 ´ 50 ´ 103 mg = 15 ´ 107 mg Suspended solids removed = 100 (50 – 10)103 = 0.4 ´ 107 mg Suspended solids after one hour = (15 ´ 107 – 0.4 ´ 107) mg Concentration of suspended solids = (15 ´ 107 – 0.4 ´ 107)/(3000 ´ 103) = 48.667 mg/L Similar iterations can be performed, which are tabulated below. Time G h

0 1 2 3 4 5 6 7 8 9 10 11

suspended solids in outlet water from filter, mg/L 10 10 10 10 10 10 10 10 10 9 9 9

suspended solids in CW at the end of G h, mg/L 50.0 48.667 47.378 46.131 44.927 43.763 42.637 41.549 40.598 39.448 38.433 37.452

Time G h

12 13 14 15 16 17 18 19 20 21 22 23

suspended suspended solids in solids in outlet CW at the water from end of filter, mg/L G h, mg/L 9 9 9 9 9 9 9 9 8 8 8 8

36.503 35.587 34.700 33.844 33.016 32.215 31.441 30.693 29.927 29.206 28.499 27.815

54 Solutions Manual—Stoichiometry

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

8 7 7 7 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5

27.155 26.483 25.833 25.205 24.565 23.946 23.348 22.770 22.211 21.671 21.148 20.643 20.155 19.683 19.194 18.721 18.263 17.821 17.394 16.981 16.582 15.822 15.462

47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

15.113 14.776 14.450 14.135 13.831 13.536 13.231 13.143 12.710 12.453 12.205 11.965 11.733 11.508 11.291 11.082 10.879 10.683 10.493 10.310 10.133 9.962

Thus time required for bringing down the suspended solids to 10 mg/L level = 68 h Ans

EXERCISE 3.43 Basis: 3000 m3 hold-up DG = 1 h Suspended solids at G = 0 h = 3000 ´ 50 ´ 107 = 15 ´ 107 mg Suspended solids removed in the filter = 4 ´ 106 mg/h (from Exercise 3.42 after first hour of filtration.) Dust (turbidity) added in the basin = 16 ´ 10–9 ´ 106 ´ 106 kg/h º 1.696 ´ 106 mg/h Net suspended solids after 1 h = 15 ´ 107 – 4 ´ 106 + 1.696 ´ 106 = 147.696 ´ 106 mg/h Suspended solid in CW = 147.696 ´ 106/(3000 ´ 103) = 49.232 mg/L

Material Balances without Chemical Reaction 55

Time G h

suspended solids in outlet water from filter, mg/L

suspended solids in CW at the end of G h, mg/L

0

–

50.0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 9 9 9 9 9 9 9 8 9 9 9 9 9 9 9 9 9 9 9 9 9

49.232 48.490 47.772 47.078 46.407 45.759 45.132 44.526 43.941 42.810 42.282 41.771 41.277 40.800 40.339 39.891 39.427 38.978 38.544 38.124 37.718 37.326 36.947 36.581 36.227 35.885 35.554 35.234 34.925 34.626 34.337 34.058 33.788 33.527 33.275 33.031 32.795

Time G h

38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74

suspended suspended solids in solids in outlet CW at the water from end of filter, mg/L G h, mg/L 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7

32.567 32.347 32.134 31.929 31.730 31.537 31.352 31.172 30.998 30.831 30.668 30.511 30.359 30.213 30.071 29.900 29.736 29.577 29.423 29.274 29.130 28.991 28.857 28.727 28.601 28.480 28.363 28.249 28.140 28.034 27.931 27.832 27.737 27.644 27.555 27.468 27.351 (Contd.)

56 Solutions Manual—Stoichiometry

(Contd.) 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

27.238 27.129 27.023 26.921 26.822 26.722 26.635 26.546 26.460 26.380 26.296 26.218 26.149 26.070 26.000 25.932 25.866 25.803

93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

25.741 25.682 25.624 25.569 25.515 25.464 25.413 25.335 25.318 25.273 25.229 25.187 25.146 25.106 25.068 25.031 24.999

Time, required to bring down the turbidity level to 25 mg/L = 109 hAns.

EXERCISE 3.44 Material balance of components A after time G: Input – output = accumulation 0 – F × CAG = V × When

DG ® 0,

DC DG

F dCA =– dG V dCAG

Integrating on both the sides,

z

CAG

CA 0

dCA F =– dG dCAG V F q V CAG = CA0 e–(FG/V)

ln CAG – ln CAO = – or

EXERCISE 3.45 Basis: V m3 volume of the receiver Input – output = accumulation

Ans.

Material Balances without Chemical Reaction 57

Output = 0 Actual volume swept = v = v0 × Dv

LM MN

v = v0 1 + c -

Swept volumetric flow rate =

Input =

LM MN

FG IJ OP H K PQ

3600 s v0 p0 kmol/h ZRT

RS UV OP in 1 h T W PQ 1/ C

3600 s v0 p0 p p -c 1+ cZRT 100 p0 p0

After time G,

1/ C

p p -c 100 p0 p0

dn dG n = no. of kmol present at time G

Accumulation in the receiver = where

= Therefore,

V dp × ZRT dG

LM MN p dp 3600 s v p L = 1+ cM V 100 p dG MN L RpU p dq = dp/ M1 + c - cS V 100 p MN Tp W

RS UV T W RpU - cS V Tp W OP PQ

3600 s v0 p0 p p V dp × = -c 1+ cZRT dG ZRT 100 p0 p0 0

or

3600 s v0 p0 V

Substitute

1/ C

0

0

z z

LM p LM1 + c - p 100 p MN MN

p1

G

dG =

dp

0

p0

0

0

0

0

z

R1

1

LM N

dr / 1 + c -

V q = 3600 s v0

z

R1

1

RS p UV Tp W

1/ C

-c 0

r = p/p0 and R1 = p1/p0 3600 s v0 q = ZRT

or

OP PQ OP PQ

1/ C

Integrating on both the sides, 3600 s v0 V

1/ C

r - c(r )1 / C 100

0

OP Q

dr 1 + c - (r /100) - c(r )1/ C

OP O PQ PPQ

58 Solutions Manual—Stoichiometry

Note: For a double acting compressor, V q = 7200 s v0

z

R1

dr 1 + c - (r /100) - c(r )1/ C

1

EXERCISE 3.46 Basis: Air receiver having 60 m3 capacity is to be from 101.325 kPa a to 0.7 MPa g Geometric volume of cylinder v0 = (p/4) (0.312) (0.15) = 1.132 ´ 10–2 m3 3 V = 60 m , C = 1.4, c = 0.07, R1 = ? Each stage has equal compression ratio. Overall compression ratio = (700 + 101.325)/101.325 = 7.9085 R 1 = R2 = 7.9085 = 2.8122

L The integral M MN

z

R1

1

dr 1 + c - (r /100) - c(r )1 / C

OP of Exercise 3.45 can be solved by PQ

Numerical integration using Simpson’s rule: r

[1 + c – (r/100) – c(r)1/C]

1.000 00 1.181 22 1.362 44 1.543 66 1.724 88 1.906 10 2.087 32 2.268 54 2.499 76 2.630 98 2.812 20

0.990 00 0.979 34 0.969 07 0.959 11 0.949 42 0.939 97 0.930 72 0.921 65 0.912 75 0.903 99 0.895 37

f = 1/[1 + c – (r/100) – c(r)1/C] f1 = f2 = f3 = f4 = f5 = f6 = f7 = f8 = f9 = f10 = f11 =

1.010 00 1.021 09 1.031 92 1.042 63 1.053 27 1.063 86 1.074 44 1.085 01 1.095 59 1.106 20 1.116 85

Value of the integral = (h/3) [f1 + 4f2 + 2f3 + 4f4 + 2f5 + K + f11] where h = interval = 0.181 22 Value of the integral = (0.181 22/3) [1.010 00 + 4´ 1.021 09 + 2 ´ 1.031 92 + 4 ´ 1.042 63 + 2 ´ 1.053 27 + 4 ´ 1.063 86 + 2 ´ 1.074 44 + 4 ´ 1.085 01 + 2 ´ 1.095 59 + 4 ´ 1.106 20 + 1 ´ 1.116 85]

Material Balances without Chemical Reaction 59

= 0.060 41(1.010 00 + 4.084 36 + 2.063 84 + 4.170 52 + 2.106 54 + 4.255 44 + 2.148 88 + 4.340 04 + 2.191 18 + 4.424 80 + 1.116 85 = 0.060 41 ´ 31.912 55 = 1.927 84 For double acting compressor, stroke s = 2 ´ 9.2 s–1 which will replace the term 7200 s in the equation for finding the time in Exercise 3.45. Therefore, q = =

V ´ integral s v0 60 ´ 10 2 ´ 1.927 84 = 555.3 s . 2 ´ 9.2 ´ 1132

The above calculated time refers to (2.8122 ´ 101.325) at the discharge of first stage or 801.325 kPa a at the discharge of second sage. time required = 555.3 s º 0.154 h Ans. Evaluation of the integral by a numerical method is laborious. Mathcad can be quite useful in quick evaluation of such an integral.

4

Material Balances Involving Chemical Reactions EXERCISE 4.1 Basis: 1000 kg ammonia production Molar mass of ammonia = 17.0305 NH3 Production =

1000 = 58.7182 kmol 17.0305

3 ´ 58.7182 2 = 88.0773 kmol CH4 requirement = 88.0773/4 = 22.0193 kmol º 493.541 Nm3 Ans. (a) When air is used, 21% O2 is also fed alongwith 79 % N2. For 58.7182 kmol NH3 production, N2 requirement = 58.7182/2 = 29.3591 kmol O2 fed alongwith N2 = (21/79) 29.3591 = 7.9044 kmol H2 consumed by O2 = 2 ´ 7.9044 = 15.8088 kmol Total H2 requirement = 88.0773 + 15.8088 = 103.8861 kmol CH4 requirement = (103.8861/4) = 25.9715 kmol º 528.126 Nm3 Ans. (b)

H2 requirement =

Material Balances Involving Chemical Reactions 61

EXERCISE 4.2 Basis: 1000 kg methanol production Molar mass of methanol = 32.0419 CH3OH production = 1000/32.0419 = 31.2091 kmol CO2 requirement = 31.2091 kmol H2 requirement = 3 ´ 31.2091 = 93.6273 kmol CO2 and H2 are required in 1 : 3 molar proportions. However, reforming reaction yields CO2 and H2 in 1 : 4 molar proportions. Hence H2 is in excess and therefore methane requirement will be fixed by CO2 requirement. CH4 requirement = 31.2091 kmol º 699.521 Nm3/t Ans.

EXERCISE 4.3

Basis: N2 flow of 0.025 L/s at NTP. This basis is same as that of Exercise 3.24. Reaction: CH0.6 + 1.6 O2 = CO2 + 0.6 H2O 12.616 1.6 ´ 31.9988 O2 consumed = 4.096 ´ 10–3 kg

(Refer Exercise 3.24)

12.616 ´ 4.096 ´ 10-3 = 1.0093 ´ 10–3 kg/s 1.6 ´ 31.9988 Mass of coke Coke present in catalyst = ´ 100 Mass of catalyst

Coke burnt =

1.0093 ´ 10-3 ´ 100 0.1 = 1.0093% say 1%

=

(by mass)

Ans.

EXERCISE 4.4 Basis : Charge to reactor : 50 kg naphthalene (N) 200 kg 98% H2SO4 C10H8 + H2SO4 = C10H7SO3H + H2O (1) N MSN 128 98 208 18 C10H8 + 2 H2SO4 = C10H6(SO3H)2 + 2 H2O N DSN 128 2 ´ 98 288 2 ´ 18 New basis: 100 kg product The product will contain 18.6 kg MSN and 81.4 kg DSN. N requirement for 18.6 kg MSN = 128 ´ 18.6/208 = 11.45 kg

(2)

62 Solutions Manual—Stoichiometry

N requirement for 81.4 kg DSN = 128 ´ 81.4/288 = 36.18 kg Total N requirement = 11.45 + 36.18 = 47.63 kg Original basis assumed the charge of N as 50 kg. MSN produced = 18.6 ´ 50/47.63 = 19.53 kg DSN produced = 81.4 ´ 50/47.63 = 85.45 kg Total H2SO4 consumed = (98 ´ 19.53/208) + (2 ´ 98 ´ 85.45/288) = 67.36 kg Unconsumed H2SO4 = 196 – 67.36 = 128.64 kg Total water produced = (18 ´ 19.53/208) + (2 ´ 18 ´ 85.45/288) = 12.371 kg Component

Mass, kg

mass %

MSN DSN H2SO4 H2O

19.53 85.45 128.64 12.37

7.94 34.74 52.29 5.03

Total

245.99

100.00

Ans.

EXERCISE 4.5 Basis: 100 kg product Compound

mass, kg

SBP Nitric acid DNOSBP DNPSBP

65 15 18 2

150 63 240 240

0.4333 0.2381 0.0750 0.0083

57.41 31.55 9.94 1.10

100

—

0.7547

100.00

Total

Molar mass

kmol

mole %

Nitric acid being the limiting component, conversion has to be based on nitric acid consumption. New basis: 100 kmol product Nitric acid input = 31.55 + 2(9.94 + 1.10) = 53.63 kmol Conversion = 2(9.94 + 1.10)100/53.63 = 41.2% Yield of DNOSBP = 2 ´ 9.94 ´ 100/[2(9.94 + 1.10)] = 90.04% Yield of DNPSBP = 100 – 90.04 = 9.96% Ans.

EXERCISE 4.6 Basis: 1000 kg molasses Monosaccharides present = 1000 ´ 0.45 = 450 kg

Material Balances Involving Chemical Reactions 63

450 ´ 2 ´ 46 ´ 2 342 = 242.105 kg º 308.41 L

Theoretical ethyl alcohol production =

Ans.

EXERCISE 4.7 Basis: 1 kmol fresh i-butane (combined feed) Recycle hydrogen feed to the reactor = 0.75 kmol H2 0.75 ´ 0.1 Methane recycled with H2 = = 0.083 kmol 0.9 Total recycle = 0.75 + 0.083 = 0.833 kmol i-butane converted = 0.5 kmol Yield of i-butylene = 0.5 ´ 0.88 = 0.44 kmol [reaction (1)] H2 produced = 0.44 kmol Total H2 in product stream = 0.75 + 0.44 = 1.19 kmol Yield of propylene = 12% as per reaction (2) Propylene produced = 0.5 – 0.44 = 0.06 kmol Methane produced = 0.06 kmol Total methane in product stream = 0.083 + 0.06 = 0.143 kmol Reactor outlet stream Component

kmol

vol. %

i–C4H10 i–C4H8 C3H6 H2 CH4

0.5 0.44 0.06 1.19 0.143

21.43 18.86 2.57 51.01 6.13

Total

2.333

100.00

Ans.

EXERCISE 4.8 Basis: 100 kmol creacked gas Composition of cracked gas: Component H2 CH4 C2H4 C2H2 C3H6

kmol 56.5 5.2 0.3 7.5 0.5

Carbon kmol — 5.2 0.6 15.0 1.5

Hydrogen kmol 56.5 10.4 0.6 7.5 1.5

Oxygen — — — — —

64 Solutions Manual—Stoichiometry

CO CO2 O2 Total

25.8 4.0 0.2

25.8 4.0 —

— — —

12.9 4.0 0.2

100.0

52.1

76.5

17.1

Entire carbon atom comes from methane. Methane fed = 52.1 kmol Hydrogen from methane = 52.1 ´ 2 = 104.2 kmol Hydrogen unaccounted = 104.2 – 76.5 = 27.7 kmol This hydrogen is burnt with oxygen to produce water. Water produced = 27.7 kmol O2 used for water production = 27.7/2 = 13.85 kmol O2 consumed = 13.85 + 17.1 = 30.95 kmol Conversion of methane = (52.1 – 5.2) 100/52.1 = 90.02% Yield of acetylene = 15 ´ 100 (52.1 – 5.2) = 31.98%

Ans. (a)

Ans. (c) Ans. (b) Ans. (d) Ans. (c)

EXERCISE 4.9 Basis: 1 litre mixture of N2 + O2 Mass of water, absorbed on silica gel = 0.0362 g = 0.002 01 mol Let P = Phenolphthalein reading, mL, M = Methylorange reading (i.e. total alkalinity), mL and N = Normality Total alkalinity =

M ´ N ´ 50 000 mg/L as CaCO3 vol.of sample, mL

P = 35.4 mL and 1 M 2 Therefore, the solution contains hydroxide and carbonate alkalinies. Normally of titrating acid, N = 0.012 (2P – M) N ´ 50 000 ppm as CaCO3 Hydroxide alkalinity = vol. of sample in mL

M = 38 mL. Hence P >

(2 ´ 35.4 – 38)0.012 ´ 50 000 10 = 1956 ppm (mg/L) as CaCO3 38 ´ 0.012 ´ 50 000 Total alkalinity = 10

=

Material Balances Involving Chemical Reactions 65

= 2280 ppm (mg/L) as CaCO3 KOH is solution = 1956 ´ 56/50 = 2190.7 ppm or mg/L K2CO3 in solution = (2280 – 1956) 69/50 = 447.1 ppm or mg/L K2CO3 formation takes place due to the reaction of CO2 with KOH. 2 KOH + CO2 = K2CO3 + H2O 2 ´ 56

44

138

18

CO2 absorbed = 44 ´ 447.1/138 = 142.6 mg/L Since the solution’s volume is 1 litre, CO2 absorbed = 142.6 mg º 0.1426 g º 0.003 24 mol Water in the flue gas = 0.0362 g º 0.002 01 mol Molar volume at 101.325 kPa a and 298.15 K (25 °C), V = 8.314 ´ 298/101.325 = 24.464 L/mol Let x = mol of O2 and y = mol of N2 in flask 32x + 28y = 1.16 24.464 (x + y) = 1 Solving the equations, x = 0.003 87 mol, y = 0.037 mol Component

mol

mole % (dry)

(1) (2)

Molar mass

Mass, g

mass %

CO2 O2 N2 H2O

0.003 24 0.003 87 0.037 0.002 01

7.34 8.78 83.88 —

44 32 28 18

0.1426 0.1238 .1.036 0.0362

10.65 9.25 77.39 2.71

Total

0.046 12

100.00

—

1.3386

100.00

Ans.

EXERCISE 4.10 Basis: 500 kg limestone CaCO3 content of limestone = 500 ´ 0.60 = 300 kg MgCO3 content of limestone = 500 ´ 0.335 = 167.5 kg Inerts = 500 – 300 – 167.5 = 32.5 kg Reactions are: CaCO3 + H2SO4 = CaSO4 + H2O + CO2

(1)

MgCO3 + H2SO4 = MgSO4 +

(2)

100

84.3

98

98

136

120.3

18

44

H2O + CO2

18

44

Stoichiometric H2SO4 requirement = (98 ´ 300/100) + (167.5 ´ 98/84.3) = 294 + 194.7 = 488.7 kg Amount of acid charged = 488.7 ´ 1.15 = 562.0 kg (100%) Ans. (a) Strength of acid = 12% (by mass) Quantity of aqueous acid = 562.0/0.12 = 4683.3 kg

66 Solutions Manual—Stoichiometry

Products of reaction (1): CaSO4 produced = 136 ´ 300/100 = 408 kg H2O produced = 18 ´ 300/100 = 54 kg CO2 produced = 44 ´ 300/100 = 132 kg º 3 kmol Products of reaction (2): MgSO4 produced = 120.3 ´ 167.5/84.3 = 239.0 kg H2O produced = 18 ´ 167.5/84.3 = 35.8 kg CO2 produced = 44 ´ 167.5/84.3 = 88 kg º 2 kmol Total CO2, escaped from reactor = 132 + 88 = 220 kg º 5 kmol Total H2O at the end of reactions = 4683.3 – 562 + 54 + 36 = 4219.1 kg Compound CaCO3 MgCO3 Inerts H2SO4 H2O CaSO4 MgSO4 Total

Mass of reactants, kg 300.0 167.5 32.5 562.0 4121.3 Nil Nil 5183.3

Mass of products, kg Nil Nil 32.5 73.3 4211.3 408.0 239.0 4964.1

mass % in product — — 0.65 1.48 84.84 8.21 4.82 100.00 Ans. (b) and (c)

EXERCISE 4.11 Basis: 100 kmol of hydrogen chloride gas 4 HCl + O2 = 2 Cl2 + 2 H2O Theoretical oxygen requirement = 100/4 = 25 kmol Actual oxygen supply = 25 ´ 1.3 = 32.5 kmol This oxygen is supplied in the form of air. N2, entering with O2 = 79 ´ 32.5/21 = 122.3 kmol The reaction goes to 80% completion. HCl reacted = 100 ´ 0.80 = 80 kmol O2 reacted = 80/4 = 20 kmol Cl2 produced = 2 ´ 80/4 = 40 kmol Component HCl O2

kmol 20.0 12.5

Ans.(d)

mole % (dry basis) 10.27 6.42

Material Balances Involving Chemical Reactions 67

Cl2 N2

40.0 122.3

20.53 62.78

Total

194.8

100.00

Ans.

EXERCISE 4.12 Basis: 100 kmol of (A + inerts) In a constant pressure reactor, the volume doubles by doubling total number of moles (assuming that Ideal Gas Law holds). At the end of reaction, total no. of moles = 200 kmol A in the input mixture = 100 ´ 0.75 = 75 kmol Let x kmol A are reacted. Unreacted A = 75 – x kmol Product = 3x kmol Total gas mixture at the end of reaction = 75 – x + 25 + 3x = 100 + 2x kmol 100 + 2x = 200 x = 50 kmol Conversion = 50 ´ 100/75 = 66.67% Ans.

EXERCISE 4.13 Basis: 100 kmol dry inlet gas CO present in ingoing gas mixture = a kmol Let nCO kmol CO are reacted in the shift converter. CO in outlet gas mixture = (a – nCO) kmol CO2 and H2 produced = nCO (each) Total dry gas mixture at the outlet = (100 + nCO) kmol % CO, present in outcoming gas mixture = b a - nCO b = 100 + nCO 100

Simplifying, nCO =

100( a - b) 100 + b

Q.E.S.

EXERCISE 4.14 Basis: 100 kmol of gases, entering secondary converter. The mixture contains 4 kmol SO2, 13 kmol O2 and 83 kmol N2. Let x kmol SO2 are reacted to SO3 as per reaction: 1 SO2 + O2 = SO3 2 Outgoing gas mixture from the converter will contain (4 – x) kmol SO2, (13 – 0.5x) kmol O2 and 83 kmol N2 on SO3–free basis.

68 Solutions Manual—Stoichiometry

( 4 - x )100 = 0.45 ( 4 - x ) + (13 - 0.5 x ) + 83 400 - 100 x = 100 - 1.5 x 400 – 100 x = 99.325 x = x= Conversion = =

0.45 45 – 0.675x 355 3.574 kmol 3.574 ´ 100/4 89.35%

Ans.

EXERCISE 4.15 Basis: 100 kmol of outgoing gas mixture from reactor (1) Reactions are: CO2 + 4 H2 = CH4 + 2 H2O (2) CO2 + H2 = CO + H2O The outgoing gas mixture (100 kmol) contains 1.68 kmol CH4 and 0.12 kmol CO. Reaction (1) : CO2 required = 1.68 kmol H2 required = 4 ´ 1.68 = 6.72 kmol Reaction (2) : CO2 required = 0.12 kmol H2 required = 0.12 kmol Total CO2 required = 1.68 + 0.12 = 1.8 kmol Total H2 required = 6.72 + 0.12 = 6.84 kmol Unconverted CO2 = 57.1 kmol Total CO2, entering the reactor = 57.1 + 1.8 = 58.9 kmol Conversion of CO2 per gas = 1.8 ´ 100/58.9 = 3.06% Ans. (a) Ans. (b) Yield of CH4 = 1.68 ´ 100/1.8 = 93.33% Total H2, entering the reactor = 6.84 + 41.1 = 47.94 kmol CO2 content in incoming mixture = (58.9 ´ 100) / (58.9 + 47.94) = 55.13% H2 content in incoming mixture = 100 – 55.13 = 44.87% Ans.(c)

EXERCISE 4.16 Basis: 100 kmol dry outgoing gas mixture N2 content of the mixture = 69.05 kmol O2 entering with N2 through air = 21 ´ 69.05/79 = 18.36 kmol Air, entering the system = 69.05 + 18.36 = 87.31 kmol O2 unreacted = 2.55 kmol O2 reacted = 18.36 – 2.55 = 15.8 kmol Reactions: 2 CH3CHO + O2 = 2 CH3COOH (1) 2 CH3CHO + 5 O2 = 4 CO2 + 4 H2O (2)

Material Balances Involving Chemical Reactions 69

CO2 content of exit gas mixture O2 reacted as per reaction (2) CH3CHO, reacted as per reaction (2) O2 reacted as per reaction (1) Acetic acid produced Acetaldehyde, reacted as per reaction (1) Total CH3CHO reacted CH3CHO in exit gases Total CH3CHO, entering the reactor Conversion Yield of acetic acid air 29 ´ 87.41 = acetic acid 30.58 ´ 44 Water, produced as per reaction (2) CH3CHO in the exit gases Acetic acid removed with water

Mass ratio,

= 4.85 kmol = 5 ´ 4.85/4 = 6.06 kmol = 2 ´ 4.85/4 = 2.43 kmol = 15.81 – 6.06 = 9.75 kmol = 2 ´ 9.75 = 19.5 kmol = 19.50 kmol = 2.43 + 19.50 = 21.93 kmol = 8.65 kmol = 21.93 + 8.65 = 30.58 kmol = 21.93 ´ 100/30.58 = 71.7% Ans. (a) = 19.50 ´ 100/21.93 = 88.9% Ans. (b) = 1.884 kg/kg

Ans. (c)

= 4.85 kmol = 14.90 kmol = 19.50 – 14.90 = 4.60 kmol Composition of gas mixture, leaving reactor: Component CO2 CH3CHO CH3CHOH O2 H2O N2 Total

kmol

mole %

4.85 8.65 19.50 2.55 4.85 69.05

4.43 7.90 17.82 2.33 4.43 63.09

109.45

100.00

Ans. (e)

EXERCISE 4.17 Basis: 100 kg CH4, 100 kg O2 and 100 H2O Moles of methane = 100/16 = 6.25 kmol Moles of oxygen = 100/32 = 3.125 kmol Moles of steam = 100/18 = 5.555 kmol In the reactor, oxygen will be completely consumed. As a result, CO, CO2, H2 and CH4 will appear in the product mixture. Let a, b, c and d be kmol of CO2 CO, H2O and H2 in the product gas mixture, respectively.

70 Solutions Manual—Stoichiometry

Carbon balance: Hydrogen balance:

a + b = 6.25 (1) c + d = 2 ´ 6.25 + 5.555 = 18.055 (2) Oxygen balance: (c/2) + a + (b/2) = 3.125 + (5.556/2) = 5.903 or 2a + b + c = 11.806 (3) Eq. (3) – Eq. (1): 2a + b + c – a – b = 11.806 – 6.25 a + c = 5.556 (4) Eq. (2) + 2 ´ Eq. (1) – Eq. (3): c + d + 2a + 2b – 2a – b – c = 18.056 + 2 ´ 6.25 – 11.806 b + d = 18.75 (5) yCO 2 × yH 2 Based on kinetic considerations for shift reaction, K p = yCO × yH 2 O Since there is no change of number of moles during the shift reaction, mole fraction can be replaced by number of moles in the equilibrium equation. (ad)/(bc) = 0.7 (6) From Eq. (5), d = (18.75 – b) From Eq. (1), d = 18.75 – 6.25 + a = 12.5 + a From Eq. (1), b = 6.25 – a From Eq. (4), c = 5.556 – a Substituting values of b, c and d in Eq. (6):

a(12.5 + a) = 0.7 (6.25 - a)(5.556 - a ) Simplifying, 0.3a2 + 20.7642a – 24.3075 Solving the quadratic equation, a b c d

=0 = 1.151 kmol CO2 = 6.25 – 1.151 = 5.099 kmol CO = 5.556 – 1.151 = 4.405 kmol H2O = 12.5 + a = 13.651 kmol H2

Ans.

EXERCISE 4.18 Basis: 1 litre water to be treated with lime and soda ash. In this case, since only temporary hardness is present in water, only burnt lime (CaO) is required to be added for treatment. CaO required to be added = 284 ´ 56/100 = 159 mg/L CaO Ans.

EXERCISE 4.19 Basis: 1 litre water In this case, since water contains temporary and permanent hardness, lime and soda ash will be required to treat them. Requirement of CaO = 56 ´ 232.6/100 = 130.3 mg/L Requirement of Na2CO2 = 106 ´ 623.4/100 = 660.8 mg/L

Material Balances Involving Chemical Reactions 71

Permanent hardness can be due to chlorides and sulphates of Ca and Mg. Since this split-up is not known, it is not possible to give actual concentrations of all components.

EXERCISE 4.20 Basis: 1 litre water to be softened In the softener, temporary hardness is removed by passing the water through sodium based cation exchanger. Raw water contains 275.5 mg/L Ca(HCO3)2 and 329.2 mg/L Mg(HCO3)2. Equivalent NaCl requirement (theoretical) for regeneration = 58.5 ´ 2 [(257.5/162) + (329.2/146.3)] = 449.25 mg/L Total NaCl consumption (theoretical) = 449.25 ´ 50 ´ 8/1000 = 179.7 kg Actual consumption of NaCl = 60 ´ 4.24 = 254.4 kg Excess NaCl = 254.4 – 179.7 = 74.7 kg Excess NaCl over theoretical = 74.7 ´ 100/179.7 = 41.5 % Ans.

EXERCISE 4.21 Basis: 100 kg cottonseed oil Each sponification reaction will be separately dealt. (i) Saponification of oleodipalmitin:

—

CH2OOCH33C17

—

CHOOCH31C15 + 3 KOH = C17H33COOK + 2 C15H31COOK + C3H5(OH)3 CH2OOCCH31C15 832

3 ´ 56

92

Stoichiometric requirement of KOH = 3 ´ 56 ´ 8/832 = 1.615 kg Gycerine produced = 92 ´ 8/832 = 0.885 kg (ii) Saponification of Olepalmitostearin:

a a

CH OOCH !!C%

CHOOCH!# C # + 3 KOH = C#H !COOK + C%H !!COOK + C!H#(OH)! + C #H !#COOK CH OOCH ! C# 860

3 ´ 56

KOH requirement= 3 ´ 56 ´ 5/860 = 0.977 kg Gycerine produced= 92 ´ 5/860 = 0.535 kg

92

72 Solutions Manual—Stoichiometry

(iii) Saponification of palmito-oleolinolein

—

CH2OOCH33C17

—

CHOOCH31C15 + 3 KOH = C17H33COOK + C15H31COOK + C17H31COOK + C3H5(OH)3

CH2OOCH31C17 856

3 ´ 56

92

KOH required = 8.047 kg and glycerine produced = 4.407 kg (iv)

Saponification of palmitodilinolein:

—

CH2OOCH31C15

—

CHOOCH31C17 + 3 KOH = C15H31COOK + 2 C17H31COOK + C3H5(OH)3 CH2OOCH31C17 854

3 ´ 56

KOH required = 3.541 kg and glycerine produced = 1.939 kg (v) Saponification of oleodilinolein:

92

—

CH2OOCH33C17

—

CHOOCH31C17 + 3 KOH = C17H33COOK + C3H5(OH)3 + 2 C17H31COOK

CH2OOCH31C17 880

3 ´ 56

92

KOH requirement = 3 ´ 56 ´ 28/880 = 5.345 kg Glycerine produced = 92 ´ 28/880 = 2.927 kg Total theoretical requirement of KOH = 1.615 + 0.977 + 8.047 + 3.541 + 5.345 = 19.525 kg Total glycerine produced = 0.885 + 0.535 + 4.407 + 1.939 + 2.927 = 10.693 kg Ans.

EXERCISE 4.22 Basis: 100 kg castor fatty acids Composition of castor fatty acids: Fatty acid

kg

Molar mass

Palmitic acid Stearic acid Oleic acid Linoleic acid Linolenic acid

1.4 1.2 4.5 6.0 0.5

256.42 284.48 282.46 280.45 278.43

kmol 0.0055 0.0042 0.0159 0.0214 0.0018

H2 requirement, kmol

0.0159 0.0428 0.0054

Material Balances Involving Chemical Reactions 73

Ricinoleic acid

86.4

Total

298.46

100.0

0.2895

0.2895

0.3383

0.3536

100 0.3383 = 295.6 Average molar mass of castor oil = 295.6 ´ 3 + 92.09 – 3 ´ 18.02 = 924.83

Average molar mass of castor fatty acids =

924.83 ´ 100 295.6 ´ 3 = 104.29 kg H2 requirement = 0.3536 kmol/100 kg castor fatty acids = 0.3391 kmol/100 kg castor oil = 7.601 Nm3/100 kg castor oil Ans.(b) Iodine value = 0.3536 kmol/100 kg castor fatty acids = 0.3391 kmol/100 kg castor oil = 86.07 kg/100 kg castor oil Ans. (a)

Quantity of castor oil =

EXERCISE 4.23 Basis: 100 kg soybean fatty acids Fatty acid

kg

Molar mass

kmol

Palmitic acid Stearic acid Oleic acid Linoleic acid Linolenic acid

46.04 5.60 21.94 23.22 3.20

256.42 284.48 282.46 280.45 278.43

0.1795 0.0197 0.0777 0.0828 0.0115

Total

100.0

—

0.3712

100 = 269.4 0.3712 Average molar mass of used soybean oil = 269.4 ´ 3 + 92.09 – 3 ´ 18.02 = 846.23 846.23 ´ 100 Quantity of used soybean oil = 269.4 ´ 3 = 104.71 kg º 0.1237 kmol Reaction : CH3(CH2)14 COOH + CH3OH = CH3(CH2)14COOCH3 + H2O

Average molar mass of fatty acids =

Palmitic acid 256.42

Methanol 32.04

Methyl ester 270.44

Methyl ester produced equivalent to palmitic acid

Water 18.02

74 Solutions Manual—Stoichiometry

270.44 ´ 46.04 = 48.56 kg 256.42 32.04 ´ 46.04 Methanol required = = 5.75 kg 256.42 Water produced = 46.04 + 5.75 – 48.56 = 3.23 kg Similar calculations can be made for other fatty acids.

=

Fatty acid

kg

Methanol required, kg

Methyl ester produced, kg

water produced kg

Palmitic acid Stearic acid Oleic acid Linoleic acid Linolenic acid

46.04 5.60 21.94 23.22 3.20

5.75 0.62 2.49 2.65 0.37

48.56 5.88 23.03 24.38 3.36

3.23 0.34 1.40 1.49 0.21

Total

100.00

11.88

105.21

6.67

11.88 = 0.1135 kg/kg soybean oil 104.71 105.21 = 1.005 kg/kg soybean oil Specific methyl ester production = 104.71

Specific methanol requirement =

Ans.

EXERCISE 4.24 Basis: 100 kg soybean fatty acids Reaction: CH3(CH2)14COOCH3 + 2 H2 = CH3(CH2)14CH2OH + CH3OH FAME of Palmitic acid

Hydrogen

FOH of Palmitic acid

Methanol

Similar reactions can be written for other fatty acids. FAME of

Palmitic acid Stearic acid Oleic acid Linoleic acid Linolenic acid

Molar mass of FAME

kg FAME*

270.44 298.50 296.48 294.83 292.81

48.56 5.88 23.03 24.38 3.36

Total

105.21

Molar mass of corres-

ponding 242.43 270.49 268.47 266.82 264.80

kg FOH

kg CH3OH

43.53 5.33 20.85 22.06 3.04

5.75 0.63 2.49 2.65 0.36

94.81

11.89

* Reference: Exercise 4.23

FOH produced = 94.81 kg/105.21 kg FAME = 90.1 kg/100 kg FAME Ans. Note: Methanol produced = 11.89 kg = Methanol consumption in Exercise 4.23

Material Balances Involving Chemical Reactions 75

EXERCISE 4.25 Basis: Urea feed rate of 350 kg/h . 350 = 5.833 kmol/h Molar feed rate of urea, nu = 60 H2SO4 required = SO3 required = Actual H2SO4 charged = = H2O entering with H2SO4 = = Actual SO3 charged = = On completion of reaction, H2SO4 consumed = SO3 consumed by reaction (1) = SO3 consumed by reaction (2) =

5.833 kmol/h º 571.67 kg/h 5.833 kmol/h º 466.64 kg/h 1.25 ´ 571.66 714.58 kg/h 2 ´ 714.58 98

14.58 kg/h 3.5 ´ 466.64 1633.24 kg/h 571.67 kg/h 466.64 kg/h 80 ´ 14.58 18

= 64.8 kg/h H2SO4 produced by reaction (2) = 14.58 + 64.8 = 79.38 kg/h 194 ´ 350 = 1131.67 kg/h 60 . mass, mi kg/h 714.58 – 571.67 + 79.38 = 222.29 1633.24 – 466.64 – 64.8 = 1101.80 1131.67

Sulphamic acid produced = Component H2SO4 SO3 Sulphamic acid Total

2455.76

mass % 9.05 44.87 46.08 100.00

Ans.

EXERCISE 4.26 Basis: Liquid feed rate = 6.1 kg/h For 100 kmol liquid feed rate, following calculations are made which are converted for 6.1 kg/h feed rate in last column.

76 Solutions Manual—Stoichiometry

Liquid Feed Component

Molar mass

C10H20 C11H22 C12H24 C13H26 C14H28

140 154 168 182 196

Total

kmol kmol feed

kg

mass %

0.8 9.9 82.2 6.7 0.4

112.0 1 524.6 13 809.6 1 219.4 78.4

0.67 9.11 82.47 7.28 0.47

100.0

16 744.0

100.00

Average molar mass of liquid =

kg/h for 6.1 kg/h feed 0.040 0.555 5.030 0.444 0.028

87 71 67 08 67

6.100 00

16 744 100

= 167.44 Tetramer feed rate =

6.1 ´ 0.822 167.44

= 0.029 95 kmol/h Gas feed rate = 1 Nm3/h (saturated) = 0.044 61 kmol/h Vapour pressure of water at 50°C (323.15 K) = 12.335 kPa (Ref. Chatper 6) Feed gas pressure = 0.35 bar-g = 1.363 25 bar a Moisture in gas =

0.123 35 (1.363 25 - 0.123 35)

= 0.0995 kmol/kmol dry gas º 0.0905 kmol/kmol wet gas Dry feed as = 0.04461 (1 – 0.0905) = 0.04057 kmol/h Feed Gas (Dry) Component

mole fraction

kmol/h

H2S CO2 HC

0.967 0.013 0.020

0.039 23 0.000 53 0.000 81

Total

1.000

0.040 57

Material Balances Involving Chemical Reactions 77

Moisture in gas = 0.044 61 – 0.040 57 = 0.004 04 kmol/h Tetramer (C12H24) consumed = 0.029 95 kmol/h DDM produced = 0.029 95 kmol/h Molar mass of DDM = 202 DDM produced = 0.029 95 ´ 202 = 6.0499 kg/h Liquid Flowing Out of Reactor Component

kg/h

mass %

C10H20 C11H22 C13H26 C14 H28 C12H26S(DDM)

0.040 87 0.555 71 0.444 08 0.028 67 6.049 90

0.57 7.81 6.24 0.40 84.98

Total

7.119 23

100.00

Ans.(a) H2S consumed = 0.029 95 kmol/h BF3 introduced = 0.35 Nm3/h = 0.001 56 kmol/h Gas Stream Exit of Reactor Component

Kmol

mole %

H2S CO2 HC BF3

0.039 23 – 0.029 95 = 0.009 28 0.000 53 0.000 81 0.001 56

76.19 4.35 6.65 12.81

Total

0.012 18

100.00

Ans.(b) Dry gas/liquid feed = 0.040 57/(6.0/167.44) = 1.114 kmol/kmol Conversion of H2S = (0.029 95/0.039 23) 100 = 76.34 %

EXERCISE 4.27 Basis: 4 kmol FeS2 burnt Theoretical oxygen requirement = 11 kmol SO2 produced due to complete combustion (roasting) = 8 kmol Excess O2 supply = 11 ´ 0.6 = 6.6 kmol O2 is supplied from air. N2 entering with O2 = 79 (11 ´ 1.6)/21 = 66.21 kmol

Ans. (c) Ans.(d)

78 Solutions Manual—Stoichiometry

Gas

kmol

mole %

SO2 O2 N2

8.0 6.6 66.21

9.90 8.17 81.93

Total

80.81

100.00

Ans.

EXERCISE 4.28 Basis: 100 kg ZnS ore It contains 74 kg ZnS and 26% kg inerts. ZnS = [74/(65.4 + 32)] = 0.76 kmol Roasting reaction: 2 ZnS + 3 O2 = 2 ZnO + 2 SO2 Theoretical O2 requirement = 3 ´ 0.76/2 = 1.14 kmol Actual O2 supply = 1.55 ´ 1.14 = 1.767 kmol N2 entering with O2 = 79 ´ 1.767/21 = 6.647 kmol Total air = 1.767 + 6.647 = 8.414 kmol SO2 produced = 0.76 kmol Composition of burner gases: Component SO2 O2 N2 Total

kmol

mole %

0.76 (1.767 – 1.14) = 0.627 6.647

9.46 7.80 82.74

8.034

100.00

Ans. (a) SO3 produced in converter = 0.76 ´ 0.98 = 0.745 kmol Unconverted SO2 = 0.76 – 0.745 = 0.015 kmol O2 consumed = 0.745/2 = 0.373 kmol Unused O2 = 0.627 – 0.373 = 0.254 kmol Composition of converter exit gas mixture: Component

kmol

mole %

SO2 O2 SO3 N2

0.015 0.254 0.745 6.647

0.20 3.32 9.72 86.76

Total

7.661

100.00

Assume that all SO3 is absorbed. SO3 absorbed = 0.745 kmol H2SO4 produced = 0.745 kmol = 73.01 kg of 100% strength = 81.12 kg of 90% strength

Ans. (b)

Material Balances Involving Chemical Reactions 79

For 1000 kg/h 90% acid production, requirement of sulphide ore = 100 ´ 1000/81.12 = 1232.7 kg/h Ans. (c) Requirement of dry air = 8.414 ´ 1000/81.12 = 103.72 kmol/h Specific volume of air at 100 kPa and 300 K V = RT/p = 8.314 (300)/100 = 24.942 m3/kmol Volumetric flow rate of air = 103.72 ´ 24.942 = 2587 m3/h Ans. (d)

EXERCISE 4.29 Basis: 100 kg pyrites FeS2 content = 88 kg Roasting reaction: 4 FeS2 + 11 O2 = 2 Fe2O3 + 8 SO2 Theoretical O2 requirement = 11 ´ 88/(4 ´ 120) = 2.02 kmol Excess air = 150% Actual O2 supply = 2.02 ´ 2.5 = 5.05 kmol N2 entering with O2 = 79 ´ 5.05/21 = 19.0 kmol Total air supply = 19 + 5.05 = 24.05 kmol Sp. vol.of dry air to 100 kPa and 300 K = 24.942 m3/kmol Volumetric air supply = 24.05 ´ 24.942 = 600 m3 Ans. (d) New basis: 100 kg cinder Cinder contains 2.6 kg S. S present in form of FeS2 = 2.6 ´ 0.4 = 1.04 kg FeS2 content of cinder = 120 ´ 1.04/32 = 3.90 kg S present in the form of SO2 = 2.6 – 1.04 = 1.56 kg SO3 content of cinder = 80 ´ 1.56/32 = 3.90 kg (Fe2O3 + gangue) in cinder = 100 – (3.90 + 3.90) = 92.20 kg Let x be the amount of FeS2 roasted. Unburnt FeS2 = (88 – x) kg Fe2O3 produced = 160 ´ 2 ´ x/(120 ´ 4) = 0.667 x kg Sulphur free cinder = 0.667 x + 12 kg 88 – x FeS2 unburnt 3.90 = = S- free cinder 0.667 x + 12 92.20 x = 85.1 kg

80 Solutions Manual—Stoichiometry

Unburnt FeS2 Quantity of cinder Sulphur, lost in cinder Total sulphur charged % S, lost in the cinder Actual FeS2 roasted FeS2 roasted to SO2 FeS2 roasted to SO3 O2 consumed

= 88 – 85.1 = 2.9 kg = 2.9 ´ 100/3.9 = 74.4 kg Ans. (a) = 74.4 ´ 0.026 = 1.93 kg = 64 ´ 88/120 = 46.93 kg = 1.93 ´ 100/46.93 = 4.11% Ans. (b) = 85.1/120 = 0.709 kmol = 0.709 ´ 0.92 = 0.652 kmol = 0.709 – 0.652 = 0.057 kmol = [11 ´ 0.652/4] + [15 ´ 0.057/4] = 1.793 + 0.214 = 2.007 kmol SO2 produced = 8 ´ 0.652/4 = 1.304 kmol SO3 produced = 8 ´ 0.057/4 = 0.114 kmol Excess oxygen = 5.05 – 2.007 = 3.043 kmol Composition of burner exit gas mixture: Component

kmol

mole %

mole % on SO3 – free basis

SO2 SO3 O2 N2

1.304 0.114 3.043 19.000

5.56 0.49 12.97 80.98

5.59 — 13.03 81.38

Total

23.461

100.00

100.00

Ans. (c)

EXERCISE 4.30 Basis: 100 kmol fuel gases CO2 content of fuel gases = 12.8 kmol Let x kmol CO3 are added in the furnace due to decomposition of MgCO3. Total CO2 at the exit = 12.8 + x kmol Total flue gas mixture = 100 + x kmol The exit gas mixture contains 24% CO2 12.8 + x = 0.24, or x = 14.74 kmol Ans. (a) 100 + x Component CO2 O2 N2 Total

Molar mass 44 32 28

Incoming gases kmol kg 12.8 563.2 6.1 195.2 81.1 2270.8 100.0

3029.2

Outgoing gases kmol kg 27.54 1211.76 6.10 195.20 81.10 2270.80 114.74

3677.76

Avg. molar mass of dry incoming gases= 30.29 Avg. molar mass of dry outgoing gases = 3677.79/114.74 = 32.05

mole % 24.00 5.32 70.68 100.00

Ans. (b)

Material Balances Involving Chemical Reactions 81

EXERCISE 4.31 Basis: 100 kmol/h of feed to reactor There are 4 reactions and 7 reaction components. Hence total 11 equations can be written. Component balance equations: nB,in = nB,out - A1 + 2A 4 Benzene balance: nT,in = nT,out + A1 - A 2 nX,in = nX,out + A 2 - A 3

Toluene balance: Xylene balance: Pseudocumene balance: Methane balance:

nP,in = nP,out + A 3 nCH4 , in = nCH4 ,out - A1 - A 2 - A3

nH2 , in = nH2 ,out + A1 + A 2 + A3 - A 4 nDP2 , in = nDP,out - A 4 Diphenyl balance: Reaction performance equations: nC,B × nB,in = - A1 + 2A 4 nC,T × nT,in = - A1 - A 2

Hydrogen balance:

nC,X × nX,in = A 2 - A 3 nC,P × nP,in = A3

These eleven equations can be represented in matrix form as under. ù é n ù é nB,in ú é 1 0 0 0 0 0 0 -1 0 0 2 ù ê nB,0 ú ê nT,in ú ê 0 1 0 0 0 0 0 1 -1 0 0 ú ê T,0 ú ê n ú ê 0 0 1 0 0 0 0 0 1 -1 0 ú ê nX,0 ú ê X,in ú ê 0 0 0 1 0 0 0 0 0 1 0 ú ê nP,0 ú ê nP,in ê 0 0 0 0 1 0 0 -1 -1 -1 0 ú ê nCH ,0 ú ê nCH 4 ,in ú ú ê 0 0 0 0 0 1 0 -1 -1 -1 1ú ê n 4 ú = ê nH ,in ú ê 0 0 0 0 0 0 1 0 0 0 -1ú ê H2 ,0 ú ê n 2 ú ê 0 0 0 0 0 0 0 -1 0 0 2 ú ê nDP,0 ú ê DP,in f n ê 0 0 0 0 0 0 0 2 -1 0 0 ú êA1 ú ê C,B× B,in ú ê 0 0 0 0 0 0 0 0 1 -1 0 ú êA 2 ú ê f C,T × nT,in ú ê 0 0 0 0 0 0 0 0 0 1 0 ú êA 3 ú ê f C,X × nX,in ú êë úû ëêA 3 ûú êëê f C,× nP,in úûú é1 ê0 ê0 ê0 ê0 ê0 ê0 ê0 ê0 ê0 ê0 ëê

0 1 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

é n ù 0 -1 0 0 2 ù ê B,0 ú é 1 ù n T,0 0 1 -1 0 0 ú ê ê 3.67 ú ú ú 0 0 1 -1 0 ú ê nX,0 ú ê 6 0 0 0 1 0 ú ê nP,0 ú ê 7.34 ú ú 0 -1 -1 -1 0 ú ê nCH4 ,0 ú ê 0 0 1 1 1 -1ú ê nH ,0 ú = ê81.99 ú ú 1 0 0 0 -1ú ê n 2 ú ê 0 0 -1 0 0 2 ú ê DP,0 ú ê 0.2 ú ú ê 2.5 ú 0 2 -1 0 0 ú êA1 ú ê 1.02 ú 0 0 1 -1 0 ú êA 2 ú A ê ú êëê 5.138úûú 3 0 0 0 1 0 ûú êëA 4 úû V×0 = I 0 = V–1 I

82 Solutions Manual—Stoichiometry

EXERCISE 4.32 Basis: Liberation of 127 g iodine Electrochemical reactions: KI ¾ ¾¾® K + I 1 I - - e ¾ ¾¾® I 2 2 Equivalent mass of iodine = 126.9 Let I = current passed through the cell, amperes Total Faradays passed = (I ´ 3600 ´ 3)/96 485

Material Balances Involving Chemical Reactions 83

I ´ 3600 ´ 3 ´ 126.9 = 127.0 96 485 I = 8.95 amperes.

Iodine liberated =

Ans.

EXERCISE 4.33

Basis: 2 m3 oxygen liberation at NTP Molar liberation = 2/22.414 = 0.0892 kmol º 2855.4 g Electrochemical reactions:

AgNO3 ¾ ¾¾® Ag+ + NO 3Ag+ + e ¾ ¾¾® Ag NO 3- – e ¾ ¾¾® NO3

NO3 + H2O ¾ ¾¾® HNO3 + OH– 2 OH– – 2e ¾ ¾¾® H2O + 1/2 O2 1 equivalent of oxygen = 2 equivalents of Ag Ag deposited = 108 ´ 2855.4/8 = 38 548 g º 38.548 kg Ans. (a) Total Faradays passed through the solution = 1130 ´ 3600 ´ 9/96 485 = 379.457 Theoretical Ag liberated = 379.457 ´ 108 = 40 981 g Current efficiency = 38 548 ´ 100/40 981 = 94.1% Ans. (b)

EXERCISE 4.34 Let

m = mass of 0.1 N HCl solution to be added, g H = density of resulting solution, g/mL

Balance of H+ ion:

-3 é (1500 ´ 1.25 ´ 103 ) + m ù ´ m 10 û ´ 10 -4 1 1500 ´ 10 + ´ 10 = ë 1.005 H ´ 103 (1875 000 + m) 1.5 ´ 10–6 + 9.95 ´ 10–5 m = ´ 10-7 H pH of 0.1 N HCl solution = –log (0.1) = 1 1 H = w å i Hi 1 H = æ 1875 000 ö 1 æ ö 1 m çè 1875 000 + m ÷ø 1.25 + çè 1875 000 + m ÷ø 1.005

–9

=

1875 000 + m 1500 000 + 0.995 m

(1)

(2)

84 Solutions Manual—Stoichiometry

Solving Eq. (1) and Eq. (2), m = 1509 g

Ans.

EXERCISE 4.35 Basis: 2 kmol of Ca(OH)2 to be recarbonated Reactions:

Ca(OH)2 + CO2 = CaCO3 + H2O

(1)

Ca(OH)2 + 2 CO2 = Ca(HCO3)2

(2)

Ca(OH)2 + Ca(HCO3)2 = 2 CaCO3 + 2H2O

(3)

Let x be the kmol of Ca(OH)2 bypassed. Ca(OH)2 reacted as per reaction (2) = (2 – x) kmol CO2 required = 2 (2 – x) kmol Total CO2 required = 2 kmol even after reaction 2 (2 – x) = 2 or x = 1 kmol Bypass total feed = 1 ´ 100/2 = 50%

Ans.

EXERCISE 4.36 Basis: 100 kmol/s fresh feed. This is the same basis as that of Example 4.17. Case (a): Conversion per pass = 24%, ammonia separation = 65% Case (b): Conversion per pass = 25%, ammonia separation = 70% Case (a) N2 reacted, kmol/s H2 reacted, kmol/s NH3 produced, kmol/s Total gas mixture, leaving the converter, kmol/s NH3 in outlet gas mixture, kmol/s NH3 separated, kmol/s NH3 uncondensed, kmol/s Composition of gas mixture, leaving separator, kmol/s Component N2 H2 NH3 Inerts Total

0.24 a 0.72 a 0.48 a M – 0.48 a

Case (b) 0.25 a 0.75 a 0.50 a M – 0.5 a

0.9 M – 3.52 a

0.9 M – 3.5 a

0.585 M – 2.288 a 0.315 M – 1.232 a

0.63 M – 2.45 a 0.27 M – 1.05 a

0.76 a 2.28 a 0.315 M – 1.232 a 0.1 M

0.75 a 2.25 a 0.27 M – 1.05 a 0.1 M

0.415 M + 1.808 a

0.37 M + 1.95 a

Material Balances Involving Chemical Reactions 85

Case (a): Purge balance:

0.1 M P = 1.0 0.415 M + 1.808 a Recycle stream = 0.415 M + 1.808 a – P Substituting the value in Eq. (1) of Example 4.15, 100 + 0.415 M + 1.808 a – P = M Nitrogen balance: Nitrogen lost in purge =

(3)

0.76 a P kmol/s (0.415 M + 1.808 a)

Nitrogen in recycle stream = 0.76 a – Hence 0.76 a –

(2)

0.76 a P kmol/s (0.415 M + 1.808 a)

0.76 a P + 24.75 = a (0.415 M + 1.808 a)

0.415 M + 1.808 a = 0.1 MP from Eq. (2) 0.76 a P + 24.75 = 1 0.75 a – 0.1 M P 24.75 M a= 0.24 M + 7.6 4.15 M + 18.08 a From Eq. (2), P= M Substitute values of a and P in Eq. (3) and simplify. 0.1404 M2 – 63.306 M – 280.98 = 0 Mathcad solution: f (M ) := 0.140 M2 – 63.306 M – 280.98 M := 350 soln := root (f(M), M) soln = 455.293 or M = 455.293 kmol/s

(4)

Substitute

24.75 ´ 455.293 0.24 ´ 455.293 + 7.6 = 96.419 kmol/s 4.15 ´ 455.293 + 18.08 ´ 96.419 P= 455.293 = 7.971 kmol/s

(5)

a=

Case (b) Purge balance:

0.1 MP = 1.0 0.37 M + 1.95 a

Ans.

(2)

86 Solutions Manual—Stoichiometry

recycle stream = 0.37 M + 1.95 a – P or 100 + 0.37 M + 1.95 a – P = M (3) Nitrogen balance: 0.75 a P kmol/s Nitrogen lost in purge = (0.37 M + 1.95 a) 0.75 a P kmol/s Nitrogen in recycle stream = 0.75 a – (0.37 M + 1.95 a) Hence, 0.75 a –

0.75 a P + 24.75 = a (0.37 M + 1.95 a)

Substitute

(4)

0.37 M + 1.95 a = 0.1 M P from Eq. (2) 0.75 a –

0.75 a P + 24.75 = a 0.1 M P a=

24.75 M unchanged from Example 4.14 0.25 M + 7.5

3.7 M + 19.5 a M Substitute values of a and P in Eq. (3) and simplifying, 0.1575 M2 – 67.6125 M – 239.625 = 0 Mathcad solution: f (M) := 0.1575 M2 – 67.6125 M – 239.625 M := 300 soln := root (f(M), M) soln := 432.801 or M = 432.801 kmol/s

From Eq. (2),

P=

a=

24.75 ´ 432.801 0.25 ´ 432.801 + 7.5

= 92.5826 kmol/s 3.7 ´ 432.801 + 19.5 ´ 92.5826 432.801 = 7.8713 kmol/s Note: Exercise 9.3 is Mathcad exercise for this exercise.

P=

Ans.

EXERCISE 4.37 Since simultaneous equations cannot be easily solved, program, developed in exercise 9.7 is used. In this program value of i is changed, keeping all other perameters constant. (a) For i = 0.09 P = 11.951 kmol/s (b) For i = 0.11, P = 9.162 kmol/s.

Material Balances Involving Chemical Reactions 87

Using values of M, P and a from the program, ammonia productions can be calculated .

EXERCISE 4.38 Basis: Make-up hydrogen stream = 45 000 Nm3/h H2 content of make-up stream = 45 000 ´ 0.885 = 39 825 Nm3/h H2 consumed in hydrocraker unit = H2 in make-up stream – H2 lost in HP purge – H2 lost in LP purge HP purge stream = 4500 Nm3/h H2 lost in HP purge = 4500 ´ 0.745 = 3352.5 Nm3/h LP purge stream = 12 000 Nm3/h H2 lost in LP purge = 12 000 ´ 0.517 = 6462.5 Nm3/h H2 consumed in hydrocracking unit = 39 825 – 3352.5 – 6462.5 = 30 010 Nm3/h Membrane System I: Hydrocarbons (HC) in HP purge = 4500 – 3352.5 = 1147.5 Nm3/h Membrane system removes 85% HC in the reject stream I. HC in Reject I = 1147.5 ´ 0.85 = 975.375 Nm3/h HC in Permeate I = 1147.5 – 975.375 = 172.125 Nm3/h Permeate I constains (1 – 0.945 =) 0.055 mole fraction HC. Permeate I flow rate = 172.125/0.055 = 3129.545 Nm3/h H2 recycled in Parmeate I = 3129.545 – 172.125 = 2957.42 Nm3/h Since fresh make-up has 88.5 % H2, reduction in make-up stream = = 3341.71 Nm3/h

Ans.(a)

2957.42 0.885

88 Solutions Manual—Stoichiometry

% reduction =

3341.71 ´ 100 45 000

= 7.43

Ans.(b)

Membrane System II: HC in LP purge = 12 500 – 6462.5 = 6037.5 Nm3/h Membrane system removes 88% HC. HC in Reject II = 6037.5 ´ 0.88 = 5313 Nm3/h HC in Permeate II = 6037.5 – 5313 = 724.5 Nm3/h Permeate II contains (1 – 0.92 = ) 0.08 amole fraction HC. Permeate II flow rate =

724.5 = 9056.25 Nm3/h 0.08

H2 recycled in Permeate II = 9056.25 – 724.5 = 8331.75 Nm3/h Reduction in make-up stream =

8331.75 0.88

= 9414.41 Nm3/h % reduction =

9414.41 ´ 100 45 000

= 20.92 Ans.(c) Note: Reduction in make-up stream will result in same reduction in H 2 consumption.

EXERCISE 4.39 Basis: 1 litre feed water Water going to C II should be 1 litre, irrespective of the recycled water. Sodium content also should be equivalent to that of feed water. Na content of feed water = C – H meq Na recycled in the water from AI = X (C – H) meq where X = litres of water recycled/litre feed Recycled Na is in the form of bicarbonates which means that alkalinity to the extent of X (C – H) meq is added to feed water. Total alkalinity in mixed feed to CI = M + X (C – H) meq Total alkalinity should be equal to H to convert all hardness causing cations (which are strong in nature) to bicarbonates so that the same can be removed in CI.

Material Balances Involving Chemical Reactions 89

M + X (C – H) = H X = (H – M)/(C – H) Because of alkalinity recycle, CO2 availability at the exit of CI is H meq. However, it is desired to convert all the strong anions (such as SO4) to bicarbonates. Strong anions to be converted to HCO-3 = A – H meq = CO2 requirement from degasser (D) Now CO2 content of the product water from D should equal to alkalinity in the feed water, irrespective of recycle. Recycled CO2 = y × M meq yM=A–H y = (A – H)/M

F H

(1 + x) (1 + y) = 1 +

H - M C- H

I F1 + A - H I KH M K

=

(C - M )( M + A - H ) (C - H ) M

=

(C - M )(C - H ) (C - H ) M

because M + A = C

=

C- M M

because C – M = A

=

A M

EXERCISE 4.40 Let R kmol/h of dry tail gas recycled. Moisture in recycle stream = 0.1146 R kmol/h Let F kmol/h be the flow of dry ambient air. Moisture in ambient air = 0.017 72 F kmol/h Since total molar flow rate of gas mixture to the reactor is unchanged, 1.1146 R + 1.017 72 F = 1363.1 (1) Let yO2 , yUC and yN 2 be the mole fractions of oxygen, undesired components (i.e. CO + CO2 + H2 + CH4 + Ether) and nitrogen respectively in the dry recycle stream. yO 2 + yUC+ yN 2 = 1 on dry basis

(2)

In the total feed to the reactor, oxygen concentration is 10 mole %. 0.21 F + R × yO2 = 1488.1 ´ 0.1 = 148.81 Nitrogen, entering in air = 0.79 F N2 in recycle stream = R × yN 2 = R – R × yO2 – R × yUC

(3)

90 Solutions Manual—Stoichiometry

Since reaction pattern is unchanged, addition of UC = 13.365 kmol/h UC in tail gas from absorber = R × yUC + 13.365 kmol/h Total oxygen consumption = 68.559 kmol/h O2 in tail gas from absorber = 0.21 F + R × yO2 – 68.559 = 148.81 – 68.559 = 80.251 kmol/h

[from Eq. (3)]

N2 in tail gas from absorber = R – R × yO 2 – R × yUC+ 0.79 F Total tail gas from absorber = R × yUC + 13.365 + 0.21 F + R × yO 2 – 68.559 + R – R × yO2 – R × yUC + 0.79 F = R + F – 55.194 kmol/h

Balance of oxygen:

80.251 = yO 2 R + F - 55194 . Multiply Eq. (3) by

(4)

1.017 72 = 4.8463 0.21

1.017 72 F + 4.8463 R × yO2 = 721.78

(5)

Substract Eq. (5) from Eq. (1) 1.1146 R – 4.8463 R × yO2 = 641.32 or or

R – 4.348 R × yO2 = 575.381 R × yO2 = 0.23 R – 132.332

From Eq. (1), F = 1339.37 – 1.0952 R Combining Eq. (4) and Eq. (6), 80.251 0.23 R - 132.332 = R + F - 55194 . R Substitute value of F form Eq. (7) and simplify. 0.0219 R2 – 227.567 R + 169 937.6 = 0 Solution of Eq.(8) yields R = 809.9 kmol/h Substitute value of R in Eq. (6). R × yO2 = 0.23 ´ 809.9 – 132.332 = 53.94 kmol/h yO 2 = 0.0666 or 6.66% O2 in recycle.

Substitute value of R in Eq. (7). F = 1339.37 – 1.0952 ´ 809.9 = 452.37 kmol/h

(6) (7)

(8)

Material Balances Involving Chemical Reactions 91

From Eq. (2),

yN 2 + yUC = 1.0 – yO 2 = 0.9334

UC Balance: R × yUC + 13.365 = R + F - 55.194 R + F – 55.194 = = 809.9 yUC + 13.365 =

yUC

(9)

809.9 + 452.37 – 55.194 1207.076 kmol/h 1207.076 yUC

yUC = 0.0337 or 3.37% in recycle. yN 2 = 0.9334 – 0.0337 = 0.8997

or 89.97% in recycle Compositions of tail gas and recycle gas will be same. Composition of Gaseous Mixture, Entering Reactor Component CH3OH O2 N2 H2O UC

kmol 125.00 148.81 452.37 ´ 0.79 + 809.9 ´ 0.8997 = 1086.04 452.3 ´ 0.017 72 + 809.9 ´ 0.1146 = 100.84 809.9 ´ 0.0337 = 27.30

Total

1487.99

mole % (wet basis) 8.40 10.00 72.99 6.78 1.83 100.00

Composition of Gaseous Mixture Leaving Reactor Component CH3OH HCHO UC O2 N2 H2O Total

mole %

kmol

Wet

1.25 111.375 27.30 + 13.365 = 40.665 80.251 1086.04 100.84 + 129.937 = 230.779 1550.358

0.08 7.18 2.62 5.18 70.05 14.89 100.00

Dry tail gas — — 3.37 6.65 89.98 — 100.00

Purge stream must contain 13.365 kmol/h UC. Total purge stream =

13.365 = 396.6 kmol/h (dry) 0.0337

Recycle stream 809.9 = Purge stream 396.6

92 Solutions Manual—Stoichiometry

= 2.042 kmol/kmol Recycle ratio =

Ans. (a)

Recycle stream 809.9 ´ 1.1146 = 452.37 ´ 1.017 72 Fresh air stream

= 1.96 kmol/kmol

Ans. (b)

Total dry tail gas stream from absorber = 1550.358 – (1.25 + 111.375 + 230.779) = 1206.954 kmol/h 40.665 ´ 100 1206.954 = 3.37%

Concentration of UC in tail gas =

(Check !)

Ans. (c)

EXERCISE 4.41 Basis: 100 kmol benzene feed Let fresh H2 = a kmol Recycle H2 stream = b kmol Recycle cyclohexane stream = c kmol Mixed feed, entering the reactor = 100 + a + b + c kmol Inerts in the mixed feed = 0.1 (100 + a + b + c) = 10 + 0.1a + 0.1b + 0.1c kmol Benzene concentration in mixed feed = 18.5 mole % 100 = 0.185 100 + a + b + c 18.5 + 0.185a + 0.185b + 0.185c = 100 0.185a + 0.185b + 0.185c = 100 – 18.5 = 81.5 0.9a + 0.9b + 0.9c = 396.486 (1) Hydrogen in the mixed feed = 100 + a + b + c – 10 – 0.1a – 0.1b – 0.1c – c – 100 = 0.9a + 0.9b – 0.1c – 10 kmol H2/benzene = 3.3 0.9a + 0.9b + 0.1c – 10 = 3.3 ´ 100 = 330 0.9a + 0.9b + 0.1c = 340 (2) Eq. (1) – Eq. (2) yields c = 56.486 kmol 0.9b = 340 + 0.1c – 0.9a = 345.645 – 0.9a or b = 384.054 – a (3) The conversion of benzene is 100%. Therefore 100 kmol benzene and 300 kmol H2 are consumed.

Material Balances Involving Chemical Reactions 93

Cyclohexane produced = 100 kmol All the cyclohexane is condensed in the cooler. Cooler exit gas mixture: H2 = 0.9a + 0.9b – 0.1c – 10 – 300 = 0.9a + 0.9b – 0.1c – 310 kmol Inerts = 10 + 0.1a – 0.1b + 0.1c kmol Total gas mixture = a + b – 300 kmol Let P be kmol of purge gas mixture. Since the compositions of purge gas mixture and the recycle gas mixture are same, (0.1a + 0.1b + 0.1c + 10) P purge gas will also contain inerts = ( a + b - 300) This should be equal to the inerts in the fresh hydrogen stream. (0.1a + 0.1b + 0.1c + 10) P = 0.025 a ( a + b - 300) 0.025a( a + b - 300) (0.1a + 0.1b + 0.1c + 10) Recycle hydrogen stream = a + b – 300 – P =b or a – P = 300 or P = a – 300 Hydrogen balance:

P=

F 0.9a + 0.9b - 0.1c - 310 I P + 300 = (1 – 0.025) a H a + b - 300 K

(4)

= 0.975 a But from Eq. (2), 0.9a + 0.9b – 0.1c – 310 = 30 ( 0.975a - 300)( a + b - 300) P= 30 = a – 300 from Eq. (5) 2 0.975a + 0.975ab – 292.5 a – 300a – 300b + 90 000 = 30a – 9000 or a2 + ab – 638.46a – 307.69b + 101 538 = 0 (6) Substitute the value of b from Eq. (3) into Eq. (6). a2 + a(384.054 – a) – 638.46a – 307.69 (384.054 – a) + 101 538 = 0 a2 + 383.054a – a2 – 638.46a – 118 170 + 307.69a + 101 538 =0 53.284 a = 16 632 a = 312.139 kmol b = 384.054 – 312.139 = 71.915 kmol P = 312.139 – 300 = 12.139 kmol Ans.

94 Solutions Manual—Stoichiometry

EXERCISE 4.42 Hydrogen in the reactor effluent = 30 kmol Recovery of H2 = 0.9 ´ 30 = 27 kmol Inerts in the purge = 0.025 a kmol Inerts in the cooler exit gas mixture = (0.025a/0.95) = 0.0263a kmol Inerts in the recycle stream = (0.0263 – 0.025) a = 0.0013a kmol Recycle hydrogen stream = 27 + 0.0013a =b or b – 0.0013 a = 27 Make-up hydrogen requirement = 330 – 27 = 303 kmol Inerts entering with make-up H2 = 303 ´ 2.5/97.5 = 7.769 kmol a = 303 + 7.769 = 310.769 kmol b = 27 + 0.0013 ´ 310.768 = 27.404 kmol Inerts in the reactor inlet = 0.0263 a = 8.137 kmol Total mixed feed = 100/0.185 = 540.541 kmol Recycle cyclohexane = 540.541 – 330 – 8.173 – 100 = 102.367 kmol Inerts content of mixed feed = 8.173 ´ 100/540.541 = 1.51 mole % Purge, P = 30 + 8.173 – 27 – 0.404 = 10.769 kmol

EXERCISE 4.43 Basis: 3500 kg/h ethylene oxide (EO) production rate Only first two reactions are considered. Compound Ethylene (C2H4) Oxygen (O2) Ethylene oxide (C2H4O) Carbon dioxide (CO2) Water (H2O)

Molar mass 28.0538 31.9988 44.0532 44.0098 18.0153

Ethylene required for EO production = 28.0538 ´ 3500/44.0532 = 2228.86 kg/h º 79.45 kmol/h

Ans.

Material Balances Involving Chemical Reactions 95

Since the yield of EO is 70%, total ethylene reacted = 2228.86/0.7 = 3184.08 kg/h Ethylene converted to CO2 and H2O = 3184.08 – 2228.86 = 955.22 kg/h Since the conversion of ethylene is 50% per pass, ethylene fed to the reactor = 3184.08/0.5 = 6368.16 kg/h º 227.000 kmol/h Total mixed feed to the reactor = 227.0/0.1 = 2270 kmol/h Inerts in the mixed feed = 2270 ´ 0.1 = 227.0 kmol/h O2 consumed for reaction (1) = 227.0 ´ 0.5 ´ 0.7/2 = 39.725 kmol/h O2 consumed for reaction (2) = 227.0 ´ 0.5 ´ 0.3 ´ 3 = 102.15 kmol/h Total O2 consumed = 39.725 + 102.15 = 141.875 kmol/h O2 concentration of the mixed feed = 141.875 ´ 100/2270 = 6.25% Ans. (d) 2 ´ 102.15 = 68.1 kmol/h 3 = H2O produced Inerts (N2 + Ar) entering with O2 = (102.15 + 39.725) ´ 0.03/0.97 = 4.39 kmol/h

CO2 produced =

Component Ethylene Oxygen Inerts (N2 + Ar) CO2 + H2O Ethylene oxide Total

Mixed feed to reactor (M), kmol/h 227.000 141.875 227.000 1674.125 (by balance) Nil 2270.00

Reactor effluent, kmol/h 113.500 Nil 227.000 1674.125 + 68.10 ´ 2 = 1810.325 3500/44.0532 = 79.450 2230.275

Let recycle stream R1 contain x kmol/h of (CO2 + H2O). Ethylene and inerts content are 113.50 and 227.00 kmol/h, respectively. Total stream R1 will amount to (340.5 + x) kmol/h. Let purge rate be P kmol/h.

96 Solutions Manual—Stoichiometry

x (340.5 + x - P) (340.5 + x ) = 1674.125

CO2 + H2O in stream R2 =

Simplifying, x2 – 1333.625 x – x P – 570 040 = 0

(1)

227.00 P (340.5 + x ) = 4.39 kmol/h or P = 6.585 + 0.019 34x. Substitute value of P in Eq. (1). x2 – 1333.625 x – x (6.585 + 0.019 34x) – 570 040 = 0 or 0.980 66x2 – 1340.21 x – 570 040 = 0 Solving, x = 1707.14 kmol/h

Inerts in purge =

Streams, kmol/h R2

(2)

Component

R1

Purge

Ethylene Inerts CO2 + H2O Oxygen

113.50 227.00 1707.14 Nil

2.194 4.392 33.015 Nil

111.306 222.608 1674.125 Nil

Fresh feed (M – R2) 115.694 4.392 Nil 141.875

Total

2047.64

39.601

2008.039

261.961

Recycle ratio = 2008.039/261.961 = 7.665 kmol/kmol Ans. (c) Ethylene feed rate = 115.694 ´ 28.0538 = 3245.66 kg/h Ans. (a) Oxygen feed rate = (141.875 + 4.39) 22.414 = 3277.8 Nm3/h Ans. (b)

EXERCISE 4.44 Basis: 100 kmol fresh ethylene feed Feed consists of 96 kmol of ethylene and 4 kmol of non-reactive gases (NRG). Let R and C be kmol of (ethylene + NRG) in the recycle stream and combined feed, respectively. Overall materials balance at A: F+R=C (1) If x is the mole fraction of ethylene in recycle stream, ethylene balance fields: 0.96 ´ 100 + x R = 0.85 C (2) Reactor inlet stream: Component Ethylene Water

kmol 0.85 C 0.85 ´ 0.65 C = 0.5525 C

Material Balances Involving Chemical Reactions 97

NRG

0.15 C

Total

1.5525 C

Conversion per pass Ethylene reacted Water reacted Ethanol produced Reactor outlet stream: Component Ethylene NRG Ethanol Total

= 5%, based on ethylene = 0.05 ´ 0.85 C = 0.0425 C kmol = 0.0425 C kmol = 0.0425 C kmol kmol

(0.85 – 0.0425) C = 0.8075 C (0.5525 – 0.0425) C = 0.51 C 0.15 C 0.0425 C 1.51 C

In the scrubber, it is assumed that all the water and alcohol are removed: Scrubber outlet off-gases: Component

Ethylene NRG Total

kmol

0.8075 C 0.1500 C 0.9575 C

Mole fraction of NRG in off gases = x = 0.15 C/0.9575 C = 0.1567 Let P be kmol of purge gas. NRG in purge = NRG in fresh feed 0.1567 P = 4 P = 4/0.1567 = 25.526 kmol R = 0.9575 C – 25.526 kmol Substitute the value of R in Eq. (2). 96 + (0.9573 C – 25.526) (1 – 0.1567)= 0.85 C Solving the equation, C = 1752.33 kmol R = C – 100 = 1652.33 kmol Recycle ratio = R/F = 1652.33/100 = 16.52 kmol/kmol fresh feed Ans. (a) Ethylene in the purge = 25.526 – 4.0 = 21.526 kmol Unreacted ethylene = 0.8075 C = 0.8075 ´ 1752.33 = 1415.00 kmol % loss of ethylene = 21.526 ´ 100/1415 = 1.52 Ans. (c)

98 Solutions Manual—Stoichiometry

Component

C2 H 4 H2O NRG C2H5OH

Gas mixture, incoming to reactor kmol mole % 1489.41 54.83 968.12 35.64 258.92 9.53 Nil Nil

Gas mixture, outgoing from reactor kmol mole % 1414.94 53.56 893.65 33.82 258.92 9.80 74.47 2.82

Total

2716.45

2641.98

100.00

100.00

Ans. (d) and (e)

EXERCISE 4.45 Basis: 100 kmol mixed feed Carbon input to the reactor = 4 ´ 100 = 400 kmol New basis: 100 kmol reactor effluent gases Carbon content = 3.25 + 1.16 ´ 2 + 1.08 ´ 2 + 1.32 ´ 3 + 0.64 ´ 3 + 2.5 ´ 4 + 4.93 ´ 4 + 17.88 ´ 4 + 30.91 ´ 4 + 8.78 ´ 4 + 0.16 ´ 5 + 5.34 ´ 1 = 279.75 kmol for 100 kmol reactor offluent stream Therefore for 100 kmol mixed feed (i.e. original basis), reactor effluent = 400 ´ 100/279.75 = 143 kmol Hydrogen in reactor effluents = 143 ´ 0.2205 = 31.53 kmol Hydrogen is present in the fuel gas. Quantity of fuel gas = 31.53/0.7357 = 42.86 kmol Balance of butadiene: Butadiene in the reactor effluent = 143 ´ 0.0878 = 12.56 kmol Butadiene in the fuel gas = 42.86 ´ 0.0024 = 0.1 kmol Butadiene in the product mixture = 12.56 – 0.1 – (100 ´ 0.34)/100 = 12.12 kmol Total quantity of product mixture = 12.12/0.9831 = 12.33 kmol

Material Balances Involving Chemical Reactions 99

n-butylene in product mixture = 12.33 – 12.12 = 0.21 kmol i.e.1.69% Coke in reactor affluents = 5.34 ´ 143/100 = 7.64 kmol This coke will deposit on the catalyst and will provide heat to the catalyst bed during regeneration. Leftover recycle stream = 143 – 12.33 – 7.64 – 42.86 = 80.17 kmol Fresh feed = 100 – 80.17 = 19.83 kmol Recycle ratio = recycle stream/fresh feed = 80.17/19.83 = 4.043 kmol/kmol Ans. (a) Component

Fresh feed mole %

i-butane i-butylene n-butylene n-butane 1,3-butadiene Total

Mixed feed

kmol

kmol

Recycle stream kmol

mole %

1.5 Nil Nil 98.5 Nil

0.297 Nil Nil 19.533 Nil

3.84 6.94 25.29 63.59 0.34

3.543 6.940 25.290 44.057 0.340

4.42 8.66 31.55 54.95 0.42

100.0

19.830

100.00

80.17

100.00

Total n-butane consumed = = Total butadiene produced = = º

100 ´ 0.6359 – 143 ´ 0.3091 19.39 kmol 143 ´ 0.0878 – 0.34 12.215 kmol 12.215 kmol n-butane consumed for butadiene production Yield of butadiene = 12.215 ´ 100/19.39 = 63.00% Ans. (c)

EXERCISE 4.46 Basis: 265 000 m3/h at 448.15 K (175°C) flue gases SO2 content of the flue gases = 1160 mg/m3 ´ 265 000 m3/h 1 kg/mg 10 6 = 307.4 kg/h SO2 reacted = 307.4 ´ 0.9 = 276.66 kg/h Na2CO3 + SO2 = Na2SO3 + CO2 Na2CO3 reacted = 106 ´ 276.66/64 = 458.22 kg/h

´

Reaction:

100 Solutions Manual—Stoichiometry

Na2SO3 formed = 126 ´ 276.66/64 = 544.67 kg/h CO2 formed = 458.22 + 276.66 – 544.67 = 190.21 kg/h Specific volume of ingoing flue gases = RT/p = 8.314 (448.15)/106.6 = 34.94 m3/kmol Molar flow rate of flue gases = 265 000/34.94 = 7584.43 kmol/h Water vapours in flue gases = 7600 kg/h = 422.22 kmol/h Dry gas mixture flow rate = 7584.43 – 422.22 = 7162.21 kmol/h Dry gas mixture flow rates at inlet and outlet remain same as equal moles of CO2 are produced in place SO2. SO2 in outgoing gases = 307.4 – 276.66 = 30.74 kg/h Vapour pressure of water at 323 K (50°C), pw = 12.335 kPa Water vapours in outgoing gases = 12.335/(106.6 – 12.335) = 0.1309 kmol/kmol dry gas Total water vapours in outgoing flue gas = 7162.21 ´ 0.1309 = 937.53 kmol/h Water, evaporated in absorber = 937.53 – 422.22 = 515.31 kmol/h º 9275.6 kg/h Moist flue gases, leaving the absorber = 7162.21 + 937.53 = 8099.74 kmol/h Stoichiometric Na2CO3 requirement = 106 ´ 307.4/64 = 509.13 kg/h Actual Na2CO3 supply = 1.5 ´ 509.13 = 763.7 kg/h Excess Na2CO3 over actual reacted = 763.7 – 458.22 = 305.48 kg/h Total solids in purge stream = 544.67 + 305.48 = 850.15 kg/h Total solids should be 8% in purge stream. Flow rate of purge stream = 850.15/0.08 = 10 626.88 kg/h Water in purge stream = 10 626.88 – 850.15 = 9776.73 kg/h Make-up Na2CO3 solution = 763.7/0.3 = 2545.7 kg/h Water in the make-up solution = 2545.7 – 763.7 = 1782 kg/h Dilution water requirement = 9275.6 + 9776.7 – 1782 = 17 270.3 kg/h Ans.

EXERCISE 4.47 Rock phosphate contains 34.55 kg P2O5. Ca3(PO4)2 = 3 CaO + P2O5 310

3 ´ 56

142

Material Balances Involving Chemical Reactions 101

Equivalent Ca3(PO4)2 in rock = 75.43 kg HCl fed = 3 ´ 100 = 300 kg of 25% concentration = 75 kg 100% HCl = H3PO4 + 3 CaCl2 Reaction: Ca3(PO4)2 + 6 HCl Theoretical HCl requirement = [6 ´ 36.5/310] (75.43) = 53.29 kg Ans. (a) Excess acid = (75 – 53.29) 100/53.29 = 40.74% Reaction of CaF2 : CaF2 + 2 HCl = CaCl2 + 2 HF 78

2 ´ 36.5

111

HCl required for this reaction = 2 ´ 36.5 ´ 2.5/78 = 2.34 kg CaCl2 produced = 111 ´ 2.5/78 = 3.558 kg Actual conversion of the phosphate = 75.43 ´ 0.95 = 71.66 kg Ca3(PO4)2 HCl consumed = 6 ´ 36.5 ´ 71.66/310 = 50.62 kg CaCl2 produced = 3 ´ 111 ´ 71.66/310 = 76.98 kg Balance CaO in the rock = 48.62 – 40.88 = 7.74 kg CaCl2 produced by the reaction of CaO and HCl = 7.74 ´ 111/56 = 15.34 kg Total CaCl2 produced = 76.98 + 3.558 + 15.34 = 95.880 kg H3PO4 produced = 2 ´ 98 ´ 71.66/310 = 45.31 kg Recovery of H3PO4 by leaching = 45.31 ´ 0.97 = 43.95 kg Thus for 43.95 kg H3PO4 production, rock phosphate consumption is 100 kg. For 2 t/h of 80 % H3PO4 production, rock phosphate requirement = 2000 ´ 0.8/0.4395 = 3640.5 kg/h º 3.64 t/h Ans. (b) CaCl2 produced = 95.88 ´ 2.0 ´ 0.8/0.4395 = 3.491 t/h Ans. (c)

EXERCISE 4.48 Basis: Ammonia flow to burner = 3266 Nm3/h Molar flow rate = 3266/22.414 = 145.71 kmol/h Volumetric flow rate of air = 29 394 Nm3/h (dry) Molar flow rate of dry air = 29 394/22.414 = 1311.41 kmol/h O2 supply to burner = 1311.41 ´ 0.21

102 Solutions Manual—Stoichiometry

= 275.40 kmol/h N2 supply to burner = 1311.41 – 275.4 = 1036.01 kmol/h Let x be kmol of NH3 reacted as per reaction (1). NO formed = x kmol/h O2 consumed = 1.25x kmol/h H2O produced = 1.5x kmol/h NH3 reacted as per reaction (2) = (145.71 – x) kmol/h N2 formed = 0.5 (145.71 – x) = 72.86 – 0.5x kmol/h O2 consumed as per reaction (2) = (3/4) (145.71 – x) = 109.28 – 0.75x kmol/h H2O produced = 1.5 (145.71 – x) = 218.57 – 1.5x kmol/h Composition of reactor effluent: Component NO O2

kmol/h

H2O

x 275.4 – [1.25x + 109.28 – 0.75 x] = 166.12 – 0.5x 1036.01 + 72.86 – 0.5x = 1108.87 – 0.5x 218.57 – 1.5x + 1.5x = 218.57

Total

1493.56

N2

% NO in burner gases = 100 x/1493.56 = 9.27 or x = 138.45 kmol/h N2 in burner gas = 1108.87 – 138.45 ´ 0.5 = 1039.65 kmol/h O2 in burner gas = 166.12 – 138.45 ´ 0.5 = 96.9 kmol/h Material balance across absorber: Air entry to absorber = 8145 Nm3/h = 363.39 kmol/h (dry) O2, entering the absorber = 363.39 ´ 0.21 = 76.31 kmol/h N2, entering the absorber = 363.39 – 76.31 = 287.08 kmol/h In the absorber, two reactions are taking place. Combining both the reactions [i.e. reactions (3) and (4)], 3 2 NO + O2 + H2O = 2 HNO3 2 O2 consumed = (3/4) y = 0.75 y where y = NO consumed in absorber, kmol/h

Material Balances Involving Chemical Reactions 103

Composition of absorber outlet gas: Component

kmol/h

NO O2 N2

138.45 – y 96.9 + 76.31 – 0.75 y = 173.21 – 0.75 y 1039.65 + 287.08 = 1326.73

Total

1638.39 – 1.75 y

(138.45 - y)100 1638.39 - 1.75 y = 0.2 y = 135.65 kmol/h NO consumed ´ 100 Absorber efficiency = NO fed

% NO in absorber outlet gases = or

= Combustion efficiency of reactor = = = Overall efficiency = = =

135.65 ´ 100 = 98.0% Ans. (b) 138.45 NH 3 converted to NO ´ 100 Total NH 3 fed 138.45 ´ 100 145.71 95.02% Ans. (a) Combustion efficiency ´ Absorption efficiency 0.98 ´ 0.9502 0.9312 or 93.12% Ans. (c)

EXERCISE 4.49 Basis: 1000 kg pig iron Fe, available per kg ore = 112 ´ 0.9/160 = 0.63 kg Ore, required to produce 1000 kg pig iron = 950/0.63 = 1508 kg Slag contains CaO, MgO and SiO2. Silica balance: Si presents in pig iron = 0.01 ´ 1000 = 10 kg Equivalent SiO2 = 60 ´ 10/28 = 21.43 kg Silica in ore = 0.1 ´ 1508 = 150.8 kg Silica in coke = 0.1 ´ 1000 = 100 kg Let x be the amount (in kg) of limestone added to the furnace. Silica in the slag = 150.8 + 100 + 0.02 x – 21.43

104 Solutions Manual—Stoichiometry

= 229.37 + 0.02 x kg CaO in the slag = 0.95 ´ 56 x/100 = 0.532x kg MgO in the slag = 0.03 ´ 40.3 x/84.3 = 0.0143x kg Total mass of the slag = 229.37 + 0.02x + 0.532x + 0.0143x = 229.37 + 0.5663x kg CaO in the slag = 0.532x ´ 100/(229.37 + 0.5663 x) % MgO in the slag = 0.0143x ´ 100/(229.37 + 0.5663 x) However, (CaO + MgO) in slag = 45% (0.532x + 0.0143x) 100 = 45(229.37 + 0.5663x) or x = 354.1 kg/t pig iron

Ans.

5

Energy Balances EXERCISE 5.1 Basis: 10 000 Lph of thermic fluid Energy balance: E1 +

gö gö qm æ p1 q æp + Z1 ÷ + Q + W = E2 + m ç 2 + Z 2 ÷ ç gc ø gc ø J è r J è r

Z1 = p1 = p2 = p2 – p1 =

Z2 101.325 kPa a 100 + 101.325 = 201.325 kPa a 201.325 – 101.325 = 100 kPa º 0.1 MPa W = pump work = 1.1 ´ 0.5 = 0.55 kW Q = 232.6 kW × qm = m = 10 000 ´ 0.75 = 7500 kg/h E2 – E1 =

7500 ( -0.1) ´ 1000 + 0.5 + 232.6 3600 ´ 0.75 ´ 1000

= – 0.277 + 0.5 + 232.6 = 232.823 kW This change in energy is responsible for rise in temperature of the thermic fluid. Rise in temperature, T = (232.823 ´ 3600)/(7500 ´ 2.68) = 41.7 K or 41.7°C Outlet temperature of thermic fluid = 473.15 + 41.7 = 514.85 K (241.7ºC) Ans.

EXERCISE 5.2 Basis: 1 kmol oxygen T1 = 350 K and T2 = 1500 K Ref. 7:

106 Solutions Manual—Stoichiometry

z

T2

H1 =

(29.8832 – 11.3842 ´ 10–8 T + 43.3779 ´ 10–6 T 2

T1

– 37.0082 ´ 10–9 T 3 + 10.1006 ´ 10–12 T 4) dT

= 29.8832 (1500 – 350) –

11.3842 ´ 10–3 (15002 – 3502) 2

+

43.3779 37.0082 ´ 10–6 (15003 – 3503) – ´ 10–9 (15004 – 3504) 3 4

+

101006 . ´ 10–12 (15005 – 3505) 5

= 34 365.7 – 12 109.9 + 48 180.2 – 46 700.0 + 15 330.0 = 39 066 kJ Ref. 5:

z

T2

H2 =

Ans. (ii)

(26.0257 + 11.7551 ´ 10–3 T – 2.3426 ´ 10–6 T 2

T1

– 0.5623 ´ 10–9 T 3) dT 117551 . ´ 10–3 (15002 – 3502) = 26.0257(1500 – 350) + 2 2.3426 0.5623 –6 3 ´ 10 (1500 – 3503) – ´ 10–3 (15004 – 3504) – 4 3 = 29 929.6 + 12 504.5 + 2601.9 – 709.6 = 39 122.6 kJ

Ans. (i)

Based on absolute enthalpies (Table – 5.22), H3 = 49 273 – [(11 603 – 8597)(350 – 298.15)/(400 – 298.15) + 8597] = 49 273 – (8597 + 1530) = 39 146 kJ

EXERCISE 5.3 Basis: 1 kmol SO2 T1 = 300 K and T2 = 1000 K Ref. 7:

z

T2

H=

T1

(25.7725 + 57.8938 ´ 10–2 T – 38.0844 ´ 10–6 T 2 + 8.6063 ´ 10–9 T 3) dT

Ans. (iii)

Energy Balances 107

= 25.7725 (1000 – 300) +

57.8938 ´ 10–3 (10002 – 3002) 2

38.0844 8.6063 ´ 10–6 (10003 – 3003) – ´ 10–9 (10004 – 3004) 3 4 = 18 040.8 + 26 341.7 – 12 352.0 + 2134.1

+

= 34 164.6 kJ/kmol Ref. 9:

z

T2

H=

T1

Ans. (a)

(24.7706 + 62.9481 ´ 10–3 T – 44.2582 ´ 10–6 T 2 + 11.122 ´ 10–9 T3) dT

= 24.7726 (1000 – 300) +

62.9481 ´ 10–3 (10002 – 3002) 2

44.2582 11122 . ´ 10–6 (10003 – 3003) + ´ 10–9 (10004 – 3004) 3 4 = 17 340.8 + 28 641.4 – 14 354.4 + 2758.0 = 34 385.8 kJ/kmol Ans. (b)

–

Ref. 61:

z

T2

H=

(44.4586 + 10.634 ´ 10–3 T + 5.945 ´ 105/T 2) dT

T1

10.634 ´ 10–3 (10002 – 3002) 2 + 5.945 ´ 105 [(1/1000) – (1/300)] = 30 420.6 + 4838.5 – 1387.2 = 33 871.9 kJ/kmol Use of absolute enthalpies (Table 5.22): H = – 314 357 – (– 347 921) = 33 564 kJ/kmol

= 43.458 (1000 – 300) +

Ans. (c)

Ans. (d)

EXERCISE 5.4 Basis: 1 kmol dry gas mixture from absorber T1 = 343 K and T2 = 618 K ° equation constants Cmp

Component kmol ni CH4

0.0025

ai × ni

bi × ni ´ 103

ci × ni ´ 106

di × ni ´ 109

0.0481

0.1303

0.0299

– 0.0283

108 Solutions Manual—Stoichiometry

CO CO2 H2 N2 Ar H2O

0.0038 0.001 0.7462 0.2435 0.003 0.0126

0.1103 0.0214 21.3492 7.2054 0.0623 0.4094

– 0.0107 0.0643 0.7607 – 1.2518 — 0.0010

0.0443 – 0.0411 – 0.1101 3.2100 — 0.1665

– 0.0179 0.0098 0.5738 – 1.2097 — – 0.0573

Total

1.0126

29.2061

– 0.2840

3.2995

– 0.7296

For 100 kmol dry gas mixture, heat exchanged Q = 100 [29.2061 (618.15 – 343.15) – 0.2840 ´ 10–3 (618.152 – 343.152)/2 + 3.2995 ´ 10–6 (618.153 – 343.153)/3 – 0.7296 ´ 10–9 (618.154 – 343.154)/4 = 100 [8031.7 – 37.5 + 215.3 – 24.1] = 818 540 kJ Use data of Table – 5.22. Component

ni kmol

A= (Ho– H0o + DH fo)1

CH4 CO CO CO2 H2 Ar H2O

0.0025 0.0038 0.001 0.7462 0.2435 0.003 0.0126

– 55 311 – 103 981 – 382 381 + 9 709 + 9 902 — – 27 751

Total

1.0126

ni × A – – – + +

138 395 382 7245 2411 — – 2870 + 5871

Ans. (a)

B= (H º – H0o + DH of )2 ni × B – 42 622 – 95 802 – 307 401 + 17 722 + 18 029 — – 218 133

– 107 – 364 – 370 + 13 224 + 4 390 — – 2748 + 14 025

Enthalpy change for 100 kmol dry gas mixture, Q = 100 (14 025 – 5871) + 100 ´ 0.003 ´ 20.7723 (618.15 – 343.15) = 815 400 + 1714 = 817 114 kJ (including that of argon) Ans. (b) At 343.15 K (70ºC), pw = 31.162 kPa (Ref. Table 6.13) Mole fraction of water vapour in gas mixture = 0.0126/1.0126 = 0.012 44 If total system pressure is p, 31.162 = 0.012 44 p or p = 2505 kPa a º 2.505 MPa a or 25.05 bar a Ans. (c)

Energy Balances 109

EXERCISE 5.5 Basis: 5000 kg/h hot oil Heat load of oil = 5000 ´ 2.51 (423.15 – 338.15) = 1066 750 kJ/h º 296.32 kW Heat gained by water = 10000 ´ 4.1868 (T – 294.15) Where T = outlet temperature of cooling water Heat loss = Heat gain 10 000 ´ 4.1868 (T – 294.15) = 1066 750 T = 319.63 K (46.5ºC)

Ans.

EXERCISE 5.6 Basis: 1 kg Diphyl DT

z

T2

Q=

(1.436 + 0.002 18 T) dT

T1

= = Mean heat capacity = Q= Error = =

1.436 (533.15 – 453.15) + 0.002 18 (533.152 – 453.152)/2 114.9 + 86 = 200.9 kJ/kg Ans. (a) (2.03 + 2.206)/2 = 2.118 kJ/(kg × K) 2.118 (533.15 – 453.15) = 169.4 kJ/kg (169.4 – 200.9) 100/200.9 – 15.7% Ans.(b)

EXERCISE 5.7 For benzene:

C = a + bT 1.591 = a + 283b 2.018 = 1 + 338b Solving the equations, a = – 0.6051, b = 0.007 76 For toluene:

(1) (2)

C¢ = a¢ + b¢T

1.524 = a¢ + 283b¢ 2.236 = a¢ + 358b¢ Solving the equations, a ¢ = – 1.1645, b = 0.0095 Basis: 5 kmol benzene – toluene mixture (a) The mixture contains 3 kmol benzene and 2 kmol toluene. Mass of benzene = 3 ´ 78 = 234 kg Mass of toluene = 2 ´ 92 = 184 kg Let T be the temperature of the mixture.

z T

Heat gained by benzene =

303

m1 × Cl1 × dT

(3) (4)

110 Solutions Manual—Stoichiometry

0.007 76 2 (T – 3032) 2 = 0.907 92 T 2 – 141.593 T – 40 452

= 234 [– 0.6051 (T – 303)] +

z

373

Heat lost by toluene =

m2 × Cl2 × dT

T

= 184 [– 1.1645 (373 – T)] + (0.0095/2) (3732 – T2) = 416 77 + 214 268 T – 0.874 T2 Heat loss by toluene = heat gain by benzene 0.907 92 T 2 = 141.593 T – 40 452 = 41 677 + 214.268 T – 0.874 T2 1.781 92 T 2 – 355.861 T – 82 129 = 0 Solving the quadratic equation, T = 336.6 K (63.45ºC) Ans. (a) (b) Use of data given in Table – 5.3 (Ref. 7). T1 = 303 K (30ºC) Let the final temperature of the mixture after mixing is T K.

z T

Heat gained by benzene = 3

(– 7.2733 + 770.541 ´ 10–3 T

303

– 1648.18 ´ 10–6 T2 + 1897.94 ´ 10–9 T 3) = 3 [– 7.2733 (T – 303) + 770.541 ´ 10–3 (T2 – 3032)/2 – 1648.18 ´ 10–3 (T9 – 3033)/3 + 1897.94 ´ 10–3 (T4 – 3034)/4] =1423.47 ´ 10–9 T 4 – 1624.17 ´ 10–6 T 3 + 1.1559 ´ 10–3 T 2 – 21.8199 T – 65 651.7 T2 = 373 K (100ºC)

z

373

Heat given-up by toluene = 2

(1.8083 + 812.223 ´ 10–3 T

T

– 1512.67 ´ 10–6 T 2+ 1630.01 ´ 10–9 T 3)

= 2 [1.8083(373 – T ) + 812.223 ´ 10–3 (3732 – T2)/2 – 1512.67 ´ 10–6 (3733 – T3)/3 + 1630.01 ´ 10–9 (3734 – T4)/4] = – 815 ´ 10–9 T 4 + 1008.4 ´ 10–6 T 3 – 0.8122 ´ 10–3 T2 – 3.6166 T + 77 795.4

Energy Balances 111

Heat gained by benzene = heat given-up by toluene Equating the two and simplifying, 2238.47 ´ 10–9 T 4 – 2362.57 ´ 10–6 T 3 + 1.9681 T2 – 18.2033 T = 143 447.1 The equation can be solved by trial and error method or by Mathcad. Mixing temperature = 334.25 K (61.1ºC)

Ans. (b)

EXERCISE 5.8 (a) From Table – 5.4 for acetic acid, log p = 4.682 06 –

1642.540 (T - 39.764)

T = 316.15 K log p = 4.682 06 –

1642.540 (316.15 - 39.764)

= –1.260 86 or p = 0.0548 bar or 5.48 kPa (b) From Table – 5.4, for sulphur trioxide, log p = 4.205 15 –

Ans.

892.175 T - 103.564

T = 335.15 K log p = 0.352 69 p = 2.253 bar

Ans.

EXERCISE 5.9 (a) From Table – 5.5, for acetone,

LM N

lv T - To = c T lv1 c - TB

OP Q

0.38

0.38 lv é 508.1 - 313.15 ù = ê ú lv1 ë 508.1 - 329.3 û lv = 29.1 ´ 1.0334 = 30.072 kJ/mol (b) From Table – 5.5, for carbon disulphide,

é 552.0 - 413.15 ù lv = ê ú 26.74 ë 552.0 - 319.4 û

0.38

l v = 29.1 kJ/mol 1

112 Solutions Manual—Stoichiometry

l v = 26.74 ´ 0.821 97 = 21 98 kJ/mol

Ans.

EXERCISE 5.10 From Table – 5.4, for chlorobenzene 1435.675 T - 55.124 1.013 25 bar 347.292 402.42 K or tB = 129.27°C 404.9 K or tB = 131.75ºC

log p = 4.110 83 –

p= T – 55.124 = T = TB = Table – 5.5 reports TB = Riedel equation: 1.092(ln pc - 5.6182 ) lv = 0.930 - TBr R × TB Read Tc and pc values from Appendix III.2 TBr = TB /Tc = 404.9/633.0 = 0.6396 pc = 4.53 MPa = 4530 kPa lv = Table – 5.5 reports

8.31451 ´ 404.9 ´ 1.092 (ln 4530 - 5.6182) (0.930 - 0.6396)

= 35 450 kJ/kmol lv = 35 190 kJ/kmol

Ans.

EXERCISE 5.11 (a) Benzene:

Riediel equation:

pc = 4.895 MPa, Tc = 562.05 K, TB = 353.3 K Tr = 353.3/562.05 = 0.6286 at TB é1.092 (ln 4895 - 5.6182) ù l v1 = 8.314 51 ´ 353.3 ê ú (0.930 - 0.6286) ë û

= 30 628 kJ/kmol at 353.3 K NIST equation:

l v1 = 47 410 ´ e–(0.1231 ´ 0.6286) ´ (1 – 0.6286)0.3602 = 30 712 kJ/kmol at 353.3 K At t/T = 25°C/298.15 K, Tr = 298.15/562.05 = 0.530 47 l v2 = 47 410 ´ e–(0.1231 ´ 0.530 47) ´ (1 – 0.530 47)0.3602 = 47 410 ´ 0.936 79 ´ 0.761 61 = 33 826 kJ/kmol at 298.15 K

Energy Balances 113

Watson equation: é ù lv2 = 29 100 ê 562.05 - 298.15 ú ë 562.05 - 353.3 û

0.38

= 31 811 kJ/kmol at 298.15 K (b) Acetone: pc = 4.700 MPa, Tc = 508.1 K, TB = 329.3 K Tr = 329.3/508.1 = 0.6481 at TB

Riedel equation:

é1.092 (ln 4700 - 5.6182) ù lv1 = 8.314 51 329.3 ê ú (0.930 - 0.6481) ë û

= 30 091 kJ/kmol at 329.3 K NIST equation:

lv1 = 46 950 ´ e–(0.2826 ´ 0.6481) ´ (1 – 0.6481)0.2876 = 29 101 kJ/kmol at 329.3 K At t/T = 25°C/298.15 K, Tr = 298.15/508.1 = 0.5868 lv2 = 46 950 ´ e–(0.2826 ´ 0.5868) ´ (1 – 0.5868)0.2876 = 30 984 kJ/kmol at 298.15 K

Watson equation: é ù lv2 = 29 100 ê 508.1 - 298.15 ú ë 508.1 329.3 û

0.38

= 30 931 kJ/kmol at 298.15 K

Ans.

EXERCISE 5.12 Basis: 100 kg naphthalene Total heat to be supplied consists of sensible heat, supplied to the solid from 303.15 K to its melting point (353.35 K), latent heat of fusion l f at 353.35 K, sensible heat supply to liquid naphthalene from the melting point (353.35 K) to its normal boiling point (490 K) and the latent heat of vaporzation at 490 K.

z

Q2 = Cs dT = 100 [– 0.092 (353.35 – 303.15) + (0.0046/2) (353.352 – 303.152)] = – 461.8 + 7579.9 = 7118.1 kJ Q2 = m × lf = 100 ´ 150.7 = 15 070 kJ For evaluation of Q3, C1 is required to be expressed in the polynomial form of temperature. Cl = a + bT 1.738 = a + b ´ 353 (1) 2.135 = a + b ´ 473 (2)

114 Solutions Manual—Stoichiometry

z

Solving the equations, a = 0.57 and b = 0.003 31 Q3 = m

Cl dT

= 100 [0.57 (491 – 353) + (0.003 31/2) (4912 – 3532)] = 7866 + 19 276 = 27 142 kJ Q4 = m ×lv = 100 ´ 316.1 = 31 610 kJ Total heat load = S Qi = 7118.1 + 15 070 + 27 142 + 31 610 = 80 940.1 kJ Latent heat of vaporisation of the eutectic mixture at 171 kPa a = 278.0 kJ/kg (Ref. Table 5.6) Quantity of the eutectic mixture condensed = 80 940.1/278.0 = 291.15 kg

Ans.

EXERCISE 5.13 Basis:

10 000 kg/h superheated stem @ 0.44 MPa a and 543 K (270°C) is mixed with 7500 kg/h saturated steam at 0.44 MPa a. From Table AIV.2 enthalpy of saturated steam = 2741.9 kJ/kg From Table AIV.3 enthalpy of superheated steam = 3004.5 kJ/kg Total enthalpy of 7500 kg/h saturated steam = 7500 ´ 2741.9 = 20 564 250 kJ/h º 5712.29 kW Total enthalpy of 10 000 kg/h superheated steam = 10 000 ´ 3004.5 = 30 045 000 kJ/h º 8345.83 kW Total enthalpy of mixed fluids = 5712.29 + 8345.83 = 14 058.12 kW º 50 609 250 kJ/h 50 609 250 Specific enthalpy of mixture = = 2891.96 kJ/kg 17 500 From steam tables (AIV.3), temperature of mixed steam = 489.15 K (216.0°C) at 0.44 MPa a which is superheated.

EXERCISE 5.14 Basis: 100 kg superheated steam @ 0.5 MPa a and 523 K (250°C) From Steam Tables AIV.3, specific enthalpy of superheated steam = 2961.1 kJ/kg

Energy Balances 115

From Steam Tables AIV.2, specific enthalpy of saturated steam at 0.5 MPa a = 2747.5 kJ/kg at (TS) saturation temp. of 425 K (151.85°C). Enthalpy of steam to be reduced = 2961.1 – 2747.5 = 213.6 kJ/kg Enthalpy of water to be sprayed = 125.66 kJ/kg Let a be the quantity of water to be sprayed. 100 ´ 2961.1 + a ´ 125.66 = 2747.5 (100 + a) or a = 8.15 kg Ans.

EXERCISE 5.15 Basis: 1 kg condensate Operating pressure of dryer = 310 kPa a Operating pressure of flash vessel = 101.325 kPa a At 310 kPa a, Ts = 407.8 K (134.65°C), h = 566.23 kJ/kg and l v = 2159.9 kJ/kg At 101.3 kPa a, Ts = 373.15 K (100°C), h = 419.06 kJ/kg, l v = 2256.9 kJ/kg and H = 2676.0 kJ/kg Let the flash quantity be a kg per kg of condensate at 310 kPa. (1 – a) 419.06 + 2676.0a = 566..23 a = 0.065 kg/kg condensate Enthalpy of saturated steam at 780 kPa a= 2767.5 kJ/kg Let y and y¢ be the quantities of water to be sprayed before modification and after the modification per kg of saturated steam at 310 kPa, respectively. y ´ 419.06 + (1 – y) 2767.5 = (566.23 + 2159.9) = 2726.13 y = 0.0176 kg/kg condensate After the modification, y¢ ´ 419.06 + 0.065 ´ 2676.0 + (1 – y¢ – 0.065) 2767.5 = 2726.13 y¢ = 0.015 kg/kg condensate Make-up 0.8 MPa a steam = 1 – 0.065 – 0.015 = 0.92 kg Reduction in make-up steam = [(0.9824 - 0.92)/0.9824] 100 = 6.35%

EXERCISE 5.16 Basis: 1 kmol dry gas mixture. T1 = 1473.15 K, T2 = 573.15 K

Ans.

116 Solutions Manual—Stoichiometry

Component

kmol ni

° equation constants Cmp ai × ni

bi × ni ´ 103

ci × ni ´ 103

di × ni ´ 103

H2 CO CO2 CH4 N2 H2O

0.557 0.345 0.028 0.005 0.065 1.850

15.936 10.015 0.598 0.096 1.923 60.110

0.568 – 0.972 1.800 0.261 – 0.334 0.147

– 0.082 4.017 – 1.149 0.060 0.857 24.440

0.428 – 1.624 0.274 – 0.057 – 0.323 – 8.413

Total

2.850

88.678

1.470

28.143

– 9.715

Heat lost in WHB, DH = 88.678 (1473 – 573) + 1.470 ´ 10–3 (14732 – 5732)/2 + 28.143 ´ 10–6 (14733 – 5733)/3 – 9.715 ´ 10–9 (14734 – 5734)/4 = 79 810.2 + 1353.4 + 28 216.9 – 11 172.1 = 98 208.4 kJ/kmol dry gas mixture º 9820 840 kJ/100 kmol dry gas mixture Steam pressure p = 4.0 MPa g = 4.1013 MPa a Ts = 524.8 K (251.8°C) (Ref. A IV.2) Water enters WHB at 504.8 K (231.8°C), 20 K lower than 524.8 K. Enthalpy of water at 504.8 K, h = 998.7 kJ/kg Enthalpy of saturated steam at 4.1013 MPa a, Hs = 2800 kJ/kg Enthalpy supplied to water in WHB = 2800 – 998.7 = 1801.3 kJ/kg Steam generated in WHB = 9820 840/1801.3 = 5452 kg per 100 kmol dry gas Ans.

EXERCISE 5.17 Basis: 100 kg saturated liquid ammonia at 705 kPa a Let y be the quantity of NH3 vapours flashed. Liquid NH3 at 101.325 kPa a = 100 – y kg Heat balance: Enthalpy of liquid NH3 at 705 kPa a = Enthalpy of liquid NH3 at 101.3 kPa a + Enthalpy of flash vapours at 101.3 kPa a 100 ´ 265.56 = (100 – y) 49.1 + y ´ 1418.7 kJ/h (º 44.85 kW) Solving the equation, y = 15.8 kg or 15.8 % flashing Ans.

EXERCISE 5.18 Basis: 100 kg NH3 vapours, entering 1st stage of compressor

Energy Balances 117

Enthalpy H1 = 100 ´ 1632.68 = 163 268 kJ/h (º 45.35 kW) of gas leaving 1st stage. Enthalpy of 100 kg liquid NH3 at 276 K = 100 ´ 213.92 = 21 392 kJ/h (º 5.94 kW) Heat, removed in flash cooler = 163 268 – 21 392 = 141 876 kJ/h (º 39.41 kW) Let y be the quantity of liquid NH3, obtained from the condenser and is utilised in flash cooler. Out of this, a part will be flashed due to pressure reduction. Another basis: 1 kg liquid NH3 at 1930.3 kPa a and 321 K (48°C) Enthalpy of saturated liquid at 1930.3 kPa a = Enthalpy of saturated liquid at 276 K (3°C) + Enthalpy of saturated vapours at 276 K (3°C) 100 ´ 431.07 = (100 – z) 213.92 + z ´ 1464.92 where z = amount of flash vapours, produced due to pressure reduction z = 17.36 kg Thus 17.36% liquid, entering the flash cooler (i.e. y kg) will be flashed. Balance liq. NH3, available for cooling 1st stage vapours = (1 – 0.1734) y = 0.8266y kg/h This ammonia should cater for the heat load of 141 876 kJ/h. 0.8266 y ´ 1251.0 = 141 876 or y = 137.13 kg/h Ans. (a) Heat load of condenser = 1663.91 – 431.07 = 169 059 kJ/h º 46.96 kW Cooling water to condenser = 169 059/[3600 ´ 4.1868 ´ (313 – 305)] = 1.402 kg/s Ans. (b)

EXERCISE 5.19 Basis: 65 kW refrigeration load in chiller From Table 5.7, Enthalpy of saturated R-134a liquid at 40°C (313.15 K) = 256.35 kJ/kg Saturation temperature of R-134a at 101.325 kPa = –26°C Superheat = 10°C Temperature of R-134a gas, leaving chiller = –26 + 10 = –16°C Enthalpy of superheated R-134a gas at 101.325 kPa and –16°C = 391.0 kJ/kg (Ref. Table 5.71) Heat picked up by R-134a in chiller = 391.0 – 256.35 = 134.65 kJ/kg

118 Solutions Manual—Stoichiometry

Evaporation rate = 60

kJ 1 kg × s 134.65 kJ

= 0.4456 kg/s Ans.(a) With the incorporation of the economizer, R-134a gas heats upto 10°C at 101.325 kPa. Heat picked up in the economizer = 412 – 391 = 21 kJ/kg In exchange, R-134a liquid under saturation pressure (10.165 bar) at 40°C (313.15 K) is cooled. Enthalpy of sub-cooled R-134a liquid = 256.35 – 21 = 235.35 kJ/kg Heat picked up in chiller = 391 – 235.35 = 155.65 kJ/kg Evaporation rate =

EXERCISE 5.20

65 = 0.4176 kg/s with 155.65 economizer Ans. (b)

(a) Locate a point representing 273 K (0°C) on p-H diagram for CO 2 (Fig. 5.11). Follow constant enthalpy line (i.e. vertical line) which intersects the pressure line representing 1.0 MPa g (= 1.101 MPa a). Read dryness fraction of 0.277. Thus vapours produced by pressure reduction is 27.7%. Ans. (b) Enthalpy of CO2 at 1.101 MPa a and 313 K(40°C) = 811 kJ/kg Enthalpy of vapour-liquid mixture at 1.101 MPa a = 500 kJ/kg Enthalpy required to produce CO2 gas at 1.101 MPa a and 313 K = 811.0 – 500 = 311 kJ/kg

Ans.

EXERCISE 5.21 Basis: 1 kmol dry gas T1 = 403.15 K and T2 = 313.15 K Component kmol yi CO2 O2 N2 H2

0.947 0.008 0.030 0.015

Total

1.000

Molar mass Mi

ni × M i

41.677 0.256 0.840 0.030

Critical pressure yi ×pci p ci MPa 7.375 6.983 5.042 0.040 3.394 0.102 1.297 0.019

Critical temperature Tci yi ×Tci K 304.12 288.00 154.59 1.24 126.09 3.78 33.2 0.48

44.01 31.9988 28.0134 2.016

42.803

7.144

293.50

Energy Balances 119

Average molar mass = 42.803 of incoming gas mixture Pseudo critical pressure pc = 7.144 MPa Pseudo critical temperature Tc = 293.5 K (20.35°C) New Basis: 100 kmol dry gas Component

kmol ni

C° mp equation constants ai × ni

bi × ni ´ 103

ci × ni ´ 106

di × ni ´ 109

CO2 O2 N2 H2

94.7 0.8 3.0 1.5

2023.313 20.821 88.773 42.916

6087.704 9.404 – 15.423 1.529

– 3887.492 – 1.874 39.549 – 0.221

928.051 – 0.450 – 14.904 1.154

Total

100.0

2175.823

6083.214

– 3850.038

913.851

Tr1 = 403.15/293.5 = 1.374,

Tr2 = 313.15/293.5 = 1.067 Avg. Tr = (1.374 + 1.067)/2 = 1.220 pr = 20/7.144 = 2.8 Fig. 5.1 does not provide value of C Rmp for the required pr and Tr. From Lee at el article (ref. 12). R Cmp

 

R = Cmp

0

 

+ w CmRp

1

R At pr = 2.8 and Tr = 1.22

C  R mp

0

= 5.681

Acentric factor (wi) for each component is to be considered and S (yi × wi) is required to be calculated for the mixture, However, since CO2 is 94.7% in the mixture, its wCO value is taken for calculation. 2 w = 0.225 for CO2 (Ref. 3)

C  R mp

R Cmp

R

1

= 7.222 = 5.681 + (0.225 ´ 7.222) = 7.306

C Rmp = 7.306 ´ 8.314 51 = 60.746 Cmp = C omp + C Rmp = C omp + 60.746 kJ/(kmol × K) Therefore, corrected heat capacity equation will become Cmp = (2175.823 + 60.746 ´ 100) + 6083.214 ´ 10–9 T – 3850.038 ´ 10–3 T2 + 913.851 ´ 10–3 T 3 Heat removed in water cooler = 8250.423 (403.15 – 313.15) + 6083.214 ´ 10–3 (403.152 – 313.152)/2 – 3850.038 ´ 10–6 (403.153 – 313.153)/3 + 913.851 ´ 10–9 (403.154 – 313.154)/4 = 742 538 + 196 087 – 44 680 + 3838

120 Solutions Manual—Stoichiometry

= 897 783 kJ/h º 249.38 kW For p = 4.0 MPa a pr = 4.0/7.144 = 0.56, Tri = 313.15/293.5 = 1.067 0

R ö æ Cmp ç ÷ = 1.2062 è R ø R ö æ Cmp ç ÷ è R ø

From Fig. 5.1,

1

= 2.3857

R ö æ Cmp ç ÷ = 1.2062 + (0.225 ´ 2.3857) = 1.743 è R ø = 1.743 ´ 8.314 14 = 14.491 R C mp = 14.5

Both are closely matching. Since incoming gas mixture has 5.3 mole % inerts, liquefaction temperature will have to be found by trails such that at the liquefaction temperature, ratio of uncondensed CO2 to inerts satisfy the vapour pressure data. At 40 bar a pressure, saturation temperature of CO2 is +5.32°C (278.47 K). Presence of inerts will require liquefaction temperature to be lower than 278.47 K. Trial 1: Assume liquefaction temperature of 273.15 K. Consider enthalpy of liquid CO2 = 0 kJ/kg at 273.15 K Heat given-up by gas from 313.15 K to 273.15 K = (2175.823 +1449.1) (313.15 – 273.15) + 6083.214 ´ 10–3 (313.152 – 273.152)/2 – 3850.038 ´ 10–6 (313.153 – 273.153)/3 + 913.851 ´ 10–9 (313.154 – 273.154)/4 = 144 997 + 71 332 – 13 255 + 925 = 203 999 kJ/h º 56.666 kW l v of CO2 at 273.15 K = 231.00 kJ/kg = 10 166 kJ/kmol (Ref. 24) Let x be the amount of uncondensed CO2 (kmol/h) (94.7 – x) 10 166 = 203 999 94.7 – x = 20.067 x = 74.633 kmol/h CO2 condensed = 20.067 kmol/h Inerts = 0.8 + 3.0 + 1.5 = 5.3 kmol/h

Energy Balances 121

74.633 Uncondense CO 2 = = 14.083 kmol / kmol 5.3 inerts At T = 273.15 K, saturation pressure of CO2, pS1 = 3485.9 kPa total pressure = 3000 kPa Applying Raoult's law,

(Ref. 24)

Moles of CO 2 3485.9 = = 6.781 Moles of inerts (4000 – 3485.9) Thus calculated ratios do not match. Trial 2: Assume liquefaction temperature (T2 ) of 275.15 K. Enthalpy of gas mixture from 313.15 K to 275.15 K = 193 955 kJ/h º 53.876 kW lv at 275.15 K = 225.05 kJ/kg = 9904.3 kJ/kmol (94.7 – x) 9904.3 = 193 955 94.7 – x = 19.532 x = 74.1168 kmol/h CO 2 74.168 = = 14.183 5.3 Inerts

CO2 condensed = 20.555 kmol/h At T = 275.15 K pS2 = 3673.3 kPa 3673.3 Moles of CO 2 = = 11.244 — no match (4000 - 3673.3) Moles of inerts Trial 3: Assume T3 = 276.15 K or t3 = 3°C. lv = 221.53 kJ/kg = 9749.4 kJ/kmol Heat given up by gas from 313.15 K to 276.15 K = 188 926 kJ/h º 52.479 kW (94.7 – x) 9749.1 = 188 926 x = 75.322 kmol/h Uncondensed CO 2 = 75.322 = 14.211 5.3 Inerts Saturation pressure of CO2 at 276.15 K, p53 = 3768.5 kPa 3768.5 Uncondensed CO 2 = = 16.279 (4000 - 3768.5) Inerts Since ratio is reversed, the liquefaction temperature is between 275.15 K and 276.15 K. Trial 4: Assume T4 = (275.15 + 276.15)/2 = 275.65 K or t4 = 2.5°C lv = (225.05 + 221.53)/2

122 Solutions Manual—Stoichiometry

= Heat given-up by gas = = º (94.7 – x) 9826.9 = x= Uncondensed CO 2 = Inerts At T4 =

223.29 kJ/kg º 9826.9 kJ/kmol (193 955 + 188 926)/2 191 441 kJ/h 53.178 kW 191 441 75.219 kmol/h 75.219 = 192 5.3 275.65 K, p54 = 3720.9 kPa 3720.9 Uncondensed CO 2 = = 13.332 (4000 - 3720.9) Inerts Further trials are unwarranted as both ratios are close enough. CO2 condensed = 94.7 – 75.219 = 19.481 kmol/h Hence liquefaction temperature = 269.85 K (–3.3 °C) CO2 liquefied = (94.7 – 75.219) 100/94.7 = 20.57 % Ans. (b) Enthalpy of saturated liquid CO2 at 37.21 bar a and 2.5°C (275.65) = 206.13 kJ/kg (Ref. 24) Liquefied gas is precooled in the revert gas exchanger before dry gas production. Note that liquid CO2 temperature after the revert gas exchanger but before the snow tower will be at higher then – 11°C. Refer Example 5.19, 100 ´ 206.13 = (100 – y) 422.61 + (–148.39 y) 571.0 y = 21 648 y = 37.91 kg dry ice/100 kg liquid CO2 For 100 kmol raw gas input, dry ice production = 19.481 ´ 0.3791 = 7.385 kmol/h º 325.2 kg/h For 100 kg dry ice production, raw gas requirement = 100 ´ 100 ´ 42.803/325.02 = 1316.93 kg/h Ans. (c) º 30.767 kmol/h Duty of water cooler = (30.767 ´ 191.441)/100 = 58 901 kJ/h º 16.361 kW Ans. (a) Note: Application of Raoult's law may not be fully justified as pressure of liquid CO2 is high (40 bar a). Use of modified Raoult's law or pure component pressures (Sec. 2.6.3) is more appropriate.

Energy Balances 123

EXERCISE 5.22 Basis: 32 000 kg of total ammonia (liquid + vapour) in each tank wagon Temperature of ammonia in wagon = –29°C (244.15 K) Let x = quantity of ammonia vapours, contained in the wagon (in kg) Liquid ammonia = 32 000 – x kg Total volume of each wagon = 60.663 m3 x (32 000 – x ) + = 60.663 676.25 1.0844 x = 14.49 kg vapours of ammonia Liquid NH3 = 320 00 – 14.49 = 31 985.51 kg Total enthalpy, H1 = 14.49 ´ 1425.06 + 31 985.51 ´ 68.32 = 2205 899.1 kJ Similar values of liquid ammonia and vapour ammonia at different temperatures can be calculated. Temperature, °C(K)

NH3 vapours, kg

–29 (244.15) –27 (246.15) –25 (248.15) –23 (250.15) –21 (252.15) –19 (254.15) –17 (256.15)

14.49 15.62 16.85 18.12 19.44 20.84 22.27

liquid NH3, kg

total enthalpy, kJ

31 985.51 31 984.38 31 983.15 31 981.88 31 980.56 31 978.16 31 377.73

2205 899.1 2493 097.8 2781 056.0 3069 691.4 3359 336.6 3649 276.3 3940 422.8

Temperature change from 244.15 K to 246.15 K Average temperature, T = 246.15 K Ambient temperature Ta = 298.15 K Heat gain, f1 = 1.3 (298.15 – 245.15) + 3.76 = 72.66 kW º 261 576 kJ/h Enthalpy gain between two temperatures, H1¢ = 2493 097.8 – 2205 899.1

= 287 198.7 kJ Time, required for increase in temp. = 287 198.7/261 576 = 1.098 h Similar calculations are listed below: Rise in temperature K From to 244.15 246.15 246.15 248.15

Enthalpy change, kJ 287 198.7 287 958.2

Heat transfer rate, f, kJ/h

Time, h

261 576 252 216

1.098 1.142

124 Solutions Manual—Stoichiometry

248.15 250.15 252.15 254.15

250.15 252.15 254.15 256.15

288 635.4 289 645.2 289 939.7 291 146.5

Total

1734 523.7

242 856 233 496 224 136 214 776

1.189 1.240 1.294 1.356 7.319

Ans. (a) 244.15 + 256.15 = 250 K 2 f = 1.3 [298.15 – 250.15] + 3.76 = 66.16 kW º 238 176 kJ/h Time q = 1734 523.7/238 176 = 7.283 h

Average T =

Ans. (b)

EXERCISE 5.23 Basis: 1650 Nm3/h air flow rate Molar flow rate = 1650/22.414 = 73.615 kmol/h Mass flow rate = 73.615 ´ 28.84 = 2123.06 kg/h Temperature drop from 463.15 K to 453.15 K: Heat loss rate at 463.15 K (190°C) = 575.8 W/m [Ref. Fig. (5.35)] Heat loss rate at 459.15 K (180°C) = 532.3 W/m [Ref. Fig. (5.35)] Average heat loss rate = (575.8 + 532.3)/2457.5 = 554.05 W/m Enthalpy to be removed from air = 2123.06 ´ 1.006 ´ 10 = 21 358 kJ/h º 5.9328 kW Length of pipe, required for cooling, L1 = 5.9328 ´ 1000/554.05 L1 = 10.71 m Similar calculations are listed below, keeping enthalpy removal quantity as same (i.e. 5.9328 kW) for 10 K drop. Temp, °C(K)

Heat loss rate, ref. Fig. (5.35), W/m

Avg. heat loss rate, W/m

190 (463.15) 180 (453.15) 170 (443.15) 160 (433.15) 150 (423.15) 140 (413.15) 130 (403.15) 120 (393.15) 110 (383.15) 100 (373.15)

575.8 532.3 481.1 434.4 400.0 364.8 324.6 292.6 266.4 244.3

— 554.05 506.7 457.75 417.2 382.4 344.7 308.6 279.5 255.35

Total

—

—

Length of pipe m — 10.71 11.71 12.96 14.22 15.51 17.21 19.22 21.23 23.23 146.00

Energy Balances 125

Balance length of pipe = 150 – 146 = 4 m Heat loss rate at 363.15 K (90°C) = 216.4 W/m Avg. heat loss rate between 373.15 K (100°C) and 363 K(90°C) = (244.3 + 216.4)/2 = 230.35 W/m Length of pipe, required to cool to 363 K (90 °C) = 5932.8/230.35 = 25.76 m However, actual length to be cooled = 4 m Drop in temperature = (4/25.76) 10 = 1.55 K Temperature of air at the end of 150 m pipe = 373.15 – 1.55 = 371.6 K (98.45°C) Ans.

EXERCISE 5.24 Basis: 1 kmol mixture of benzene and toluene containing 0.6 kmol benzene. Antoine equation: 1203.835 for benzene: log p = 4.018 14 – (T – 53.226) 1343.943 for toluene: log p = 4.078 27 – (T – 53.773) At p = 101.325 kPa, = 1.013 25 bar Ts1 = 353.25 K, Ts2 = 383.77 K ts2 = 110.62 °C ts1 = 80.1°C Initial guess, T1 = 353.60 ´ 0.6 + 383.77 ´ 0.4 = 365.67 K say 366 K Bubble point: component 1 = benzene and component 2 = toluene. All pressures are kPa. Temperature, K x1

T1 = 366 p si

0.6 0.4

147.653 59.416

1.0

T2 = 360 yi

p si

T3 = 364 yi

p si

T5 = 362.75 yi

p si

yi

0.8743 124.155 0.2346 48.928

0.7352 139.468 0.1932 55.737

0.8259 134.533 0.7966 0.2200 53.132 0.2121

1.1089

0.9284

1.0459

1.0087

At Syi = 1.00. Bubble point, TBP = 362.75 K or 89.6°C

Ans. (a–i)

Dew point: Temperature, K y1 0.6

T1 = 366 p si xi 147.653

T2 = 368 p si xi

0.4117 156.206

T3 = 370 p si xi

0.3892 165.137

0.3681

p si

T4 = 369 xi

160.62 0.3785

126 Solutions Manual—Stoichiometry

0.4 1.0

59.416

0.6821

63.283

0.6405

1.0938

67.350

1.0297

0.6018

65.29 0.6208

0.9699

0.9993

At Sx i = 1.00 dew point, TDP= 369 K or 95.85 °C Enthalpy (sensible heat) of liquid mixture at T: Reference temperature: T0 = 273.15 K (0 °C) H1 =

Ans. (a – ii)

TBP

ò Cl1 × dT

273.15

= – 484.365 (362.75 – 273.15) + 5056.235 ´ 10–3 (362.752 – 273.152)/2 – 14 292.204 ´ 10–6 (362.753 – 273.153)/3 + 14 419.754 ´ 10–8 (362.754 – 273.154)/4 = – 43 471.8 + 144 250.8 – 130 473.5 + 42 394.1 = 12 699.6 kJ/kmol benzene H2 =

TBP

ò Cl2 × dT

273.15

= — 56.3627 (362.75 – 273.15) + 1768.423 ´ 10–3 (362.752 – 273.152)/2 – 5192.623 ´ 10–3 (362.753 – 273.153)/3 + 5497.39 ´ 10–4 (362.754 – 273.154)/4 = – 5058.6 + 50 451.8 – 47 403.5 + 16 162.3 = 14 152.0 kJ/kmol toluene Hsol = S (Hi × xi) = 0.6 ´ 12 699.6 + 0.4 ´ 14 152.0 = 7619.8 + 5660.8 = 13 280.6 kJ/kmol liquid mixture From Fig. 6.2, Hsol = 13 733 kJ/kmol liquid mixture Enthalpy of vapour mixture at TDP : l v1

é 562.05 – 369.0 ù = ê ú ë 562.05 – 353.3 û

0.38

´ 30 720

= 29 820.7 kJ/kmol of benzene at TDP é 591.75 – 369.0 ù l v2 = ê ú ë 591.75 – 383.8 û

0.38

´ 33 180

= 34 058.3 kJ/kmol of toluene at TDP H1¢ =

TDP

ò Cl1 × dT

273.15

= – 484.365 (369.0 – 273.15)

Energy Balances 127

+ 5056.235 ´ 10–3 (369.02 – 273.152)/2 – 14 292.204 ´ 10–6 (369.03 – 273.153)/3 + 14 419.754 ´ 10–9 (369.04 – 273.154)/4 = – 46 547.5 + 155 999.5 – 142 626.2 + 46 883.5 = + 13 709.3 kJ/kmol benzene H 2¢ =

TDP

ò Cl2 × dT

273.15

= – 56.3627 (369.0 – 273.15) + 1768.423 ´ 10–3 (369.02 – 273.152)/2 – 5192.623 ´ 10–6 (369.03 – 273.153)/3 + 5497.39 ´ 10–9 (369.04 – 273.154)/4 = – 5416.5 + 54 561.0 – 51 818.7 + 17 873.9 = + 15 199.7 kJ/kmol toluene Enthalpy of benzene vapour at TDP = 13 709.3 + 29 820.7 = 43 530 kJ/kmol benzene vapour Enthalpy of toluene vapour at TDP = 15 199.7 + 34 058.3 = 49 258 kJ/kmol toluene vapour

b

Enthalpy of vapour mixture = S Hi¢ × yi

g

= 0.6 ´ 43 530 + 0.4 ´ 49 258 = 45 821 kJ/kmol vapour mixture Ans. (b) From Fig. 6.2, enthalpy of vapour mixture = 45 666.7 kJ/kmol vapour mixture

EXERCISE 5.25 Basis: 1 kmol natural gas Temperature

ni

CH4 CO2 C2 H 6 C3 H 3 i-C4H10 n-C4H10 C5H12 C6H14 C7H18

0.8957 0.0112 0.0526 0.0197 0.0068 0.0047 0.0038 0.0031 0.0024

Total

1.0000

Ki 2.70 0.90 0.38 0.098 0.038 0.024 0.0075 0.0019 0.0007

Values of Li for trial value of L L = 0.5

0.1

0.01

0.041

0.2421 0.0059 0.0381 0.0179 0.0065 0.0046 0.0038 0.0031 0.0024

0.0354 0.0012 0.0019 0.0105 0.0051 0.0039 0.0035 0.0030 0.0024

0.0033 0.0001 0.0014 0.0019 0.0014 0.0014 0.0022 0.0026 0.0022

0.0140 0.0005 0.0053 0.0060 0.0036 0.0030 0.0032 0.0030 0.0024

0.3244

0.0769

0.0165

0.041

V = 1 – 0.041 = 0.959 kmol and L = 0.041 kmol

Ans.

128 Solutions Manual—Stoichiometry

EXERCISE 5.26 Basis: 1 kmol crude gas mixture from demethanizing unit Cryogenic temp. in cold box = 111 K (– 162.15 °C) Component

yi

K*i

H2 CH4 C2H6 N2

0.750 0.200 0.045 0.005

49.73 0.057 9.45 ´ 10–5 1.158

1.000

—

L1 = 0.5 0.0148 0.1892 0.0450 0.0022

L2 = 0.25 0.0050 0.1708 0.0450 0.0011

L3 = 0.22 0.0042 0.1664 0.0450 0.0010

L4 = 0.215 0.0041 0.1655 0.0450 0.0010

V 0.7459 0.0345 Nil 0.0040

0.2512

0.2219

0.2166

0.2156

0.7844

*

Ki values are calculated from Table 5.15 of the text at T = 111 K. New basis: crude gas stream flow = 18 720 Nm3/h = 835.19 kmol/h Ingoing gas stream: ni kmol/h

Molar

( m i )

Hi , kJ/kg

mi × Hi

mass

kg/h

at 2.75 MPa a and 300.15 K

kW

H2 CH4 C2H6 N2

626.4 167.03 37.58 4.18

2 16 30 28

1252.8 2672.5 1127.4 117.0

3972.1 888.43 607.82 305.67

1382.29 659.54 190.35 9.93

Total

835.19

components

5169.8

2242.11

Product gas stream: component

kmol/h ni

Molar mass

kg/h ( m i )

H2 CH4 C2H8 N2

622.98 28.81 Nil 3.34

2 16 30 28

1246.0 461.0 Nil 93.5

Total

655.13

Hi, kJ/kg at 2.75 MPa a, and 297.65 K 3935.7 883.44 604.90 303.17

1800.5

mi× Hi kW 1362.19 113.13 Nil 7.83 1483.19

Tail gas stream: component

H2

ni kmol/h 3.42

Molar

( m i )

Hi, kJ/kg

mi× Hi

mass

kg/h

at 140 KPa a, and 297.15 K

kW

2

6.8

3917.6

7.40

Energy Balances 129

CH4 C2H6 N2

138.22 37.58 0.84

Total

180.06

16 30 28

Net heat change, f1 = = Heat infiltration, f2 = Refrigeration requirement, f3 = º Recovery of hydrogen = Purity of product hydrogen stream = =

2211.5 1127.4 23.5

907.34 665.26 308.14

557.38 208.34 2.01

3369.3

775.13

775.13 + 1483.19 – 2242.11 16.21 kW (endothermic) 20.9 kW 20.9 – 16.21 = 4.69 kW 1.334 TR (622.98 ´ 100)/626.4 = 99.45% 622.98 ´ 100/655.13 95.1% Ans.

EXERCISE 5.27 (a) C7H16(g) + 11 O2(g) ¾ ¾¾® 7 CO2(g) + 8 H2O(g) Heat of combustion of n-heptane (g), DH°c = – 4501.46 kJ/kmol 7 C(s) + 7 O2(g) ¾ ¾¾® 7 CO2(g) Heat of combustion of carbon (s), DH°c = – 393.51 kJ/kmol 8 H2(g) + 4 O2(g) ¾ ¾¾® 8 H2O(g) Heat of combustion of hydrogen (g), DH°c (g) = – 241.82 kJ/kmol Heat of formation of n-heptane (g) = Heat of combustion of carbon + Heat of combustion of hydrogen – Heat of combustion of n-heptane DH°f = – 7 ´ 393.51 – 8 ´ 241.82 + 4501.46 = – 187.67 kJ/kmol Ans. From Table AV.2 DH°f = – 187.78 kJ/kmol (b) C2H5OH (g) ¾ ¾¾® 2 CO2(g) + 3 H2O(g) Heat of combustion of ethyl alcohol (g), DH°c = – 1277.53 kJ/kmol 2 C(s) + 2 O2(g) ¾ ¾¾® 2 CO2(g) 3 H2(g) + 1.5 O2(g) ¾ ¾¾® 3 H2O(g) D H°f of C2H5OH (g) = – 2 ´ 393.51 – 3 ´ 241.82 + 1277.53 = – 234.95 kJ/kmol D H° From Table AV.2, = – 234.95 kJ/kmol f (c) C4H6(l) + 5.5 O2 (g) ¾ ¾¾® 4 CO2(g) + 3 H2O(g) Heat of combustion of butadiene (1,3), DH°c = – 2386.69 kJ/kmol 4 C(s) + 4 O2(g) ¾ ¾¾® 4 CO2(g) 3 H2(g) + 1.5 O2(g) ¾ ¾¾® 3 H2O(g) D H°f of C4H6(l) = – 4 ´ 393.51 – 3 ´ 241.82 + 2386.69

Ans.

130 Solutions Manual—Stoichiometry

= 87.19 kJ/kmol From Table AV.2, D H°f = 87.19 kJ/kmol

Ans.

EXERCISE 5.28 (a)

(b)

DH°c (g) for hydrogen = lv at 298.15 for water = º D H°f (l) =

– 241.82 kJ/kmol 2442.5 kJ/kg 44.002 kJ/mol – 241.82 – 44.002

= – 285.82 kJ/mol D H°f (g) for methanol = – 200.94 kJ/mol

Ans.

Watson equation: é 512.5 – 298.15 ù lv at 298.15 K for methanol = 35 210 ê ú ë 512.5 – 337.7 û = 38 048 kJ/kmol D H°f (l) = – 200.94 – 38.05 = – 238.99 kJ/mol

0.38

Ans.

NIST equation: For methand, pc = 80.84 bar, Tc = 512.5 K At T = 298.15 K Tr = 0.5818 lv at 298.15 K= 45.3 ´ e–(–0.31)(0.5818) ´ (1 – 0.5818)0.4241 = 37.485 kJ/mol DH°f (l) = –200.94 – 37.485 = –238.43 kJ/mol (c)

Ans.

D H°f (g) for CS2 = 117.36 kJ/mol

Watson equation: é 552 – 298.15 ù l v = 26 740 ê ú ë 552 – 319.04 û = 27 627 kJ/kmol

0.38

D H°f (l) = 117.36 – 27.64 = 89.72 kJ/mol Ans. NIST equation: Tc = 552.0 K For CS2, pc = 73.00 bar At T = 298.15 K Tr = 298.15/552 = 0.5786 lv at 298.15 K = 37.07 ´ e–(0.2264 ´ 0.5786) ´ (1 – 0.5786)0.2264 = 26.739 kJ/mol DH°f (l) = +117.36 – 26.739 = 90.62 kJ/mol Ans.

Energy Balances 131

EXERCISE 5.29 Basis: 1 kg NaHCO3 D H ro = (– 1130.68 – 393.51 – 241.82) – 2(– 950.81) = + 135.61 kJ/2 mol NaHCO3 º + 67.805 kJ/mol NaHCO3 (endothermic) Molar mass of NaHCO3 = 84.0066 Moles of NaHCO3 = 1000/84.0066 = 11.904 mol/kg Heat to be supplied for dissociation = 67.805 ´ 11.904 Ans. = 807.15 kJ/kg NaHCO3

EXERCISE 5.30

d

D H ro = D Hfo

i

products

d

– D Hfo

i

reactants

= (– 1412.2 – 393.51) – (– 1130.68 – 824.2) = 149.17 kJ/mol Na2CO3 or mol Fe2O3 Ans.

EXERCISE 5.31 Basis: 100 kg CH4 reformed Input heat balance: Gas

kg

Molar kmol

Tempera-

mass

ture K(°C)

ni

(H° – DH°o + DH°f )i ni × (H° – DH°o + DH°f ), kJ

kJ/kmol

100 100 100

16 32 18

6.250 3.125 5.55

698 (425) 698 (425) 1253 (980)

– 38 416 + 2 102 – 192 447

– 240 100 + 65 691 – 1068 081

Total 300

—

14.925

—

—

– 1242 490

CH4 O2 H2O

Outlet Heat balance: Molar kmol

Tempera- (H° – DH°o – DH°f )i ni × (H° – DH°o

mass

ni

ture K(°C)

kJ/kmol

H2 27.3 H2O 79.2 CO 149.8 CO2 50.7

2 18 28 44

13.65 4.40 5.10 1.15

1198 (925) 1198 (925) 1198 (925) 1198 (925)

+ 35 134 – 194 882 – 76 942 – 339 706

+ 479 579 – 857 481 – 392 404 – 390 662

Total 300.0

—

24.30

—

—

1160 968

Gas

kg

Heat of reaction,

+ DH°f )i, kJ

D H r° = 1160 968 – (1242 490) = 81 522 kJ (endothermic)

Ans.

132 Solutions Manual—Stoichiometry

EXERCISE 5.32 Basis: 0.1 kg coke and N2 flow of 0.025 L × s–1. These are same bases as those of Exercises 3.24 and 4.1. Reaction: CH0.6 + 1.6 O2 = CO2 + 0.6 H2 – 101.1 1.6 ´ 0 – 393.51 0.6 ´ (– 241.82) DH°f D H °r = – 393.51 – 0.6 ´ 241.82 + 101.1 = – 393.51 – 145.09 + 101.1 = – 437.5 kJ/1.6 mol oxygen (exothermic) Actual oxygen consumed = 0.128 mol Heat generated = 437.5 ´ 0.128/1.6 = 35 kJ Ans.

EXERCISE 5.33 Many data are taken from Example 4.20. Heat capacity Data of RII Exit Gas stream Component CH3OH

. ni kmol/h

Heat capacity (C°mpr2 ) equation constants . . . . ai× ni bi× ni ´ 103 ci× ni ´ 106 di× ni ´ 109

1.415

35.2

72.0

83.0

– 63.9

218.267

10 520.5

—

—

—

CO2

16.481

352.1

1059.5

– 676.6

161.5

CO

1.857

53.9

– 5.2

21.6

– 8.7

H2

3.714

106.3

3.8

– 0.5

2.9

CH4

1.16

22.3

60.5

13.9

– 13.1

1.858

122.3

—

—

—

O2

159.270

4145.1

1872.2

– 373.1

– 89.6

N2

1082.930

32 044.9

– 5567.3

14 276.2

– 5380.0

H2O

268.04

8709.2

21.3

3541.0

– 1218.9

Total

1754.852

56 111.8

– 2483.2

16 885.5

– 6609.8

HCHO

(CH3)2O

In the gas to gas heat exchanger, air is preheated from 308.15 K to 523.15 K. Air flow rate is 1395.1 kmol/h against 1363.1 kmol/h in Example 4.10. From Example 5.37 (Table 5.31),

z

523.15

Heat duty of heat exchanger f6 = (1395.1/1363.1)

C°mpa

308.15

Energy Balances 133

= 8935 667 kJ/h º 2482.13 kW In the static mixer (I), 1395.1 kmol/h of air at 523.15 K is mixed with 93 kmol/ h methanol at 351.47 K (i.e. at the outlet of evaporator). From Table 5.31,

z

T1

(93/125)

z

T1

C°mpme dT = (1395.1/1363.1)

351. 47

C°mpa dT

523.15

Solving by Mathcad, T1 = 504.9 K or 231.75 °C Enthalpy of reactor input stream over 298.15 K,

z

504.9

f2 = (93/125)

z

504.9

C°mpme dT + (1395.1/1363.1)

298.15

C°mpa

298.15

= 9546 895 kJ/h º 2651.92 kW Heat of reactions at 298.15 K in RI, D H°rI = 20320 839 ´ 92.07/123.75

= 15118 704 kJ/h º 4199.64 kW Heat Capacity Data for RI Exit Gas stream . Component ni, kmol/h

. ai × ni

Heat capacity (Cpmr1 ° ) equation constants . . . bi × ni ´ 103 ci × ni ´ 106 di × ni ´ 109

CH3OH HCHO

0.93 92.07

23.1 4437.7

47.3 —

54.5 —

– 42.0 —

CO2 CO H2 CH4 (CH3)2O O2

6.537 0.737 1.474 0.460 0.737 236.865

139.7 21.4 43.2 8.9 48.5 6164.6

420.2 – 2.1 1.5 24.0 — 2784.4

– 268.3 8.6 – 0.2 5.5 — – 554.9

64.1 – 3.5 1.1 – 5.2 — – 133.2

N2 H2O

1082.930 120.974

32 044.9 3930.7

– 5567.3 9.6

14 276.2 1598.2

– 5380.0 – 550.1

Total

1543.714

46 861.7

– 2282.4

15 119.6

– 6048.8

Enthalpy of RI exit stream, at 613.15 K over 298.15 K,

134 Solutions Manual—Stoichiometry

z

613.15

f ¢3 =

Cmpr1 ° dT

298.15

= 15 260 251 kJ/h º 4238.96 kW Heat transfer in RI = 9546 895 + 15 118 704 – 15 260 251 = 9405 348 kJ/h º 2612.60 kW Enthalpy of additional methanol vapours added to RI exit stream 140.548 f 3¢¢ = 125

z

351. 47

Cmpa ° dT

298.15

= 344 999 kJ/h º 95.83 kW Total enthalpy of gas mixture, entering RII = 15 260 251 + 344 999 = 15 605 250 kJ/h Let T2 be the temperature of stream, entering RII. By Mathcad, T2 = 582.3 K or 309.15 °C Enthalpy of RII exit gas stream,

z

613.15

f4 =

C°mpr2 dT

298.15

= 18 246 584 kJ/h º 5068.50 kW 140.063 ´ 20 320 639 123.75 = 22 999 351 kJ/h

Heat of reaction in RII =

º 6388.71 kW Heat transfer in RII = 15 260 251 + 22 999 351 – 18 246 584 = 20 013 018 kJ/h º 5559.17 kW RII exit gases leave the gas heat exchanger at 383.15 K (110°C). Heat transfer in BFW heater and heat exchanger

z

613.15

=

C°mpr2 = 13 404 122 kJ/h

383.15

Energy Balances 135

Heat transfer in BFW heater, f5 = 13 404 122 – 8935 667 = 4468 455 kJ/h º 1241.24 kW Total heat transfer taken place for steam generation = 9405 348 + 20 013 018 + 4468 455 = 33 886 821 kJ/h º 9413.01 kW Steam generated = = Total methanol evaporated = = Heat duty of evaporator f1 = = º

33 886 821 2629.3 12 888.2 kg/h 93 + 140.548 233.548 kmol/h 233.548 ´ 4551 263 125 8503 506 kJ/h 2362.01 kW

8503 506 2119.7 = 4011.7 kg/h

Steam consumption in evaporator, ms =

Ans.

EXERCISE 5.34 (a) Basis: 1 kmol CH4 Reaction: CH4 + C2H4 = C3H8 (gaseous) Gas

D H°f kJ/kmol

CH4 C2H4 C3H8

– 74 520 + 52 550 – 104 680

Constants of C°mp equation a 19.2494 4.1261 – 4.2227

b ´ 103

c ´ 106

d ´ 109

52.1135 155.0213 306.264

11.973 – 81.5455 – 158.6316

– 11.3173 16.9755 32.1455

D H °r r= – 104 680 – (– 74 520 + 52 550) = – 82 710 kJ/kmol

D a = – 4.2227 – 4.1261 – 19.2494 = – 27.5982 Db ´ 103 = 306.264 – (155.0213 + 52.1135) = 99.1292 Dc ´ 106 = – 158.6316 + 81.5455 – 11.973 = – 89.0591 Dd ´109 = 32.1455 – 16.9755 + 11.3173 = 26.4873

136 Solutions Manual—Stoichiometry

DH°o = D H°RT – DaT – (Db/2)T 2 – (Dc/3)T 3 – (Dd/4)T 4

For T = 298.15 K, D H r°T = D H °r = – 82 710 kJ/kmol D H °0 = – 82 710 + 27.5982 ´ 298.15 – (99.1297 ´ 10–3/2) (298.15)2

+ (89.0591 ´ 10–6/3)(298.15)3 – (26.4873 ´ 10–9/4)(298.15)4 = – 82 710 + 8828 – 4406 + 787 –52 = – 77 553 kJ/kmol – heat of reaction at 0 K D Hr°T = – 78 154 – 27.598 T + 49.565 ´ 10–3 T2 – 29.686 ´ 10–6 T 3

+ 6.622 ´ 10–9 T4 Ans. CO(g) + H2O(g) = CO2(g) + H2(g)

(b) D H°f

– 110.53

– 241.82

– 393.51

0

all in kJ/mol

D Hr° = – 393.51 – (– 110.53 – 241.82)

= – 41.16 kJ/mol Gas

Constants of C°mp equation b ´ 103

a CO2 H2 CO H2O

21.3655 28.6105 29.0277 32.4921

64.2841 1.0194 – 2.8165 0.0796

c ´ 103

d ´ 103

– 41.0506 – 0.1476 11.6437 13.2107

9.7999 0.769 – 4.7063 – 4.5474

D a = 21.3655 + 28.6105 – 29.0277 – 32.4921 = – 11.5438 Db ´ 103 = 64.2841 + 1.0194 – (– 2.8165 + 0.0796) = 68.0404 Dc ´ 103 = – 41.0506 – 0.1476 – (11.6437 + 13.2107) = – 66.0526 Dd ´ 103 = 9.7999 + 0.769 – (– 4.7063 – 4.5474) = 19.8226 D H 0° = – 41 160 + 11.5438 (298.15) + 68.0404 ´ 10–3 (298.15)2/2

+ 66.0526 ´ 10–6 (298.15)3/3 – 19.8226 ´ 10–9 (298.15)4/4 = –41 160 + 3442 – 3024 + 583 – 39 = –40 378 kJ/kmol at 0 K D H °r = –40 378 – 11.544 T + 34.02 ´ 10–3 T 2 – 22.018 ´ 10–6 T 3

+ 4.956 ´ 10–9 T 4 (c) Basis = 1 kmol CO consumed or CH3OH produced CO(g) + 2 H2(g) = CH3OH(g)

Ans.

Energy Balances 137

Gas

DHºf kJ/mol

CH3OH CO H2

– 200.94 – 110.53 O

Constants of C mp º equation a 24.8692 29.0277 28.6105

b ´ 103

c ´ 103

d ´ 103

50.8755 – 2.8165 1.0194

58.6278 11.6437 – 0.1476

– 45.1266 – 4.7063 0.769

DHºr = – 200.94 – (– 110.53) = – 90.41 kJ/mol CO consumed Da = 24.8692 – (29.0277 + 2 ´ 28.6105) = – 61.3795 Db ´ 103 = 50.8755 – (– 2.8165 + 2 ´ 1.0194) = 51.6532 Dc ´ 106 = 58.6278 – (11.6437 – 2 ´ 0.1476) = 47.2793 Dd ´ 109 = – 45.1266 – (– 4.7063 + 2 ´ 0.769) = – 41.9583 DHº0 = – 90 410 + 61.3795 ´ 298.15 – 51.6532 ´ 10–3 (298.15)2/2 – 47.2793 ´ 10–6 (298.15)3/3 + 41.9583 ´ 10–9 (298.15)4/4 = –90 410 + 18 300 – 2296 – 418 + 83 = – 74 741 kJ/kmol DHºrT = – 74 741 – 61.38 T + 25.827 ´ 10–3 T2 + 15.76 ´ 10–6 T3 – 10.49 ´ 10–9 T 4 Ans. At T = 593.15 K DHºrT = – 78 741 – 61.38 ´ 593.15 + 25.827 ´ 10–3 (593.15)2 + + 15.76 ´ 10–6 (593.15)3 – 10.49 ´ 10–9 (593.15)4 = – 74 748 – 36 408 + 9087 + 3289 – 1298 = – 100 071 kJ/kmol CH3OH Ans.

EXERCISE 5.35 Let H¢ be enthalpy in kJ/kg, T be temperature in K and p be pressure in bar a H=

H¢ 1.055 056 ´ 2.204 623

H¢ 2.326 t = 1.8T – 459.67 p and p¢ = or p = 14.503 77 p¢ 14.503 77 Substitute these values in equation. H = – 34.38 + 0.7209 (1.8T – 459.67) 2.326 ´ 105 1075 . + 7.763 ´10–6 (1.8T – 459.67)2 + 14.503 77 p Simplifying 17.2403 ´ 103 H = – 846.934 + 2.9884 T + 58.5038 ´ 10–6T2 + Ans. p =

138 Solutions Manual—Stoichiometry

EXERCISE 5.36 Basis: 100 kmol outgoing gas mixture Gas

kmol ni incoming

outgoing

CO2 H2 CH4 CO

58.9 47.94 — —

57.10 41.10 1.68 0.12

Total

107.84

100.00

Constants of Cºmp equation

a 21.3655 28.6105 19.2494 29.0277

b ´ 103

c ´ 106

d ´ 109

64.2841 1.0194 52.1135 – 2.8165

– 41.0506 – 0.1476 11.973 11.6437

9.7999 0.769 – 11.3173 – 4.7063

T1 = 298.15 K, T2 = 588.15 K H1 =

T2

ò (ni × c pi ) × dT

T1

= (21.3655 ´ 58.9 + 28.6105 ´ 47.94) (588.15 – 298.15) + (64.2841 ´ 58.9 + 1.0194 ´ 47.94) 10–3 (588.152 – 298.152)/2 – (41.0506 ´ 58.9 + 0.1476 ´ 47.94) 10–6 (588.153 – 298.153)/3 + (9.7999 ´ 58.9 + 0.769 ´ 47.94) 10–9 (588.154 – 98.154)/4 = 762 704 + 492 875 – 143 032 + 17 157 = 1129 704 kJ Heat of reactions: CO2(g) + 4 H2(g) = CH4(g) + 2 H2 (g) (1) DHºr1 = 2 (– 241.82) + (– 74.52) – [(– 392.51) + 0] = – 164.65 kJ/mol CO2 reacted (exothermic) Actual CO2 consumed as per reaction (1) = 1.68 kmol Heat evolved = 164 650 ´ 1.68 = 276 612 kJ CO2(g) + H2(g) = CO(g) + H2O(g) (2) DH ºr2 = – 241.82 – 110.53 + 393.51 = 41.16 kJ/mol CO2 (endothermic) Actual CO2 consumed as per reaction (2) = 0.12 kmol Heat absorbed = 41 160 ´ 0.12 = 4939 kJ Total enthalpy of product gas mixture = 1129 704 + 276 612 – 4939 = 1401 377 kJ Let T3 be the final temperature of product gas mixture. H1 =

T3

ò

(ni × c pi ) × dT

298.15

= (57.1 ´ 21.3655 + 41.1 ´ 28.6105 + 1.68 ´ 19.2494 + 0.12 ´ 29.0277) (T3 – 298.15) + (64.2841 ´ 57.1

+ 41.1 ´ 1.0194 + 1.68 ´ 52.1135 + 0.12 ´ (– 2.8165)) ´ 10–3

Energy Balances 139

(T23 – 298.152)/2 + (– 41.0506 ´ 57.1 – 0.1476 ´ 41.1 + 11.973 ´ 1.68 + 11.973 ´ 1.68 + 11.6437 ´ 0.12) ´ 10–6 (T32 – 298.152)/3 + (9.7999 ´ 57.1 + 0.769 ´ 41.1 – 11.3173 ´ 1.68 – 4.7063 ´ 0.12) ´ 10–9 (T3 – 298.15)/4 H2 = 2431.68 (T3 – 298.15) + 3799.7 ´ 10–3 (T32 – 2982.15)/2 – 2328.54 ´ 10–6 (T33 – 2983.15)/3 + 571.6 ´ 10–9 (T 43 – 2984.15)/4 = 2431.68 T3 + 1899.87 ´ 10–9 T34 – 776.23 ´ 10–6 T 39 – 142.9 ´ 10–9 T34 – 724 641 – 168 716 + 20 542 – 1127 = 1401 377 Using Mathcad T3 = 669.61 K or 396.46°C Ans.

EXERCISE 5.37 Basis: 100 kmol dry gas mixture, entering the shift converter. Refer Exercise 5.32 (b). T1 = 618.15 K DH ºrT1 = – 41 160 – 7134 + 12 993 – 5197 + 723 = – 39 775 kJ/kmol CO T2 = 783.15 K DH ºrT2 = – 41 160 – 9039 + 20 857 – 10 570 + 1863 = – 38 049 kJ/kmol CO Average heat of reaction, DHºrT = [– 39 775 + (– 38 049)]/2 = – 38 912 kJ/kmol CO Gas CO H2 CO2 N2 Ar H2O

(Hº – Hº + DHºf ), kJ/kmol 618 K

783 K

– 95 804 17 723 – 370 400 18 029 — – 218 133

– 90 644 22 585 – 362 262 23 126 — – 211 933

Average Cºmp between 618 K and 783 K, 31.273 29.467 49.321 30.891 39.946 (constant) 37.576

Partial oxidation of NG: Sni × Cºmpi = ( 0.37 ´ 31.272 + 0.6 ´ 29.467 + 0.2 ´ 49.321 + 0.01 ´ 39.948) 100 = 3063.7 kJ/(100 kmol dry gas × K)

140 Solutions Manual—Stoichiometry

Partial oxidation of FO: Sni × Cºmpt = ( 0.46 ´ 31.273 + 0.48 ´ 29.467 + 0.05 ´ 49.321 + 0.01 ´ 39.948) 100 = 3139.5 kJ/(100 kmol dry gas × K) Reforming of NG: Sni × Cºmpt = (0.15 ´ 31.273 + 0.56 ´ 29.467 + 0.07 ´ 49.321 + 0.217 ´ 30.891 + 0.003 ´ 39.948) 100 = 3116.8 kJ/(100 kmol dry gas × K) Average Sni × Cºmpt = (3063.7 + 3139.5 + 3146.8)/3 = 3116.7 kJ/(100 kmol dry gas × K) For steam, accompanying the gas mixture, ni × Cºmpt = 37.576 ´ 100 ´ x = 3757.6 x kJ/(100 kmol × K) Rise in temperature in shift converter, DT = »

38912 a 12.485 a = K 3116.7 + 3757.6 x 1 + 1.206 x 12.5 a K 1 + 1.2 x

Ans.

EXERCISE 5.38 Basis: Flowrate of offgases from uranium oxide dissolver = 38.3 Nm3/h Flow rate = 919.2 Nm3/day º 41.01 kmol/day NO2 entering the reactor = 41.01 ´ 0.025 = 1.025 kmol/day Reaction: 8 NH3(g) + 6 NO2(g) = 7 N2(g) DHºf : kJ/mol – 45.94 33.18 0 o DH r = – 241.82 ´ 12 – (45.94 ´ 8 + 6 ´ 33.18) = – 2733.4 kJ/6 mol NO2 º – 455 567 kJ/kmol NO2 (exothermic) Total heat generated = 455 567 ´ 1.025 = 466 956 kJ/d º 5.405 kW Theoretical NH3 required = 4 ´ 1.025/3 = 1.367 kmol/d Actual NH3 fed = 2 Nm3/h º 0.089 kmol/h º 2.141 kmol/d Excess NH3 = [(2.141 – 1.367)/1.367]/100 = 56.6%

+ 12 H2O(g) –241.82

Ans. (a)

Energy Balances 141

Reactants: Component

. n i, kmol/d

N2 O2 NO2 H2O NH3

11.278 2.051 1.025 26.656 2.141

. n i× a i 337.73 53.38 24.17 866.11 54.92

Total

41.099

1336.31

Constants of Cºmp equation . . . n i× bi ´ 103 n i× ci ´ 106 n i× di ´ 109 – 57.98 148.68 – 56.03 24.11 – 4.80 – 1.15 54.93 – 32.28 6.70 2.12 352.14 – 121.22 71.68 0.75 – 6.6 94.86

464.49

– 178.3

Enthalpy of reactants at 393.15 K (120ºC) over 298.15 K (25ºC): f1 = 1336.31 (393.15 – 298.15) + 94.86 ´ 10–3 (393.152 – 298.152)/2 + 464.49 ´ 10–6 (393.153 – 298.153)/3 – 178.3 ´ 10–9 (393.154 – 298.154)/4 = 126 949 + 3114 + 5301 – 712 = 134 652 kJ/d º 1.558 kW Total enthalpy of product gas mixture = f1 + DHºr = 134 652 + 466 956 = 601 608 kJ/d º 6.963 kW Products: Component

. n i, kmol/d

N2 O2 H2O NH3

12.474 2.051 28.706 0.774

. n i× a i 369.12 53.38 932.72 19.85

Total

44.005

1375.07

Constants of Cºmp equation . . . n i× bi ´ 109 n i× ci ´ 106 n i× di ´ 109 – 64.13 164.44 – 61.97 24.11 – 4.80 – 1.15 2.28 379.23 – 130.54 25.91 0.27 – 2.39 – 11.83

539.14

– 196.05

Enthalpy of products at T K over 298.15 K, f2 = 1375.07 (T – 298.15) – 11.83 ´ 10–3 (T2 – 298.152)/2 + 539.14 ´ 10–8 (T3 – 298.153)/3 – 196.05 ´ 10–3 (T4 – 298.154)/4 = 1375.07 – 5.92 ´ 10–3 T2 + 179.71 ´ 10–6 T3 – 49.01 ´ 10–9 T4 – 409 771 + 525 – 4756 + 387 = 601 608 kJ/d º 6.963 kW By trial and error or by Mathcad, T = 703.78 K or 430.63ºC Ans. (b)

EXERCISE 5.39 Basis: 38.3 Nm3/h offgases Flow rate = 41.01 kmol/d (Ref. Exercise 5.38) NO2 entering the reactor = 41.01 ´ 0.035 = 1.435 kmol/d

142 Solutions Manual—Stoichiometry

Theoretical NH3 required = 1.435 ´ 4/3 = 1.913 kmol/d Excess NH3 = [(2.141 –1.913)/1.913]100 = 11.92% Total heat generated = 1.435 ´ 455 567 = 653 739 kJ/d º 7.566 kW Reactants: . Components n i kmol/d

. n i× a i

Constants of Cºmp . . n i× bi ´ 103 n i× ci ´ 106

. n i× di ´ 109

N2 O2 NO2 H2O NH3

10.868 2.051 1.435 26.656 2.141

321.59 53.38 33.84 866.11 54.92

– 55.87 24.11 76.91 2.12 71.68

143.27 – 4.80 – 45.19 352.14 0.75

– 53.99 – 1.15 9.38 – 121.22 – 6.6

Total

43.151

1329.84

118.95

446.17

– 173.58

Enthalpy of reactants at 393.15 K (120ºC) over 298.15 K (25ºC), f1 = 1329.84 (393.15 – 298.15) + 118.95 ´ 10–3 (393.152 – 298.152)/2 + 446.17 ´ 10–6 (393.153 – 298.153)/3 – 173.58 ´ 10–9 (393.154 – 298.154)/4 = 126 335 + 3904 + 5092 – 693 = 134 638 kJ/d º 1.558 kW This enthalpy does not include enthalpy of nitrogen (x kmol/d) at 303.15 K (30ºC). Total enthalpy of product gas mixture (excluding that of N2 stream), f¢1 = 134 638 + 653 739 = 788 377 kJ/d º 9.125 kW NH3 consumed= 1.913 kmol/d N2 produced = 1.913 ´ 7/8 = 1.674 kmol/d Total N2 in product stream = 10.868 + 1.674 = 12.542 kmol/d H2O produced = 1.913 ´ 12/8 = 2.87 kmol/d Total H2O in product stream = 26.656 + 2.87 = 29.526 kmol/d Products (excluding extra nitrogen supply): . Components ni kmol/d

. n i× a i

Constants of Cºmp equation . . n i× bi ´ 103 n i× ci ´ 106

. n i× di ´ 109

N2 O2 H2O NH3

12.544 2.051 29.526 0.228

371.13 53.38 959.36 5.85

– 64.48 24.11 2.35 7.63

165.34 – 4.80 390.06 0.08

– 62.31 – 1.15 – 134.27 – 0.7

Total

44.347

1389.72

– 30.39

550.68

– 198.43

Enthalpy of product gas stream (excluding extra N2 supply) at 753.15 K over 298 K, f¢2 = 1389.72 (753.15 – 298.15) – 30.39 ´ 10–3 (753.152 – 298.152)/2

Energy Balances 143

+ 550.68 ´ 10–6 (753.153 – 298.153)/3 – 198.43 ´ 10–9 (753.154 – 298.154)/4 = 632 323 – 7266 + 73 515 – 15 558 = 683 014 kJ/d º 7.905 kW Enthalpy carried by extra nitrogen in feed stream f¢1 – f¢2 = 788 377 – 683 014 = 105 363 kJ/d º 1.219 kW Enthalpy of extra nitrogen at 753.15 K over 303.15 K = 29.5909 (753.15 – 303.15) – 5.141 ´ 10–3 (753.152 – 303.152)/2 + 13.1829 ´ 10–6 (753.153 – 303.153)/3 – 4.968 ´ 10–9 (753.154 – 303.154)/4 = 133 16 – 1222 + 1754 – 389 = 13 459 kJ/kmol N2 Flow of nitrogen required to control the temperature = 105 363/13 459 = 7.828 kmol/d º 7.31 Nm3/h Ans.

EXERCISE 5.40 Basis: 1 kmol toluene O2 requirement for benzaldehyde production = 1 kmol Excess O2 = 100% Actual O2 supply = 2 kmol N2 supply = 2 ´ 79/21 = 7.524 kmol Air supply = 2 + 7.524 = 9.524 kmol DHor = – 36.7 – 241.82 – (50.17) = – 328.69 kJ/mol Side reaction is the standard combustion reaction. DHoc = –3772.0 kJ/mol (NHV) Toluene burnt to CO2 and H2O = 0.005 kmol Total toluene converted = 0.13 kmol Total oxygen consumed = 0.125 + 0.005 ´ 9 = 0.17 kmol Toluene reacted to benzaldehyde = 0.13 – 0.005 = 0.125 kmol Total heat produced at 298.15 K (25ºC) = 0.125 ´ 328 690 + 0.005 ´ 377 200 = 42 972 kJ Inlet gas mixture is at 448.15 K (175ºC):

144 Solutions Manual—Stoichiometry

Constants of Cºmp equation Components kmol ni C7H8 1.0 2.0 O2 N2 7.524 Total

10.524

mole %

ni×ai

ni×bi ´ 103

9.5 19.0 71.5

– 35.19 52.05 222.64

563.18 23.51 – 38.68

– 349.80 – 4.69 99.19

82.59 – 1.12 – 37.38

548.01

– 255.30

44.09

100.0

239.500

ni×ci ´ 106

ni×di ´ 109

Enthalpy of ingoing gas mixture at 448.15 K over 298.15 K, H1 = 239.5 (448.15 – 298.15) + 548.01 ´ 10–3 (448.152 – 298.152)/2 – 255.3 ´ 10–6 (448.153 – 298.153)/3 + 44.09 ´ 10–9 (448.154 – 298.154)/4 = 35 925 + 30 673 – 5404 + 357 = 61 551 kJ Outlet gas mixture at 468.15 K (195ºC): Constants of Cºmp equation Components kmol ni C7H8 0.87 O2 1.83 7.524 N2 C7H8O 0.125 0.035 CO2 H2O 0.020 Total

10.404

mole %

n i× a i

ni× bi ´ 103

ni× ci ´ 106

ni× di ´ 109

8.36 17.59 72.32 1.20 0.34 0.19

– 30.62 47.63 222.64 16.25 0.75 0.65

489.97 21.51 – 38.68 — 2.25 —

– 304.32 – 4.29 99.19 — 1.44 0.26

71.85 – 1.03 – 37.38 — 0.34 – 0.09

100.00

257.30

475.05

– 210.60

33.69

Enthalpy of outcoming gas mixture at 468.15 K over 298.15 K, H2 = 257.30 (468.15 – 298.15) + 475.05 ´ 10–3 (468.152 – 298.152)/2 – 210.6 ´ 10–6 (468.153 – 298.153)/3 + 33.69 ´ 10–9 (468.154 – 298.154)/4 = 43 741 + 30 943 – 5342 + 338 = 69 680 kJ Heat liberated = 61 551 + 42 972 – 69 680 = 34 843 kJ Ans.

EXERCISE 5.41 Basis: 1 kmol S2(g) Stoichiometric CH4 requirement = 0.5 kmol Actual supply of natural gas = 2 kmol Actual CH4 supply = 2 ´ 0.6 = 1.2 kmol Excess CH4 = 1.2 – 0.5 = 0.7 kmol % Excess = 0.7 ´ 100/05 = 140 Conversion = 80% S2 reacted = 0.8 kmol

Ans. (a)

Energy Balances 145

CS2 produced = 0.8/2 = 0.4 kmol Ratio =

CS2 produced 76 ´ 0.4 = = 30.4 kg/kmol S2 fed 1

Ans.(b)

Unreacted S2 = 1.0 – 0.8 = 0.2 kmol Unreacted CH4 = 1.2 – 0.4 = 0.8 kmol H2S produced = 0.8 kmol N2 entering with natural gas = 0.4 ´ 2.0 = 0.8 kmol Reactants: Gas

kmol, ni

mole %

CH4 S2 N2

1.20 1.00 0.80

40.0 33.3 26.7

Total

3.00

100.0

(Hº – Hºo + DH fº )i, kJ/kmol at 720 K (427ºC) – 38 305 22 951 20 537 —

ni× (Hº – H oº + DHfº)i, kJ 45 966 22 951 16 430 –

6 585

Products: Gas

kmol, ni

mole %

CH4 S2 CS2 H2S N2

0.8 0.2 0.4 0.8 0.8

26.67 6.67 13.33 26.66 26.67

Total

3.0

100.00

(Hº – Hºo + DHºf ), at 977 K (704 ºC) – 20 257 + 33 118 + 32 118 – 45 137 + 29 322 —

ni× (Hº – Hºo + DHºf ), kJ – 16 206 + 6 624 + 12 843 – 36 110 23 458 – 9 391

Heat change in the reactor = – 9391 – (– 6585) = – 2806 kJ/kmol S2 Antoine constants for CS2 A = 6.0677, B = + 1169.1 and C = – 31.55 at T = 298.15 K Log p = 4.066 83 – p = 0.481 bar or p = 48.1 kPa

Ans. (c and d)

1168.62 (298.15 - 31.616)

Ans. (e)

EXERCISE 5.42 Basis: 3 kmol H2S supply, including the bypass Out of 3 kmol, 1 kmol H2S enters the furnace and 2 kmol bypass the furnace. 3 H2S(g) + O2(g) = SO2(g) + H2O(g) (1) 2 (2) SO2(g) + 2 H2S(g) = 3 S(l) + 2 H2O(g)

146 Solutions Manual—Stoichiometry

Theoretical oxygen requirement = 1.5 kmol Oxygen supply = 40% excess N2, entering with air = 79 ´ 2.1/21 = 7.9 kmol N2, entering with H2S = 10 ´ 3/90 = 0.333 kmol Composition of gases, leaving the converter: Gas O2 N2 H2O

kmol 2.1 – 1.5 = 0.6 0.333 + 7.9 = 8.233 1.0

Total

9.833

mole % 6.11 83.73 10.16 100.00

Ans. (a) Heat balance across furnace boiler: DHºr = – 241.82 – 296.83 – (– 20.63) = – 518.02 kJ/mol H2S Since the reactants are at 298.15 K (25 ºC), DHRectants = 0 Product stream from furnace boiler: Component

ni kmol mole %

Constants of Cºmp equation

O2 H2O SO2 N2

0.6 1.0 1.0 8.233

5.54 9.23 9.23 76.00

n i× a i 15.615 32.492 24.771 243.622

Total

10.833

100.00

316.500

ni× bi ´ 103 7.053 0.08 62.948 – 42.326 27.755

ni× ci ´ 106 ni× di ´ 109 – 1.406 – 0.337 13.211 – 4.547 – 44.258 11.122 108.535 – 40.902 76.082

– 34.664

Enthalpy of product stream at 593.15 K over 298.15 K, DHProducts = 316.5 (593.15 – 298.15) + 27.775 10–3 (593.152 – 298.152)/2 + 76.082 ´ 10–6 (593.153 – 298.153)/3

– 34.664 ´ 10–9 (593.154 – 298.154)/4 = 93 368 + 3648 + 4617 – 1003 = 100 630 kJ Heat transfer at Q1 = 100 630 – 518 020 = – 417 390 kJ/kmol H2S Ans.(b) New basis: 9000 kg/h liquid sulphur production Production rate = 9000/32 = 281.25 kmol/h From both the reactions when 3 kmol H2S are consumed. 3 kmol liquid sulphur are produced. Of these 3 kmol H2S, one kmol enters the furnace boiler. H2S entering the furnace = 281.25/3 = 93.75 kmol/h

Energy Balances 147

Heat transfer at Q1 = 417 390 ´ 93.75 kmol/h = 39 130 313 kJ/h (heat liberation) º 10 869.5 kW

Ans. (c)

EXERCISE 5.43 Basis: 100 kmol fresh ethylene feed (96% pure) Heat capacity constants for combined (mixed) feed: Component

ni kmol

Constants of Cºmp

ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

C2H4 H2O N2

0.5483 0.3564 0.0953

2.262 11.58 2.82

84.998 0.028 – 0.49

– 44.711 4.708 1.256

9.308 – 1.621 – 0.473

Total

1.0000

16.662

84.536

– 38.747

7.214

Ref. Temperature T0 = 298.15 K, T1 = 573.15 K Enthalpy of feed gas = 2716.45 [16.662 (573.15 – 298.15) + 84.536 ´ 10–3 (573.152 – 298.152)/2 – 38.747 ´ 10–6 (573.153 – 298.153)/3 + 7.214 ´ 10–9 (573.154 – 298.154)/4] = 2716.45 [4582.05 + 10 127.73 – 2089.45 + 180.37] Hº = 2716.45 ´ 12 800.7 = 34 772 462 kJ Real gas enthalpy, H = 34 772 462 + 2716.45 ´ 962 = 37 385 687 kJ DHrº = – 234.95 – (– 241.82 + 52.50) = – 45.63 kJ/mol enthanol (exothermic) Total heat produced = 45 630 (1489.41 – 1414.94) = 3398 066 kJ Ideal gas conditions: Enthalpy of products= 34 772 462 + 3 398 066 = 38 170 528 kJ Heat capacity constants for reactor exit gas mixture: Constants of C mp º equation

Component

ni kmol

C2 H 4 H2O N2 C2H5OH

0.5356 0.3382 0.0980 0.0282

2.210 10.989 2.9 0.294

83.029 0.027 – 0.504 5.908

– 43.676 4.468 1.292 – 2.326

9.092 – 1.538 – 0.487 0.111

1.0000

16.393

88.46

– 40.242

7.178

Total

ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

148 Solutions Manual—Stoichiometry

Enthalpy of product gas mixture = 2641.98 [16.393 (T – 298.15) + 88.46 ´ 10–3 (T2 – 298.152)/2 – 40.242 ´ 10–6 (T3 – 298.153)/3 + 7.178 ´ 10–9 (T 4 – 298.154] = 38 170 528 Solving by Mathcad, T = 605.02 K or 331.87°C Real enthalpy of product stream

= 37 385 687 + 3398 066 = 40 783 753 kJ

Ideal gas enthalpy of product stream = 40 783 753 – 914 ´ 2641.98 = 38 368 983 kJ Solving by Mathcad, T = 606.42 K or 333.27 °C Ans. There is hardly any effect of residual enthalpy on reaction exit gas mixture temperature.

EXERCISE 5.44 Standard heat of reaction at 25 °C (298.15 K), DH°r = – 479.8 – 92.31 – (–484.2) – 0 (Appendix AIV.2) = –87.91 kJ/mol reactant Average temperature = (40 + 100)/2 = 70°C Mean Heat Capacity at 100°C (373.15 K) Component

State

Acetic acid Hydrogen chloride MCA Chloride

Liquid Gas Liquid Gas

Cpmi, kJ/(kg × K) 1.686 0.7977 1.88 0.45

Cpmi values are calculated using, equations given in Tables 5.1 and 5.3. DHr at 100°C (373.15 K) = –87.99 ´ 1000 + (1.88 + 0.7977 – 1.686 – 0.485) (373.15 – 298.15) = –87 842 kJ/kmol reacant º –87.842 kJ/mol reactant Ans.(a) . Let nt = Molar flow rate of gas-vapour mixture, leaving the reactor, kmol/h Basis: 100 kg/h MCA production rate molar production rate = kmol/h Molar feed rate of chlorine = 1.2 ´ 10.582 = 12.6984 kmol/h Chlorine gas, leaving the reactor = 12.6984 – 10.582 = 2.1164 kmol/h

1000 = 10.582 94.5

Energy Balances 149

Molar flow rate of HCl = 10.582 kmol/h nt = nCl2 + nHCl + nAA + nMCA

and

pt = pCl + pHCl + PAA + PMCA 2 = pCl2 + pHCl + pvAA × xAA + pvMCA × xMCA

At 100°C (373.15 K),

log pvAA = 4.682 06 –

1642.540 373.15 - 39.764

pvAA = 0.569 14 bar º 426.9 Torr log pvMCA = 4.690 87 –

pCl

2

pvMCA = º pt = = + pHCl =

pCl2 + pHCl pt

=

1733.960 373.15 - 92.154

0.0331 bar 24.8 Torr 1.1 atm = 836 Torr pCl + pHCl + 0.5 ´ 426.9 + 0.5 ´ 24.8 2 610.15

nCl2 + nHCl nt

10.582 + 2.1164 610.5 = nt 836 nt = 17.3888 kmol/h

nAA =

426.9 ´ 0.5 ´ 17.3888 = 4.4398 kmol/h 836

nMCA =

24.8 ´ 0.5 ´ 17.3888 = 0.2579 kmol/h 836

This gas-vapour mixture, is cooled to 40°C (313 K) with the help of cooling water in overhead condenser. Trial and error method is adopted to find the condensation rates. Trial - I: Assume total condensation of acetic acid and MCA. x¢AA = Mole fraction of AA in the condensed liquid =

4.4398 = 0.9451 (4.4398 + 0.2579)

x¢MCA = 1 – 0.9451 = 0.0549

150 Solutions Manual—Stoichiometry

Assume a pressure drop of 50 Torr in the heat exchanger. Gases leave the overhead condenser at p¢t = 836 – 50 = 786 Torr and 40°C At T = 40 + 273.15 = 313.15 K, p¢vAA = 0.0472 bar º 35.4 Torr p¢vMCA = 0.0007 bar º 0.52 Torr 786 = p¢HCL + p¢Cl + 0.9451 ´ 35.4 + 0.0549 ´ 0.52 2 p¢CL + p¢HCl = 752.51 Torr 2

10.582 + 2.1164 752.51 = nt¢ 786 t = 13.2635 kmol/h n¢ AA = n¢ MCA = n¢

13.2635 ´ 0.9451 ´ 35.4 = 0.5646 kmol/h 786 0.0549 ´ 0.52 ´ 13.2635 786

= 0.482 ´ 10–3 kmol/h Trial II: For this trial, composition of the condensed liquid is calculated as follows. 4.4398 - 0.5646 x¢¢AA = (4.4398 - 0.5646) + (0.2579 - 0.482 ´ 10-3 )

= x¢¢MCA = p¢¢Cl + p¢¢HCl = 2 = t = n¢¢

0.9387 0.0613 786 – 0.9387 ´ 35.4 – 0.0613 ´ 0.52 752.74 Torr (10.582 + 2.1164) ´ 786 752.74

= 13.2595 kmol/h AA = n¢¢ MCA = n¢¢

13.2595 ´ 0.9387 ´ 35.4 = 0.5606 kmol/h 786 13.2595 ´ 0.0613 ´ 0.52 = 0.538 ´ 10–3 kmol/h 786

4.4398 - 0.5646 = x¢¢¢ AA (4.4398 - 0.5646) + (0.2579 - 0.538 ´ 10-3 )

Energy Balances 151

= 0.9377 x¢¢¢ MCA = 0.0623

Both these values are close enough to x¢¢AA and x¢¢MCA and hence new trial is not justified. Acetic acid condensed = (4.4398 – 0.5646) 60 = 232.51 kg/h MCA condensed = (0.2579 – 0.538 ´ 10–3) 94.5 = 24.32 kg/h Heat duty of overhead condenser: fc = sensible heat transfer of gas-vapour mixture + Latent heat transfer for condensation of AA and MCA vapours + Subcooling of condensed liquid





fc = m Cl2 × CpCl2 + m HCl × CpHCl + m AA × CpAA + m MCA × CpMAA DTv +  m AA ¢ × lAA + m MCA ¢ × lMCA  + m AA ¢¢ × ClAA + m MCA ¢¢ ClMCA  DTL m Cl2 = 2.1164 ´ 71 = 150.26 kg/h m HCl = 10.528 ´ 36.5 = 386.24 kg/h æ 4.4398 + 0.5646 ö ÷ø 60 m AA = çè 2

= 150.13 kg/h m MCA

(0.2579 + 0.538 ´ 10-3 ) 94.5 = 2

= 12.21 kg/h  AA = 232.75 kg/h m¢  MCA = 24.32 kg/h m¢

 AA = m¢¢  MCA = m¢¢

0 + 232.75 = 116.38 kg/h 2 0 + 24.32 = 12.16 kg/h 2

Specific Heats of Liquids and Gases at 70°C (143.15 K) Component Chlorine Hydrogen chloride Acetic acid MCA

Cpi, kJ/(kg × °C) 0.4864 0.7977 1.2254 1.0467

CLi, kJ/(kg × °C) – – 1.7 1.9

152 Solutions Manual—Stoichiometry

AT 70°C (143.15 K), following latent heats of vaporization are calculated using Watson equation. lAA = 25 668 kJ/kmol = 427.8 kJ/kg lMCA = 33 633 kJ/kmol = 355.9 kJ/kg fc = (150.26 ´ 0.4864 + 386.24 ´ 0.7977 + 150.13 ´ 1.2254 + 12.21 ´ 1.0467) (100 – 40) + (232.51 ´ 427.8 + 24.32 ´ 355.9) + 116.38 ´ 1.7 + 12.16 ´ 1.9) (100 – 40) = 34676.3 + 108 123.3 + 13 248.6 = 156 048.2 kJ/h º 43.347 kW Ans.(b) Energy balance around reactor: Heat to be removed in jacket = Heat produced in reaction – Heat removed in overhead condenser. = 87 842 ´ 10.582 – 156 579 = 772 965 kJ/h º 214.713 kW Ans.(c)

EXERCISE 5.45 The rate of chemical reaction is R =

dM = M {e ( 0.080 73 T -40.1841) } /183 672 dG

The equation can be rewritten as DM = Dq × M {e ( 0.080 73 T-40.1841) } /183 672 where DM = M0 – M = initial mass of reactant – final mass of reactant M 0 = 250 kg T1 = 535 K Energy balance: M1Cp (T1 – T0) = DHr (M0 – M1) where Cp = average heat capacity of total mass = 1.675 kJ/(kg × k) T1 = temperature of mass after Dq s For Dq = 1 s DM = = M1 = = =

250 {e ( 0.080 73 ´ 535 - 40.1841) } /183 672 0.027 52 kg M0 – D M 250 – 0.027 52 249.972 48 kg

Energy Balances 153

Energy balance: 249.972 48 ´ 1.675 (T1 – 535) = 607 (0.027 52) T1 – 535 = 0.04 T1 = 535.04 K Similar iterations can be performed for 24 seconds which are tabulated below. Time s

Mass of reactants, kg

Rate of recion, kg/s

Reaction temperature, K

0

250

0.027 52

535.04

1

249.972 48

0.027 60

535.08

2

249.944 88

0.027 69

535.12

3

249.917 19

0.027 77

535.16

4

249.889 42

0.027 86

535.204

5

249.861 56

0.027 96

535.245

6

249.833 60

0.028 05

535.286

7

249.805 55

0.028 14

535.327

8

249.777 41

0.028 23

535.368

9

249.749 18

0.028 32

535.409

10

249.720 86

0.028 41

535.450

11

249.692 45

0.028 50

535.491

12

249.663 95

0.028 59

535.533

13

249.635 36

0.028 68

535.575

14

249.606 68

0.028 78

535.617

15

249.577 90

0.028 87

535.659

16

249.549 03

0.028 97

535.701

17

249.520 06

0.029 06

535.743

18

249.491 00

0.029 16

535.785

19

249.461 84

0.029 25

535.828

20

249.432 59

0.029 35

535.871

21

249.403 24

0.029 45

535.914

22

249.373 79

0.029 55

535.957

23

249.344 24

0.029 65

536.000

24

249.314 59

At q = 24 s, M = 249.319 59 kg

Ans.

154 Solutions Manual—Stoichiometry

EXERCISE 5.46 Heat of crystallisation is just opposite the heat to solution. Heat of crystallisation = – 17.58 kJ/mol (exothermic) º – 17 580/280.7 º – 62.63 kJ/kg NiSO4 × 7 H2O where molar mass of NiSO4 × 7 H2O = 280.7

Ans.

EXERCISE 5.47 Heat of crystallisation = –11.95 kJ/mol (exothermic) CuSO4(s) + 5 H2O(l) ¾ ¾¾® CuSO4 × 5 H2O(s)

– 771.36

5 (– 285.83)

?

DH°f

D H°r = – 11.95 = (DH°f of CuSO4 × 5 H2O) – (– 771.36 – 5 ´ –285.83) DH°f of CuSO4 × 5 H2O = – 11.95 – 2200.51 = – 2212.46 kJ/mol

Ans.

EXERCISE 5.48 Basis: 1 mol KCIO3 Molar mass of KClO3 = 122.5 Water, present in KClO3 = 5.56 mol º 5.56 ´ 18 º 100.08 kg Mass % KClO3 = 122.5 ´ 100/(122.5 + 100.08) = 55.04 For preparation of 1000 kg solution, KClO3 dissolved = 0.55 ´ 1000 = 550 kg DH = 37 260 kJ/kmol KClO3 (endothermic) Heat absorbed = 37 260 ´ 550/122.5 = 167 290 kJ Ans.

EXERCISE 5.49 (a) Basis: 200 kg NaClO3 dissolved to produce 40% solution From Fig. 5.35, integral heat of solution for 40% solution = + 157 kJ/kg NaClO3 (endothermic) For dissolution of 200 kg NaClO3, heat absorbed = 157 ´ 200 = 31 400 kJ (b) Basis: 500 kg of 30% NaClO3 solution NaClO3 dissolved = 500 ´ 0.3 = 150 kg From Fig. 5.35, integral heat solution for 30% solution, = 176.3 kJ/kg NaClO3 (endothermic) For 150 kg NaClO3 dissolution, heat absorbed = 150 ´ 176.3 = 26 445 kJ (c) Integral heat of solution of 40% soln. = + 157 kJ/kg NaClO3 Integral heat of solution of 20% soln. = + 190 kJ/kg NaClO3

Ans.

Ans.

Energy Balances 155

Heat of dilution = 190 – (157) = 33 kJ/kg NaClO3 (endothermic) Heat added for dilution = 33 kJ/kg NaClO3

Ans.

EXERCISE 5.50 Molar mass of KOH = 56.1056 Molar mass of H2O = 18.0153 mol of Molar mass mass mole Enthalpy HE, Excess HE¢, Excess H2O/mol of 1 mol fraction of fraction of enthalpy, enthalpy, of KOH KOH and KOH in of KOH formation kJ/mol kJ/kmol number of solution in at 298.15 K, KOH solution mol H2O wi solution kJ/mol KOH

0 3 3.5 4 4.5 5 6 8 10 12 15 20 25 30 40 50 75 100 150 200

110.1515 119.1592 128.1668 137.1745 146.1821 164.1974 200.2280 236.2586 272.2892 326.3351 416.4116 506.4881 596.5646 776.7176 956.8706 1407.2531 1857.6356 2758.4006 3659.1656

0.5093 0.4708 0.4378 0.4090 0.3838 0.3417 0.2802 0.2375 0.2061 0.1719 0.1347 0.1108 0.0940 0.0722 0.0586 0.0399 0.0302 0.0203 0.0153

0.2500 0.2222 0.2000 0.1818 0.1667 0.1429 0.1111 0.0909 0.0769 0.0625 0.0476 0.0385 0.0323 0.0244 0.0196 0.0132 0.0099 0.0066 0.0050

–424.764 –469.462 –471.302 –472.955 –474.503 –475.712 –477.093 –478.775 –479.725 –480.306 –480.825 –481.189 –481.302 –481.369 –481.461 –481.520 –481.595 –481.637 –481.700 –481.742

–44.698 –11174.50 –46.538 –10341.78 –48.191 –9638.20 –49.739 –9043.45 –50.948 –8491.33 –52.329 –7475.57 –54.011 –6001.22 –54.961 –4996.45 –55.542 –4272.46 –56.061 –3503.81 –56.425 –2686.90 –56.538 –2174.54 –56.605 –1825.97 –56.697 –1382.85 –56.756 –1112.86 –56.831 –747.78 –56.873 –563.10 –56.936 –377.06 –56.978 –283.47

A plot of HE vs mole fraction of KOH is prepared in Excel and equation is fitted. HE = –386.583x13 + 311.412x12 – 4.513x1 – 56.9851 where HE = Excess enthalpy, kJ/mol KOH at 298.15 K R 2 = 0.9994 Another plot of H E¢ vs x1 x2 is prepared in Excel and equation is fitted. HE¢ = Excess enthalpy, kJ/kmol solution at 298.15 K HE¢ = 12 280x12 x22 – 62 695x1 x2 + 80.196 where HE¢ = Excess enthalpy, kJ/kmol solution at 298.15 K R 2 = 0.9998

156 Solutions Manual—Stoichiometry

Another plot of HE¢¢vs x1x2 is prepared in Excel and equation is fitted. HE¢¢ = –2687.7 x12 x22 – 43.552 x1x2 + 0.6235 where HE¢¢ = Excess enthalpy, kJ/kg solution at 298.15 K R 2 = 0.9999 x1 = mole fraction of KOH in aqueous solution Range of x1 = 0 to 0.25, wi = 0 to 0.51 Heat of dilution chart: Mass fraction of KOH, wi

HE¢¢, Excess enthalpy, kJ/kg solution

0.5093 0.4708 0.4378 0.4090 0.3838 0.3417 0.2802 0.2375 0.2061 0.1719 0.1347 0.1108 0.0940 0.0722 0.0586 0.0399 0.0302 0.0203 0.0153

–101.45 –86.79 –75.20 –65.93 –58.09 –45.53 –29.97 –21.15 –15.69 –10.74 –6.45 –4.29 –3.06 –1.78 –1.16 –0.53 –0.30 –0.14 –0.08

Usong above data, following diagram is prepared. Heat of dilution, kJ/kg solution

Heat of dilution chart for KOH-H2O system 0 –20

0

0.1

0.2

0.3

–40 –60 –80 –100 –200

mass fraction of KOH

0.4

0.5

0.6

Energy Balances 157

EXERCISE 5.51 (a) Locate two points on Fig. 5.16. One point corresponds to 60% NaOH at 373 K (100°C). Another point corresponds to pure water (i.e. 0% NaOH) at 308 K (35°C). Draw a line joining the two points. The line intersects vertical axis of 10% NaOH at 335 K (62°C). (b) H1 = Enthalpy of 10% NaOH soln. at 335 K (62°C) = 231.3 kJ/kg solution (from Fig. 5.16) H2 = Enthalpy of 10% NaOH soln. at 298.15 K (25°C) = 100 kJ/kg solution (from Fig. 5.16) Enthalpy to be removed = 231.3 – 100 = 131.3 kJ/kg solution Ans. b (i) Specific heats of 10% NaOH solution at 335 K (62°C) and 298.15 K (25°C) are 3.84 and 3.78 kJ/(kg solution × K) as read from Fig. 5.4. Average specific heat = (3.84 + 3.78)/2 = 3.81 kJ/(kg soln × K) Heat removed = 3.81 (335 – 298) = 141 kJ/kg solution Ans. b (ii)

EXERCISE 5.52 Basis: 100 kg 32% N solution Urea dissolved = 50 kg N from urea = 50 ´ 0.4665 = 23.325 kg Required N from NH2NO3 = 32 – 23.325 = 8.675 kg NH4NO3 dissolved =

80.0434 ´ 8.675 2 ´ 14,0067

= 24.787 kg º 0.3097 kmol º 309.7 mol H2O in solution = 100 – 50 – 24.787 = 25.213 kg Associated urea = 1.1758 ´ 0.3097 = 0.3641 kmol º 21.869 kg — o m = 0.3097 ´ 1000/25.213 = 12.2833 molality of NH4NO3 × 1.1758 NH2CONH2 Additional urea dissolved =

50 – 0.3641 60.0553

= 0.8326 – 0.3641

158 Solutions Manual—Stoichiometry

= 0.4685 kmol º 468.5 mol Water = 25.213 kg = 1.3995 kmol For additional urea dissolution, Waber 1.3995 = = 2.99, say 3 Urea 0.4685

DH°f of urea crystals = –333.51 kJ/mol urea DH°f of NH2CONH2, 3 H2O = –320.98 kJ/mol urea Heat of solution, H Ei = –320.98 – (–333.51) = + 12.346 kJ/mol urea For NH4NO3× 1.1758 NH2CONH3, o o 2 o 3 HE2 = 40.3044 – 2.5962 m— + 0.1582 (m— ) – 3.4782 (m— ) E 2 H 2 = 40.3044 – 2.5962 ´ 12.2833 + 0.1582 ´ (12.2833) – 3.4782 ´ (12.2833)3 ´ 10–3 = 25.838 kJ/mol NH4NO3× 1.1758 NH2CONH2 Total heat effect, HE = 12.346 ´ 468.5 + 25.838 ´ 309.7 = 5784.1 + 8002.0 = 13 786.1 kJ at 298.15 K (endothermic) Ans.

EXERCISE 5.53 Basis: 100 kg 20% N solution NH4NO3 dissolved = 30 kg º 0.3748 kmol N from NH4NO3 =

28.0134 ´ 30 80.0434

= 10.5 kg Required N from urea = 20 – 10.5 = 9.5 kg Urea dissolved =

9.5 = 20.364 kg 0.4665

º 0.3391 kmol H2O in solution = 100 – 30 – 20.364 = 49.636 kg º 2.755 kmol Urea/NH4NO3 = 0.3391/0.3748 = 0.9048

Energy Balances 159

Since ratio of 1.1758 is required, NH4NO3 is dissolved additionally. Associated NH4NO3 = o = m—

0.3391 = 0.2884 kmol º 288.4 mol 1.1758

0.2884 ´ 1000 = 5.81 49.636

Additional NH4NO3 dissolved = 0.3748 – 0.2884 = 0.0864 kmol º 86.4 mol Water 2.755 NH 4 NO3 = 0.0864 = 31.887, say 30

DH°f of NH4NO3 crystals = –365.56 kJ/mol NH4NO3 DH°f of NH4NO3 × 30 H2O = – 342.105 kJ/mol NH4NO3 Heat of solution, HE1 = –342.105 – (–365.56) = + 23.455 kJ/mol NH4NO3 For dissolution of NH4NO3× 1.1758 NH2CONH2, HE2 = 40.3044 – 2.5962 (5.81) + 0.1582 ´ (5.81)2 –3.4782 ´ 10–3 (5.81)3 = 29.8784 kJ/mol NH4NO3× 1.1758 NH2CONH2 Total heat effect, HE = 23.455 ´ 288.4 + 29.878 ´ 86.4 = 6764.4 + 2581.5 = 9345.9 kJ (endothermic) Ans.

EXERCISE 5.54 When mole fraction of ester is 0.95 in solution, enthalpy of mixing = 326.6 kJ/kg solution Molar mass of ester (CH3COOC2H5) = 88 Mass of CH3COOC2H5 per kmol mixture, m2 = 88 ´ 0.95 Integral heat to solution of ester = 326.6/(88 ´ 0.95) = 3.907 kJ/kg ester In a similar manner, following table is prepared. Mole fraction of C2H5OH

0.05

Mass of C2H5OH

mole fraction

Mass of CH3COOC2H5

per kmol of per kmol mixture CH3COOC2H5 mixture (m1) kg (m2) kg 2.3

0.95

83.6

Mass fraction of ester 0.973

Internal heat of solution kJ/kmol kJ/kg soln. ester (H) (H/m2) 326.6

3.907 (Contd.)

160 Solutions Manual—Stoichiometry

(Contd.) 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

4.6 9.2 13.8 18.4 23.0 27.6 32.2 36.8 41.4 43.7

0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.05

79.2 70.4 61.6 52.8 44.0 35.2 26.4 17.6 8.8 4.4

0.945 0.884 0.817 0.742 0.657 0.561 0.451 0.324 0.175 0.091

596.1 1032.3 1319.8 1467.0 1486.1 1385.8 1176.3 859.1 531.3 245.8

7.527 14.663 21.425 27.784 33.775 39.369 44.557 48.813 60.375 55.864

Integral heat of mixing (in kJ/kg ester) and mass fraction of ester are plotted below. Ans. 70.0

INTERGRAL HEAT OF MIXING kJ/kg ESTER

60.0

50.0

40.0

30.0

20.0

10.0

0.0 0.091

0.175

0.324

0.451

0.561

0.657

0.742

0.817

0.884

0.945

MASS FRACTION OF ESTER

EXERCISE 5.55 Basis: 1 kmol mixture Enthalpy of 0.18 kmol ethanol at 318.15 K over 273.15 K: H1 = 0.18

318.15

ò Cl1 dT

273.15

= 0.18 [– 325.137 (318.15 – 273.15) + 4137.87 ´ 10–3 (318.152 – 273.152)/2 – 14 030.7 ´ 10–3 (318.153 – 273.153)/3 + 17 035.4 ´ 10–3 (318.154 – 273.154)/4] = 0.18 [– 14 631 + 55 051 – 55 295 + 19 925] = 909 kJ

0.973

Energy Balances 161

Enthalpy of 0.82 kmol n-hexane at 318 K over 273 K: H2 =

318.15

ò Cl2 dT

273.15

= 0.82 [31.421 (318.15 – 273.15) + 976.058 ´ 10–3 (318.152 – 273.152)/2 – 2353.68 ´ 10–3 (318.159 – 273.159)/3 + 3092.73 ´ 10–9 (318.154 – 273.154)/4] = 0.82 [1414 + 12 986 – 9276 + 3617] = 7167.6 kJ Enthalpy of mixture at 318 K = 909 + 7167.6 – 561.4 = 7515.2 kJ/kmol Ans.

EXERCISE 5.56 Basis: 100 kg mixed acid Refer Exercise 3.14. For producing 100 kg mixed acid having 39% HNO3, 42% H2SO4 and 19% H2O, 68.3% HNO3 requirement = 57.1 kg 97.9% H2SO4 requirement = 42.9 kg Component (Acid)

68. 3 % HNO3 97.9% H2SO4 Mixed acid

(Ref. Fig. 5.21) Enthalpy at 273 K (0 °C) kJ/kg – 191.8 – 10.5 – 226.9

(Ref. Fig. 5.20) specific heat kJ/(kg × K) 2.6 1.42 2.05

Enthalpy at 303 K (0 °C) kJ/kg – 113.8 + 32.1 —

Enthalpy at 313 K (0 °C) kJ/kg — — – 144.9

Heat of mixing = –144.9 ´ 100 – (–113.8 ´ 57.1 + 32.1 ´ 42.9) = –14 490 – (– 6498 + 1377) = –9377 kJ (exothermic) Ans.

EXERCISE 5.57 Basis: 100 kg spent acid Let x = amount of 98% H2SO4 to be added, kg and y = amount of 69% HNO3 to be added, kg Total acid after mixing = 100 + x + y H2SO4 balance: 44.4 + 0.98 x = 0.4 (100 + x + y) (1) HNO3 Balance: 11.3 + 0.69 y = 0.25 (100 + x + y) (2) Solving the equations, x = 22.84 kg 98% H2SO4 y = 44.11 kg 69% HNO3 Ans. (a)

162 Solutions Manual—Stoichiometry

Amount, kg mi

Acid

spent acid 69% HNO3 98% H2SO4 fortified mixed acid

(Ref. Fig. 5.21) Enthalpy at 273 K (0 °C), hi kJ/kg

100 44.11 22.84 166.95

– 215.8 – 191.8 – 10.5 – 284.7

mi × hi kJ

– 21 580 – 8 460 – 600 – 47 531

Heat of mixing = – 47 531 – (– 21 580 – 8460 – 600) = – 16 891 kJ/100 kg or 168.91 kJ/kg spent acid

Ans.(b)

EXERCISE 5.58 Basis: 100 L solution containing 4.5% NH3 Use Fig. 5.22. Locate a point representing liquid ammonia (w = 1.0) at 240 K (–33°C). Locate another point representing water (w = 0) at 298 K (25°C). Join both the points. Line crosses 4.5% NH3 at 305 K (32°C) having an enthalpy value of 93.16 kJ/kg solution. Enthalpy of 4.5% NH3 at 298.15 K(25°C) = 67.17 kJ/kg solution Change in enthalpy = 93.16 – 67.17 = 25.99 kJ/kg solution (exothermic) Total solution = 100 ´ 0.98 = 98 kg Heat evolved = 25.99 ´ 98 = 2547.0 kJ Ans.(i) Ammonia dissolved = 98 ´ 0.045 = 4.41 kg º 0.259 kmol Use data, contained in Table 5.59. For use of the Data of Table 5.59, heat of formation of liquid NH3 is required at 240 K (– 33°C). From Fig. 5.22, Enthalpy of anhydrous NH3 at 298.15 K (25°C) = 464.9 kJ/kg NH3 Enthalpy of anhydrous NH3 at 240 K (– 33°C) = 211.1 kJ/kg NH3 Heat evolved by cooling anhydrous NH3 = 464.9 – 211.1 = 253.8 kJ/kg º 4322.5 kJ/kmol NH3 D H°f at 298.15 K (25°C) of vapour NH3 = – 46.11 kJ/mol NH3 Use Eq. (5.25). l v = 23 346 kJ/kmol at T = 239.7 K n = 0.38, Tc = 405.6 K (Ref. Table 5.5) 0.33

LM N

OP Q

0.33

107.5 é 405.5 – 298.15 ù lv = ê = ú 165.8 ë 405.5 – 239.8 û 23.325 l v = 23.35 ´ 0.848 = 19 775 kJ/kmol at 298.15 K D H°f of liquid NH3 at = – 46 110 – 4322.5 – 19 775 = – 70 207.5 kJ/kmol

Energy Balances 163

D H°f of 4.51% NH3 solution (Ref. Table 5.60) = 80 073 kJ/kmol Heat change = – 80 073 – (–70 236.5) = 9836.5 kJ/kmol NH3 Heat evolved = 9836.5 ´ 0.259 = 2547.7 kJ Ans. (ii)

EXERCISE 5.59 Basis: 1000 kg 15% solution NH3 content of the solution = 1000 ´ 0.15 = 150 kg º 8.8075 kmol From Table 5.59, D H°f at 298.15 K (25°C) of 32% NH3 solution = – 77.707 kJ/mol NH3 15.9% NH3 solution = – 79.319 kJ/mol NH3 4.5% NH3 solution = – 80.073 kJ/mol NH3 When 32% NH3 solution is diluted to 15.9% NH3 soln., heat evolved = – 79.319 – (– 77.707) = – 1.612 kJ/mol NH3 When 4.5% NH3 soln. is concentrated to 15.9% NH3 soln., heat absorbed = – 79.319 – (– 80.073) = + 0.754 kJ/mol NH3 Now on Fig. 5.22, locate points representing 32% NH3 (mass) at 298.15 K (25°C) and 4.5% (mass) NH3 at 298.15 K (25°C). Join both the points. Measure distances from intersection point on 15.9% NH3. Distances to both the points are measured to be 24.4 and 17.6 linear units, respectively. 24.4 Requirement of 4.5% soln. = ´ 1000 = 580.6 kg (24.4 + 17.6) Ammonia content of 4.5% soln. = 580.6 ´ 0.045 = 26.13 kg º 1.534 kmol Heat absorbed = 1.534 ´ 0.754 ´ 1000 = 1156.8 kJ Requirement of 32% soln. = 1000 – 580.6 = 419.4 kg NH3 content of 32% soln. = 419.4 ´ 0.32 = 134.2 kg º 7.88 kmol Heat evolved = 7.88 ´ 1.612 ´ 1000 = 12 702.6 kJ Net heat change = 12 702.6 – 1156.8 = 11 545.8 kJ (evolved) Ans.

EXERCISE 5.60 Basis: 100 kmol/h gas mixture entering the absorber Gas mixture enters the absorber at 318.15 K (45°C) and leaves from the top at 303.15 K (30°C). Total pressure at absorber top = 6.85 kPa g = 108.175 kPa a

164 Solutions Manual—Stoichiometry

Vapour pressure of water at 303 K (30°C), pw = 4.2415 kPa [Table – 6.13] Moisture content of gas mixture, leaving the absorber = 4.2415/(108.175 – 4.2415) = 0.0408 kmol/kmol dry gas Total water leaving the absorber in the gas mixture = 0.0408 (1 – 0.2445) 100 = 3.082 kmol/h º 55.5 kg/h HCl absorbed = 24.45 kmol/h º 891.5 kg/h Aqueous acid produced = 891.5/0.3 = 2971.5 kg/h Water content of the acid = 2971.5 – 891.5 = 2080.0 kg/h Total water fed to absorber = 2080.0 + 55.5 = 2135.5 kg/h From Table 5.79, standard heats of formation of 30% aqueous HCl and 5% aqueous HCl solutions are – 155.1 and – 165.0 kJ/mol HCl (by interpolations), respectively. Heat evolved due to absorption at 298.15 K = standard heat of formation of 30% acid – standard heat of formation of HCl gas = – 155.1 – (– 92.31) = – 62.79 kJ/mol HCl (exothermic) Total heat evolved f1 = 62 790 ´ 24.45 = 1535 216 kJ/h º 426.449 kW Feed gas temperature = 318.15 K (45°C) Component

.

ni kmol/h

.

.

heat capacity equation constants .

.

ni × bi ´ 103 2803.71 1922.50

ni × ci ´ 106 644.15 – 867.44

ni × di ´ 109 – 608.87 143.86

CH4 CH3Cl

53.80 18.30

ni × ai 1035.62 247.83

CH2Cl2 CHCl3 CCl4 HCl

2.65 0.65 0.15 24.45

44.81 19.67 7.52 741.05

372.20 98.20 22.15 – 186.04

– 249.17 – 77.14 – 19.79 324.23

63.04 21.55 5.90 – 105.96

100.00

2096.50

5032.72

– 245.16

– 480.48

Total

Outgoing gas-mixture from absorber: Temp. = 30 K (30°C)

Energy Balances 165

Component

.

ni kmol/h

.

ni × ai

.

heat capacity equation constants

ni × bi ´ 103

.

.

ni × ci ´ 106

ni × di ´ 109

CH4 CH3Cl CH2Cl3 CHCl3 CCl4 H2O

53.80 18.30 2.65 0.65 0.15 3.082

1035.62 247.83 44.81 19.67 7.52 104.93

2803.71 1922.50 372.20 98.20 22.15 – 29.74

644.15 – 867.44 – 249.17 – 77.14 – 19.79 101.70

– 608.87 143.86 63.04 21.55 5.90 – 63.02

Total

78.632

1460.38

5189.02

– 467.76

– 437.54

Reference temperature, T0 = 298.15 K (25°C) Enthalpy of feed gas f2 = 2096.50 (318.15 – 298.15) + 5032.72 ´ 10–3 (318.152 – 298.152)/2 – 245.16 ´ 10–6 (318.153 – 298.153)/3 – 480.48 ´ 10–9 (318.154 – 298.154)/4 = 41 930 + 31 017 – 466 – 281 = 72 200 kJ/h º 20.056 kW Enthalpy of absorber off-gas, f3 = 1460.38 (303.15 – 298.15) + 5189.02 ´ 10–3 (303.152 – 298.152)/2 – 467.76 ´ 10–6 (303.153 – 298.153)/3 – 437.54 ´ 10–9 (303.154 – 298.154)/4 = 7302 + 7800 – 211 – 59 = 14 832 kJ/h º 4.12 kW Enthalpy of feed water f4 = 2135.5 ´ 4.1868 (303.15 – 298.15) = 44 705 kJ/h º 12.418 kW Enthalpy of outgoing acid (30%) H5 = 0 kJ/h at 298.15 K Total heat of to be removed = f1 + f2 – f3 + f4 = 426.449 + 20.055 – 4.12 + 12.418 = 454.802 kW Ans. (a) If 5% acid is produced, Total aqueons acid produced = 891.5/0.05 = 17 830 kg/h Water content of acid = 17 830 – 891.5 = 16 938.5 kg/h Total water fed = 16 938.5 + 55.5 = 16 994 kg/h

166 Solutions Manual—Stoichiometry

Enthalpy of feed water f4 = 16 994 ´ 4.1868 (303.15 – 298.15) = 355 752 kJ/h º 98.82 kW Heat evolved due to absorption at 298.15 K = – 165.0 – (– 92.31) = – 72.69 kJ/mol HCl f¢1 = 72 690 ´ 24.45 = 1777 271 kJ/h º 493.686 kW Enthalpy of outgoing acid (5% cocn.) = f¢1 + f2 – f3 + f¢4 = 493.686 + 20.055 – 4.12 + 98.82 = 608.441 kW Temperature rise for outgoing acid = 608.441 ´ 3600/(17 830 ´ 4.19) = 29.31 K Temperature of outgoing acid = 29.31 + 298.15 = 327.46 K (54.3°C) Ans. (b)

EXERCISE 5.61

HE = H – Hid [Eq. (5.74)] Hid = x1 H1 + x2 H2 Reference temperature, t0/T0 = 0°C/273.15 K Enthalpy of cyclohexane (1) at 50°C over 0°C, H1 = 155.78 (323.15 – 273.15) = 7789 kJ/kmol Enthalpy of cyclohexanone (2) at 50°C over 0°C, H2 = 176.84 (323.15 – 273.15) = 8842.0 kJ/kmol HE can be calculated for x1 = 0.1 and x2 = 0.9 as HE = 0.1 ´ 0.9 [3836.7 – 959.3 (0.9 – 0.1) + 815.2 (0.9 – 0.1)2 – 485.6 (0.9 – 0.1)3] = 300.81 kJ/kmol mixure (endothermic) In a similar manner, HE can be calculated for x1 = 0.1 to 1 in steps of 0.1. x1

HE (endothermic)

Hid

H

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0 300.81 551.95 746.00 881.66 959.18 975.61 920.21 769.70 483.70 0

8842.0 8736.8 8631.5 8526.2 8420.9 8315.6 8210.2 8104.9 7999.6 7894.3 7789.0

8842.0 9037.6 9183.5 9272.2 9302.6 9274.8 9185.8 9025.1 8769.3 8378.0 7789.0

All enthalpy values are in kJ/kmol.

Energy Balances 167

H1 = H + x2

dH dx1

Similarly, H 2 = H – x1

(1) dH dx1

(2)

H = Hid + HE = 7789 x1 + 8842.0 x2 + x1x2 [3836.7 – 959.3 (x2 – x1) + 815.2 (x2 – x1)2 – 485.6 (x2 – x1)3] Substituting x2 = 1 – x1 and simplifying, H = –3884.8 x15 + 6451.2 x14 – 4137.8 x13 – 1635.6 x12 + 2153.9 x1 + 8842.1

dH = –19424 x14 + 25 804.8 x13 –12 413.4 x12 – 3271.2 x1 + 2153.9 dx1 Substituting x2 = 1 – x1, H and

dH in Eq. (1), dx1

H1 = 15 539.2 x15 –38 777.6 x14 + 34 080.4 x13 –10 777.8 x12 –3271.2 x1 + 10 996.0 (3) or (4) H 2 = 15 539.2x15 – 19 353.6 x14 + 8275.6 x13 + 1635.6 x12 + 8842.1 Using Eq. (3) and Eq. (4), following table is prepared x1

H

H1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

8842.0 9037.6 9183.5 9272.2 9302.6 9274.8 9185.8 9025.1 8769.3 8378.0 7789.0

– 10 591.5 10 126.2 9 688.5 9 310.6 8 988.0 8 697.4 8 415.8 8 139.0 7 900.3 7 789.0

H2 8 842.0 8 865.0 8 947.7 9 093.7 9 297.1 9 561.5 9 918.6 10 447.0 11 290.6 12 677.7 –

Following is the plot of H vs x1, H1 and H 2 are evaluated at x1 = 0.4 and x1 = 0.7.

168 Solutions Manual—Stoichiometry 10 500 H2 at x1 = 0.7 10 000 9500

Enthalpy (H) kJ/kg solution

H2 at x1 = 0.4

H1 at x1 = 0.4

9000

H1 at x1 = 0.7

8500 8000 7500

0

0.2

0.4

x1

0.6

0.8

1

Enthalpy of cyclohexane (x1)-Cyclohexanone (x2) Solutions at 50°C (325.15 K)

EXERCISE 5.62 t0/T0 = 0°C/273.15 K Enthalpy of benzene (1) at 298.15 K over 273.15 K, H1 = 129.8 (298.15 – 273.15) = 3245 kJ/kmol Entalpy of cyclohoxane (2) at 298.15 K over 273.15 K, H2 = 150.83 (298.15 – 273.15) = 3770.75 kJ/kmol 4 3 H = –490.664 x1 + 645.318 x1 – 3309.315 x12 + 2628.911 x1 + 3770.75 H1 = 1471.992 x14 – 3253.292 x13 + 5245.269 x12 – 6618.63 x1 + 6399.661 H 2 = 1471.992 x14 – 1290.636 x13 + 3309.315 x12 + 3770.75 x1

HE (endothermic)

H

H1

H2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

– 282.97 502.94 662.01 761.11 800.01 777.24 690.24 535.19 307.67 –

3770.75 4001.14 4168.54 4275.03 4321.56 4307.88 4232.54 4092.96 3885.34 3604.74 3245.0

– 5787.15 5262.08 4810.23 4420.92 4087.00 3804.84 3574.35 3398.97 3285.69 3245.0

3770.75 3802.70 3895.15 4045.66 4255.32 4528.75 4874.10 5303.05 5830.83 6476.20 –

All enthalpy values are in kJ/kmol.

Energy Balances 169

Following is the plot of H vs x1, H1 and H 2 are evaluated at x1 = 0.4 and x1 = 0.7. H2 at x1 = 0.7

5400 5000 4600

Enthalpy (H) kJ/kg solution

H2 at x1 = 0.4 4200

H1 at x1 = 0.4

3800

H1 at x1 = 0.7

3400 3000

0

0.2

0.4

x1

0.6

0.8

1

Enthalpy of Benzene(1)-Cyclohexane(2) Solutions at 25°C (298.15 K)

EXERCISE 5.63

t0/T0 = 0°C/273.15 K Enthalpy of 1,2-dichloroethane(1) at 40°C (313.15 K) over 0°C (273.15 K) H1 = 123.9 (313.15 – 273.15) = 4956 kJ/kmol Enthalpy of dimethyl carbonate(2) at 40°C (313.15 K) over 0°C (273.15 K) H2 = 173.33 (313.15 – 273.15) = 6933.2 kJ/kmol H = –1040 x14 + 2161.41 x13 – 598.17 x12 – 2500.41 x1 + 6933.2 4 3 2 H1 = 3120 x1 – 8482.82 x1 + 7082.4 x1 – 1196.34 x1 + 4432.79 H 2 = 3120 x14 – 4322.82 x13 + 598.17 x12 + 6933.2 x1

HE (exothermic)

H

H1

H2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0 –56.25 –112.95 –160.87 –193.30 –205.99 –197.21 –167.71 –120.76 –62.11 0

6933.2 6679.24 6424.82 6179.18 5949.04 5738.69 5549.69 5381.47 5230.70 5091.64 4956.03

– 4375.81 4413.95 4507.54 4624.41 4739.87 4836.71 4905.23 4943.20 4955.88 4956.03

6933.2 6935.17 6927.54 6895.59 6832.12 6737.39 6619.16 6492.69 6380.70 6313.41 –

All enthalpy values are in kJ/kmol.

170 Solutions Manual—Stoichiometry

Following is the plot of H vs x1 × H1 and H 2 are evaluated at x1 = 0.3 and x1 = 0.6.

Enthalpy (H) kJ/kg solution

7000 H2 at x1 = 0.3 H2 at x1 = 0.6 6500 6000 5500 5000 4500

0

0.2

0.4

x1

0.6

0.8

1

H1 at x1 = 0.3 H1 at x1 = 0.6

Enthalpy of 1,2-Dichlorothane(1) - Dimethyl Carbonate(2) solution at 40°C (313.15 K)

EXERCISE 5.64 Basis: 100 Nm3/h vent stream Molar flow rate =

100 = 4.4615 kmol/h 22.414

Bromine in the vent stream = 4.4615 ´ 0.0125 = 0.0558 kmol/h NaOH required = 2 ´ 0.0558 = 0.1116 kmol/h º 4.464 kg/h Excess NaOH used is 20% Actual NaOH used = 4.464 ´ 1.2 = 5.357 kg/h Excess NaOH = 5.537 – 4.464 = 0.893 kg/h º 0.0223 kmol/h Aqueous solution rate =

5.357 = 107.14 kg/h 0.05

Water in solution = 107.14 – 5.357 = 101.783 kg/h º 5.650 kmol/h Water produced = 0.0558 kmol/h 2 NaOH(sol) + Br2(g) = NaOBr(sol) + NaBr(sol) + H2O(l) DH°f – 470.11 30.91 – 334.3 – 361.665 – 285.83 DH°r = –334.3 – 361.665 – 285.83 + 30.91] DH°r = 72.485 kJ/mol Br2

Energy Balances 171

Heat generated = 72.485 ´ 1000 ´ 0.0558 = 4044.66 kJ/h º 1.1235 kW Component

kmol/h

mole %

kg/h

mass %

NaOBr NaBr NaOH H2O Total

0.0558 0.0558 0.0223 5.7058 5.8397

0.96 0.96 0.38 97.70 100.00

6.634 5.741 0.893 102.792 116.060

5.72 4.95 0.77 88.56 100.00

Solid concentration of solution = 100 – 88.56 = 11.44% by mass

Ans.(a)

Ans.(b)

EXERCISE 5.65 Basis: Flow rate of moist flue gas = 265 000 Nm3/h Refer solution of Exercise 4.46. Molar flow rate of ingoing gas mixture = 7584.43 kmol/h at 106.6 kPa and 175°C (448.15 K) Moisture, entering with gas mixture = 422.22 kmol/h Dry flue gas flow rate = 7584.43 – 422.22 = 7162.21 kmol/h Molar flow rates of incoming and outgoing flue gas (dry) remain constant as SO2 reacted equals CO2 generation. Moisture in outgoing gas mixture = 937.53 kmol/h Water evaporated in absorber = 937.53 – 422.22 = 515.31 kmol/h º 9275.6 kg/h Composition of solution purged from system Component

kg/h

mass %

kmol/h

mole %

Na2SO3 Na2CO3 H2O Total

544.67 305.48 9 776.73 10 626.88

5.13 2.87 92.00 100.00

4.3213 2.8822 542.6904 549.8939

0.79 0.52 98.69 100.00

Molar mass of solution being purged =

10 626.88 = 19.325 kg/kmol 549.8939

Fresh aqueous 30 % NaOH is made-up and its flow rate is 2545.7 kg/h. It enters at 32°C (305.15 K). 1 kg solution contains 0.3 kg NaOH or 0.002 83 kmol NaOH and 0.7 kg H2O or 0.038 86 kmol H2O. In this solution,

172 Solutions Manual—Stoichiometry

0.038 86 H2O = = 13.73 kmol/kmol 0.002 83 NaOH In the circulating solution, H2O 98.69 Na 2SO3 = 0.79 = 124.92 say 125 H2O 98.69 = = 189.79 say 190 Na 2 CO3 0.52

In all three cases, water to chemical ratio is quite high and hence standard heat of formation of solution having 1 molality concentration (repressented by 'ai' in Annexure V.1) values can be taken for calculations. Reaction: Na2CO3(sol) + SO2(g) = Na2SO3(sol) + CO2(g) DH°f –1157.38 – 296.83 – 1115.87 – 393.51 DH°r = –1115.87 – 393.51 – [–1157.38 – 296.83] = –55.17 kJ/mol SO2 reacted SO2 reacted = 276.66 kg/h º 4.3185 kmol/h Heat generated, f1 = 55.17 ´ 1000 ´ 4.3185 = 238 252 kJ/h º 66.181 kW Enthalpy of water at 32°C (305.15 K), same as that of make-up solution = 134.0 kJ/kg Enthalpy of water vapour at 50°C (323.15 K) = 2592.2 kJ/kg Heat picked up by water = 2592.2 – 134.0 = 2458.2 kJ/kg Heat absorbed by water, f2 = 9275.6 ´ 2458.2 = 22 801 280 kJ/h º 6333.689 kW Mean heat capacity of flue gas, C°mp = 30.7 kJ/(kmol dry gas × K) Heat given up by flue gas f3 = 7162.21 ´ 30.7 (448.15 – 323.15) = 27 484 981 kJ/h º 7634.717 kW Reference temperature, t0/T0 = 32°C/305.15 K

Energy Balances 173

Enthalpy of make-up solution = 0 kJ/h Heats of formation of make-up solution and that of diluted solution are nearly same and hence heat of dilution is neglibigle. Final solution can be assumed to leave at 50°C (323.15 K). Enthalpy of purge solution at 50°C (323.15 K) over 32°C (305.15 K), f4 = 10 626.88 ´ 4 (323.15 – 305.15) = 765 135 kJ/h º 212.538 kW Net heat exchange in the heat exchanger, f = f1 + f3 – f2 – f4 = 66.181 + 7634.717 – 6333.689 – 212.538 = 1154.671 kW (exothermic) Heat, amounting to 1154.671 kW is required to be removed in the heat exchanges. Rise in cooling water temperature is 8°C (8 K). Required cooling water flow =

1154.671 ´ 3600 4 ´8

= 129 900 kg/h º 129.9 m3/h

Ans.

EXERCISE 5.66 Refer solution of Exercise 5.44. Final vent gases have following composition Component

kmol/h

mole %

Cl2 HCl Acetic acid (AA) Total

2.1164 10.5820 0.5646 13.2630

15.96 79.79 4.25 100.00

kg/h

mass %

150.065 385.829 33.905 569.799

26.34 67.71 5.95 100.00

Reactions: 2 NaOH(sol) + Cl2(g) = NaOCl(sol) + H2O(l) + NaCl(sol) (1) NaOH(sol) + HCl(g) = NaCl(sol) + H2O(l) (2) NaOH(sol) + CH3COOH(g) = CH2COONa(sol) + H2O(l) (3) Aqueous solution of 10% NaOH is used for scrubbing. Consider standard enthalpies of formation of NaOH, NaOCl, NaCl and CH3COONa in 'ai' state for calculation of heats of reaction. From Appendix V.1 values are taken. DH°r1 = (–347.3) + (–407.27) + (–285.83) – 2[–470.11] = –100.18 kJ/mol Cl2

174 Solutions Manual—Stoichiometry

DH°r2 = –407.27 – 285.83 – [–470.11 – 92.31] = –130.68 kJ/mol HCl DH°r3 = –726.13 – 285.83 – [–470.11 – 432.6] = –109.25 kJ/mol AA In the absorber, absorption of three components is assumed 100%. Total heat effect, f = –100.18 ´ 1000 ´ 2.1164 – 130.68 ´ 1000 ´ 10.582 – 109.25 ´ 1000 ´ 0.5646 = –1656 559 kJ/h (exothermic) º –460.155 kW Total NaOH consumed for all three reactions = 2 ´ 2.1164 + 10.582 + 0.5646 = 15.3794 kmol/h º 615.176 kg/h For completion of reactions, 10% excess is used. NaOH used = 615.176 ´ 1.1 = 676.69 kg/h NaOH leftover in final solution = 676.69 – 615.176 = 61.5 kg/h Solution introduced at top =

676.69 0.1

= 6766.9 kg/h Water in the solution = 6766.9 – 676.69 = 6090.2 kg/h NaOCl produced = 2.1164 kmol/h º 157.55 kg/h Water produced by reactions (2) and (3) = 10.582 + 0.5646 + 2.1164 = 13.263 kmol/h º 238.937 kg/h NaCl produced = 10.582 + 2.1164 kmol/h = 12.6984 kmol/h º 724.13 kg/h CH3COONa produced = 0.5646 kmol/h = 46.32 kg/h

Energy Balances 175

Component

kg/h

mass %

NaOCl NaCl CH3COONa H2O Total

157.55 742.13 46.32 6329.14 7275.14

2.16 10.20 0.64 87.00 100.00

Solids in final solution = 100 – 87.0 = 13.0 % Heat capacity of solution = 3.5 kJ/(kg × K) Rise in temperature of solution =

1656 559 7275.14 ´ 3.5

= 65.06 K or °C Aqueous 10% solution is introduced at 32°C (305.15 K). Temperature of outgoing solution = 32.5 + 65.06 = 97.56°C say 97.6°C or 370.75 K Ans.

EXERCISE 5.67

Basis: 100 kg 19.1% H2SO4 SO3 in the acid = 15.6 kg º 0.195 kmol (ref. Table 5.64) H2O in the acid = 100 – 15.6 = 84.4 kg º 4.689 kmol Strong acid: oleum having 29 mass % free SO3 From Table 5.64, it can be inferred that 100 kg oleum will contain 87 kg SO3 and 13 kg water. Resultant acid has 53.8% H2SO4 strength. Thus again from Table 5.64, this resultant 100 kg acid will contain 44 kg SO3 and 56 kg H2O. Let x = Amount of oleum, required to be added, kg SO3 balance: 15.6 + 0.87 x = (100 + x) 0.44 x = 66.05 kg SO3 from oleum = 66.05 ´ 0.87 = 57.46 kg º 0.718 kmol SO3 in resultant acid = 15.6 + 57.46 = 73.06 kg º 0.913 kmol H2O from oleum = 66.05 – 57.46 = 8.59 kg º 0.477 kmol H2O in resultant acid = 166.05 – 73.06 = 92.99 kg º 5.166 kmol

176 Solutions Manual—Stoichiometry

Energy balance: Enthalpy of original weak solution + Enthalpy of oleum + Heat of dissolution = Enthalpy final solution 0.195 (– 157.005) 1000 + 4.689 (– 0.084) 1000 + 0.718 (– 10.886) 1000 + 0.477 (– 81.643) 1000 + DHm = 0.913 (– 125.981) 1000 + 5.166 (– 3.559) 1000 Solving the equation, DHm = – 55 636 kJ/100 kg weak acid = – 556.36 kJ/kg week acid Ans.

EXERCISE 5.68 From Fig. 5.5, heat capacity of 53.8% acid at 291 K (18°C) = 2.404 kJ/(kg × K) heat capacity of 53.8% acid at 373 K (100°C) = 2.550 kJ/(kg × K) Average heat capacity of 53.8% acid = 2.477 kJ/(kg × K) Rise in temperature = 55 636/(166.05 ´ 2.477) = 135.3 K From Fig. 5.18, boiling point of 53.8% acid is 405 K (132°C). Therefore aqueons acid will boil.

EXERCISE 5.69 Basis: 100 kg oleum containing 41.2% free SO3 SO3 conent = 89.2 kg (Ref. Table 5.64) º 1.115 kmol Water content of oleum = 100 – 89.2 = 10.8 kg º 0.6 kmol Heat balance: (1.115 kmol SO3 is 0.6 kmol H2O) + ¥ kmol H2O + DHm = (1.115 kmol SO3 in ¥ kmol H2O) DHm + 1.115 (– 10 886) + 0.6 (– 81 643) = 1.115 (– 181 079) or DHm = – 140 779 kJ/100 kg oleum diluted

Ans.

EXERCISE 5.70 Let M be the kmol of the binary mixture in the flask after time q having the mole fraction x of the components A. The fraction dM kmol boil off during the time interval dq. Consequently let the composition of the binary drp by dx. Overall material balance: Q q = lv (M0 – M) or Q dq = – lv dM (1) Component A, lost from the liquid mixture during time dq = M x – (M – d M) (x – dx) (2)

Energy Balances 177

Component A, gained by the vapour mixture, produced during time interval dq = dM × y

Equating Eq. (2) and Eq. (3),

é ù ax = dM ê ú ë1 + (a - 1) x û

é ù ax M x – (M – dM) (x – dx) = dM ê ú ë1 + (a - 1) x û Neglecting second order term dM × dx, é ù ax x dM + M dx= dM ê ú ë1 + (a - 1) x û Dividing Eq. (5) by dM and substituting value of dM from Eq. (1), é M lv ù é d x ù é ù ax x– ê ú ê ú = ê ú ë1 + (a - 1) x û ë Q û ë dq û Eq. (1) can be rewritten as M = M0 – (Q q/lv) Substituting this value of M in Eq. (6), æ Q ×q ö l d x ax x - ç M0 × = ÷ l v ø Q dq è 1 + (a - 1) x

–

(3)

(4)

(5)

(6) (7) (8)

Qq ù d x M 0 lv é ax = –x ê1 ú M l Q d 1 ( q + a - 1) x 0 v û ë

=

(a - 1)(1 - x) x 1 + (a - 1) x

(9)

Let K = Q/(M0lv) – a constant dx (a - 1)(1 - x) x 1 [1 – Kq] = K dq 1 + (a - 1) x

(10)

dq [1 + (a - 1) x] d x =– (1 Kq ) (a - 1)(1 - x) x K

(11)

dq [(1 - x) + x + (a - 1) x]d x =– (1 Kq) (a - 1)(1 - x) x K

(12)

dq dx dx dx + + =– (1 - Kq ) K (a - 1) x K (a - 1)(1 - x) K (1 - x)

(13)

édx a dx ù dq 1 ê + ú =– (1 - Kq ) K (a - 1) ë x (1 - x) û

(14)

–

178 Solutions Manual—Stoichiometry

Integrating Eq. (14), 1 K (a - 1)

x dx ù q é x dx dq ê ò x + a ò (1 - x ) ú = – ò (1 - K q ) x 0 ë x0 û

(15)

é x ln(1 - Kq ) (1 - x) ù 1 – a ln êln ú = K (1 - x0 ) û K (a - 1) ë x0

(16)

x (1 - x0 )a 1 ln = ln (1 – K q) (a - 1) x0 (1 - x)a

(17)

or (1 – Kq)(a – 1) =

x (1 - x0 )a x0 (1 - x)a

for Kq < 1

Q.E.A.

EXERCISE 5.71 Basis:

5 kg n-heptane/n-octane mixture, containing 50 mole % each Mass of 1 kmol mixture = 0.5 ´ 100 + 0.5 ´ 114 = 107 kg 5 5 kg mixture = = 0.0467 kmol 107 Constant K =

Q 0.5 ´ 3600 = = 1.1506 h–1 M 0 lv 0.0467 ´ 33 500

x0 = 0.5 mole fraction of n-heptane

(a) for x = 0.3

x (1 - x0 )a x0 (1 - x)a

Therefore (1 – Kq)(a – 1) (1 – Kq) Kq q Overall material balance:

=

0.3(1 - 0.5) 2.16 = 0.290 08 0.5(1 - 0.3) 2.16

= 0.290 08 [Ref. Eq. (17) of Exc. 5.70] = 0.290 08(1/1.16) = 0.3441 = 1 – 0.3441 = 0.6559 = 0.6559/1.1506 = 0.57 h º 34.2 min Ans.(a-i)

Qq = lv (M0 – M) [Ref. Eq. (1)] 0.5 ´ 3600 ´ 0.57 = 33 500 (0.0467 – M) or M = 0.016 07 kmol 1 kmol final mixture will contain (0.3 ´ 100 + 0.7 ´ 114) = 109.8 kg material. 0.016 07 kmol mixture = 0.016 07 ´ 109.8 = 1.764 kg Ans.(a-ii)

Energy Balances 179

(b) : q = 15 min = 0.25 h (1 – Kq) = 1 – 1.1506 ´ 0.25 = 0.712 35 (1 – Kq)(a – 1) = (0.712 35)1.16 = 0.674 72 x(1 - 0.5)2.16 x (0.5)2.16 ´ 2.16 = 0.5(1 - x) 0.5 (1 - x) 2.16

= 0.447 57 ´

x (1 - x) 2.16

= 0.672 72 or x = 1.5077 (1 – x)2.16 By trial and error, x = 0.4366 or 43.66 mole % n-heptane Ans. (b-i) Overall material balance 0.5 ´ 3600 ´ 0.25 = 33 500 (0.0467 – M) or M = 0.0333 kmol Molar mass of final mixture = 0.4366 ´ 100 + 114 ´ 0.5635 = 107.89 Left over quantity in the flask = 107.89 ´ 0.0333 = 3.59 kg Ans. (b-ii)

EXERCISE 5.72 Basis: Ammonia feed to burner = 3266 Nm3/h = 145.71 kmol/h This basis is choosen to be same as that assumed in Exercise 4.48. Total feed = Ammonia + Air (dry) = 145.71 + 1311.41 = 1457.12 kmol/h Ammonia content of the feed = 145.71 ´ 100/1457.12 = 10% (check !) Yield of NO = Combustion efficiency as derived in Exercise 4.48 = 95% Since both the conditions of the example are satisfied, the calculations of Exercise 4.48 can be utilised for the balance calculations. Heat of reactions: There are two reactions taking place in burner. Reaction (1): 4 NH3 + 5 O2 = 4 NO + 6 H2O 4 (– 46.11) 5 (0) 4 (90.25) 6 (– 241.82) DH¢f , kJ/mol DH°r1 = 6(–241.82) + 4(+90.25) – [4(–46.11)] = – 905.48 kJ/4 mol NH3 º – 226.37 kJ/mol NH3

180 Solutions Manual—Stoichiometry

Ammonia reacted as per reaction (1) = 138.45 kmol/h Heat generated as per reaction (1) f1 = 226.37 ´ 1000 ´ 138.45 = 31 340 927 kJ/h at 298.15 K º 8705.813 kW Reaction (2): 4 NH3 + 3 O2 = 2 N2 + 6 H2O 4 (– 46.11) 3 (0) 2 (0) 6(– 241.82) DH°f, kJ/mol DH°r2 = 6 (– 241.82) – 4 (– 46.11) = – 1266.48 kJ/4 mol NH3 º –316.62 kJ/mol NH3 Ammonia reacted as per reaction (2) = 145.71 – 138.45 = 7.26 kmol/h Heat generated as per reaction (2) f2 = 316.62 ´ 1000 ´ 7.26 = 2 298 661 kJ/h = 638.517 kW Moisture, entering with air = 1311.41 ´ 0.016 ´ 29/18 = 33.805 kmol/h Component NH3 O2 N2 H2O Total

.

ni

Heat capacity equation constants .

.

kmol/h

ai × n i

bi× n i ´ 103

145.71 275.40 1 036.01 33.805 1 490.925

3 737.5 5 658.1 30 656.5 1 098.4 41 150.5

4 878.5 22 058.1 –5326.1 2.7 21 613.2

.

ci× n i ´ 106

.

di× n i ´ 109

51.3 –17 195.1 13 657.6 446.6 –3 039.6

–449.3 4 674.1 5 146.9 –153.7 9 218.0

C°mp = 37 413.0 + 16 734.7 ´ 10–3 T – 3090.9 ´ 10–6 T2 + 9667.3 ´ 10–9 T 3 air C°mp = 3737.5 + 4878.5 ´ 10–3 T + 51.3 ´ 10–6 T2 – 449.3 ´ 10–9 T 3 am

Air is preheated to 260°C (533.15 K) and introduced to the reactor. Reference temperature, t0/T0 = 25°C/298.15 K Enthalpy of air at 533.15 K over 298.15 K, f3 =

553.15

ò C°mp dT

298.15

air

= 11 416 807.1 kJ/h = 3171.335 kW

Energy Balances 181

Product stream from reactor .

Component

ni

Heat capacity equation constants .

kmol/h NO O2 N2 H2O Total

.

n i × ai

138.450 96.895 1039.645 252.375 1 527.365

4 082.4 1 990.7 30 764.0 8 200.2 45 037.3

.

.

n i × bi ´ 103

n i × ci ´ 106

n i × di ´ 109

– 284.2 7 760.8 –5 344.8 20.1 2 151.9

1 569.7 –6 049.8 13 705.5 3 334.1 12 559.5

–667.3 1 644.5 –5 165.0 –1 147.7 –5 335.5

Product stream temperature is to be limited to 910°C (1183.15 K). Enthalpy of reactor exit gas stream at 1183.15 K over 298.15 K.

f4 =

1183.15

ò 45 037.3 + 2151.9 ´ 10–3 T + 12 559.5 ´ 10–6 T2 – 5335.5

298.15

´ 10–9 T3) dT = 45 488 086.1 kJ/h º 12 635.579 kW Let enthalpy of superheated ammonia be f5 at T K. f3 + f5 + f1 + f2 = f4 f5 = 45 488 086.1 – 11 416 807.1 – 31 340 927 – 2298 661 = 431 691 kJ/h = Solving by mathcad,

T

ò Comp

298.15

am

dT

T = 378.45 K or t = 105.3°C

Ans.

6

Stoichiometry and Unit Operations EXERCISE 6.1 Basis:

Feed rate = 4000 kg/h Ethanol content = 4000 ´ 0.3 = 1200 kg/h Water content = 4000 – 1200 = 2800 kg/h Molar feed rate, F =

1200 2800 + 46 18

= 26.087 + 155.556 = 181.643 kmol/h Ethanol with NBP = 78.25 °C (351.4 K) is a more volatile component. Ethanol content in feed, zF =

26.087 181.643

= 0.1436 at bubble point Mole fraction of ethanol in distillate, (92 / 46) xD = (92 / 46) + (8 /18)

= 0.8182 Bottom product is nearly pure water. xB = 0 Ethanol balance: F × z F = D × xD + xB × W 181.643 ´ 0.1436 = 0.8182 D + 0 ´ W D = 31.88 kmol/h W= F– D = 181.643 – 31.88 = 149.763 kmol/h

Stoichiometry and Unit Operations 183

Using data, contained in Table 6.21 enthalpy-concentration diagram for ethanolwater binary is prepared as under. 95

P = 1.01 325 bar a

373 368

90

363

85

358

80

353

Temperature, K

Temperature, °C

100

75 348 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mole Fraction of Ethanol t-x-y Diagram For Ethanol & Water 49100 49000 48900 48800 48700

Enthalpy of Liquid, kJ/kmol Mixture

48600 10200

48500

9800

48400

9400

48300

9000

48200

8600

48100

8200

48000

Enthalpy of Vapour, kJ/mol Mixture

P = 1.01 325 bar a

7800 7400 7000 6600

0.0

01 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Mole Fraction of Ethanol Enthalpy-Concentration Diagram For Ethanol & Water

Reflux ratio = 2 kmol/kmol Overhead vapour flow = (2 + 1)D = 3 ´ 31.88 = 95.64 kmol/h From the enthalpy-concentration diagram, following values are read. For feed, ZF = 0.1436 HL = 6915 kJ/kmol mixture

184 Solutions Manual—Stoichiometry

For distillate,

xD = 0.8182, HV = 48 900 kJ/kmol mixture H L = 9000 kJ/kmol mixture For bottom, xB = 0, HL = 7550 kJ/kmol water Heat duty of overhead condenser, fC = 95.64 (48 900 – 9000) = 3816 036 kJ/h º 1060.01 kW Enthalpy of feed, fF = 181.643 ´ 69 15 = 1256 061 kJ/h º 348.906 kW Ethalpy of bottom product, fB = 149.763 ´ 7550 = 1130 711 kJ/h º 314.086 kW Enthalpy of distillate product, fD = 31.88 ´ 9000 = 286 920 k/h º 79.7 kW Heat duty of reboiler, fRB = fC + fB + fD – fF = 286 920 + 1130 711 + 3816 036 – 1256 061 = 3977 606 kJ/h º 1104.891 kW Ans.

EXERCISE 6.2 Basis: 10 000 kg/h feed Let D and B be the amount of distillate and bottom products, respectively. Overall material balance: D + B = 10 000 (1) Balance of DEA: 0.5 ´ 10 000 = 0.99 D + 0.05 B 19.8 D + B = 100 000 (2) Solving the equations, B = 5212.8 kg/h D = 4787.2 kg/h Ans. (a) Enthalpy balance: Enthalpy of distillate product, f1 = 4787.2 ´ 372.2 = 1781 795 kJ/h º 494.94 kW Reflux ratio = 0.8 kmol/kmol product Total vapours, condensed in the overhead condenser = 4787.2 (1 + 0.8) = 8617 kg/h

Stoichiometry and Unit Operations 185

Heat (latent heat only) duty of the condenser, f2 = 8617 (1572.6 – 372.2) = 10 343 847 kJ/h º 2873.29 kW Ans. (b) Enthalpy of feed, f3 = 10 000 ´ 418.7 = 4187 000 kJ/h = 1163.06 kW Enthalpy of bottoms, f4 = 5212.8 ´ 558.1 = 2909 263.7 kJ/h º 809.13 kW Heat load of reboiler, f5 = Enthalpy of distillate + Enthalpy of bottoms + Heat removed in condenser – Enthalpy of feed = 494.94 + 808.13 + 2873.29 – 1163.06 = 3013.3 kW Ans. (c)

EXERCISE 6.3 Basis: Fresh feed rate of 10 000 kg/h of 50:50 mixture of toluene and methylcyclohexane (MCH) by mass. Let T1 and T2 be the extractive distillation column and solvent recovery column, respectively. Toluene feed to T1 = 5000 kg/h It is desired to recover 95% toluene as an overhead product of 90% purity from T2. Material balance over T2: Toluene in overhead product = 5000 ´ 0.95 = 4750 kg/h Overhead product rate = 4750/0.99 = 4798 kg/h MCH in bottom product = 4798 – 4750 = 48 kg/h Bottom product of T2 tower is the recycled solvent. Recycled solvent = 3.3 ´ 10 000 = 33 000 kg/h Presence of toluene in the bottom product = 0.9% Phenol in the recycled solvent = 33 000 ´ 0.991 = 32 703 kg/h Toluene in the recycled solvent = 33 000 – 32 703 = 297 kg/h Composition of F2: Component Toluene MCH Phenol Total

kg/h

mass %

4750 + 297 = 5 047 48 32 703

13.35 0.13 86.52

37 798

100.00

186 Solutions Manual—Stoichiometry

Material balance over T1: Toluene unrecovered = 5000 – 4750 = 250 kg/h This unrecovered toluene finds its ways into the overhead product from T1. MCH in the overhead product = 5000 – 48 = 4952 kg/h Since phenol content of overhead product is 0.2%, Overhead product rate = (250 + 4952)/0.998 = 5212.42 kg/h Phenol in the overhead product = 5212.42 – (250 + 4952) = 10.42 kg/h Make-up phenol rate will be same as loss of phenol in the overhead product. Ans.

EXERCISE 6.4

Basis: Fresh feed, F = 100 kg Let recycle stream be V3 kg. Total feed to C 1 = F + V3 = 100 + V3 kg Both streams are azeotropes and contain 96% (by mass) ethanol. Overhead vapour from C1 is a ternary azeotrope. Component

Molar mass

mole %

kg

mass %

46 78 18

22.8 53.0 23.3

1048.8 4204.2 419.4

18.48 74.12 7.40

100.0

5672.4

100.00

C2H5OH C6H6 H2O Total

Bottom product W1 from C 1 is pure ethanol. Hence all water appear in the ternary azeotrope. Ethanol entering C 1 = 0.04 (100 + V3) kg Overhead vapour =

0.04(100 + V3 ) = 54.054 + 0.540 54 V3 0.074

Material balance across C 2: Entire benzene of D1 comes out in V2. Benzene balance:

Ethanol balance: Ethanol in

0.751 V2 = 0.11 D1 D1 = 6.827 V2 W2 = 0.53 D1 – 0.125 V2

Stoichiometry and Unit Operations 187

= 0.53 (6.827 V2) – 0.125 V2 = 3.4933 V2 kg

Water balance: Water in W2

= 0.35 D1 – 0.124 V2 = 0.35 (6.827 V2) – 0.124 V2 = 2.2655 V2 kg Quantity of W2 = 3.4933 V2 + 2.2655 V2 = 5.7588 V2 kg Material balance across C 3: Water, entering in fresh feed (F), must be purged out as W3 stream. W2 = V3 + 4 5.7588 V2 – V3 = 4 Water in V3 = 2.2655 V2 – 4 But V3 contains 4% water. 0.04 V3 = 2.2655 V2 – 4 or 56.6375 V2 – V3 = 100 Solving Eq. (1) and Eq. (2), V 2 = 1.887 kg and V3 = 6.8669 kg Component

V3

W3

(1)

(2)

W2

kg

mass %

kg

mass %

kg

mass %

C2H5OH H2O

6.5922 0.2747

96.00 4.00

0 4.000

0 100.00

6.5922 4.2747

60.66 39.34

Total

6.8669

100.00

4.0000

100.00

10.8669

100.00

Component

V2

D1

kg

mass %

0.2359 1.4171

12.50 75.10

H2O

0.2340

12.40

Total

1.8870

100.00

C2H5OH C6H6

kg

V1 mass %

kg

mass %

6.8281 1.4171

53.54 11.11

10.6751 42.8160

18.48 74.12

4.5087

35.35

12.7539

100.00

4.2747 57.7658

Reflux R = V1 + V2 – D1 = 57.7658 + 1.887 – 12.7539 = 46.8989 kg/100 kg fresh feed

7.40 100.00

188 Solutions Manual—Stoichiometry

Composition of Reflux Component

kg

mass %

C2H5OH C6H6 H2O

4.0829 42.8160 0

8.71 91.29 0

Total

46.8989

100.00

EXERCISE 6.5 (a) From Fig. 6.26, Stream, leaving reaction section= R + P kmol Light component balance: R xR = (R+ P) xE – P xP or

xR =

( R + P) xE - Pxp R

Since the above equation does not involve the term Fp, it is evident that the recycle composition is independent of Fp, i.e. feed to the distillation column. Ans. (i) Light component balance across the distillation column: D × xD + P × xP = FP xE and D = FP – P xD =

FP × xE - P × xP FP - P

From the above relationship, it can be inferred that as FP decreases, xD increases for a given xE and xP. Maximum value of xD can be unity. For xD = 1, Fpm – P = Fpm × xE – P × xP or F pm = P (1 – xP)/(1 – xE) Ans. (ii) (b) Fig. 6.27, Light component balance across the distillation column: FP = B + P xE × FP = B × xB + P × xP xE =

FP xE - P × xP Fp - P

From the above relationship, it can be seen that as FP decreases, xB decreases for a given xE and xP. Minimum value of xB can be zero. For xB = 0,

Stoichiometry and Unit Operations 189

Fpm xE = P × xP F pm = P × xP /xE

Ans.

EXERCISE 6.6 Molar mass of ACN = 41.0519 Molar mass of H2O = 18.0153 Feed contains 50% ACN by mass. 1 kg feed solution contains 0.5 kg or 0.0122 kmol ACN and 0.5 kg or 0.0278 kmol H2O. Mole fraction of ACN in feed =

0.0122 (0.0122 + 0.0278)

= 0.305 Mole fraction of H2O in feed = 1 – 0.305 = 0.695 Basis: 100 kmol fresh feed Mixed feed to CI = 100 + R kmol Recycle stream contains 0.6 mole fraction ACN. ACN in mixed feed = 100 ´ 0.305 + 0.6 R = 30.5 + 0.6 R kmol Bottom product from CI is pure water. Bottom product from CII is 99.0 mole % ACN. ACN of fresh feed must be purged out as the bottom product of CII. 30.5 = 30.81 kmol 0.99 Water in bottom product = 30.81 – 30.5 = 0.31 kmol ACN, contained in mixed feed is distilled as an azeotrope, containing 69.0 mole % ACN.

Quantity of bottom product from CII =

30.5 + 0.6 R 0.69 = 44.203 + 0.8696 R Water in D1 = 44.203 + 0.8696 R – 30.5 – 0.6 R = 13.703 + 0.2696 R kmol

Quantity of D1 =

D1 is fed to CII.

Water in D2 = 13.703 + 0.2696 R – 0.31 = 13.393 + 0.2696 R kmol D2 is the azeotrope containing 40% water. Quantity of R =

(13.393 + 0.2696 R ) 0.4

= 33.4825 + 0.674 R R = 102.7 kmol/100 kmol fresh feed

Ans.

190 Solutions Manual—Stoichiometry

EXERCISE 6.7 Basis: 100 m3/h of feed Specific volume of feed gas mixture at 100 kPa and 318.15 K = 26.455 m3/kmol (Ref. Table 7.8) Feed gas rate = 100/26.455 = 3.78 kmol/h HCl in the feed gas = 3.78 ´ 0.35 = 1.323 kmol/h Nitrogen in the feed = 3.78 – 1.323 = 2.457 kmol/h HCl absorbed = 1.323 ´ 0.96 = 1.27 kmol/h HCl unabsorbed = 1.323 – 1.27 = 0.053 kmol/h HCl content of outgoing dry gas mixture =

0.053 ´ 100 (2.457 + 0.053)

= 2.11% Vapour pressure of H2O at 308 K (35 °C) = 5.6216 kPa

Ans. (b) (Table 6.13)

kmol 5.6216 = 0.0596 kmol dry gas 100 - 5.6216 Total outgoing gas mixture = (2.457 + 0.053)(1 + 0.0596) = 2.66 kmol/h Specific volume of outgoing gas mixture at 98 kPa and 308.15 K = 26.144 m3/kmol (Ref. Table 7.8) Volumetric flow rate = 2.66 ´ 26.144 = 69.543 m3/h Ans. (a) Water content of gas mixture =

D H°f of 3.9% HCl solution = – 165.36 kJ/mol HCl (Ref. Table 5.79) D H°f of gaseous HCl at 298.15 K = – 92.31 kJ/mol HCl Heat of solution at 298.15 K = – 165.36 – (– 92.31) = – 73.05 kJ/mol HCl Heat generated, f2 = 73.05 ´ 1000 ´ 1.27 = 92 774 kJ/h at 298.15 K º 25.770 kW Heat balance: Component

Inlet kmol

C°mp equation constants ni × ai

ni × bi ´ 10

3

ni × ci ´ 106

ni × di ´ 109

N2 HCl

2.457 1.323

72.705 40.099

– 12.631 – 10.067

32.39 17.544

– 12.206 – 5.733

Total

3.780

112.804

– 22.698

49.934

– 17.939

Stoichiometry and Unit Operations 191

HCl absorbed = 1.27 kmol/h = 46.355 kg/h Final solution flow =

46.355 = 1188.59 kg/h 0.039

Assume water input at 25°C (298.15 K). Component Outlet kmol ni

C mp º equation constants ni × ai

ni × bi ´ 103

ni× ci ´ 106

ni × di ´ 109

N2 H2O HCl

2.457 0.15 0.053

72.705 4.874 1.606

– 12.631 0.012 – 0.403

32.39 1.982 0.703

– 12.206 – 0.682 – 0.230

Total

2.660

79.185

– 13.022

35.075

– 13.118

Enthalpy of incoming gas mixture, f1 = 112.804 (318.15 – 298.15) – 22.698 ´ 10–3 (318.152 – 298.152)/2 + 49.934 ´ 10–6 (318.153 – 2983)/3 – 17.939 ´ 10–9 (318.154 – 298.154)/4 = 2256.1 – 139.8 + 94.8 – 10.5 = 2200.6 kJ/h º 0.611 kW Enthalpy of outgoing gas mixture, f3 = 79.185 (303.15 – 298.15) – 13.022 ´ 10–3 (303.152 – 298.152)/2 + 35.075 (303.153 – 298.153)/3 – 13.118 ´ 10–9 (303.154 – 298.154)/4 = 396 – 19.6 + 15.9 – 1.8 = 390.5 kJ/h º 0.108 kW Heat content of final solution, f4 = 2200.6 + 92 774 – 390.5 = 94584.1 kJ/h º 26.273 kW 94 584.1 1188.59 ´ 4.19 = 19 K over 298.15 K

Rise in temperature of final solution =

Final temperature = 298.15 + 19 = 317.15 K or 44°C

Ans. (c)

EXERCISE 6.8 Basis: 1425 m3/h of gas at 303 K and 101.325 kPa Specific volume V = 24.876 m3/kmol (Table 7.8) Flow rate of gas = 1425/24.876 = 57.28 kmol/h SO2 content of ingoing gas = 57.28 ´ 0.148 = 8.48 kmol/h Inerts in the feed gas = 57.28 – 8.48 = 48.8 kmol/h

192 Solutions Manual—Stoichiometry

SO2 content of outgoing gas mixture = 1% Flow rate of outgoing gas mixture = 48.8/0.99 = 49.29 kmol/h SO2 in outgoing gas mixture = 49.29 – 48.8 = 0.49 kmol/h SO2 absorbed = 8.50 – 0.49 = 8.01 kg mol/h º 512.64 kg/h Water flow rate = 16.5 L/s = 59 400 kg/h Flow rate of final solution = 59 400 + 512.64 = 59 912.64 kg/h SO2 content of final solution = 512.64 ´ 100/59 912.64 = 0.86% Ans.

EXERCISE 6.9 Basis: 100 kg fresh feed Refer Fig. E6.3 Point F on the ternary diagram represents 76% C9, 18% AcOH and 6% Ac2O by mass. This point lies above the top binodal curve. Since distillation boundary passes between the two binodal curves, one fractionating column will be sufficient to separate total C9 from the feed mixture and the top product from column–1 will be a ternary azeotrope. Point N represents pure C9. Join NS which will intersect base line (0% C9) at S. Point R2 represents an azeotrope of C9 and AcOH (63.4% by mass). Quantity of mixture (represented by) S FN = Quantity of N FS

Since FN + FS = 100 kg (i.e. fresh feed) Quantity of S = [FN/(FN + FS)] = 4.9 units ´ 100/20.4 units = 24 kg Quantity of N = 100 – 24 = 76 kg of C9 Simple mass balance will also give the same result. Feed A lies on line SR2 and also on the binodal curve.

AR2 Quantity of mixture (represented by) S 5.6 units = = 7 units Quantity of mixture (represented by) A SR2 Quantity of mixture (represented by) A = 7 ´ 24/5.6 = 30 kg Composition of A is read from the ternary diagram as AcOH: 0.728, Ac2O: 0.2 and C9: 0.072 (mass fractions). Quantity of R2 = Quantity of A – Quantity of S = 30 – 24 = 6 kg

Ternary Diagram for HOAc – Ac20 – C9:Naphtha Naphtha (C9) Naptha 0.9

F 0.7 B IN

OD

AL CU RV R1 ILLA E TION BOU NDA 0.5 RY

E

E

TIF

0.5

C9

R2

6.0

LIN

1.0

LIN

3

Ac20

s%

0.5

1.0

0.1 BINODAL CURVE

Mass % HOAc ¾ ¾¾® Fig. E6.3 Solutionof Exercise 6.8

A S

6.0

HOAc

Stoichiometry and Unit Operations 193

0.

6.0

as

D

0.3

0.3 0.7

M

TIF

s%

Ac

2O

DIST

as M

M

194 Solutions Manual—Stoichiometry

AD is a tieline, drawn in proportion to other sideby tielines, which intersects top binodal curve at R1

AD1 Quantity of mixture (represented by) D 9.7 units = = 2.7 units Quantity of mixture (represented by) A R1 D Quantity of D = 9.7 ´ 30/2.7 = 107.8 kg Quantity of R1 = 107.8 – 30 = 77.8 kg Compositions of R1 and D are calculated from the diagram as follows. Stream AcOH 0.372 0.47

R1 D

Composition, mass fraction Ac2O C9 0.072 0.559 0.108 0.422

Join ND. Point M lies on ND. Quantity of N 76 units DM = = 107.8 units Quantity of D NM

DM 76 DM 76 = = = 76 + 107.8 1838 . DM + TN ND ND = 12.4 units DM = 12.4 ´ 76/183.8 = 5.13 units Composition of M is read from the diagram as: AcOH: 27.6%, Ac2O: 6.3% and C9: 66.1% (mass %) All the resuls are summarised for N = 1000 kg/h in Table – 6.24 in the text. Ans.

EXERCISE 6.10 Basis: 6500 m3/h of feed gas in terms of H2 Actual flow of feed gas = 6500/0.78 = 8333.3 m3/h This flow is rated at 101.3 kPa and 300.15 K (27 ºC). However, the actual conditions are 1.6 MPa g and 473.15 K(200 ºC). Assuring ideal gas law, actual flow = (101.3 ´ 8333.3 ´ 473.15)/(300.15 ´ 1701.3) = 782.3 m3/h Ans. (a) Rest calculations will be based on 101.3 kPa and 300.15 K (27 ºC). CO2 in inlet gas mixture = 8333.3 ´ 0.165 = 1375 m3/h (H2 + inerts) = 8333.3 – 1375 = 6958.3 m3/h

Stoichiometry and Unit Operations 195

Outgoint gas mixture = 6958.3/0.995 = 6993.3 m3/h CO2 in outgoing gas mixture = 6993.3 – 6958.3 = 35 m3/h CO2 absorbed = 1375 – 35 = 1340 m3/h Specific vol. at 101.3 kPa and 298.15 K (25 ºC), V = 24.465 m3/kmol CO2 absorbed = 1340/24.465 = 54.77 kmol/h Ans. (b) Initial concn. of CO2 in lean TEA solution = 2420 mL/L Flow of lean TEA aqueous solution = 900 L/min CO2 in inlet solution = 2.42 ´ 900/1000 = 2.18 m3/min CO2 in rich solution = 2.18 + (1340/60) = 24.51 m3/min CO2 concn. in outgoing solution = 24.51 ´ 1000/0.9 = 27 237 mL/L Ans. (c)

EXERCISE 6.11 Basis: 3000 m3/h of gas mixture at 101.3 kPa and 300.15 K (27 ºC) Actual flow rate at 4 bar g and 523.15 K (250 ºC) = 101.325 ´ 3000 ´ 523.15/(300.15 ´ 501.325) = 1056.8 m3/h Ans. (a) Rest other calculations are based on 101.325 kPa and 298.15 K (25 ºC). CO2 in inlet gas = 3000 ´ 0.18 = 540 m3/h CO2 free gas mixture = 3000 – 540 = 2460 m3/h Flow of outgoing gas mixture = 2460/0.998 = 2464.9 m3/h CO2 in outgoing gas mixture = 2464.9 – 2460 = 4.9 m3/h CO2 absorbed = 540 – 4.9 = 535.1 m3/h º 535.1/24.465 º 21.87 kmol/h Ans. (b) Change in CO2 content of DAPOL solution = 5800 – 2200 = 3600 mL/L Flow rate of aqueous DAPOL solution =

5351 . ´ 106 60 ´ 3600

= 2477.3 L/min º 41.3 L/s

EXERCISE 6.12 Basis: 0.625 L/s (37.5 L/min = 2250 L/h) MEA solution flow rate Gas inlet flow = 1000 m3/h of dry gas mixture

Ans. (c)

196 Solutions Manual—Stoichiometry

Gas enters the absorber at 101.3 kPa a and 333.15 K (60 ºC). Vapour pressure of water at 333.15 K, pw = 19.92 kPa (Ref. Table 6.13) Moisture content of incoming gas mixture =

19.92 = 0.2447 kmol/kmol dry gas 101.325 - 19.92

Flow of dry incoming gas mixture = 38.3 kmol/h Moisture, entering the absorber = 38.3 ´ 0.2447 = 9.372 kmol/h Vapour pressure of water at 318.15 K (45 ºC) = 9.582 kPa (Ref. Table 6.13) Moisture content of outgoing gas mixture = 9.582/(90 – 9.582) = 0.119 kmol/kmol dry gas Flow of dry outgoing gas mixture = 34.75 kmol/h Water vapours, leaving the absorber = 34.75 ´ 0.119 = 4.135 kmol/h Water vapours condensed = 9.372 – 4.135 = 5.237 kmol/h º 94.27 kg/h Inlet gas mixture: Component

ni kmol

Cºmp equation constants

N2 H2 CO CO2 CH4 H2O

8.50 25.51 0.19 3.98 0.12 9.37

ni × ai 251.52 729.85 5.52 85.03 2.31 304.45

Total

47.67

1378.68

ni × bi ´ 103 – 43.70 26.00 – 0.54 255.85 4.25 0.75 244.61

ni × ci ´ 106 112.05 – 3.77 2.21 – 163.38 1.44 123.78 72.33

ni × bi ´ 109 – 42.23 19.62 – 0.89 39.00 – 1.36 – 42.61 – 28.47

Enthalpy of inlet gas at 333.15 K over 298.15 K, f1 = 1378.68 (333.15 – 298.15) + 244.61 ´ 10–3 (333.152 – 298.152)/2 + 72.33 ´ 10–6 (333.153 – 298.153)/3 – 28.47 ´ 10–9 (333.154 – 298.154)/4 = 48 254 + 2702 + 252 – 31 = 51 177 kJ/h º 14.216 kW Outlet gas mixture: Component kmol ni N2

8.50

Cmp º equation constants ni × ai 251.52

ni × bi ´ 103 – 43.70

ni × ci ´ 106 112.05

ni × bi ´ 109 – 42.23

(contd.)

Stoichiometry and Unit Operations 197

(contd.) H2 CO CO2 CH4 H2O

25.51 0.19 1.61 0.12 4.14

729.85 5.52 34.40 2.31 139.07

26.00 – 0.54 103.50 6.25 0.34

– 3.77 2.21 – 66.09 1.44 56.54

19.62 – 0.89 15.78 – 1.36 – 19.46

Total

40.07

1162.67

91.85

102.38

– 28.54

Enthalpy of outlet gas at 318.15 K over 298.15 K, f2 = 1162.67 (318.15 – 298.15) + 91.85 ´ 10–3 (318.152 – 298.152)/2 + 102.38 ´ 10–6 (318.153 – 298.153)/3 – 28.47 ´ 10–9 (318.154 – 298.154)/4 = 23 253 + 566 + 197 – 17 = 23 996 kJ/h º 6.667 kW Assume that water is condensed at an average temperature of (333 + 318)/2 = 325.5 K lv of water at 325.5 K = 2376 kJ/kg (Table AIV.1) Heat removed due to condensation f3 = 94.27 ´ 2376 = 223 986 kJ/h º 62.218 kW Enthalpy of feed solution = 2250 ´ 4.19 (318.15 – 298.15) = 188 550 kJ/h º 52.375 kW Flow of outgoing solution = 2250 + 2.35 ´ 44 + 94.27 = 2447.67 kg/h Heat of absorption = 2.35 ´ 44.0098 ´ 1675 = 173 234 kJ/h = 48.12 kW (exothermic) Enthalpy of feed gas + Enthalpy of feed solution + Heat of absorption + Heat of condensation = Enthalpy of outgoing gas + Enthalpy of outgoing solution Let T = temperature of outgoing solution, K 3600 (14.216 + 52.375 + 48.12 + 62.218) = 6.667 ´ 3600 + 2447.67 ´ 4.19 (T – 298.15) 10 255.7 (T – 298.15) = 3600 ´ 170.262 T – 298.15 = 59.77 T = 357.9 K or 84.75°C Ans.

EXERCISE 6.13 Basis: 25 000 m3/h feed gas

198 Solutions Manual—Stoichiometry

Molar Mole feed rate = 25 000/22.414 = 1115.4 kmol/h Gas

Molar mass

kmol/h

kg/h

H2 HCl N2 CCl4

2 36.5 28 154

921.3 12.3 129.4 52.4

1 842.6 449.0 3 623.2 8 069.6

Total

—

1115.4

13 984.4

13 984.4 1115.4 = 12.54 HCl absorbed = 12.3 ´ 0.999 = 12.29 kmol/h Reaction: HCl (g) + NaOH(ai) = NaCl(ai) + H2O(l) DHºf – 92.31 – 470.11 – 407.27 – 285.83 DHºR = – 407.27 – 285.83 – (– 92.31 – 470.11) = – 130.68 kJ/mol HCl NaOH required = 12.29 kmol/h º 491.6 kg/h Pressure of gas at oulet = 3.38 bar g = 4.393 25 bar a Gas mixture will leave saturated at 311.15 K (38ºC) at the top. Total pressure = 338 kPa a Vapour pressure of H2O at 311.15 K (38ºC), pw = 6.624 kPa (Table 6.13) Water content of outgoing gas mixture = 6.624/(439.325 – 6.624) = 0.0153 kmol/kmol dry gas Molar flow rate of outgoing dry gas mixture = 1115.4 – 12.29 = 1103.11 kmol/h Water vapours, accompanying gas mixture = 1103.11 ´ 0.0153 = 16.878 kmol/h º 303.8 kg/h Composition of solution, leaving the absorber:

Average molar mass of gas mixture =

Component NaCl NaOH H2O Total

kg/h

mass %

12.29 ´ 58.5 = 719.0 3930 ´ 0.15 – 491.6 = 97.9 12.29 ´ 18 + 3930 ´ 0.85 – 303.8 = 3257.9

17.65 2.40 79.95

4074.8

100.00

Ans. (a)

Stoichiometry and Unit Operations 199

Total heat generated = 12.29 ´ 1000 ´ 130.68 = 1606 057 kJ/h = 446.127 kW Ans. (b) Use Fig. 5.16 for heat of solution calculations: Locate a point, representing 50% NaOH solution at 303 K (38ºC). Locate another point, representing 0% NaOH at 327 K (30ºC). Join the points. The line interesects 15% axis at 327 K (54ºC). Enthalpy of 15% solution at 327 K (54ºC) = 187.5 kJ/kg soln. Enthalpy of 15% solution at 311 K (38ºC) = 118.8 kcal/kg soln. Total heat to be removed = (187.5 – 118.8)3930 = 269 991 kJ/h º 75 kW Ans. (c)

EXERCISE 6.14 Basis: Ammonia flow to burner = 3266 Nm3/h Following data are tabulated from Excercise 4.48 Components

Dry inlet gas kmol/h

mole %

Dry outlet gas kmol/h

mole %

NO O2 N2

138.45 96.9 1039.65

10.86 7.60 81.54

2.80 71.47 1326.73

0.20 5.10 94.70

Total

1275.00

100.00

1401.00

100.00

Pressure of inlet gas mixture = 0.15 MPa g = 251.325 kPa a Vapour pressure of H2O at 40ºC = 7.375 kPa (Ref. Table 6.13) Water content = 7.375/(251.325 – 7.375) = 0.0302 kmol/kmol dry gas Water, accompanying ingoing gas = 0.0302 ´ 1275 = 38.505 kmol/h º 693.1 kg/h Pressure of outgoing gas mixture = 10 kPa g = 111.325 kPa a Vapour pressure of H2O at 323 K (50ºC) = 12.335 kPa Water content =

12.335 = 0.1246 kmol/kmol dry gas - 12.335) (111325 .

Water, accompanying outgoing gas = 0.1246 ´ 1401 = 174.56 kmol/h º 3142.2 kg/h

200 Solutions Manual—Stoichiometry

Total water in reactor outlet gas mixture = 252.375 kmol/h (Ref.Exercise 5.72) Water, entering with secondary air = 363.39 ´ 29 ´ 0.016 = 168.6 kg/h º 9.37 kmol/h Water condensed in the cooler = 252.375 + 9.37 – 38.505 = 223.24 kmol/h = 4018.32 kg/h at 313 K (40ºC) Acid produced = 135.65 kmol/h (same as NO consumed) = 8546 kg/h at 100% concn. of HNO3 Final strength of acid = 58% (by mass) Final acid quantity = 8546/0.58 = 14 734.5 kg/h Water content of acid = 14 734.5 – 8546 = 6188.5 kg/h Water consumed in the reaction = 135.65/2 = 67.825 kmol/h º 1220.9 kg/h Water balance across absorber: Let a be the amount of demineralised water added, 4018.32 + a + 693.1 = 6188.5 + 3142.2 + 1220.9 a = 5840.18 kg/h Reaction: 2 NO(g) + 1.5 O(g) + H2O(l) = 2 HNO3(l) DHºf

2 ´ 90.309

0

– 286.021

2 (– 174.213)

DHºr = 2 (–174.213) – (2 ´ 90.309 – 286.021) = 243.023 kJ/2 mol NO Heat of reaction at 298.15 K (25 ºC) = 243.023 ´ 135.65 ´ 1000/2 = 16 483 035 kJ/h (exothermic) º 4578.62 kW Air introduced at 311 K (40 ºC) = 363.39 kmol/h (secondary) O2 feed = 76.31 kmol/h N2 feed = 287.08 kmol/h Base temperature T0 = 298.15 K (25 ºC) Enthalpy of water vapour at 313 K (40 ºC) over 298.15 K (25 ºC) = 2574.4 – 104.77 = 2469.63 kJ/kg Enthalpy of water vapour at 323 K (50ºC) over 298.15 K (25ºC) = 2592.2 – 104.77 = 2487.43 kJ/kg

Stoichiometry and Unit Operations 201

For finding the cooling load, enthalpy balance across the absorber has to be made. Absorber inlet gas enthalpy: Temperature of inlet gas = 313.15 K (40ºC) Component

C mp º equation constants

ni kmol

ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

NO

138.45

4 082.4

– 284.2

1 569.7

– 667.3

O2

96.90

2 521.9

1139.1

– 227.0

– 54.5

N2

1039.65

30 764.2

– 5344.8

13 705.6

– 5165.0

Total

1275.00

37 368.5

– 4489.9

15 048.3

– 5886.8

Enthalpy of inlet gas, f1 = 37 368.5 (313.15 – 289.15) – 4489.9 ´ 10–3 (313.152 – 298.152)/2 + 15 048.3 ´ 10–6 (313.153 – 298.153)/3 – 5886.8 ´ 10–9 (313.154 – 298.154)/4 = 558 511 kJ/h º 155.142 kW Enthalpy of moist air at 313.15 K over 298.15 K =

313.15

o ò Cmpair dT

298.15

(use data from Exercise 5.72

= 186 690 kJ/h

(by Mathcad)

º 51.858 kW Absorber outlet gas enthalpy: Temperature of outlet gas = 323.15 K (50°C) C mp º equation constants ni × ai

ni × bi ´ 103

ni × ci ´ 106

NO O2 N2

ni kmol 2.80 71.47 1326.73

82.6 1 860.1 39 259.1

– 5.7 840.1 – 6820.7

31.7 – 167.4 17 490.2

– 13.5 – 40.2 – 6591.2

Total

1400.00

41 201.8

– 5986.3

17 354.5

– 6644.9

Component

ni × di ´ 109

Enthalpy of outgoing gas mixture, f2 = 41 201.8 (323.15 – 298.15) – 5586.3 ´ 10–3 (323.152 – 298.152)/2

202 Solutions Manual—Stoichiometry

+ 17 354.5 ´ 10–6 (323.153 – 298.153)/3 – 6644.9 ´ 10–9 (323.154 – 298.154)/4 = 1020 458 kJ/h º 283.461 kW Heat of dilution : Ref. Table 6.25 DHºf of 58% HNO3 liquid = – 197.00 kJ/mol HNO3 (by interpolation) DHºf of 100% HNO3 liquid = – 174.1 kJ/mol HNO3 Enthalpy change = – 197.00 – (– 174.1) = – 22.9 kJ/mol HNO3 Heat evolved at 298.15 K, f3 = 135.65 ´ 22 900 = 3106 385 kJ/h º 862.885 kW Heat capacity of 58% HNO3 = 2.64 kJ/(kg × K) from Fig. 5.20 Enthalpy of liquid 58% HNO3 over 298.15 K (25 °C) = 2.64 (313.15 – 298.15) = 39.6 kJ/kg at 313.15 K Total enthalpy of 58% acid = 14 734.4 ´ 39.6 = 583 482 kJ/h º 162.078 kW Heat evolved at 313.15 K (40°C) (net) = 862.885 – 162.078 = 700.807 kW Enthalpy of condensate from cooler = 4018.32 ´ 4.1868 (313.15 – 298.15) = 252 359 kJ/h º 70.1 kW Enthalpy of condensate + Enthalpy of make-up water + Enthalpy of inlet gas + Enthalpy of secondary air + Heat of reaction at 298.15 K (25°C) + Heat of dilution at 313 K (40°C) = Enthalpy of tail gas + Heat removed by cooling water (DH) + Enthalpy of 58% acid 70.1 + 0 + 155.142 + 51.858 + (2469.63 ´ 693.1)/3600 + 4578.62 + 700.807 = 283.461 + (2487.43 ´ 3142.2)/3600 + DH D H = 3809.029 kW æ 3809.029 ´ 3600 ö Cooling water circulation rate = ç è 8 ´ 4.1868 ´ 1000 ÷ø

= 409.6 m3/h

EXERCISE 6.15 Basis: 1000 kg/h oil-free solid meal Solvent feed = 0.665 ´ 1000 = 665 kg/h Oil content of solvent = 665 ´ 0.015 = 9.975 kg/h Benzene in the solvent = 665 – 9.975 = 655.025 kg/h

Ans.

Stoichiometry and Unit Operations 203

Underflow, rate U = 0.507 ´ 1000 = 507 kg/h Oil content of underflow = 507 ´ 0.1183 = 60 kg/h Benzene content of underflow = 507 – 60 = 447 kg/h Total solution, entering with meal charge: Oil flow = 0.4 ´ 1000 = 400 kg/h Benzene = 0.025 ´ 1000 = 25 kg/h Total solution, C = 400 + 25 = 425 kg/h Overall material balance: C + S = U + O Overflow, O = C + S – U = 425 + 665 – 507 = 583 kg/h Balance of oil: C xo + S yo = O yo + U xu 425

FG 0.4 IJ + 665 ´ 0.015 = 583 y H 0.4 + 0.025 K

o

+ 507 ´ 0.1183

Solving the equation, yo = 0.6 or 60%

Ans.

EXERCISE 6.16 Basis: 7500 kg 29.6% Na2SO4 solution Na2SO4 in the solution = 7500 ´ 0.296 = 2220 kg Water in the solution = 7500 – 2220 = 5280 kg Water lost by evaporation = 5280 ´ 0.05 = 264 kg Water in the crystals and mother liquor = 5280 – 264 = 5016 kg Let x be the amount of crystals produced. Quantity of mother liquor = 5016 + 2220 – x = 7236 – x kg Balance of Na2SO4: Na2SO4 in mother liquor = 0.183 (7236 – x) = 1324.19 – 0.183 x kg x kg crystals will contain = (142 x/322) = 0.441 x kg Na2SO4 where molar mass of Na2SO4 = 142 and Molar mass of Na2SO4 × 10 H2O crystals = 322 0.441 x + 1324.19 – 0.183 x = 2220 x = 3472.1 kg crystals Mother liquor quantity = 7236 – 3472.1 = 3763.9 kg Ans.

204 Solutions Manual—Stoichiometry

EXERCISE 6.17 Basis: 100 kg initial solution FeSO4 in solution = 28 kg Associated water in crystals (FeSO4 × 7 H2O) = (126 ´ 28)/152 = 23.21 kg FeSO4 × 7 H2O dissolved in water = 28 + 23.21 = 51.21 kg Copper as dissolved = 51.21/0.96 = 53.34 kg Free water in solution = 100 – 53.34 = 46.66 kg Free water and impurities remain constant during crystallization and the total quantity of the two figures should be taken for calculation purpose. By cooling, solubility of FeSO4 is reduced to 20.51 kg per 100 kg water. Associated water = (20.51 ´ 126)/152 = 17.0 kg Free water = 100 – 17 = 83 kg FeSO4 × 7H2O quantity = 20.51 + 17 = 37.51 kg Crystals of FeSO4 × 7H2O per 1000 kg free water = 37.51 ´ 100/83 = 45.19 kg In the original solution, crystals of FeSO4 × 7H2O per 100 kg free water = (23.21 + 28) 100/(46.66 – 5) = 122.92 kg Yield per 100 kg free water = 122.92 – 45.19 = 77.73 kg For 500 kg yield, original solution requirement = (109.75 + 100) 500/77.73 = 1349.2 kg Copperas to be charged = 1349.2 ´ 0.5334 = 719.7 kg Ans.

EXERCISE 6.18 Basis: 1000 kg/h of feed (fresh) containing 20% NaNO3 Since the mother liquor of constant composition is recycled, whatever NaNO3 enters the system, the same must go out. NaNO3 in fresh feed = 0.2 ´ 1000 = 200 kg/h Crystals of NaNO3 contain 4% water. Crystals’ quantity = 200/0.96 = 208.33 kg/h Ans. (a) Chemically bonded water = 208.33 – 200 = 8.33 kg/h Let the quantity of recycled mother liquor = W kg/h Solution fed to crystalliser = W + 208.33 kg/h Feed to the crystallizer contains 58% NaNO3. NaNO3 in the solution = 0.58 (W + 208.33)

Stoichiometry and Unit Operations 205

= 0.58 W + 120.83 kg/h Water in the solution = 0.42 (W + 208.33) – 8.33 = 0.42 W + 79.17 kg/h NaNO3 in the mother liquor = 0.5 (0.42 W + 79.17) = 0.21W + 39.59 kg/h NaNO3 balance over crystalliser: 0.58 W + 120.83 = 200 + 0.21 W + 39.59 W = 320.97 kg/h Ans. (b) NaNO3 in mother liquor = (0.21 ´ 320.97) + 39.59 = 106.99 kg/h Water in mother liquor = 320.97 – 106.99 = 231.98 kg/h Mixed feed: NaNO3 content = 106.99 + 200 = 306.99 kg/h Water content = 800 + 213.98 = 1013.98 kg/h Mixed feed = 306.99 +1013.98 = 1320.97 kg/h Ans.(c) NaNO3 content of mixed feed = 306.99 ´ 100/(306.99 + 1013.98) = 23.2% Ans. (d)

EXERCISE 6.19 Basis: 675 kg/h liquid paraffin Sensible heat lost by paraffin, f1 = 675 ´ 2.93 (332 – 320) = 23 733 kJ/h º 6.59 kW Mechanical energy = 17 kW Total heat gained by water = 1.92 ´ 3600 ´ 4.1868 ´ 5.8 = 167 847 kJ/h º 46.62 kW Heat taken up by water owing to crystallisation = 46.62 – 17 – 6.59 = 23.03 kW Crystals produced = 23.03 ´ 3600/168.7 = 491.5 kg/h Crystallisation = 491.5 ´ 100/675 = 72.8%

EXERCISE 6.20 Basis: 100 kg feed Water in the feed = 17.5 kg Water in mother liquor = 17.5/2 = 8.75 kg Urea in mother liquor = 8.75 ´ 205/100 = 17.94 kg Quantity of mother liquor = (8.75 + 17.94)/(1 – 0.016) = 27.12 kg

Ans.

206 Solutions Manual—Stoichiometry

Biuret in mother liquor = 27.12 – 26.69 = 0.43 kg Biuret in crystals = 0.5 – 0.43 = 0.07 kg Urea crystallised = 82 – 17.94 = 64.06 kg Yield of urea crystals = (64.06 ´ 100)/82 = 78.12% Biuret content of urea crystals =

0.07 ´ 100 (64.06 + 0.07)

= 0.11%

Ans.

EXERCISE 6.21 Basis: 1000 kg mixture NaCl content of the mixture = 400 kg NH4Cl content = 1000 – 400 = 600 kg Since NaCl is less soluble than NH4Cl (at 323 K), all NH4Cl will dissolve and a definite quantity of NaCl will remain undissolved. Quantity of NaCl which will dissolve with 600 kg NH4Cl = 14.26 ´ 600/22.50 = 380.3 kg NaCl undissolved = 400 – 380.3 = 19.7 kg Ans. (a-i) Quantity of saturated solution = 380.3 + 600 + (600 ´ 100/22.50) = 3647 kg Ans. (a-ii) Water content of saturated solution = 3647 – 380.3 – 600 = 2666.7 kg NaCl which can be present in saturated solution at 283 K (10 °C) = 18.25 ´ 2666.7/100 = 486.7 kg Additional NaCl which can be dissolved = 486.7 – 380.3 = 106.4 kg NH4Cl remaining undissolved from the added mixture = 106.4 ´ 60/40 = 159.6 kg NH4Cl in saturated solution at 283 K = 12.49 ´ 2666.7/100 = 333.1 kg NH4Cl crystallized out = 600 – 333.1 = 266.9 Total NH4Cl separated = 266.9 + 159.6 = 426.5 kg Ans. (b – i) Additional salt treated = 106.4 + 159.6 = 266 kg Ans. (b – ii) NH4Cl which can be present in saturated solution at 373 K (100 °C) = 33.98 ´ 2666.7/100 = 906.1 kg Additional NH4Cl which can be dissolved= 40 ´ (906.1 – 600)/60

Stoichiometry and Unit Operations 207

= NaCl remaining undissolved = = NaCl in saturated solution at 373 K = = NaCl crystallized out = Total NaCl separated = Additional salt which can be treated = =

204.1 kg 40 ´ 306.1/60 204.1 kg 10.77 ´ 2666.7/100 287.2 kg 380.3 – 287.2 = 93.1 kg 204.1 + 93.1 = 297.2 kg 204.1 + 306.1 510.2 kg Ans. (c)

EXERCISE 6.22 (a) Basis: 100 kg aqueous 20 % urea solution Refer given below figure. Solubility isotherms Isoconcs of nitrogen Phase boundaries

100% H2O –5°C A

O 4N

40%N 100% CO(Nh2)2

C C

0° 0°

–1 C

0°

–26.5°C

0°C

–2

35%N

3

30%N

–20°C

°C –20

H 2O

20%N 25%N

NH

–10°C ICE (H2O)

C

°C 1020°C C 30° NH4NO3

CO(Nh2)2

CO(Nh2)2

100% B NH4NO3

Fig. E6-4(a) Ternery Solubility Diagram of Ammonium Nitrate-Urea-Water System

Point A represents 20% urea while point B represents 100 % NH4NO3. Join AB. It intersects 25% N (isocon) at C. AC ´ 100 BC 2.031 units = ´ 100 1.285 units = 158.05 kg

Requirement of NH4NO3 =

208 Solutions Manual—Stoichiometry

Salt-out temperature is read as 17.9°C (291.05 K). Algebraic method: Let ma kg a NH4NO3 is required. Nitrogen balance: 0.4596 ´ 0.2 ´ 100 + 0.35 ma = (100 + ma) 0.25 ma = 158.08 kg (b) Basis: 100 kg aqueous 40% NH4NO3 solution Refer given below figure.

Ans.

Solubility isotherms Isoconcs of nitrogen Phase boundaries

100% H2O –5°C

35%N 40%N 100% D CO(Nh2)2

C C

0° 0°

–1 C

0°

–26.5°C

F

0°C

30%N

–20°C

°C –20

25%N

E

–2

H 2O

20%N

O3 4N NH

–10°C ICE (H2O)

°C 10 0°C 2 °C 30 NH4NO3

CO(Nh2)2

CO(Nh2)2

100% NH4NO3

Fig. E6-4(b) Ternery Solubility Diagram of Ammonium Nitrate-Urea-Water System

Point D represents 100% urea and point E represents 40% NH4NO3. Join DE. It intersects 30% N isocon at F. EF ´ 100 DF 1.578 units = 1.574 units ´ 100 = 100.25 kg Salt-out temperature is read as 24.4°C (297.55 K) Algebraic method: Let mb kg urea is required. Nitrogen balance: 0.35 ´ 0.4 ´ 100 + 0.4596 mb = (100 + mb) 0.3 mb = 100.25 kg

Requirement of urea =

Ans.

Stoichiometry and Unit Operations 209

(c) Let mc kg aqueous ammonium nitrate solution of 50% strength is required to be mixed with 100 kg of 70% urea solution. Refer given below figure. Solubility isotherms Isoconcs of nitrogen Phase boundaries

100% H 2O –5°C

G

C 0°

C

–1

0°

C

H

0°

–26.5°C

35%N

40%N 100% CO(Nh2)2

I

0°C

30%N

–20°C

–2

25%N

°C –20

H 2O

20%N

O3 4N NH

–10°C ICE (H2O)

°C 1020°C C 30° NH4NO3

CO(Nh2)2

CO(Nh2)2

100% NH4NO3

Fig. E6-4(c) Ternery Solubility Diagram of Ammonium Nitrate-Urea-Water System

Point G represents 70% urea while point H represents 50% NH4NO3. Join GH. It intersects 25% N isocon at I. NH4NO3 required =

HI ´ 100 GI

1.104 units = 1.155 units ´ 100

mc = 95.58 kg Salt-out temperature is read as –8.3°C (264.85 k) Algebraic method: Nitrogen balance: 100 ´ 0.7 ´ 0.4596 + mc ´ 0.5 ´ 0.35= (100 + mc) 0.25 mc = 95.6 kg

Ans.

EXERCISE 6.23 In example 5.48, solution contains 42.02 kg NH4NO3, 37.63 kg urea and 20.35 kg H2O.

210 Solutions Manual—Stoichiometry

Point representing this concentration on Fig. 6.32 reads salt out temperature to be -7.9°C (265.25 K) Ans(a) When the above solution is cooled to –20°C, it will cool on the phase boundary line between NH4NO3 and urea. At –20°C, solution concentration is read as 37% NH4NO3, 30% urea and 33% H2O (by mass). Basis: 100 kg original solution. It contains 20.35 kg water. Final solution contains 33% water. 20.35 = 61.67 kg 0.33

Final solution = NH4NO3 in final solution = Urea in final solution = Water in final solution = NH4NO3 crystallized out = = Urea crystallized out = = Total crystallized mass = NH4NO3 in the mass = = Urea in the mass =

61.67 ´ 0.37 = 22.82 kg 61.67 ´ 0.3 = 18.50 kg 61.67 – 22.82 – 18.5 = 20.35 kg 42.02 – 22.82 19.2 kg 37.63 – 18.50 19.13 kg 19.2 + 19.13 = 38.33 kg (19.2 ´ 100)/38.33 50.09% 100 – 50.09 = 49.91 % Ans.

EXERCISE 6.24

(a)

Moisture content = 5 ppm (v/v) = 5 moles/106 moles Total pressure = 101.325 kPa If p = vapour pressure of water/ice at DP p 5 = 6 101.325 ´ 1000 10

or p = 0.5066 Pa From Table 6.12 DP = – 65.5°C (207.65 K) (b)

Total pressure = 0.8 MPa = 0.8 ´ 106 Pa p 5 = 6 6 0.8 ´ 10 10

Or p = 4 Pa

From Table 6.12, DP = – 48.05°C (225.1 K)

Ans

Stoichiometry and Unit Operations 211

EXERCISE 6.25 Purge gas rate = 8000 Nm3/h = 356.9198 kmol/h Vapour pressure of ice at –40°C, p w = 12.841 Pa Total pressure = 45 bar a = 45 ´ 105 Pa Basis:

Moisture in purge gas =

(dry) (Ref. Table 6.12)

12.841 (45 ´ 105 - 12.841)

= 2.8536 ´ 10–6

kmol kmol dry gas

Moisture, condensed in cold box in 300 days = 356.9198 ´ 24 ´ 300 ´ 2.8536 ´ 10–6 = 7.3332 kmol º 132.1 kg Ans.

EXERCISE 6.26 Vapour pressure of water of 308.15 K (35°C) = 5.6216 kPa (ps) % RH = 80 Partial pressure of water, p = 0.8 ´ 5.6216 = 4.4973 kPa Vapour pressure of water at 304.15 K (31°C)= 4.4911 kPa (Table 6.13) Hence DP of air is 304.15 K (31 °C). Ans. (b) Absolute humidity H =

18 4.4973 ´ 29 (100 – 4.4973)

= 0.0292 kg/kg dry air Saturation humidity Hs =

Ans. (a)

18 5.6216 ´ 29 (100 – 5.6216)

= 0.036 97 kg/kg dry air Saturation = (0.0292 ´ 100)/0.036 97 = 79.0% Humid heat, CH = 1.006 + (1.84 ´ 0.0292) = 1.06 kJ/(kg × K) Latent heat of evaporation (lv) at 304.15 K (31 °C) = 2428.3 kJ/kg Enthalpy, i = 1.06 (304.15 – 273.15) + (2428.3 ´ 0.0292) = 103.8 kJ/kg dry air

Ans. (c) Ans. (d)

Ans. (e)

212 Solutions Manual—Stoichiometry

EXERCISE 6.27 Basis: 1 kg dry air From Fig. 6.15: At 367 K DB and 300.5 K WB, H1 = 20.7 g/kg dry air and DP1 = 298.5 K (25.5°C) At 303 K DB and 50% RH, H2 = 13.2 g/kg dry air and DP2 = 291.2 K (18.2°C) Moisture to be removed = 20.7 – 13.2 = 7.5 g/kg dry air At DP1, lv1 = 2441.35 kJ/kg and at DP2, lv2 = 2458.54 kJ/kg Average lv = (2441.35 + 2458.54)/2 = 2450 kJ/kg Since the air will be fully saturated after the spray chamber, from Fig. 6.15, i1 = 87.5 kJ/kg dry air i2 = 51.5 kJ/kg dry air Cooling load on the spray chamber = (87.5 – 51.5) + 0.0075 (307.15 – 291.2) ´ 4.1868 = 36.5 kJ/kg dry air Humid volume of air at 307 K DB and 300.5 K WB. VH = 0.899 m3/kg dry air 1

Mass flow rate of air = 60 000/0.899 = 66 740.8 kg dry air/h Cooling load of the spray chamber = 66 740.8 ´ 36.5 = 2436 039 kJ/h º 676.67 kW º 192.53 TR

Ans. (a)

Humid heat of saturated air, CHs = 1.006 + (1.84 ´ 0.0132) = 1.03 kJ/(kg dry air × K) Heating load on the heater = 66 740.8 ´ 1.03 (303 – 291.2) = 811 168 kJ/h º 225.32 kW Latent of evaporation of water at 400 kPa a, lv = 2133 kJ/kg Steam consumption = 811 168/2133 = 380.3 kg/h (c) Desired NCL = 10 g moisture/kg dry air At this humidity level let vapour pressure of water be pv.

(Table AIV.2) Ans. (b)

Stoichiometry and Unit Operations 213

pv ´ 0.622 = (101.325 - pv ) pv = From Table 6.13, DP = i¢2 at 14.02°C = Cooling load for NCL =

0.01

1.604 kPa 14.02°C (287.17 K) 39 kJ/kg dry air 87.5 – 39 + (0.0207 – 0.01) (307.15 – 287.17) ´ 4.186 8 = 48.5 + 0.9 = 49.4 kJ/kg dry air Total cooling load = 66 704.8 ´ 49.4 = 329 522 kJ/h º 915.34 kW º 260.45 TR At H = 0.01 kg/kg dry air and RH = 50%, DB requirement can be read as 25.2°C (298.35 K) from Fig. 6.15. C¢Hs = 1.006 + 1.84 ´ 0.01 = 1.0244 kJ/(kg dry air × K) Heating load = 66 704.8 ´ 1.0244 (298.35 – 287.17) = 763 956 kJ/h º 212.20 kW Steam consumption for NCL = 763 956/2133 = 358.2 kg/h Ans.(c) Note: Calculations show that for achieving NCL conditions, refrigeration requirement is 35.3% higher.

EXERCISE 6.28 Vapour pressure of water at 313.15 K (40 °C) = 7.375 kPa Absolute humidity, H =

7.375 18.015 ´ (117.3 – 7.375) 44.01

= 0.0275 kg/kg dry CO2

Ans.

EXERCISE 6.29 Vapour pressure of water 291.15 K (18°C) = 2.0624 kPa Absolute humidity in chlorine gas =

2.0624 18.015 ´ (101.325 – 2.0624) 70.906

= 0.005 279 kg/kg dry Cl2

214 Solutions Manual—Stoichiometry

On wet chlorine basis, humidity =

18.015 2.0624 ´ = 0.005 171 kg/kg wet Cl2 70.906 101.325

º 5171 ppm by mass

Ans.

EXERCISE 6.30 Basis: 4 kmol HCl gas Based on Table 5.53, following data are tabulated. Component HCl O2

Cl2 N2 H2O Total

kmol

Molar mass

kg

0.8 0.55

36.5 32

29.2 17.6

1.6 5.079 (1.6)

71 28 18

113.6 142.21 (28.8)

302.61

(331.41)

8.029 (9.629)

Average molar mass of dry gas= 302.61/8.029 = 37.69 At the outlet of Trombone cooler, part of HCl will have been absorbed in condensed water. Average molar mass of dry gas without HCl 302.61 – 29.2 = 8.029 – 0.8 Average of both molar mass = = Vapour pressure of water at 323 K (50°C) = Water vapours in the outgoing gas mixture

=

=

37.821 (37.69 + 37.821)/2 37.756 12.335 kPa (Table 6.13)

12.335 18.015 ´ = 0.066 08 kg/kg dry air (101.325 – 12.335) 37.756

Water content of gas mixture, exit of cooler = 302.61 ´ 0.066 08 = 20.0 kg Water content of ingoing gas mixture = 28.8 kg Water condensed in the cooler = 28.8 – 20.0 = 8.8 kg Aqueous acid = 8.8/0.67 = 13.134 kg/4 kmol HCl fed º 3.284 kg/kmol HCl fed HCl in aqueous acid = 13.134 – 8.8 = 4.334 kg

Stoichiometry and Unit Operations 215

Note: HCl to the tune of 4.334 kg is reduced from the exit gas quantity of 302.61 kg. Therefore one more trial is attempted. Water content of exit gas mixture = (302.61 – 4.334) ´ 0.066 08 = 19.71 kg º 1.09 kmol Water condensed = 28.8 – 19.71 = 9.09 kg Aqueous acid = 9.09/0.67 = 13.567 kg/4 kmol HCl feed º 3.39 kg/kmol HCl feed – close to previous value HCl absorbed = 9.09 ´ 0.333/0.667 = 4.538 kg º 0.124 kmol Components ni kmol

C°mp equation constants ni × bi ´ 103

ni × ai

ni × ci ´ 106

ni × di ´ 109

HCl

0.68

20.610

– 5.174

9.017

– 2.947

O2 Cl2 H2O N2

0.55 1.60 1.09 5.079

14.314 45.674 35.416 150.292

6.465 38.207 0.087 – 26.111

– 1.288 – 34.181 14.400 66.956

0.309 10.356 – 4.96 – 25.232

Total

8.999

266.306

13.474

54.904

– 22.474

Enthalpy of gas mixture at 323.15 above 298.15 K = 266.306 (323.15 – 298.15) + 13.474 ´ 10–3 (323.152 – 298.152)/2 + 54.904 ´ 10–6 (323.153 – 298.153)/3 – 22.474 ´ 10–9 (323.154 – 298.154)/4 = 6657.7 + 104.6 + 132.4 – 16.8 = 6877.9 kJ Enthalpy to be removed to condense water = 9.09 ´ 2382.9 = 21.660.6 kJ From Table 5.79, DH°f of 33% aqueous acid = – 152.9 kJ/mol HCl DH°f of gaseous HCl = – 92.31 kJ/mol HCl Net heat change = – 152.9 – (– 92.31) = – 60.59 kJ/mol HCl HCl dissolved in water = 0.8 – 0.68 = 0.12 kmol Heat of dilution = 60 590 ´ 0.12 = 7270.8 kJ (exothermic) at 298.15 K (25 °C) Sensible heat of HCl solution = (9.09 + 4.538) ´ 2.6 (50 – 25) = 886 kJ over 298.15 K (25 °C)

216 Solutions Manual—Stoichiometry

Enthalpy of inlet gas at 599.5 K = 91.520.2 kJ Enthalpy to be removed in Trombone cooler

(Ref. Example 5.39)

= 91 520 + 21 660.6 + 7270.8 – 6877.9 – 886 = 112.688 kJ/4 kmol HCl fed º 28 172 kJ/kmol HCl

EXERCISE 6.31 Refer solution of exercise 4.23. Basis: Urea feed rate = 350 kg/h Total CO2 liberated = 5.833 kmol/h CO2 leave at 7 bar and 80°C (353.15 K). Antoine equation for SO3:

892.175 log p = 4.205 15 – T - 103.564 At T1 = 353.15 K, log p1 = 4.205 15 –

892.175 (353.15 - 103.564)

or p1 = 4.27 bar 4.27 SO3, carried with CO2 = 7 - 4.27

= 1.5641 kmol/kmol CO2 SO3 quantity = 5.833 ´ 1.5641 = 9.1234 kmol/h at reactor outlet In the condenser,mixture is cooled to 30°C (303.15 K). p2 of SO3 = 346.1 Torr = 0.4614 bar

0.4614 SO2, leftover in CO2 = (6.7 - 0.4614) = 0.074 kmol/kmol CO2 SO3 quantity in CO2 at the outlet of condenser, = 5.833 ´ 0.074 = 0.4316 kmol/h SO3 condensed (refluxed) = 9.1234 – 0.4316

Ans.

Stoichiometry and Unit Operations 217

= 8.6918 kmol/h º 695.344 kg/h Latent heat removal, f3 = 695.344 ´ 42.55 = 29 587 kJ/h º 8.219 kW Gas mixture ingoing to condenser . Component ni Heat capacity equation constants kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 CO2 SO3

5.8330 9.1234

124.62 201.06

374.97 1109.62

–239.45 –838.14

57.16 222.33

Total

14.9 564

325.68

1484.59

–1077.59

279.49

Enthalpy of ingoing gas mixture at 353.15 K over 298.15 K, f1 =

353.15

ò

298.15

(325.68 + 1484.59 ´ 10–3 T – 1077.59 ´ 10–6 T2 + 279.49

´ 10–9 T 3) dT = 38 737 kJ/h º 10.760 kW Gas mixture outcoming from condenser . Component ni Heat capacity equation constants kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 CO2 SO3

5.8330 0.4316

124.62 9.51

374.97 52.49

–239.45 –39.65

57.16 10.52

Total

6.2646

134.13

427.46

–279.10

67.68

Enthalpy of outgoing gas mixture from condenser at 303.15 K over 298.15 K, f2 =

303.15

ò

298.15

(134.13 + 427.46 ´ 10 –3 T – 279.10 ´ 10 –6 T 2 + 67.68

´ 10–9 T 3) dT = 1196.4 kJ/h º 0.332 kW Heat load of condenser, fc = 38 737 + 29 587 – 1196.4 = 67 127.6 kJ/h º 18.647 kW

Ans.

218 Solutions Manual—Stoichiometry

EXERCISE 6.32 Basis: 1103.1 kmol/h HCl free gas, entering the compressor (Ref. Excercise 6.13) Antoine constants: A = 4.02291 B = 1221.781 and C = – 45.739 (Ref. Table 5.4) Use Antoine equation 5.24. Vapour pressure of CCl4 at 313.15 K (40 °C), pv is given by log p = 4.022 91 –

1221.781 (313.15 – 45.739)

or pv = 0.2844 bar º 28.44 kPa Total p = 1130 kPa a CCl4–free gas = 1103.1 – 52.4 = 1050.7 kmol/h CCl4 content of gas mixture exit affercooler 28.44 æ ö ´ 1050.7 = ç è 1130 – 28.44 ÷ø

= 27.127 kmol/h CCl4 condensed = 52.4 – 27.127 = 25.273 kmol/h º 3887.6 kg/h

Ans.

EXERCISE 6.33 Basis: 780 kmol of benzene-nitrogen mixture, entering the absorber Vapour pressure of benzene at 313.15 K (40°C) = 24.39 kPa This vapour pressure is calculated using Antoine constants (Ref. Table 5.4); A = 4.01814 B = 1203.835 and C = – 53.226 and using Eq. (5.24). Benzene content of incoming gas mixture = (24.39 ´ 780)/104 = 182.9 kmol N2 entering with benzene = 780 – 182.9 = 597.1 kmol Vapour pressure of benzene at 283.15 K (10°C) = 6.06 kPa (Calculated) Benzene content of outgoing gas mixture = 6.06/(101.325 – 6.06) = 0.0636 kmol/kmol dry gas Benzene unadsorbed = 0.0636 ´ 597.1 = 37.976 kmol Benzene removal in adsorber = 182.9 – 37.976 = 144.924 kmol º 11 320.3 kg For removal of 1000 kg/h of benzene, the mass flow rate of bone dry nitrogen = 597.1 ´ 28 ´ 1000/11 320.3 = 1476.9 kg/h Ans.(a) Change in benzene content of BD activated carbon = 0.35 – 0.05 = 0.3 kg/kg

Stoichiometry and Unit Operations 219

Required mass flow rate of BD activated carbon = 1000 ´ 1.0/0.3 = 3333 kg/h Ans.(b) For removing 1000 kg/h benzene at 333.15 K (60°C), the initial mixture of benzene-nitrogen needs to be compressed. This means the gas will be compressed to saturation pressure at 333 K. Vapour pressure of benezene at 333.15 K (60°C) = 52.187 kPa (calculated) Let total pressure after compression be p kPa. After compression and cooling to 333 K (60°C), benzene concentration should be 38.03 kmol/597.2 kmol dry N2. 52.187 37.976 = 597.1 p - 52.187

or p = 872.7 kPa a

Ans. (c)

EXERCISE 6.34 Basis: 1 kg bone dry soybean flakes C6H14 removed = 0.61 – 0.025 = 0.585 kg/kg BD flakes º 0.0068 kmol/kg BD flakes New basis: 760 kmol gas mixture, entering the vessel. For calculating vapour pressures of n-Hexane at various temperatures, Antoine constants from Table 5.5 are used. A = 4.00 266, B = 1171.530, C = – 48.784 Use Eq. (5.24). Temperature K (°C)

Vapour pressure, bar (kPa)

283.15 (10) 293.15 (20) 343.15 (70)

0.1009 (10.09) 0.1616 (16.16) 1.0539 (105.39)

n-Hexane in the inlet mixture = (10.09 ´ 760)/101.325 = 75.68 kmol N2 in the mixture = 760 – 75.68 = 684.32 kmol Partial pressure of n-hexane in outgoing mixture = 105.39 ´ 0.65 = 68.504 kPa Quantum of n-hexane in the outgoing stream =

68.504 ´ 684.32 = 1428.31 kmol (101.325 – 68.504)

n-Hexane evaporated in N2 stream = 1428.31 – 75.68 = 1352.63 kmol º 116 326 kg

220 Solutions Manual—Stoichiometry

Wet gases, leaving desolventiser = (1428.31 + 684.32) 0.0068/1352.63 = 0.0106 kmol/kg BD flakes Ans. (a) Specific volume at 101.325 kPa and 343.15K (70°C), V = RT/p = (8.314 ´ 343.15)/101.325 = 28.156 m3/kmol Total wet gases, leaving the desolventiser = 25 000 ´ 0.0106 = 265 kmol/h Volumetric flow rate of gases = 265 ´ 28.156 = 7461.34 m3/h Ans. (b) Let the pressure after compression be p kPa. 16.16 10.09 = 101.325 p

p = 162.3 kPa a

Ans. (c)

EXERCISE 6.35 Basis: 100 kg/h wet pigment Water in the pigment = 50 kg/h Water in the dried pigment = 50 ´ 0.03/0.97 = 1.55 kg/h Water evaporated = 50 – 1.55 = 48.45 kg/h Air required for evaporation of water = 48.45 ´ 100/2.08 = 2329.3 m3/h Ans. (c) Vapour pressure of H2O at 294 K (21°C) = 2.485 kPa (Ref. Table 6.13) Moisture content of incoming air = 2.485/(114 – 2.485) = 0.022 28 kmol/kmol dry air º 0.0218 kmol/kmol wet air Specific volume of air at 114 kPa and 373.15 K(100°C), V = RT/p = (8.314 ´ 373.15)/114 = 27.213 m3/kmol Molar flow rate of air = 2329.3/27.213 = 85.6 kmol moist air/h Moisture, entering with air = 85.6 ´ 0.0218 = 1.866 kmol/h BD air = 85.63 – 1.866 = 83.764 kmol/h BD air requirement = 83.764/48.45 = 1.7289 kmol/kg moisture removed Ans. (a) H2O evaporated = 48.45/18 = 2.692 kmol/h Total air, leaving the drier = 85.63 + 2.692 = 88.322 kmol/h (wet) Moisture in outgoing air = 1.867 + 2.692 = 4.559 kmol/h

Stoichiometry and Unit Operations 221

If p = partial pressure of H2O in outgoing air, p/108 = 4.559/88.322 or p = 5.575 kPa From Table 6.13, dew point = 308 K (34.85°C) Ans. (b)

EXERCISE 6.36 Miscella contains 40% edible oil (having 872 molar mass) and 60% n-hexane.

(60 / 86) Mole fraction of n-hexane = (60 / 86) + (40 / 872) = 0.9383 Absolute pressure in vessel = 760 – 197.5 = 562.5 Torr º 0.7401 atm º 0.75 bar At t/T = 60°C/333.15 K, using Antoine constants vapour pressure of n-hexane = 0.7636 bar. Partial pressure of n-hexane over miscella = 0.9383 ´ 0.7636 (Raoult's law) = 0.7165 bar n-Hexane content of vapour from vessel 0.7165 = (0.75 - 0.7165)

= 21.388 kmol/kmol inerts Basis: Ejector pulling out 25 m /h at 310 Torr and 40°C (313.15 K) Specific volume of ideal gas at 310 Torr (0.4133 bar) and 3

313.15 K = 22.414

313.15 760 ´ 273.15 310

= 62.9974 m3/kmol Gas mixture being ejected from the cooler (with cooling water) =

25 62.9974

= 0.3968 kmol/h Vapour pressure of n-hexane at 313.15 K = 0.3726 bar n-Hexane content of gas mixture

0.3762 = (0.4133) = 0.9015 kmol/kmol mixture

222 Solutions Manual—Stoichiometry

n-Hexane loss = 0.9015 ´ 0.3968 = 0.3577 kmol/h º 30.7635 kg/h Specific n-Hexane loss =

30.7635 ´ 24 200

= 3.692 kg/t soya flakes º 5.723 L/t soya flakes Ans.(a) After incorporation of the chiller, temperature of gas mixture, leaving the chiller is –10°C (263.15 K). Vapour pressure of n-hexane at 263.15 K = 0.0293 bar 0.0293 ´ 0.3968 0.4133 = 0.0281 kmol/h º 2.419 kg/h º 58.061 kg/d º 0.2903 kg/t soya flakes º 0.45 L/t soya flakes

n-Hexane loss =

Ans.(b)

Enthalpy balance: (a) System with only cooler: Gas mixture leaving cooler . Component ni Heat capacity equation constants kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 O2 N2 n-C6H14

0.0082 0.0309 0.3577

0.2134 0.9144 –1.5793

0.0964 –0.1589 208.1540

–0.0192 0.4074 –111.5517

–0.0046 –0.1535 23.2216

Total

0.3968

–0.4515

208.0915

–111.1635

23.0625

t0/T0 = 25°C/298.15 K Enthalpy of gas mixture, leaving cooler, at 313.15 K, f2 =

313.15

ò

298.15

(–0.4515 + 208.0915 ´ 10–3T – 111.1635 ´ 10–6 T 2 + 23.0625

´ 10–9T 3) dT = –6.8 + 954.05 – 155.81 + 9.9 = 801.34 kJ/h º 0.223 kW n-Hexane in gas mixture entering condenser = (21.388 ´ 0.0391) = 0.8363 kmol/h Gas mixture entering condenser

Stoichiometry and Unit Operations 223

Component

. ni

Heat capacity equation constants ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

kmol/h

ni × ai

O2 N2 n-C6H14

0.0082 0.0309 0.8363

0.2134 0.9144 –3.6924

0.0964 –0.1589 487.0695

–0.0192 0.4074 –259.7882

–0.0056 –0.1535 52.6225

Total

0.8754

–2.5646

487.0070

–259.4000

52.4634

Eenthalpy of gas-mixture entering condenser at 333.15 K f1 =

333.15

ò

298.15

(–2.5646 + 487.007 ´ 10 –3 T – 259.4 ´ 10 –6 T 2 + 52.4634

´ 10–9 T 3) dT = –89.8 + 5380.3 – 905.5 + 57.9 = 4442.9 kJ/h º 1.234 kW n-Hexane condensed = 0.8363 – 0.3577 = 0.4786 kmol/h Average temperature of condensation = Latent heat of n-hexane at 323.15 K,

40 + 60 = 50°C or 323.15 K 2 0.38

é 507.6 - 323.15 ù lv2 = 28.85 ê ú ë 507.6 - 341.9 û = 30.05 kJ/mol Heat given up during condensation f3 = 0.4786 ´ 30.050 = 14 381.9 kJ/h º 3.995 kW Heat duty of cooler = f1 + f3 – f2 = 4442.3 + 14 381.9 – 801.3 = 18 022.9 kJ/h º 5.006 kW (b) System with cooler and chiller: Gas mixture, leaving chiller . Component ni Heat capacity equation constants kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 O2 N2 n-C6H14

0.0082 0.0309 0.0281

0.2134 0.9144 –0.1241

0.0964 –0.1589 16.3520

–0.0192 0.4074 –8.7632

–0.0056 –0.1535 1.8242

Total

0.0692

1.0039

16.2895

–8.3750

1.6651

224 Solutions Manual—Stoichiometry

Enthalpy of gas mixture at 263.15 K over 298.15 K f4 =

263.15

ò

298.15

(1.0039 + 16.2895 ´ 10–3 T – 8.3750 ´ 10 –6 T2 + 1.665 ´

10–9 T 3) dT = –35.1 – 160.0 + 23.1 – 1.3 = –173.3 kJ/h º –0.048 kW 40 - ( -10) 2 = 25°C or 298.15 K

Average temperature of condensation =

0.38

é 507.6 - 298.15 ù lv = 28.85 ê ú 3 ë 507.6 - 341.9 û = 31.537 kJ/mol at 298.15 K n-Hexane condensed in chiller = 0.3577 – 0.0281 = 0.3296 kmol/h Heat given-up during condensation, f5 = 0.3296 ´ 31 537 = 10 394.6 kJ/h Heat load of chiller = f2 + f5 – f4 = 801.3 + 10 394.6 – (–173.3) = 11 369.2 kJ/h º 3.158 kW º 0.899 TR

Ans.

EXERCISE 6.37 Basis: 5 m/s paper speed Production rate = 5 ´ 3600 ´ 3.8 ´ 0.081 ´ 24/1000 = 133 t/d on BD basis Ans.(a) Water content of paper, leaving the drier = 0.05 ´ 133 ´ 1000/24 = 277 kg/h Ans.(b–i) Filler Content = 0.2 ´ 0.95 ´ 133 ´ 1000/24 = 1052.92 kg/h Ans.(b–ii) Fiber in paper = 0.8 ´ 0.95 ´ 133 ´ 1000/24 = 4211.7 kg/h Ans.(b–iii) Water in the paper, leaving the presses = (1052.92 + 4211.72) 0.6/0.4 = 7897 kg/h Evaporation in the drier = 7897 – 277 = 7620 kg/h Ans.(c)

Stoichiometry and Unit Operations 225

EXERCISE 6.38 Basis: 1000 kg/h dried product X1 =

0.3 = 0.4286 kg moisture/kg dry solid 1 - 0.3

0.02 º 0.0204 kg moisture/kg dry solid 1 - 0.02 . mc = mass flow rate of dry carbon pellets = 1000 (1 – 0.02) = 980 kg/h . Let msi = mass flow rate of ingoing superheated steam, kg/h . mso = mass flow rate of outcoming superheated steam, kg/h Reference temperature, t0/T0 = 0°C/273.15 K Energy balance: . . . . mc × is1 + msi × Hsi = mc × is2 + mso × Hso + fL where is1 = enthalpy of dry solid = Cs(Ti – T0) + XiCl(Ti – T0) = 1.3146 (303.15 – 273.15) + 0.4286 ´ 4.1868 ´ (303.15 – 273.15) = 93.272 kJ/kg dry solid is2 = Cs (T – t0) + X2Cl(T – T0) = 1.3146 (383.15 – 273.15) + 0.0204 ´ 4.1868 ´ (383.15 – 273.15) = 154 kJ/kg dry solid From Appendix IV.2, Hsi = 3074.5 kJ/kg at 300°C (573.15 K) Hso = 2776.3 kJ/kg at 150°C (423.15 K) These values are at atmospheric pressure. . . 980 ´ 93.272 + msi ´ 3074.5 = 980 ´ 154 + mso ´ 2776.3 + fL Heat loss fL is 5% of total heat input . . 0.95 (980 ´ 93.272 + msi ´ 3074.5) = 980 ´ 154 + mso ´ 2776.3 (1) Material balance: . . . mo(X1 – X2) = msi – mso . . msi – mso = 980 (0.4286 – 0.0204) = 400 kg/h (2) Solving Eq. (1) and Eq. (2), . msi = 8130.15 kg/h . mso = 8530.15 kg/h Ans.

X2 =

226 Solutions Manual—Stoichiometry

EXERCISE 6.39 Basis: Solution feed rate = 970 kg/h . Let ma = mass flow rate of air through the drier on dry basis, kg/s H1 = Humidity of ingoing air = 0.036 kg moisture/kg dry air (Fig. 6.15) X1 =

0.75 = 3 kg moisture/kg dry solid at inlet (1 - 0.75)

X2 =

0.03 = 0.030 93 kg moisture/kg dry solid at outlet (1 - 0.03)

. ms = mass flow rate of dry solid = 970 ´ 0.25 = 242.5 kg/h º 0.067 36 kg/s Moisture balance: . . ms(X1 – X2) = ma (H2 – H1) . ma(H2 – 0.036) = 0.067 36 (3 – 0.030 93) (1) Enthalpy of solid at inlet above 0°C (273 K), is1 = Cs (ti – 0) + X1 CL (ti – 0) = 1.4 (60 – 0) + 3 ´ 4.1868 (60 – 0) = 837.6 kJ/kg dry solid Enthalpy of solid at outlet above 0°C (273 K) is2 = Cs(t0 – 0) + X2CL(t0 – 0) = 1.4 (70 – 0) + 0.030 93 (70 – 0) = 100.165 kJ/kg dry solid Enthalpy of at inlet of dryer above 0°C (273.15 K), ia1 = 1.006 (TDP – 273.15) + H1.iws1 + CH (TDB1 – TDP) 2 TDB1 = 280 + 273.15 = 553.15 K TDP = 35 + 273.15 = 308.15 K CH1 = 1.006 + 1.84 H1 = 1.006 + 1.84 ´ 0.036 = 1.072 24 kJ/(kg dry air × K) iws1 = Enthalpy of saturated water vapour at 35°C (308.15 K) = 2565.4 KJ/kg (Appendix IV.1) ia1 = 1.006 (308.15 – 273.15) + 0.036 ´ 2565.4 + 1.072 24 (553.15 – 308.15) = 390.26 kJ/kg dry air Enthalpy of air, leaving the dryer, ia2 = 1.006 (TDP – 273.15) + H2.iws2 + CH (TDB2 – TDP) 2 TDB2 = 80 + 273.15 = 353.15 K

Stoichiometry and Unit Operations 227

Enthalpy balance of dryer: . . . . msis1 + ma × ia1 = ms × ias2 + ma × ia2 + fL . . 0.067 36 ´ 837.6 + ma × 390.26 = 0.06736 ´ 100.65 + ma × ia2 + fL . . 56.421 + 390.26 ma = 6.7798 + ma × ia2 + fL fL = Heat loss . = 0.1 (56.421 + 390.26 ma) 44 = ma (ia2 – 351.23) (2) ia2 = 1.006 (TDP – 273.15) + H2 × iws2 + (1.006 + 1.84 H2) (353.15 – TDP) (3) Assume TDP = 54 + 273.15 = 327.15 K iws2 = 2599.2 kJ/kg (Appendix IV.1) pv2 = 15.002 kPa

15.002 18 H2 = (101.325 - 15.002) ´ 29 = 0.107 87 kg/kg dry air Ans(b) ia2 = 1.006 (327.15 – 273.15) + 0.107 87 ´ 2599.2 + (1.006 + 1.84 ´ 0.107 87) (353.15 – 327.15) = 366.02 kJ/kg dry air . From Eq. (1), ma (0.107 87 – 0.036) = 0.2 . ma = 2.7828 kg/s º 10 018 kg/h of dry air Ans.(a) From Eq. (2), ia2 – 351.23 = 44/2.7828 ia2 = 367.04 kJ/kg dry air This value is close enough to ia2, calculated earlier. Hence tDP = 54°C or TDP = 327.15 K Heat duty of air heater, . f = ma(ia1 – iaa) Enthalpy of air at the heater inlet, iaa = 1.006 (308.15 – 273.15) + 0.036 ´ 2565.4 = 127.56 kJ/kg dry air f = 2.7828 (390.26 - 127.56) = 731.042 kW 731.357 Fuel oil consumption = 41 300 ´ 0.7

= 0.025 29 kg/s º 91.03 kg/h

Ans.(c)

228 Solutions Manual—Stoichiometry

EXERCISE 6.40 Basis: Feed flow rate of 50% solution = 50 000 kg/h Caustic soda content of feed = 50 000 ´ 0.5 = 25000 kg/h Concentrated solution leaves the evaporator with 75% concentration. Flow of 75% solution =

25 000 = 33 333.33 kg/h 0.75

Evaporation rate = 50 000 – 33 333.33 = 16 666.67 kg/h Energy balance: Ws× ls = Wc× Hc + Wv × Hv – WF × HF Boiling point of pure water at 100 Torr a pressure = 51.1°C (324.25 K) Boiling point of 75% solution = 51.1 + 81 = 132.1°C or 405.25 K Enthalpy of feed (50% solution), HF = 320 kJ/kg solution (Fig. 5.16) Enthalpy of final concentrated 75% solution, H C = 1050 kJ/kg solution (Fig.5.16) Enthalpy of superheated vapour at 100 Torr (13.33 kPa) abs pressure and 132.1°C (405.25 K) = 2748.5 kJ/kg (Appendix AIV.3) Latent heat of steam at 6 bar a, l s = 2085 kJ/kg (Appendix A IV.2) Ws ´ 2085 = 33 333.33 ´ 1050 + 16 666.67 ´ 2748.5 – 50 000 ´ 320 Ws = 31 083.94 kg/h º 8.63 kg/s Water evaporated Steam consumed 16 666.67 = 31 083 = 0.5362 kg/kg Note: High boiling point elevation results in lower economy.

Economy of evaporator =

EXERCISE 6.41 Basis: Weak liquor flow = 50 L/s Mass flow rate = 1.08 ´ 24 ´ 50 ´ 3600 = 4665 600 kg/d Solids in the weak liquor = 4665 600 ´ 0.15 = 699 840 kg/d Flow of concentrated liquor = 699 840/0.55

Stoichiometry and Unit Operations 229

= 1272 436.4 kg/d Evaporation = 4665 600 – 1272 436.4 = 3393 163.6 kg/d º 141 381.8 kg/h Steam economy = 141 381.8/28 500 = 4.961 kg evaporation/kg steam

Ans. (a) Ans. (b)

EXERCISE 6.42 Basis: Weak liquor flow rate = 530 L/min (= 8.85 L/S) Mass flow rate = 530 ´ 1.05 ´ 60 = 33 390 kg/h Solids in the weak liquor = 33 390 ´ 0.1 = 3339 kg/h Mass flow rate of concentrated liquor, leaving the final (4th) Stage = 3339/0.5 = 6678 kg/h Evaporation = 33 390 – 6678 = 26 712 kg/h (total in four effects) Ans. (a) Effect No.

temperature, K (°C)

saturation pressure, kPa a

latent heat of evaporation, kJ/kg

1st

376

(103)

112.5

2249.1 (lv1)

2nd

367

(94.0)

81.5

2272.8 (lv2)

3rd

353.5

(80.5)

48.0

2308.0 (lv3)

4th

324.6

(51.6)

13.3

2379.2 (lv4)

Steam pressure in the shell of 1st stage = 103 kPa g Saturation temperature = 394 K (121°C) Latent heat of evaporation of steam (lv5) = 2199.7 kJ/kg (Appendix IV.2) Heat balance across each effect: 1st effect: Wa × lv5 = WF cpF (Tl – TF) + (WF – Wl) lv1 2199.7 Wa = 33 390 (376 – 313) ´ 4.1868 + (33390 – W1) 2249.1 Wa = 38 143.69 – 1.0225 W1 (1) 2nd effect: (WF – W1) lv1 = W1 cp2 (T2 – T1) + (W1 – W2) lv2 (33 390 – W1) 2249.1 = W1 (367 – 376) ´ 4.1868 + (W1 – W2) 2272.8 75 097 449 – 2249.1 W1 = – 37.68 W1 + 2272.8 W1 – 2272.8 W2 W1 = 16 747.05 + 0.5068 W2 (2)

230 Solutions Manual—Stoichiometry

3rd effect: (W1 – W2) lv2 = W2 cp3 (T3 – T2) + (W2 – W3) lv3 (W1 – W2) 2272.8 = W2 (353.5 – 367) ´ 4.1868 + (W2 – W3) 2308.0 W1 – 1.991 W2 = – 1.0155 W3 (3) 4th effect: (W2 – W3) lv3 = W3 cp4 (T4 – T3) + (W3 – W4) lv4 (W2 – W3) 2308 = W3 (324.6 – 353.5) ´ 4.1868 + (W3 – 6678) 2379.2 W2 – 1.978 W3 = – 6884.01 (4) Rewriting Eq. (4), 1.978 W3 = W2 + 6884.01 or W3 = 0.5056 W2 + 3480.3 (5) Substitute the value of W3 in Eq. (3). W1 – 1.991 W2 = – 0.5134 W2 – 3534.24 or W1 – 1.4776 W2 = – 3554.24 (6) W1 – 0.5068 W2 = 16 747.05 (2) (–) (+) (–) W2 = – 20 281.29 or –0.9708 W2 = 20 895.6 kg/h W1 = 27 337 kg/h, W3 = 14 045.1 kg/h and Ws = 10 191.6 kg/h Steam economy = 26 712/10 191.6 = 2.621 kg evaporation/kg steam Ans. Simultaneous equations (1) to (4) can be conveniently solving with Mathcad.

EXERCISE 6.43 Basis: Weak liquor flow rate = 530 L/min (same as Excercise - 6.33) Effect No.

temperature, K(°C)

saturation pressure, kPa a

1st 2nd 3rd 4th

362 (89.0) 347.5 (74.5) 337.5 (64.5) 324.6 (51.6)

67.5 37.6 24.5 13.3

latent heat of evaporation, kJ/kg 2287.4 (lv1) 2323.0 (lv2) 2347.6 (lv3) 2379.1 (lv4)

Heat balance across each effect: 1st effect: Ws lv = W2 cp2 (T1 – T2) + (W2 – W1) lv1 2199.7 Ws = W2 (362 – 347.5) 4.1868 + (W2 – 6678) 2287.4 Ws = 1.0675 W2 – 6944.2 (1)

Stoichiometry and Unit Operations 231

2nd effect: (W2 – W1) lv1 = W3 cp3 (T2 – T3) + (W3 – W2) lv2 (W2 – 6678) 2287.4 = W3 (347.5 – 337.5) ´ 4.1868 + (W3 – W2) 2323.0 1.9495 W2 – W3 = 6459.24 (2) 3rd effect: (W3 – W2) lv2 = W4 cp4 (T3 – T2) + (W4 – W3) lv3 (W3 – W2) 2323.0 = W4 (337.5 – 324.6) ´ 4.1868 + (W4 – W3) 2347.6 2.0106 W3 – W2 = 1.0318 W4 (3) 4th effect: (W4 – W3) lv3 = WF cpF (T4 – T1) + (WF – W1) lv4 (W4 – W3) 2347.6 = 33 390 (324.6 – 313) 4.1868 + (33 390 – W4) 2379.1 2.0134 W4 – W3 = 34 528.8 (4) From Eq. (2), W2 = 3313.28 + 0.513 W3 (5) Substitute W2 in Eq. (3). 2.0106 W3 – 3313.28 – 0.513 W3 = 1.034 W4 1.4976 W3 – 1.034 W4 = 3313.28 W3 – 0.69 W4 = 2211.39 (6) Add Eq. (4) and Eq. (6) which yield W4 = 27 762 kg/h Eq. (6) yields, W3 = 21 367 kg/h Eq. (5) yields, W2 = 14 275 kg/h Eq. (1) yields, Ws = 8294 kg/h Steam economy = 26 712/8294 = 3.221 kg evaporation/kg steam Ans. Use of Mathcad is recommended for solving simultaneous equations.

EXERCISE 6.44 Basis: Pulp handling rate = 175 t/d Feed rate of pulp slurry = (175 ´ 1000)/(3600 ´ 24 ´ 0.015) = 135.03 kg/s Salt in the slurry = 135.03 ´ 0.0925 = 12.49 kg/s Liquor in the feed (with dissolved salt) = 135.03 ´ 0.985 = 133.0 kg/s Flow of washed cake = 175 ´ 1000/(3600 ´ 24 ´ 0.18) = 11.253 kg/s Salt in the washed cake = 11.253 ´ 0.0163 = 0.183 kg/s

232 Solutions Manual—Stoichiometry

Liquor in washed cake (with dissolved salt) = 11.253 ´ 0.82 = 9.227 kg/s Wash water, entering 2nd stage = 590 L/min = 9.833 kg/s Strong liquor, leaving 1st stage = 135.03 + 9.833 – 11.253 = 133.61 kg/s Salt in strong liquor = 12.49 – 0.183 = 12.307 kg/s Let amount of salt in mud, leaving 1st stage = a kg/s and amount of salt in wash liquor, entering 1st stage = b kg/s. Salt balance around 1st stage: 12.49 + b = 12.307 + a a – b = 0.183 (1) Liquor balance around 1st stage: Liquor in washed liquor will be in equilibrium with the washed cake. Liquor in washed cake = 9.227/0.183 Salt = 50.421 Liquor in washed liquor = 50.421 × b

Ratio =

133.0 + 50.421 b = 133.61 +

F 133.61 I a H 12.307 K

– a + 4.644 b = 0.0562 Solving the equations, a = 0.249 kg/s b = 0.066 kg/s

(2)

(12.307 – 0.249) ´ 100 12.307 = 97.98% say 98%

Efficiency of 1st stage =

(0.249 – 0.183) ´ 100 0.249 = 26.51%

Efficiency of 2nd stage =

(12.307 – 0.183) ´ 100 12.307 = 98.51 %

Efficiency (overall) =

EXERCISE 6.45 Basis: 100 00 kg/h feed to the fractionator Overall material balance: F+S=W+Q+O

Ans.

Stoichiometry and Unit Operations 233

where S, W, Q and O are unknowns. Balance of CH2Cl2: 0.0168 ´ 10 000 + S ´ 0 = W ´ 0 + Q + 0.0763 + O ´ 0.907 11.89 O + Q = 2201.83

(2)

Balance of CH3OH: 0.0202 ´ 10 000 + S ´ 0 = W ´ 0 + Q ´ 0.1606 + O ´ 0.093 0.58 O + Q = 1257.78

(3)

Solving Eq. (2) and Eq. (3) O = 83.47 kg/h and Q = 1209.37 kg/h

Ans. (a)

Reflux ratio = 0.5 Components CH2Cl2 CH3OH H2O Total

Upper layer (Q)

Lower layer (O)

Molar mass

kg/h

kmol/h

mole %

kg/h

kmol/h

85 32 18

92.28 194.22 922.87

1.086 6.069 51.271

1.86 10.39 87.75

75.71 7.76

0.891 0.243

1209.37

58.426

100.00

83.47

1.134

Total products, drawn from the fractionator = 58.426 + 1.134 = 59.56 kmol/h Reflux stream = 0.5 ´ 59.56 = 29.78 kmol/h Reflux stream will have the same composition at as that of Q. Composition of overhead vapours from fractionator: Component CH2Cl2 CH3OH H2O Total

kmol/h

mole %

1.086 + 29.78 ´ 0.186 + 0.891 = 2.531 6.069 + 29.78 ´ 0.1039 + 0.243 = 9.400 51.271 + 29.78 ´ 0.8775 = 77.403 89.34

kg/h

2.83 10.53 86.64

215.14 300.94 1393.25

100.00

1909.38

Partial pressure of CH2Cl2 = 101.325 ´ 0.0283 = 2.867 kPa Partial pressure of CH3CH = 101.324 ´ 0.1053 = 10.669 kPa Partial pressure of H2O = 101.325 ´ 0.8664 = 87.788 kPa

234 Solutions Manual—Stoichiometry

Latent heat of evaporation: Components

Antoine constants (Ref. Table 5.4)

CH2Cl2 CH3OH H2O H2O

A B C 4.536 91 1327.016 –20.474 5.204 09 1581.341 – 33.500 4.6543 1435.264 – 64.848 From Appendix (A IV.2)

Calculated Saturation Temperature

Latent heat of evaporation (lv)

K (°C) kJ/kmol kJ/kg 238.75 (–34.4°C) 31.72 373.5 289.55 (16.4°C) 38.62 1205.3 369.52 (96.37°C) – – 369.5 K 2267.3

Heat removed in overhead condenser = 378.5 ´ 215.14 + 1205.3 ´ 300.99 + 1393.25 ´ 2267.3 = 80 355 + 362 783 + 3158 916 = 3602 054 kJ/h º 1000.571 kW Heat content of bottom product = W (373 – 318) ´ 4.1868 = 259.58 W kJ/h º 0.072 W kW Heat supplied to feed = 0 kJ/h on the basis of ref. temp. Heat to be supplied by steam = Heat removed in the condenser + Heat of products + Heat in bottom product – Heat of feed = 1000.571 + 0.072 W kW Enthalpy of saturated steam at 451.3 kPa = 2743 kJ/kg (Table AIV.2) Heat supplied by steam = 2743 – 158.5 = 2584.5 kJ/kg Total heat supplied = 2584.5 S kJ/h From Eq. (1), W = F + S – Q – O = 10 000 + S – 1209.37 – 83.47 = 8707.16 + S kg/h Heat supply rate = Heat loss rate = 1.02 (enthalpy to be supplied) 2584.5 S = [1000.571 + 0.072 (8707.16 + S)] 1.02 ´ 3600 2584.5 S = 5976 131 + 264.38 S 2320.12 S = 5976 131 S = 2575.8 kg/h Ans.(b) W = 8707.16 + 2575.8 = 11 282.9 kg/h Ans.(c)

7

Combustion EXERCISE 7.1 GCV = 24 070 kJ/kg on dry ash-free basis Moisture in coal = 100 – 24.6 – 49.8 – 20.3 = 5.3% GCV on as-received basis = 24 070 ´ (1 – 0.203 – 0.053) = 24 070 ´ 0.744 = 17 908 kJ/kg º 7699 Btu/lb % C = 5.88 + 0.005 12 [7699 – (40.5) (0.7)] ± 0.0053 [80 – 100 (24.6/49.8)]1.55 [Ref. Eq. (7.3)] Since 100 (24.6)/49.8 < 80 % C = 5.88 + 39.27 + 1.06 = 46.21 Ans.

EXERCISE 7.2 Basis: 100 kg air dried coal GCV = 27 235 kJ/kg on dry ash-free basis O2 content of coal = 18.6 kg = 0.581 kmol Equivalent H2 = 2 ´ 0.581 = 1.162 kmol Net H = 3.5 – (1.162 ´ 2) = 1.176 kg or 1.176% Ans. (a) Moisture in coal = 4 kg Combined water = 1.162 ´ 18 – 4 = 20.916 – 4 = 16.916 kg or 16.916% Ans. (b) GCV on as-received basis = 27 235 ´ (1 – 0.142 – 0.04) = 22 278 kJ/kg º 9578 Btu/lb Dülong’s formula: GCV = 33 950 ´ 0.615 (1 – 0.04) + 144 200 ´ 0.011 76 + 9400 ´ 0.004 = 20 044 + 1696 + 37.6 = 21 777.6 kJ/kg on as-received basis Ans. (c) Calderwood equation: Since 100(26.7/55.1) < 80 % C = 5.88 + 0.005 12(9578 – 40.5 ´ 0.4) + 0.0053 [80 – (100 ´ 26.7/55.1)]1.55

236 Solutions Manual—Stoichiometry

= 5.88 + 48.96 + 1.11 = 55.96% Mass of ash-free, moisture-free coal = 26.7 + 55.1 = 81.8 kg VM = 26.7 ´ 100/81.8 = 32.64% FC = 55.1 ´ 100/81.8 = 67.36% Oxygen requirement:

Ans. (d)

Ans. (e)

Element

Mass, kg

Molar Mass

kmol

O2 requirement kmol

Carbon Hydrogen Sulpher Oxygen

59.04 3.36 0.38 17.86

12 2 32 32

4.92 1.68 0.012 0.558

4.92 0.84 0.012 (–) 0.558

Total

84.0 (dry)

7.170

5.214

Theoretical O2 requirement = 5.214 kmol Specific air requirement = 5.214/(0.21 ´ 100) = 0.248 kmol/kg dry coal Theoretical air requirement = 5.214 ´ 29/(0.21 ´ 100) = 7.2 kg/kg dry coal Ans. (f) Excess air = 60% Total O2 supply = 5.214 ´ 1.6 = 8.342 kmol N2 from air = 8.342 ´ 79/21 = 31.383 kmol Total N2 in flue gases = 31.383 + (1.8 ´ 0.96)/28 = 31.445 kmol Total moisture in flue gas = (4.0/0.96)/18 + 1.68 = 1.911 kmol Analysis of flue gases: Gas

kmol

Actual wet analysis

Orsat analysis

CO2 SO2 O2 N2 H2O

4.92 0.012 3.128 31.445 1.911

11.88 0.03 7.55 75.92 4.62

12.48 — 7.92 79.60 —

Total

41.416 (wet) 39.505 (dry)

100.00

100.00

Ans.(g)

EXERCISE 7.3 Basis: 100 kg fuel oil (Carbon + Hydrogen) content of fuel oil = = Carbon in the fuel oil = =

100 – 1.37 98.63 kg 9.33 ´ 98.63/10.33 89.08 kg

Combustion

237

Hydrogen in fuel oil = 98.63 – 89.08 = 9.55 kg º 4.775 kmol H2O produced by combustion = 4.775 kmol º 85.95 kg [Ref. Eq. (7.1)] NCV = GCV – 2442.8 (H2O production) = 41 785 – 2442.8 ´ 85.95/100 = 39 685.4 kJ/kg Ans.

EXERCISE 7.4 (a) Gaseous n-propanol: Formula: C3H8O Water formed = 4 mol/mol n-propanol = 72.06 kg/kmol n-propanol NCV = 2068.65 – 2442.8 ´ 72.06/1000 = 1892.72 kJ/mol (b) Liquid acetone: Formula : C3H6O Water formed = 3 mol/mol acetone = 54.045 kg/kmol acetone NCV = GCV – 2442.8(H2O produced) = 1790.02 – 2442.8 ´ 54.045/1000 = 1658.00 kJ/mol

Ans.

Ans.

EXERCISE 7.5 NCV of CH4 = 802.62 kJ/mol at 298.15 K (a) Specific energy consumption = 22.0193 ´ 802.62 ´ 103/106 = 17.673 GJ/t NH3 (b) Specific energy consumption = 25.9715 ´ 802.62 ´ 103/106 = 20.845 GJ/t NH3

(Ref. Appendix V.2) Ans. Ans.

EXERCISE 7.6 Specific energy consumption = 31.2091 ´ 802.62 ´ 103/106 = 25.049 GJ/t CH3OH

Ans.

EXERCISE 7.7 NCV of motor spirit = 44 050 kJ/kg at 298.15 K (ref. Example 5.29) NCV of ethanol liquid = 1235.49 kJ/mol Molar mass of ethanol = 46.0684 kg/kmol

238 Solutions Manual—Stoichiometry

NCV of ethanol liquid =

1234.97 ´ 1000 46.0684

= 26 807.3 kJ/kg at 298.15 K NCV of the blend = 44 050 ´ 0.9 + 26 807.3 ´ 0.1 = 42 325.7 kJ/kg at 298.15 K CO2 generated by motor spirit alone

Ans.

0.8486 ´ 44.0098 12 = 3.112 kg/kg

=

CO2 generated by blend = 0.8486 ´ 44.0098 ´ 0.9 + 2 ´ 44.0098 ´ 0.1 12 46.0684 = 2.9919 kg/kg Comparison on NCV basis:

3.112 ´ 106 44 050 = 70.65 kg/GJ

CO2 generated by motor spirit =

2.9919 ´ 106 42 325.7 = 70.68 kg/GJ In both cases, CO2 generated are practically same on heat release basis. CO2 generated by blend =

EXERCISE 7.8 Basis: 1 mol associated gas Component mol ni

Molar mass Mi

ni × Mi g

Calorific values ni × GCVi ni × NCVi kJ/mol kJ GCVi NCVi 890.65 1560.69 2219.17 2868.20 2877.40 3528.83 3535.77 — —

CH4 C2 H 6 C3 H 8 i-C4H10 n-C4H10 i-C5H12 n-C5H12 N2 CO2

0.744 0.084 0.074 0.017 0.020 0.005 0.004 0.043 0.009

16.0425 30.0690 44.0956 58.1222 58.1222 72.1488 72.1488 28.0135 44.0095

11.936 2.526 3.263 0.988 1.162 0.361 0.289 1.205 0.396

Total

1.000

—

22.126

802.62 1428.64 2043.11 2648.12 2657.32 3264.73 3271.67 — —

GCV = 1096.05 ´ 1000/22.126 = 49 536.7 kJ/kg NCV = 995.93 ´ 1000/22.126 = 45 011.8 kJ/kg

662.64 131.10 164.22 48.76 57.55 17.64 14.14 — —

597.15 120.01 151.19 45.02 53.15 16.32 13.09 — —

1096.05

995.93

Combustion

239

At 101.325 kPa and 298.15 K (25 °C), specific volume = 24.465 m3/kmol GCV = 1096.05 ´ 1000/24.465 = 44 800.7 kJ/m3 NCV = 995.93 ´ 1000/24.465 = 40 708.4 kJ/m3 Ans.

EXERCISE 7.9 Basis: 100 mol refinery gas Refer tabular calculations given below: Component H2 CH4 C2H6 C3H8 n-C4H10 n-C5H12 Total

mole ni

ni × Mi, kg

ni × GCVi, kJ

ni × NCVi, kJ

74.0 13.5 7.4 3.6 1.2 0.3

148 216 222 158.4 69.6 21.6

21 151.4 12 023.8 11 549.1 7 989.0 3 452.9 1 060.7

17 894.7 10 835.4 10 571.9 7 355.2 3 188.8 981.5

100.0

835.6

57 226.9

50 827.5

3

GCV = 572.269 ´ 1000/24.465 = 23 391 kJ/m NCV = 508.275 ´ 1000/24.465 = 20 776 kJ/m3 Average molar mass = 835.6/100 = 8.356 GCV = 572 269/8.356 = 68 486 kJ/kg NCV = 508 275/8.356 = 60 828 kJ/kg

Ans.

EXERCISE 7.10 Basis: 100 mol purge gas: Component H2 N2 Ar CH4

mol, ni 69.0 23.0 2.7 5.3

Total

100.0

Molar mass, Mi 2 28 40 16

ni × Mi, g 138 644 108 84.8

ni × GCVi kJ 19 722.3 — — 4 720.4

ni × NCVi kJ 16 685.6 — — 4 253.9

974.8

24 442.7

20 939.5

GCV = 244.42 kJ/mol NCV = 209.4 kJ/mol Average molar mass = 9.748 GCV = 244.427 ´ 1000/9.748 = 25 075 kJ/kg NCV = 209.518 ´ 1000/9.748 = 21 493 kJ/kg Theoretical O2 requirement = (69/2) + (5.3 ´ 2) = 45.1 mol N2, entering with O2 = 79 ´ 45.1/21 = 169.66 mol Theoretical air requirement = 45.1 + 169.66 = 214.76 mol (dry)

Ans. (a)

240 Solutions Manual—Stoichiometry

Specific dry air requirement = 214.76 ´ 29/974.8 = 6.39 kg/kg purge gas Actual O2 supply = 1.2 ´45.1 = 54.12 mol Actual N2 entry = 54.12 ´ 79/21 = 203.59 mol Composition of dry flue gases: Component

mol

Ans.(b)

mole % (dry)

5.3 2.7 203.59 + 23.0 = 226.59 54.12 – 45.1= 9.02

CO2 Ar N2 O2 Total

2.18 1.11 93.01 3.70

243.61 100.00

Ans.(c)

EXERCISE 7.11 Basis: 100 mol SNG Component

ni mol 96.59 1.29 0.22 2.00

CH4 H2 CO CO2 Total

100.00

GCV = NCV = or GCV = NCV =

Molar mass Mi 16 2 28 44

ni × Mi g 1545.44 2.58 6.16 88.00

ni × GCVi , kJ 86 027.9 368.7 62.3 —

ni × NCVi , kJ 77 525.1 311.9 62.3 —

—

1642.18

86 458.9

77 899.3

864.589 ´ 1000/16.4218 = 52 649 kJ/kg 778.993 ´ 1000/16.4218 = 47 437 kJ/kg 864.589/22.414 = 38 574 kJ/Nm3 778.993/22.414 = 34 755 kJ/Nm3

Ans.

EXERCISE 7.12 Basis: 100 kmol tail gas (wet). Data in the following table are extracted from Table 4.20 and Table 4.21. Component CO2 CO H2 CH4 (CH3)2O O2 N2 H2O Total

ni kmol 1.17 0.13 0.26 0.08 0.13 11.28 76.67 10.28 100.00

(GCV)i kJ/kmol — 282 980 285 830 890 650 2107 400 — — —

ni × GCVi kJ — 36 787.4 74 315.8 71 252.0 273 962.0 — — — 456 317.2

(NCV)i kJ/kmol — 282 980 241 820 802 620 1931 380 — — —

ni × (NCV)i kJ — 36 787.4 62 873.2 64 209.6 251 079.4 — — — 414 949.6

Combustion

GCV = 456 317.2(22.414 ´ 100) = 203.6 kJ/Nm3 NCV = 414 949.6(22.414 ´ 100) = 185.1 kJ/Nm3

241

Ans.

EXERCISE 7.13 Basis: 100 kmol dry flue gases O2, accounted for = 10.6 + 8.7 = 19.3 kmol O2 supplied from air = 21 ´ 80.7/79 = 21.452 kmol O2 unaccounted = 21.452 – 19.3 = 2.152 kmol Hydrogen burnt = 2.152 ´ 2 = 4.304 kmol Water formed = 4.304 kmol Carbon burnt = 10.6 katom º 127.2 kg Total air supplied = 80.7 + 21.452 = 102.152 kmol New Basis: 100 kg coal Carbon content of coal = 65 kg Carbon in the refuse = [12.7 ´ 0.086/(1 – 0.086)] = 1.195 kg Carbon burnt = 65 – 1.195 = 63.805 kg Dry air supply = 102.152 ´ 63.805/127.2 = 51.241 kmol Dry air 51.241 ´ 29 = = 14.86 kg/kg Ans. (a) Ratio Coal 100 (100 + 4.304)63.805 Flue gas quantity = 127.2 = 52.32 kmol (wet) For 100 kmol dry flue gas basis: Component

Molar mass, Mi

ni × Mi, kg

CO2 O2 N2 H2O

kmol, ni 10.6 8.7 80.7 4.304

44 32 28 18

466.4 278.4 2259.6 77.5

Total

104.304

—

3081.9

Average molar mass = 3081.9/104.304 = 29.547 (wet) Ratio

flue gas 52.32 ´ 29.547 = = 15.46 kg/kg fuel (wet basis) coal 100

Ans.(b)

Hydrogen burnt = 4.304 ´ 2 ´ 63.805/127.2 = 4.32 kg/100 kg coal Hydrogen content of coal = 4.32%

Ans. (c)

242 Solutions Manual—Stoichiometry

EXERCISE 7.14 Basis: 100 kmol of dry flue gases N2 in flue gas = 100 – 10.6 – 6.0 = 83.4 kmol O2 accounted = 10.6 + 6.0 = 16.6 kmol O2 supplied through air = 21 ´ 83.4/79 = 22.17 kmol Theoretical oxygen requirement = 22.17 – 6.0 = 16.17 kmol Excess air = 6.0 ´ 100/16.17 = 37.1%

Ans. (a)

Carbon content of fuel = 10.6 katom = 127.2 kg Oxygen, consumed for burning of hydrogen = 22.17 – 16.6 = 5.57 kmol Hydrogen burnt = 5.57 ´ 2 = 11.14 kmol º 22.28 kg Ratio C:H = 127.22/22.28 = 5.71 kg/kg

Ans. (b)

EXERCISE 7.15 Basis: 100 kmol fuel gas mixture Combustion reactions: CH4 + 2 O2 = CO2

+ 2 H2O

C2H4 + 3 O2 = 2 CO2 + 2 H2O C6H6 + 7.5 O2 = 6 CO2 + 3 H2O Component

kmol

CO2 C2H4 C6H6 O2 CO H2 CH4 N2

3.4 3.7 1.5 0.3 17.4 36.8 24.9 12.0

Theoretical O2 requirement, kmol — + 11.10 + 11.25 – 0.30 + 8.70 + 18.40 + 49.80 —

Total

100.00

+ 98.95

Total carbon, kmol 3.4 7.4 9.0 — 17.4 — 24.9 —

Total CO2, produced by complete combustion = 62.1 kmol Total dry flue gas quantity = 62.1/0.10 = 621 kmol (N2 + O2) in fuel gas = 621 – 62.1 = 558.9 kmol

62.1

Combustion

243

N2 from fuel = 12 kmol (N2 + O2) from air = 558.9 – 12 = 546.9 kmol N2 corresponding to theoretical O2 = 79 ´ 98.95/21 = 372.24 kmol Since the mixture (N2 + O2) of the flue gases (i.e. 546.9 kmol) does not contain (total) theoretical oxygen, excess air = 546.9 – 372.24 = 174.66 kmol O2 in flue gases = 174.66 ´ 0.21 = 36.68 kmol N2 in flue gases = 174.66 – 36.68 = 137.98 kmol Total N2 in flue gas = 12 + 372.2 + 137.98 = 522.18 kmol Excess air = 36.68 ´ 100/98.95 = 37.07% Ans. (a) Composition of dry flue gas: Component

kmol

Orsat analysis, mole %

CO2 O2 N2

62.1 36.68 522.18

10.00 5.91 84.09

Total

620.96

100.00

Ans.(b)

EXERCISE 7.16 Basis: 100 kmol dry flue gas mixture Component

kmol

kmol of C

kmol of O2

CO2 CO O2 N2

12.4 3.1 5.4 79.1

12.4 3.1 — —

12.4 1.55 5.40 —

Total

100.00

15.5

19.35

O2 supplied through air = 21 ´ 79.1/79 = 21.03 kmol Unaccounted O2 = 21.03 – 19.35 = 1.68 kmol This O2 is used for hydrogen burning. Unburnt C = 15.5 ´ 0.15/0.85 = 2.735 kmol O2, required for complete combustion = 15.5 + 2.735 + 1.68 = 19.915 kmol Excess O2 = 21.03 – 19.915 = 1.115 kmol Excess air = 1.115 ´ 100/19.915 = 5.6% Ans.

244 Solutions Manual—Stoichiometry

EXERCISE 7.17 Comparison of the fuels is to be made on NCV basis. (A): Basis 100 kmol bottled LPG Gas C2 H 6 C3 H 2 i-C4H10 n-C4H10 Total

kmol ni 1.2 25.2 23.9 49.7

Molar mass Mi 30 44 58 58

100.0

—

Mass, ni × Mi kg 36.0 1108.8 1386.2 2882.6

LCVi, kJ/kmol 1428 640 2043 110 2648 120 2657 320

5413.6

ni × LCVi kJ 1 714 368 51 486 372 63 290 068 132 068 800 248 559 610

LCV = 248 559 610/5413.6 = 45 914 kJ/kg Cost of bottled LPG on LCV basis = 30 ´ 106/45 914 = Rs. 653.4 per GJ (B) Basis: 1 kmol associated gas Volume of 1 kmol gas is 22.414 Nm3. Gas

ni, kmol

CH4 C2 H 8 C3 H 8 n-C4H10 i-C4H10

0.866 0.086 0.039 0.007 0.002

Total

1.000

LCVi, kJ/kmol

ni × LCVi, kJ

802 1428 2043 2657 2648

695 122 79 18 5

620 640 110 320 120

069 863 681 601 296

921 510

LCV = 921 510/22.414 = 41 113 kJ/Nm3 Cost of associated gas on LCV basis = 15 ´ 106/41 113 = Rs. 364.85 per GJ Thus associated gas is cheaper than bottled LPG.

Ans.

EXERCISE 7.18 Basis: 100 kmol blast furnace (fuel) gas Gas

kmol, ni

H2 CO CO2 N2

3.2 26.2 13.0 57.6

Total

100.0

Molar mass Mi 2 28 44 28 —

Mass ni × Mi kg 6.4 733.6 572.0 1612.8

O2 requirement, kmol 1.6 13.1 Nil Nil

2924.8

14.7

Average molar mass of fuel gas = 2924.8/100 = 29.25 Theoretical N2 entry = 14.7 ´ 0.79/0.21 = 55.3 kmol from air

Combustion

245

Total dry air entry = 14.7 + 55.3 = 70 kmol Absolute humidity of air at 298.15 K (25 °C) DB and 291 K (18°C) WB = 10.2 g/kg dry air (Ref. Fig. 6.15) º 0.0164 kmol/kmol dry air Moisture, entering with air = 0.0164 ´ 70 = 1.148 kmol Total wet air = 14.7 ´ 32 + 55.3 ´ 28 + 1.148 ´ 18 = 2039.5 kg Theoretical air requirement = 2039.3/2924.8 = 0.7 kg/kg fuel Ans. (a) Excess air = 20% Actual dry air supply = 70 ´ 1.2 = 84 kmol Moisture entry = 84 ´ 0.0164 = 1.378 kmol O2 in the flue gas = (84 – 70)0.21 = 2.94 kmol N2 from air = 84 ´ 0.79 = 66.36 kmol Composition of flue gas: Gas

kmol

CO2 O2 N2 H2O

26.2 + 13.0 = 39.2 2.94 66.36 + 57.6 = 123.96 3.2 + 1.378 = 4.578

Total

170 678 (wet) 166 100 (dry)

Actual analysis (wet) mole % 22.97 1.72 72.63 2.68

Orsat analysis mole % 23.60 1.77 74.63 —

100.00

100.00

Total heat released by the combustion of 100 kmol blast furnace gas = 3.2 (241 820) + 26.2 (282 980) = 773 824 + 7414 076 = 8187 900 kJ at 298.15 K Take reference temperature = 298.15 K (25°C) Enthalpy of fuel gas: Component

ni, kmol

C° mp constants

H2 CO CO2 N2

3.2 26.2 12.0 57.6

ai × ni 91.554 760.526 256.386 1704.436

Total

100.0

2812.902

ni × bi ´ 103 32.621 – 73.792 771.409 – 296.122

ni × ci ´ 106 – 0.472 305.065 – 492.607 759.335

ni × di ´ 109 2.461 – 123.305 117.599 – 286.157

434.116

571.321

– 289.402

Enthalpy of fuel gas at 333.15 K over 298.15 K, H1 = 2812.902 (333.15 – 298.15) + 434.116 ´ 10–3 (333.152 – 298.152)/2 + 571.321 ´ 10–6 (333.153 – 298.153)/3 – 289.402 ´ 10–9 (333.154 – 298.154)/4

246 Solutions Manual—Stoichiometry

= 98 452 + 4794 + 200 – 319 = 103 127 kJ Enthalpy of flue gases (total) = 8187 900 + 103 127 = 8291 027 kJ Flue gas: Heat capacity constants

CO2

O2

N2

H2O

Total

ni, kmol

39.2

2.94

123.96

4.578

170.678

148.749 115.175 0.364 97.428 60.479 – 24.447 – 20.818 2.209

1771.791 2372.181 2431.683 1148.769 – 1239.730 – 347.821 242.652 37.857

(ni × ai)1 (ni × ai)2 (ni × bi)1 ´ 103 (ni × bi)2 ´ 103 (ni × ci)1 ´ 106 (ni × ci)2 ´ 106 (ni × di)1 ´ 109 (ni × di)2 ´ 109

837.528 76.516 1457.221 54.193 2519.937 34.560 910.894 63.849 – 1609.184 – 6.887 – 289.249 – 24.416 384.156 – 1.653 32.195 3.398

708.998 745.592 – 123.178 76.598 315.862 – 9.709 – 119.033 0.055

Suffixes 1 and 2 refer to the constants for temperature ranges 298.15 to 1500 K and 1500 to 4000 K, respectively. Enthalpy of flue gas at 1500 K over 298.15 K = 1771.791 (1500 – 298.15) + 2431.683 ´ 10–3 (15002 – 298.152)/2 – 1239.83 ´ 10–6 (15003 – 298.153)/3 + 242.652 ´ 10–9 (15004 – 298.154)/4 = 2129 693 + 2627 672 – 1383 760 + 306 628 = 3680 233 kJ Enthalpy of flue gas above 1500 K at T K = 2372.181 (T – 1500) + 1148.769 ´ 10–3 (T2 – 15002)/2 – 347.821 ´ 10–6 (T3 – 15003)/3 + 37.857 ´ 10–9 (T4 – 15004)/4 = 8291 027 – 3680 233 = 4610 794 kJ

By trial and error or by Mathcad, T = 2782.71 K or 2509.56°C

Ans.

Combustion

247

EXERCISE 7.19 Basis: 100 kg coal Carbon content of coal = 46.21 kg (refer Exercise 7.1) º 3.85 katom or kmol Sulphur content of coal = 0.7 kg º 0.022 katom or kmol Considering SO2 as CO2 in the flue gases, CO2 content of flue gases = 3.85 + 0.022 = 3.872 kmol Total dry flue gases mixture = 3.872/0.088 = 44 kmol O2 content of flue gases = 44 ´ 0.092 = 4.05 kmol N2 content of flue gases = 44 – 3.872 – 4.05 = 36.078 kmol O2 from air = (21 ´ 36.078)/79 = 9.59 kmol O2 accounted = 3.872 + 4.05 = 7.922 kmol O2 unaccounted = 9.59 – 7.922 = 1.668 kmol This O2 is used to burn the net hydrogen of the coal. Net H2 in flue = 1.668 ´ 2 = 3.336 kmol Water vapours in flue gases: Moisture of coal = 5.3 kg Moisture, sprinkled over coal = 5 kg Total moisture due to coal firing = 10.2 kg º 0.567 kmol Absolute humidity of air at 307 K (34°C) DB and 297 K(24°C) WB = 14.8 g/kg dry air (Ref. Fig. 6.15) º 0.0239 kmol/kmol dry air Moisture, entering through air = 0.0239 (9.590 + 36.078) = 1.092 kmol Water produced by burning net H2 = 3.276 kmol Total moisture in flue gases = 0.567 + 1.092 + 3.336 = 4.995 kmol Composition of flue gases: Gas

kmol

mole % (wet basis)

CO2 SO2 O2

3.85 0.022 4.05

7.85 0.04 8.27

Orsat analysis 8.80 — 9.20

248 Solutions Manual—Stoichiometry

N2 H2O

36.078 4.995

Total

48.995 (wet) 44.000 (dry)

73.65 10.19

82.00 —

100.00

100.00

Partial pressure of water vapours in flue gases, p = 0.1019 ´ 100 = 10.19 kPa From Table 6.13, dew point = 46.2°C (319.35 K) Temperature of water, entering the economiser is 311.15 K(38°C) which is lower than the dew point. This is undesirable. Corrosion is envisaged. Refuse contains 5.9% combustibles. It is assumed that this is carbon. Rest 94.1% is ash. Carbon in the refuse = 20.3 ´ 0.059/0.941 = 1.273 kg º 0.106 kmol O2 required for burning carbon of refuse = 0.106 kmol Theoretical oxygen requirement = 3.872 + 1.668 + 0.106 = 5.648 kmol Excess O2 = 9.59 – 5.648 = 3.942 kmol Excess air = (3.942 ´ 100)/5.648 = 69.8% say 70% New basis: 2500 kg/h firing rate Dry air supplied = (9.59 + 36.078) 2500/100 = 1141.7 kmol/h Moist air supplied = (9.59 + 36.078 + 1.092) 2500/100 = 1168.88 kmol/h Volumetric flow rate = 1168.88 ´ 8.314 ´ 307.15/100 = 29 849 m3/h Molar flow rate of flue gases = 48.995 ´ 2500/100 = 1224.88 kmol/h Volumetric flow rate = 1224.88 ´ 8.314 ´ 463.15/100 = 47 165 m3/h Heat Balance: Heat supplied by burning fuel, f1 = 2500 ´ 17 908 = 44 770 000 kJ/h º 12 436.11 kW Heat gain: Economiser: Heat gained by water, f4 = 12 150 (368 – 311)4.1868 = 2899 568 kJ/h º 805.436 kW Boiler pressure: At 0.9 MPa g, i.e. 10.013 bar a Ts = 453 K (180°C), H = 2776.3 kJ/kg Enthalpy of water at 368 K = 397.99 kJ/kg

Combustion

249

Heat supplied in the boiler, f5 = (2776.3 – 397.99) ´ 12 150 = 28 896 467 kJ/h º 8026.8 kW

Superheater: Enthalpy of steam at 0.9 MPa g and 473 K (180°C) = 2826.6 kJ/kg Specific enthalpy gain in the superheater = 2826.6 – 2776.3 = 50.3 kJ/kg Total heat gained in superheater, f6 = 50.3 ´ 12 150 = 611 145 kJ/h º 169.763 kW Total moisture, evaporated in the boiler = 0.567 + 3.336 = 3.903 kmol/100 kg coal º 0.7031 kg/kg coal Total moisture evaporated = 0.7031 ´ 2500 = 1757.75 kg/h lv at DP [46.2°C)] = 2392.8 kJ/kg Heat lost in evaporation, f8 = 1757.75 ´ 2392.8 = 4205 944 kJ/h º 1168.318 kW Ash content of the refuse = 2500 ´ 0.203 = 507.5 kg/h Combustibles (as C) = 507.5 ´ 0.059/0.941 = 31.82 kg/h Heat lost in refuse, f11 = 32 762 ´ 31.82 = 1042 487 kJ/h º 289.58 kW Heat loss in flue gas: Basis: 100 kg coal C°mp equation constants Component

ni × kmol

ai × ni

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

CO2 O2 N2 H2O

3.872 4.050 36.078 4.995

82.727 117.932 1067.581 162.333

248.908 4.031 – 185.477 0.398

– 158.948 – 49.647 475.613 66.027

37.945 162.503 – 179.236 – 22.728

Total

48.995

1430.573

67.860

333.745

– 1.516

Heat loss in the flue gas for 2500 kg/h firing rate 2500 [1430.333 (463.15 – 298.15) + 67.860 ´ 10–3 (463.152 – 298.152)/2 100 + 333.745 ´ 10–6 (463.153 – 298.153)/3 – 1.516 ´ 10–9 (463.154 – 298.154)/4] = 25 ´ 248 355

f7 =

250 Solutions Manual—Stoichiometry

= 6208 875 kJ/h º 1724.688 kW enthalpy, kW Heat input, burning of fuel Heat output: 1. Useful heat gain: Economiser Boiler Superheater 2. Heat loss in flue gases 3. Heat loss due to evapn. 4. Heat loss in refuse 5. Unaccounted heat loss Total

12 436.11

% 100.0

805.44 8 026.80 169.76 1 724.688 1 168.318 289.58 251.524

6.48 64.54 1.37 13.87 9.39 2.33 2.02

12 436.11

100.00

(805.44 + 8026.8 + 169.76) ´ 100 = 72.4% Ans. 12 436.11 = 6.48 + 64.54 + 1.37 = 72.39% on GCV basis Specific steam production 12 150 = 4.86 kg/kg coal at actual conditions Ans. = 2500

Boiler efficiency =

EXERCISE 7.20 Basis: 100 kg fuel oil Component Carbon Hydrogen Sulphur Oxygen Nitrogen Total

Mass, kg

kmol

O2 requirement, kmol

84.0 12.7 0.4 1.2 1.7

7.000 6.350 0.013 0.038 0.061

7.000 3.175 0.013 (–) 0.038 —

100.0

13.462

10.15

N2 entering with O2 = 79 ´ 10.15/21 = 38.183 kmol Total dry air = 10.15 + 38.183 = 48.333 kmol Specific theoretical dry air requirement = 48.333 ´ 29/100 = 14.02 kg/kg fuel Ans. (a) Moisture entering with air = 0.0155 kg/kg dry air (Ref. Fig. 6.15) º 0.025 kmol/kmol dry air Incoming moisture = 48.333 ´ 0.025 = 1.208 kmol

Combustion

251

Composition of flue gases with theoretical air: Gas CO2 SO2 H2O O2 N2 Total

kmol 7.00 0.013 6.35 + 1.208 = 7.558 Nil 38.183 + 0.061 = 38.244 52.815 (wet) 45.257 (dry)

mole % (dry basis) 15.50 (Orsat) — — — 84.50 100.00

Theoretical CO2 in flue gases = 15.50% Ans.(c) Total dry gas gases = 45.257 kmol/100 kg fuel oil º 0.4526 kmol/kg fuel oil Ans. (b-i) º 10.14 Nm3/kg fuel oil Total wet flue gases = 52.815 kmol/100 kg fuel oil º 0.528 16 kmol/kg fuel oil Ans. (b-ii) º 11.84 Nm3/kg fuel oil Let x = actual O2 supply, kmol N2, entering with O2 = 79x/21 = 3.762x kmol Total dry air = x + 3.762x = 4.762x kmol Excess O2 = x – 10.15 kmol Total flue gas mixture = 7.013 + x – 10.15 + 3.762x + 0.061 = 4.762x – 3.076 kmol (dry) CO2 content in flue gas = [7.013/(4.762x – 3.076)] = 0.065 x = 23.268 kmol Excess air = (23.268 – 10.15) 100/10.15 = 129.24% Ans. (d-i) Total dry flue gases = (4.762 ´ 23.268) – 3.076 = 107.726 kmol/100 kg fuel oil º 1.077 26 kmol/kg fuel oil Moisture, entering with air = 0.025 ´ 4.762 ´ 23.268 = 2.77 kmol Total moisture in flue gases = 6.35 + 2.77 = 9.12 kmol Total wet flue gases = 107.726 + 9.12 = 116.846 kmol/100 kg fuel oil º 1.1685 kmol/kg fuel oil Volume of flue gases at 100.7 kPa and 643.15 K (370 °C) = 1.1685 ´ 8.314 ´ 643.15/100.7 Ans. (d-ii) = 62.047 m3/kg fuel oil Mole fraction of moisture in flue gases = 9.12/(107.726 + 9.12) = 0.078 05 Partial pressure of water vapours, p = 100.7 ´ 0.078 05 = 7.86 kPa Dew point = 314.35 K (41.2°C) (Ref. Table 6.13) Ans. (e)

252 Solutions Manual—Stoichiometry

Component

ni × kmol

C°mp equation constants ai × ni

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

CO2 O2 N2 H2O

7.013 13.118 87.595 9.120

149.836 381.982 2592.015 296.328

450.824 13.550 – 450.326 0.726

– 287.888 – 160.808 1154.756 120.482

68.727 526.348 – 435.172 – 41.472

Total

116.846

3420.161

14.774

826.542

118.431

Heat lost in flue gases = 3420.161 (643.15 – 298.15) + 14.774 ´ 10–3 (643.152 – 298.152)/2 + 826.542 ´ 10–6 (643.153 – 298.153)/3 + 118.431 ´ 10–9 (643.154 – 298.154)/4 = 1179 956 + 2398 + 65 954 + 4528 = 1253 136 kJ Specific enthalpy loss = 1253 136/100 = 12 531.4 kJ/kg fuel oil Ans.( f )

EXERCISE 7.21 Basis: 100 kg furnace oil Oxygen requirement: Element Carbon Sulphur Hydrogen Total

Mass, kg

kmol

O2 requirement, kmol

85.65 3.00 11.35

7.138 0.094 5.675

7.138 0.094 2.838

100.00

12.907

10.070

Total (Orsat) CO2 in flue gases = 7.138 + 0.094 = 7.232 kmol Total dry flue gas mixture = 7.232/0.115 º 62.887 kmol Let x = kmol of oxygen supplied Total flue gas mixture = 7.232 + (x – 10.07) + 3.762x = 4.762x – 2.838 kmol = 62.887 kmol or x = 13.802 kmol Excess air = (13.802 – 10.07) 100/10.07 = 37.06% N2 in flue gases = 3.762 ´ 13.802 = 51.923 kmol Total dry air = 13.802 + 51.923 = 65.725 kmol Humidity of air at 308 K DB and 302.5 K WB = 0.0242 kg/kg dry air (ref. Fig. 6.15) Moisture, entering with air = 0.0242 ´ 29 ´ 65.725/18.015 = 2.563 kmol Total moisture in flue gases = 5.675 + 2.563 = 8.238 kmol

Combustion

Gas

kmol, ni

Molar mass, Mi 44 64 18 32 28

CO2 SO2 H2O O2 N2

7.138 0.094 8.238 3.732 51.923

Total

71.125 (wet) — 62.887 (dry)

253

ni × Mi, kg 314.07 6.02 148.29 119.42 1453.84

mole % (wet basis) 10.04 0.13 11.58 5.25 73.00

Orsat analysis, % 11.50 — — 5.93 82.57

2041.64

100.00

100.00

Partial pressure of water vapours, p = (100 – 0.12) ´ 0.1158 = 11.57 kPa Dew point = 321.85 K (48.7 °C) (Ref. Table 6.13) Heat balance: Refer Table – 7.13 for nomenclature in the text. Steam pressure = 14 bar g = 15.013 bar a Ts = 471.4 K (198.4°C), H = 2790 kJ/kg, h = 844.7 kJ/kg Production of steam = 76.2 ´ 1000/6.25 = 12 192 kg/h Continuous blow down = 1000 kg/h hsw = 482.5 kJ/kg and hfw = 343.31 kJ/kg Total boiler feed water input = 12 192 + 1000 = 13 192 kg/h f4 = 13 192 (482.5 – 343.31) = 1836 195 kJ/h º 510.05 kW Useful heat in economiser, f 4 = 12 192(482.5 – 343.31) = 1697 005 kJ/h = 471.39 kW Heat lost in blowdown water, f²12 = 1000(482.5 – 343.31) = 139 190 kJ/h º 38.66 kW f5 = (2790.0 – 482.5) 12 192 = 28 133 040 kJ/h º 7814.73 kW Heat lost in blow-down, f¢12 = (844.7 – 482.5) 1000 = 362 200 kJ/h º 100.61 kW Total heat loss in blow-down, f12 = 38.66 + 100.61 = 139.27 kW Enthalpy of steam of 15 bar, a and 553.15 K (280°C) = 2993.7 kJ/kg f6 = (2993.7 – 2790) 12 192 = 2483 510 kJ/h º 869.86 kW Absolute humidity of flue gases = 148.29/(2041.68 – 148.29) = 0.0783 kg/kg dry gas CH = 1.006 + 1.84 ´ 0.0783 = 1.15 kJ/(kg × K) Fuel firing rate = 6630/6.25 = 1060.8 kg/h

254 Solutions Manual—Stoichiometry

f7 = 2041.68 ´ 1060.8 ´ 1.15(550.15 – 298.15)/100 = 6276 529 kJ/h º 1743.48 kW (approximate calculations) Component

ni × kmol

CO2 O2 N2 H2O Total

C°mp equation constants ai ni

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

7.232 3.732 51.923 8.238

154.515 108.672 1536.448 267.670

464.903 3.714 – 266.936 0.656

– 296.878 – 45.749 684.496 108.830

70.873 149.743 – 257.953 – 37.461

71.125

2067.305

202.337

405.699

– 74.798

1060.8 [2067.305 (550.15 – 298.15)] 100 + 202.337 ´ 10–3 (550.152 – 298.152)/2 + 450.699 ´ 10–6 (550.153 – 298.153)/3 – 74.798 ´ 10–9 (550.154 – 294.154)/4] = 10.608 (520 960.9 + 21 619.3 + 3488.4 – 16) = 10.608 ´ 546 067 = 5792 679 kJ/h º 1609.08 kW Moisture, produced by burning H2 of fuel oil = 5.675 ´ 18 ´ 1060.8/100 = 1084.18 kg/h lv at 321.85 K (48.7°C) = 2388.4 kJ/kg Heat loss due to evaporation of moisture = f8 = 1084.18 ´ 2388.4 = 2589 456 kJ/h º 719.29 kW Dry air supply = 65.725 ´ 29 ´ 1060.8/100 = 20 219.11 kg/h CH = 1.006 + 1.84 ´ 0.0242 = 1.0505 kJ/(kg dry air × K) f3 = 20 219.11 ´ 1.0505 (308.15 – 298.15) = 212 402 kJ/h º 59.00 kW f2 = 1060.8 ´ 1.675 (350.15 – 298.15) = 92 396 kJ/h = 25.67 kW f1 = 1060.8 ´ 43 040 = 45 656 832 kJ/h º 12 682.45 kW Thermal efficiency of the boiler = (471.39 + 7814.73 + 689.86) 100/12 682.45 = 70.77% (Ref. Table – 7.52) Specific volume of air at 100.0 kPa and 308 K = 25.623 m3/kmol (Ref. Table – 7.8) Molar flow rate of air = (65.725 + 2.563) 1060.8/100

Heat lost in flue gas, f7 =

= 724.4 kmol/h

Combustion

255

Volumetric flow rate of air = 724.4 ´ 25.623 = 18 561.3 m3/h Molar flow rate of flue gases = 71.125 ´ 1060.8/100 = 754.494 kmol/h Volumetric flowrate of flue gases = 754.494 ´ 8.314 ´ 550.15/99.82 = 34 572 m3/h

Ans.

All the above results are tubulated in Table – 7.52 in the text.

EXERCISE 7.22 Basis: 1 kmol methanol CH3OH(l) + 3/2 O2(g) = CO2(g) + 2 H2O(g) Theoretical O2 requirement = 1.5 kmol N2 entering with air = 79 ´ 1.5/21 = 5.643 kmol Stoichimetric air requirement = 1.5 + 5.643 = 7.143 kmol º 207.15 kg Ratio, stoichiometric air/fuel = 207.15/32 = 6.47 kg dry air/kg methanol Ans. (a) NCV of liquid fuel = 638.24 kJ/mol at 298.15 K (25°C) (Ref. Table – AV.2) Excess air = 40% Actual O2 supply = 1.4 ´ 1.5 = 2.1 kmol Excess O2 = 2.1 – 1.5 = 0.6 kmol Actual N2 supply = 79 ´ 2.1/21 = 7.9 kmol Component dry

kmol wet

Orsat analysis mole %

CO2

1.0

1.0

10.53

N2

7.9

7.9

83.16

O2

0.6

0.6

6.31

H2O

—

2.0

—

Total

9.5

11.5

100.00

Since liquid methanol and air are available at 298.15 K (25°C), their enthalpies are 0 kJ/kmol at reference temperature. Heat capacity constants ni × kmol (ni × ai)1* (ni × ai)2*

CO2 1.0 21.366 37.174

N2 7.9 233.768 245.834

O2 0.6 15.615 11.060

H2O

Total

2.0 64.984 50.317

11.5 335.733 344.385

256 Solutions Manual—Stoichiometry

(ni × bi)1 ´ 103 (ni × b)2 ´ 103 (ni × ci)1 ´ 106 (ni × ci)2 ´ 106 (ni × di)1 ´ 109 (ni × di)2 ´ 109

64.284 23.237 – 41.051 – 7.379 9.8 0.821

– 40.614 25.256 104.145 – 3.201 – 39.247 0.018

7.053 13.030 – 1.405 – 4.983 – 0.337 0.693

0.159 42.564 26.421 – 10.680 – 9.095 0.965

30.882 104.087 88.109 – 26.243 – 38.879 2.497

*

Suffixes 1 and 2 refer to the constants for temperature ranges 298.15 – 1500 K and 1500 – 4000 K, respectively. Enthalpy of flue gas at 1500 K above 298.15 K = 335.733 (1500 – 298.15) + 30.882 ´ 10–3 (15002 – 298.152)/2 + 88.109 ´ 10–6 (15003 – 298.153)/3 – 38.879 ´ 10–9 (15004 – 298.154)/4 = 403 551 + 33 371 + 98 345 – 49 130 = 486 137 kJ Let T be AFT. Enthalpy of flue gas above 1500 K = 344.385 (T – 1500) + 104.087 ´ 10–3 (T2 – 15002)/2 – 26.243 ´ 10–6 (T 3 – 15003)/3 + 2.497 ´ 10–9 (T 4 – 15004)/4 = 638 240 (NCV of 1 kmol CH3OH) – 486 137 = 152 103

By trial and error, T = 1833.27 K or 1560.12°C)

Ans.

EXERCISE 7.23 Basis: Firing rate = 3000 Nm3/h GCV = 38 574 kJ/Nm3 Total heat input = 3000 ´ 38 574 = 1.157 22 ´ 108 kJ/h º 32 145 kW Total enthalpy of steam at 65 bar a and 673 K = 3170.8 kJ/kg (Table AV.3) Enthalpy of water at 373 K (100 °C) = 419.06 kJ/kg (Table AV.1) Enthalpy, supplied for steam raising = 3170.8 – 419.06 = 2751.74 kJ/kg

Combustion

257

Steam generation = 1.157 22 ´ 108 ´ 0.83/2751.74 = 34 905 kg/h or 34.905 t/h Ans. New basis: 100 kmol SNG Air requirement Component

kmol

katom C

CH4 H2 CO CO2

96.59 1.29 0.22 1.90

96.59 — 0.22 2.00

193.18 1.29 — —

kmol H2

kmol O2 — — 0.11 1.90

Total

100.00

98.71

194.47

2.01

194.47 0.22 – 2.01 + 2 2 194.045 kmol 194.045 ´ 1.1 213.345 kmol 213.345 – 194.045 = 19.4 kmol 79 ´ 213.345/21 802.584 kmol 0.018 ´ (213.345 + 802.584) 18.287 kmol 213.345 + 802.584 + 18.287 1034.186 kmol

Theoretical O2 requirement = 98.71 +

= Actual O2 supply = = Excess O2 = N2, entering through air = = Moisture entering with air = = Total moist air = = Specific combustion air requirement = 1034.186/100 = 10.342 kmol/kmol SNG Ans.(b) Total moisture in flue gas = 194.47 + 18.287 = 212.757 kmol Composition of flue gas mixture Component

kmol

CO2 O2 N2 H2O

98.710 19.400 802.584 212.757

Total

1133.451 (wet) 920.694 (dry)

Wet gas analysis mole % 8.71 1.71 70.81 18.77 100.00

Take reference temperature, T0 = 298.15 K (25°C)

Orsat gas analysis mole % 10.72 2.11 87.17 — 100.00

258 Solutions Manual—Stoichiometry

Heat capacity equation constants ni (ni × ai)1* (ni × ai)2* (ni × bi)1 ´ 103 (ni × bi)2 ´ 103 (ni × ci)1 ´ 106 (ni × ci)2 ´ 106 (ni × di)1 ´ 109 (ni × di)2 ´ 109

CO2 0.0871 1.861 3.238 5.599 2.024 – 3.576 – 0.643 0.854 0.072

N2

O2

H2O

Total

0.7081

0.0171

0.1877

1.0000

20.953 22.035 – 3.640 2.264 9.335 – 0.287 – 3.518 0.002

0.445 0.315 0.201 0.371 – 0.040 – 0.142 – 0.010 0.020

6.099 4.722 0.015 3.995 2.480 – 1.002 – 0.854 0.091

29.358 30.310 2.175 8.654 8.199 – 2.074 – 3.528 0.185

*

Suffixes 1 and 2 refer to the constants for temperature ranges 298.15 to 1500 K and 1500 to 4000 K, respectively. Total heat liberated by combustion of SNG (based on NCV) 77 899.3 ´ 1000 = 778 993 kJ/kmol SNG = 100 Enthalpy of flue gas at 1500 K above 298.15 K, H1 = 1133.451 [29.358 (1500 – 298.15) + 2.175 ´ 10–3 (15002 – 298.152)/2 + 8.199 ´ 10–6 (15003 – 298.153)/3 – 3.528 ´ 10–9 (15004 – 298.154)/4] = 42 327.4 ´ 1133.451 = 47 976 034 kJ Let T be the flame temperature. Enthalpy of flue gas above 1500 K (upto T) = 778 993 ´ 100 – 47 976 034 = 29 923 266 kJ/100 kmol SNG Hence 1133.451 [30.310 (T – 1500) + 8.654 ´ 10–3 (T2 – 15002)/2 – 2.074 ´ 10–6 (T3 – 15003)/3 + 0.185 ´ 10–9 (T4 – 15004)/4] = 29 926 186

Solving by Matcad T = 2155.25 K or 1882.1°C

Ans.(c)

Combustion

259

EXERCISE 7.24 Basis: 100 kmol SNG Let x kmol flue gas be recycled having GCV = 0 kJ/kmol Total mixed gas = 100 + x kmol GCV balance (100 + x) 288 200 = 864 589 ´ 100 or x = 200.0 kmol Let a be kmol of O2 in recycle flue gas and b be kmol of H2O in recycle flue gas. (CO2 + N2) in flue gas recycle = 200 – a – b kmol Theoretical O2 demand = 194.045 – a kmol Actual O2 supply = 1.1 (194.045 – a) = 213.345 – 1.1 a kmol Excess O2 = 213.345 – 1.1 a – (194.045 – a) = 19.4 – 0.1 a kmol 79 N2 entering with O2 = (213.345 – 1.1 a) 21 = 802.584 – 4.138 a Total dry air = 213.345 – 1.1 a + 802.584 – 4.138 a = 1015.929 – 5.238 a kmol Moisture, entering with air = 0.018 (1015.929 – 5.238 a) = 18.287 – 0.0943 a kmol Total moisture in flue gas = 194.47 + 18.287 – 0.0943 a + b = 212.757 – 0.0943 a + b kmol Flue Gas Composition: Component

kmol 98.71 802.584 – 4.138 a 212.757 – 0.0943 a + b 19.400 – 0.1 a 200 – a – b – from recycled flue gas

CO2 N2 H2O O2 CO2 + N2 Total

1333.451 – 5.3323 a

19.4 – 0.1 a a = 200 1333.451 – 5.3323 a 5.3323 a2 – 1353.451 a + 3880 = 0 Solving for a, a = 2.9 kmol (only positive root) º 1.45% in flue gas H2O balance: b 212.757 – 0.0943 a + b = 200 1333.451 – 5.3323 a Substituting value of a, O2 balance:

b 212.484 + b = 200 1317.987

260 Solutions Manual—Stoichiometry

or b = 38.01 kmol or 19.005% in flue gas (CO2 + N2) in recycle flue gas = 200 – 2.9 – 38.01 = 159.09 kmol It will be safe to assume CO2/N2 ratio, same as that of combustion gases. Check: Theoretical O2 demand = 194.045 – 2.9 = 191.145 kmol Actual O2 supply = 191.145 ´ 1.1 = 210.260 kmol Excess O2 = 210.260 – 191.145 = 19.115 kmol 79 ´ 210.260 N2 entering alongwith O2 = 21 = 790.978 kmol Total dry air = 210.260 + 790.978 = 1001.238 kmol Moisture in air = 1001.238 ´ 0.018 = 18.022 kmol Total moisture in flue gas = 194.47 + 18.022 + 38.01 = 250.502 kmol Component CO2 O2 N2 H2O

kmol 98.71 + 17.66 = 116.37 19.115 790.978 + 141.43 = 932.408 250.502

Total

Recycle ratio = Heat Capacity Data: Heat capacity constants ni, kmol (ni × ai)1* (ni × ai)2* (ni × bi)1 ´ 103 (ni × bi)2 ´ 103 (ni × ci)1 ´ 106 (ni × ci)2 ´ 106 (ni × di)1 ´ 109 (ni × di)2 ´ 109 *

1318.395 (wet) 1067.893 (dry)

Wet analysis mole % 8.83 1.45 70.72 19.00

Dry analysis mole % 10.90 1.79 87.31 —

100.00

100.00

kmol 200 = 0.1517 kmol 1318.398

Ans. (a)

CO2

N2

O2

H2O

Total

0.0883 1.887 3.282 5.676 2.052 – 3.625 – 0.652 0.865 0.073

0.7072 20.927 22.007 – 3.636 2.261 9.323 – 0.287 – 3.513 0.002

0.0145 0.377 0.267 0.170 0.315 – 0.034 – 0.120 – 0.008 0.017

0.1900 6.173 4.780 0.015 4.044 2.510 – 1.015 – 0.864 0.092

1.0000 29.364 30.336 2.225 8.672 8.174 – 2.074 – 3.520 0.184

Suffixes 1 and 2 refer to the constants for temperature ranges 298.15 – 1500 K and 1500 – 4000 K, respectively.

Combustion

261

Enthalpy of recycle flue gas at 448.15 K above 298.15 K, H¢ = 200 [29.364 (448.15 – 298.15) + 2.225 ´ 10–3 (448.152 – 298.152)/2 + 8.174 ´ 10–6 (448.153 – 298.153)/3 – 3.52 ´ 10–9 (448.154 – 298.154)/4] = 200 [4400.2 + 124.4 + 172.8 – 28.5] = 200 ´ 4668.9 = 933 780 kJ Enthalpy of total flue gas = Heat of combustion + H¢ = 77 899 300 + 933 780 = 78 833 080 kJ/1318.395 kmol flue gas º 59 794.7 kJ/kmol flue gas Enthalpy of total flue gas upto 1500 K above 298.15 K H1 = 29.364 (1500 – 298.15) + 2.225 ´ 10–3 (15002 – 298.152)/2 + 8.174 ´ 10–6 (15003 – 298.153)/3 – 3.52 ´ 10–9 (15004 – 298.154)/4 = 35 291.1 + 2404.2 + 9123.5 – 4448.0 = 42 370.8 kJ/kmol flue gas Enthalpy of flue gas above 1500 K= 59 794.7 – 42 370.8 = 17 423.9 kJ/kmol Hence 30.336 (T – 1500) + 8.672 ´ 10–3 (T2 – 15002)/2 – 2.074 ´ 10–6 (T3 – 15003)/3 + 0.184 ´ 10–9 (T 4 – 15004)/4 = 17 423.9 Solving for T by Mathcad, T = 1935.3 K or 1662.15°C Ans. (c) Combustion air requirement 1019.26 æ 1001.238 +18.022 ö =ç = ÷ è ø 100 100 = 10.1926 kmol/kmol fuel

Ans. (b)

EXERCISE 7.25 Basis: 100 kmol dry blue gas N2 content = 2.4% which comes from air. O2 entering with N2 = 21 ´ 2.4/79 = 0.638 kmol Total O2, available in blue gas, is calculated in the table given below: Gas

kmol

kmol C

kmol O2

kmol H2

H2 CO CO2 N2 O2 CH4

48.5 44.2 2.3 2.4 1.0 1.6

— 44.2 2.3 — — 1.6

— 22.1 2.3 — 1.0 —

48.5 — — — — 3.2

100.0

48.1

25.4

51.7

Total

262 Solutions Manual—Stoichiometry

O2 appeared in blue gas after decomposition of steam = 25.4 – 0.638 = 24.762 kmol Total steam decomposed = 24.762 ´ 2 = 49.524 kmol H2, available from decomposition = 49.524 kmol Net hydrogen from fuel = 51.7 – 49.524 = 2.176 kmol º 4.352 kg Carbon content of gas = 48.1 kmol º 577.2 kg New basis: 100 kg coal Total combustibles = FC + VM = 26.9 + 52.9 = 79.8 kg Refuse = 17.5/(1.0 – 0.19) = 21.605 kg Carbon content of refuse = 21.605 – 17.5 = 4.105 kg Thus coal used for production of 100 kmol blue gas = (577.2 + 4.352) ´ 100/(79.8 – 4.105) = 768.283 kg Carbon used up = 577.2 ´ 75.695/(577.2 + 4.352) = 75.129 kg Total carbon in fuel = 75.129 + 4.105 = 79.234 kg or mass % Ans.(c) Net hydrogen = 79.8 – 79.234 = 0.566 kg or 0.57 mass % Ans.(c)

steam decomposed (49.524 ´ 18.015) = coal charged 768.283 = 1.16 kg/kg Dry blue gas produced = 100 ´ 100/768.283 = 13.02 kmol/100 kg coal

Ratio,

Ans. (b) Ans. (a)

EXERCISE 7.26 Basis: 100 kmol dry blue gas Gas

kmol

kmol C

kmol O2

kmol H2

H2 CH4 CO CO2 N2

49.0 0.8 41.0 4.7 4.5

— 0.8 41.0 4.7 —

— — 20.5 4.7 —

49.0 1.6 — — —

Total

100.0

46.5

25.2

50.6

As N2 comes from air, O2 supply from air = 4.5 ´ 21/79 = 1.196 kmol Unaccounted O2 = 25.2 – 1.196 = 24.004 kmol Unaccounted O2 is available from decomposition of steam. Steam decomposed = 24.004 ´ 2 = 48.008 kmol steam decomposed 48.008 = = 0.48 kmol/kmol Ratio, coal charged 100

Combustion

263

H2 produced by decomposition of steam = 48.008 kmol Net H2 from coal = 50.6 – 48.008 = 2.592 kmol New basis: 100 kmol dry carburetted water gas Gas

kmol

kmol C

kmol O2

kmol H2

H2 CH4 CO CO2 C3H6 O2 N2

37.0 14.0 30.5 5.6 7.0 0.4 5.5

— 14.0 30.5 5.6 21.0 — —

— — 15.25 5.6 — 0.4 —

37.0 28.0 — — 21.0 — —

Total

100.0

71.1

21.25

86.0

O2 from air = 21 ´ 5.5/79 = 1.462 kmol It is assumed that fuel oil, used for carburettion, does not contain N2. O2 provided from decomposition of steam = 21.25 – 1.462 = 19.788 kmol Steam decomposed = 19.788 ´ 2 = 39.576 kmol H2, produced by decomposition = 39.576 kmol Net H2 from oil = 86.0 – 39.576 = 46.424 kmol steam decomposed 39.576 Ratio, = = 0.396 kmol/kmol dry carburetted gas 100 Let x be kmol of dry blue gas and y be kg of fuel oil sprayed for making 100 kmol dry carburetted water gas. 80% C of fuel oil appears in final gas. Carbon, available for carburettion = 0.87 ´ 0.8/12 = 0.058 kmol/kg fuel oil Similarly, hydrogen available for carburettion = 0.13 ´ 0.9/2 = 0.0585 kmol/kg fuel oil Balance of carbon: 0.465 x + 0.058 y = 71.1 Balance of net hydrogen: 0.025 92 x + 0.0585 y = 46.424 Solving the simultaneous equations, x = 57.03 kmol blue gas (dry) y = 768.51 kg fuel oil sprayed Total steam, decomposed for producing 100 kg dry carburetted gas = 39.576 kmol Steam, decomposed for producing 57.03 kmol dry blue gas = 57.03 ´ 0.480 08 = 27.374 kmol Steam, decomposed during carburettion = 39.576 – 27.374

264 Solutions Manual—Stoichiometry

= 12.202 kmol º 219.64 kg

additional steam decomposed = 219.64/100 dry carburetted gas = 2.196 kg/kmol dry carburetted water gas 100 Ratio, = dry blue water gas 57.03 = 1.754 kmol/kmol fuel oil used for carburetion 768.5 Ratio, = dry blue gas 57.03 = 13.48 kg/kmol

Ratio,

Ans. (c)

Ans. (b)

Ans. (a)

EXERCISE 7.27 Basis: 100 kmol producer gas Gas

kmol

kmol C

kmol O2

kmol H2

H2 CH4 C2H 4 CO CO2 N2

10.5 2.1 0.4 22.0 7.7 57.3

— 2.1 0.8 22.0 7.7 —

— — — 11.0 7.7 —

10.5 4.2 0.8 — — —

Total

100.0

32.6

18.7

15.5

Mass, kg 391.2 598.4 31 New basis: 100 kg coal Carbon content of refuse = [18.5/(1 – 0.062)]0.062 = 1.223 kg Carbon utilised = 62 – 1.223 = 60.777 kg Thus coal, used for 100 kmol dry producer gas = 391.2/0.607 77 = 643.64 kg N2 from coal = 643.64 ´ 0.06 = 38.62 kg º 1.38 kmol O2 from coal = 643.64 ´ 0.093 = 59.86 kg º 1.87 kmol N2 from air = 57.3 – 1.38 = 55.92 kmol O2 from air = 21 ´ 55.92/79 = 14.86 kmol O2 accounted for = 14.86 + 1.87 = 16.73 kmol O2 from decomposition for steam = 18.7 – 16.73 = 1.97 kmol Steam decomposed = 1.97 ´ 2 = 3.94 kmol H2 produced due to decomposition of steam = 3.94 kmol (Table 6.13) Vapour pressure of water 298.15 K (25°C), pw = 3.166 kPa Humidity of producer gas = 3.166/(106.7 – 3.166) = 0.0306 kmol/kmol dry gas Water vapours with producer gas = 0.306 ´ 100

Combustion

265

= 3.06 kmol H2 from fuel = 15.5 – 3.94 = 11.56 kmol º 23.12 kg Check from fuel analysis: H2 from fuel = 643.64 ´ 0.036 = 23.17 kg Moist producer gas: Molar mass, Mi

mass, kg (ni × Mi)

H2 CH4 C2H4 CO CO2 N2 H2O

Gas

kmol, ni 10.5 2.1 0.4 22.0 7.7 57.3 3.06

2 16 28 28 44 28 18

21.0 33.6 11.2 616.0 338.8 1604.4 55.08

Total

103.06

—

2680.08

moist producer gas 2680.08 = coal 643.64 = 4.16 kg/kg Ans. (a) Dry air = 14.86 ´ 32 + 55.92 ´ 28 = 2041.28 kg Absolute humidity of air at 311 K (38°C) DB and 299 K (26°C) WB, = 16.4 g/kg dry air (Ref. Fig. 6.15) º 0.0263 kmol/kmol dry air Moisture from air = 16.4 ´ 2041.28/1000 = 33.477 kg º 1.86 kmol moist air (2041.28 + 33.477) = = 3.223 kg/kg Ratio, coal 643.64 Ans. (b) Steam supplied = 3.94 + 3.06 – 1.86 = 5.14 kmol º 92.52 kg Ratio, steam/coal = 92.52/643.64 = 0.144 kg/kg Ans. (c) Energy balance: Basis: 100 kmol dry producer gas Base temperature: 298.15 K (25°C) Enthalpy of steam at 4 bar a and 623 K (350°C) = 3170 kJ/kg (Ref. Table – AV.3) Enthalpy of steam over 298.15 K (25°C) = 3170 – 104.8 = 3065.2 kJ/kg Total enthalpy of supplied steam = 92.52 ´ 3065.2 = 283 592 kJ Humid heat of air, CH = 1.006 + 1.84 ´ 0.0164 = 1.036 kJ/(kg dry air × K) Total enthalpy of air = 2041.28 ´ 1.036 (311 – 298) = 27 492 kJ Calorific value of coal on as-received basis = 27 800 (1 – 0.185) = 22 657 kJ/kg (dry)

Ratio

266 Solutions Manual—Stoichiometry

Enthalpy of coal = 22 657 ´ 643.64 = 14 582 951 kJ Component

kmol × ni

C°mp equation constants

H2 CH4 C2 H 4 CO CO2 N2 H2O

10.5 2.1 0.4 22.0 7.7 57.3 3.06

ni × ai 206.407 40.424 1.650 638.609 164.514 1695.559 99.426

Total

103.06

2846.589

ni × bi ´ 103 103.978 109.438 62.009 – 61.963 494.988 – 294.579 0.244 414.115

ni × ci ´ 106 ni × di ´ 109 3036.805 – 7786.180 25.143 – 23.770 – 32.618 6.790 256.161 – 103.539 – 316.090 75.459 755.380 – 284.666 40.425 – 13.915 3765.206

– 8129.821

Enthalpy of producer gas at 1200.15 K over 298.15 K = 2846.589 (1200.15 – 298.15) + 414.115 ´ 10–3 (1200.152 – 298.152)/2 + 3765.206 ´ 10–6 (1200.153 – 298.153)/3 – 8129.821 ´ 10–9 (1200.154 – 298.154)/4 = 2567 623 + 279 775 + 2135 545 – 4198 471 = 784 472 kJ Sensible heat of producer gas over 298.15 K (25 °C) = 784 472 kJ Latent heat of water vapours at 298.15 K (25 °C) = 2442.8 kJ/kg Heat, lost in evaporating formed water of coal = 11.56 ´ 18 ´ 2442.8 = 508 298 kJ Calorific value of producer gas: Gas

kmol, ni

GCVi, kJ/mol

ni × GCVi, kJ

H2 CH4 C2H4 CO

10.5 2.1 0.4 22.0

285.83 890.65 1411.2 282.98

3001 215 1870 365 564 480 6225 560

Total

35.0

—

11 661 620

Enthalpy balance is given in Table –7.54 in the text.

Ans.

EXERCISE 7.28 Basis: 100 kg waste liquor to be incinerated Let x kmol of natural gas be required as the stabilizing fuel. Water liquor incineration: Component

kg

kmol

O2 requirement, kmol

Carbon Hydrogen Sulphur Oxygen Water

7.5 0.5 5.7 12.5 58.8

0.625 0.250 0.178 0.391 3.267

0.625 0.125 0.178 (–) 0.391 .Nil

Total

85.0

4.711

0.537

Combustion

267

New basis: 100 kmol natural gas (Ref. Table 7.7 in the text) Components

kmol

CH4 C2H8 C3H8 C4H10 (total) CO2

89.4 5.0 1.9 1.0 0.7

Total

98.0

O2 requirement kmol 178.8 17.5 9.5 6.5 —

CO2 production kmol 89.4 10.0 5.7 4.0 0.7

212.3

109.8

H2O production kmol 178.8 15.0 7.6 5.0 — 206.4

Total theoretical O2 requirement for 100 kg waste liquor incineration = 0.537 + (212.3 x/100) = 0.537 + 2.123 x kmol Total heat liberated at 298.15 K (25 °C) = 3520 ´ 100 + (855 320 x/100) = 352 000 + 8553.20 x kJ Actual O2 supply = 1.15 (0.537 + 2.123 x) = 0.6176 + 2.4415 x kmol Excess O2, appearing in flue gas = 0.0806 + 0.3185 x kmol N2 supplied through air = 3.762 (0.6176 + 2.4415 x) = 2.3234 + 9.1849 x kmol Flue gas: Component

kmol 0.625 + 1.098 x = 0.625 + 1.098 x 3.267 + 0.25 + 2.064 x = 3.517 + 2.064 x 0.178 0.02 x + 2.3234 + 9.1849 x = 2.3234 + 9.2049 x 0.0806 + 0.3185 x

CO2 H2O SO2 N2 O2 Total

6.7240 + 12.6854 x

Absolute enthalpies: (Ref. Table 5.22 from text) Component

(H° – H°o + DH°f ), kJ/kmol at 1200 K

CO2 H2O SO2 N2 O2

– 339 593 – 194 796 – 303 331 + 36 699 + 38 399

475 K

298.15 K

– 376 906 – 223 245 – 340 659 + 13 745 + 13 900

– 384 158 – 229 252 – 347 992 + 8 613 + 8 597

Absolute enthalpy different, kJ 1200 K – 475 K 37 313 28 449 37 328 22 954 24 499

1200 K – 298.15 44 565 34 456 44 661 28 086 29 802

Enthalpy of flue gas at 1200 K above 298.15 K = (0.625 + 1.098 x) 44 565 + (3.517 + 2.064 x) 34 456 + 0.178 ´ 44 661 + (2.3234 + 9.2049 x) 28 086

268 Solutions Manual—Stoichiometry

+ (0.0806 + 0.3185 x) 29 802 = 224 642 + 388 071 x kJ However, this enthalpy rise is solely due to heat liberation due to natural gas combustion: 224 642 + 388 071 x = 352 000 + 8553.2 x or x = 0.3356 kmol NG/100 kg waste liquor º 7.522 Nm3 NG/100 kg waste liquor Flue gas analysis: Substituting the value of x, following table is obtained. Component CO2 H2O SO2 N2 O2 Total

kmol 0.9935 4.2097 0.1780 5.4126 0.1875 10.9813 (wet) 6.7716 (dry)

Wet gas analysis, % 9.05 38.34 1.62 49.29 1.70 100.00

Furnace pressure p = Partial pressure of water, pH O = 2 = From Steam Tables (Table – 6.13), Water dew point = A = ln (pH O) = 2 Partial pressure of SO2, pSO = 2 = B = ln (pSO ) = 2

Dry gas analysis, % 14.67 — 2.63 79.93 2.77 100.00

Orsat analysis, % 17.30 — — 79.93 2.77 100.00

103.5 kPa 103.5 ´ 0.3834 39.682 kPa 348.85 (75.7°C) ln (39.682) = 3.6809 103.5 ´ 0.0162 1.6767 kPa ln (1.6767) = 0.5168

Ans. (b–i)

Substituting the values of A and B, sulphurous acid (H2SO3) dew point can be calculated. (1000/TDP) = 3.5752 – 0.1845 ´ 3.6809 – 9.333 ´ 10–4 ´ 0.5168 – 9.13 ´ 10–4 ´ 0.5168 ´ 3.6809 = 3.5752 – 0.6791 – 0.0005 – 0.0017 = 2.8939 Sulphurous acid dew point TDP(SO ) = 1000/2.8939 2 = 345.55 K or 72.4°C Ans. (b-ii) For liquid fuel (i.e. waste liquor), C/H mole ratio = 0.625/0.25 Excess air = 15% From Fig. 7.3, sulphuric and dew point TDP(SO3) = 426 K = 153°C Ans. (b-iii) Enthalpy available for steam generation = Enthalpy of flue gas at 1200 K above 475 K = 0.9935 ´ 37 313 + 4.2057 ´ 28 449 + 0.178 ´ 37 328 + 5.4126 ´ 22 954 + 0.1875 ´ 24 499 = 292 311 kJ

Combustion

269

Enthalpy of saturated steam at 15 bar g (16.01 bar a) = 2791.8 kJ/kg Enthalpy of water at 350 K (77°C) = 322.33 kJ/kg Enthalpy supplied for steam generation = 2791.8 – 323.33 = 2469.47 kJ/kg steam Enthalpy to be supplied for 5000 kg/h steam generation = 2469.47 ´ 5000 = 12 347 350 kJ/h º 3429.82 kW Boiler efficiency = 85% 12 347 350 ´ 100 = 4969.5 kg/h Waste liquor firing rate = 292 311 ´ 0.85 Ans. (a-i) Stabilizing fuel (NG) firing rate = 0.3356 ´ 49.695 = 16.6775 kmol/h º 373.81 Nm3/h Ans. (a-ii)

EXERCISE 7.29 Combustion reaction:

F H

Cn Hm + n + Basis: 1 mol fuel

I K

m m O2 = n CO2 + H2O 2 4

m mol 4 m Let actual oxygen supply = n + + y mol 4

Theoretic oxygen required = n +

F H

Total N2 = 3.762 n +

I K

m + y from air and CO2 = n mol 4

==

y 4.762 y + [n + (m /4)]3.762 + n

y=

= (1 - 4.762 = )

% excess air =

100 y [n + ( m / 4)]

Where excess O2 = y mol Solving the equation for y,

By definition,

=

LMF n + m I3.762 + nOP NH 4 K Q LM N

19.048 + 3.762r 100 = 4+r (1 - 4.762 =)

OP Q

270 Solutions Manual—Stoichiometry

where

r = m/n 4 For methane, r = = 4 and = = 5/100 = 0.05 1 Substituting the values in the equation, % excess air = =

LM N

19.048 + 3.762 ´ 4 100 ´ 0.05 4+4 [1 - ( 4.762 ´ 0.05)] 5(19.048 + 15.048) = 27.97% (1 - 0.2381)8

Ans.

EXERCISE 7.30 Basis: 1 mol fuel Inerts content of fuel = a mol Combustibles = (1 – a) mol Combustion reaction:

F H

(1 – a)Cn Hm + (1 – a) n +

I K

m m O2 = (1 – a)n CO2 + (1 – a) H2O 2 4

F H

Theoretical O2 required = (1 – a) n + In flue gases,

I K

m + y mol 4

O2 = y mol i.e. excess oxygen

F H

I K

m + y mol 4 CO2 = (1 – a)n mol, Inerts = a mol Mole fraction of oxygen in flue gases, y == 4.762 y + (1 - a ) n + ( m / 4) 3.762 + (1 - a )n + a Solving the equation for y,

Total N2 = 3.762(1 – a) n +

y=

F H

% excess air =

I K

= m [3.762 (1 – a) n + + n(1 – a) + a] (1 - 4.762 =) 4

By definition,

100 y (1 - a)[n + ( m / 4)]

LM N

=

100= 19.048 + 3.762 r 4a + (1 - 4.762=) (4 + r ) n(1 - a)( 4 + r )

=

100= (1 - 4.762=)

LM19.048 + 3.762r + c OP Q N (4 + r )

where cf = correction factor

f

OP Q

OP Q

Combustion

= Now when a = cf = % excess air = = =

4a n(1 a)( 4 + r ) 0.05 4 ´ 0.05/(1 ´ 0.95 ´ 8) = 0.0263 5 19.048 + 15.048 + 0.0263 (1 - 0.2381) 8 5 ´ 34.307/(0.7619 ´ 8) 28.14

LM N

OP Q

EXERCISE 7.31 Basis: 1 mol fuel Combustion reactions can be written as under: m m x1 Cn Hm + x1 n1 + 1 O2 = x1 n1 CO2 + x1 1 H2O 1 1 2 4 m2 m x2 Cn Hm + x2 n2 + O2 = x2 n2 CO2 + x2 1 H2O 2 2 2 4 mi mi O2 = xi ni CO2 + xi H2O x1 Cni Hmi + xi ni + 4 2 mj i mol Theoretical O2 required = å xj n j + 4 1 mj i Actual O2 supply = å xj n j + + y mol 4 1 Mole fraction of O2 in flue gases, = y == ( 4.762 y + 3.762) é i æ ù i mj ö + xj ú + å xjnj êå x j ç n j + ÷ 4 ø ëê 1 è ûú 1 Solving the equation for y,

F H F H F H

I K

I K

I K

FG H FG H

IJ K IJ K

i é ù mj ö i æ = + å xi n j ú ê3.762å x j ç n j + ÷ (1 - 4.762=) ëê 4 ø 1 è 1 ûú By definition,

y=

% excess air =

271

100 y mj ö æ å xj çnj + ÷ 4 ø è 1 i

Substituting the value of y and solving the equation, i é ù xjnj å ê ú 100= 1 % excess air = ê3.762 + i ú i (1 - 4.762=) ê ånj xj + å xjmj ú 1 1 ëê ûú

Ans.

272 Solutions Manual—Stoichiometry

=

LM N

19.048 + 3.762 rm 100= 4 + rm (1 - 4.762=) i

where, rm =

åx m j

1 i

OP Q

i

j

=

å x j n j rj 1

i

å xjnj

å xjnj

and rj = mj / nj

1

1

EXERCISE 7.32 Basis: 1 mol fuel Combustion reactions can be written as follows: m x1(1 – a1) Cn Hm + x1 (1 – a1) n1 + 1 O2 1 1 4 m1 = x1 (1 – a1)n1 CO2 + x1 (1 – a1) H2O 2 m x2(1 – a2) Cn Hm + x2 (1 – a2) n2 + 2 O2 2 2 4 m2 = x2 (1 – a2)n2 CO2 + x2(1 – a2) H2O 2 m xi (1 – ai)Cni Hmj + xi (1 – ai) n4 + i O2 4 m = xi (1 – ai) ni CO2 + xi (1 – a2) i H2O 2

F H

I K

F H

I K

F H

I K

i

Theoretical O2 required = å x j (1 – aj) [nj + (mj /4)] mol 1 i

Let actual O2 supply = å x j (1 – aj) [nj + (mj /4)] + y mol 1

Mole fraction of O2 in flue gases, ==

Solving for y, y=

y

é ì üù ê4.762 y + 3.762 íå x j (1 - a j )[n j + (m j /4)]ýú î1 þú ê 1 ê i ú ê +å (1 - a j )n j x j + å x j a j ú i ë 1 û i

= é ìi ü 3.762 íå x j (1 - a j )[n j + (m /4)]ý ê (1 - 4.762=) ë î1 þ i i ù + å x j nj (1 – aj) + å x j a j ú 1 û 1

Combustion

273

By definition, 100 y

% excess air =

i

å x j (1 - a j )[n j + (m j /4)] 1

Substituting the value of y and solving the equation, % excess air

i i é ù å x j n j (1 - a j )å x j × a j ú ê = 1 1 = ê3.762 + i ú (1 - 4.762= ) ê å x j (1 - a j )[n j + (m j /4)] ú êë úû 1 i é ù xjaj å ê ú = 1 1 = + i ê3.762 + ú (1 - 4.762= ) ê 1 + rm ú + x a n m (1 )[ ( /4)] å j j j j êë úû 1 3.762 rm + 19.048 = + c fm = 4 + rm (1 - 4.762=)

LM N

i

where rm =

å x j m j (1 - a j ) i

1

å x j n j + (1 - a j ) 1

and cfm = åi 4 x a j j 1

OP Q

i

=

å x j n j rj (1 - a j ) 1

i

å x j n j (1 - a j ) 1

i

å [n j x j ][(1 - a j )(4 + rj )] 1

EXERCISE 7.33 Fuels Firing in Reformer Furnace Fuel Firing rate kmol/h mole fraction xi Natural gas 375 0.590 Tail gas 158 0.249 Naphtha 102.3 0.161 rm = = Cfm =

ai 0.028 0.820 0.005

ri 2.75 5.55 1.75

[(1 - 0.28) 0.59 ´ 2.75 + (1 - 0.82) 0.249 ´ 5.55 + (1 - 0.005) 0.161 ´ 1.75] [(1 - 0.28) 0.59 + (1 - 0.82) 0.249 + (1 - 005) 0.161] (1.1682 + 0.2488 + 0.2803) 1.6973 = 2.695 = (0.4248 + 0.0448 + 0.1602) 0.6298 [4 ´ 0.59 ´ 0.028 + 4 ´ 0.249 ´ 0.82 + 4 ´ 0.161 ´ 0.005] [0.59 (1 - 0.028)(4 + 2.75) + 0.249 (1 - 0.82) (4 + 5.55) + 0.161 (1 - 0.005) (4 + 1.75)]

274 Solutions Manual—Stoichiometry

(0.0661 + 0.8167 + 0.0032) 0.886 = (3.871 + 0.428 + 0.9211) 5.2201 = 0.1697

=

Excess air =

é (19.048 + 3.762 ´ 2.695) ù 100 ´ 0.035 + 0.1697 ú ê (1 - 4.762 ´ 0.035) ë (4 + 2.695) û

3.5 é 29.1866 ù + 0.1697 ú ê 0.8333 ë 6.695 û = 19.02 %

=

Ans.

8

Stoichiometry and Industrial Problems EXERCISE 8.1 Basis: 100 L/min 3 N potassium solution, entering the tower Concentration of K in solution = 39 ´ 3 = 117 g/L Ingoing solution: K in K2CO3 form = 0.7 ´ 117 = 81.9 g/L K2CO3 in solution = 138 ´ 81.9/78 = 144.9 g/L K in KHCO3 form = 0.3 ´ 117 = 35.1 g/L KHCO3 in solution = 100 ´ 35.1/39 = 90 g/L Outcoming solution : K in K2CO3 form = 0.32 ´ 117 = 37.44 g/L K2CO3 in solution = 138 ´ 37.44/78 = 66.24 g/L K in KHCO3 form = 0.68 ´ 117 ´ 100/39 = 204 g/L Ans. (a) Liquor rate = 1.65 L/s = 5940 L/h K2CO3 in entering solution = 144.9 ´ 5940/1000 = 860.7 kg/h KHCO3 in entering solution = 90 ´ 5940/1000 = 534.6 kg/h K2CO3 in outgoing solution = 66.25 ´ 5940/1000 = 393.5 kg/h KHCO3 in outgoing solution = 204 ´ 5940/1000 = 1211.8 kg/h Chemical reaction, accompanying absorption is as follows: K2CO3 + CO2 + H2O = 2 KHCO3 Molar mass 138 44 18 2 ´ 100 K2CO3 converted to KHCO3 = 860.7 – 393.5 = 467.2 kg/h CO2 absorbed = 467.2 ´ 44/138 = 149 kg/h Ans. (b) H2O reacted = 467.2 ´ 18/138 = 60.9 kg/h

276 Solutions Manual—Stoichiometry

Total KHCO3 produced = 467.2 + 149.0 + 60.9 = 677.1 kg/h Total KHCO3 in outgoing solution = 534.6 + 677.1 = 1211.7 kg/h This calculated total KHCO3 tallies with the figure of 1211.8 kg/h, calculated earlier. New basis: 100 kmol of incoming gas mixture CO2 content = 18 kmol Outgoing gas mixture contains 6.9% CO2. Quantity of outgoing gas mixture = 82/(1 – 0.069) = 88.08 kmol CO2 absorbed = 100 – 88.08 = 11.92 kmol CO2 recovery = 11.92 ´ 100/18 = 66.2% Ans. (c) CO2 absorption rate for 1.65 L/s flowrate = 149/44 = 3.386 kmol/h Incoming gas rate = 100 ´ 3.386/11.92 = 28.4 kmol/h Ans. (d)

EXERCISE 8.2 Basis: 100 kmol dry flue gas, leaving boiler furnace. Let x kmol dry air mixes with flue gas. Balance of dry flue gas across RAH: 100 ´ 0.14 = (100 + x) 0.125 x = 12 kmol Thus 0.12 kmol dry air mixes with one kmol dry flue gas. Ans. (a) Dew point of flue gas = 319.3 K (46.3°C) Vapour pressure of water at 319.3 K, pw = 10.242 kPa Total pressure of flue gas, leaving boiler furnace, p = 100 kPa 10.242 Moisture, present in flue gas from boiler, = (100 - 10.242) = 0.114 kmol/kmol dry flue gas Total wet flue gas = 100 + 11.4 = 111.4 kmol Humidity of air at 313 K DB and 295 K WB = 0.0134 kg/kg dry air = 0.0216 kmol/kmol dry air Dry air mixing with flue gas = 12 kmol Wet air mixing with flue gas = 12 (1 + 0.0216) = 12.259 kmol

Stoichiometry and Industrial Problems 277

Wet Flue Gas Before RAH Component

kmol

CO2 O2 N2 H2O Total

C°mp equation constants

14.0 3.0 83.0 11.4

ai 299.12 78.08 2456.04 370.41

bi ´ 103 899.98 35.27 – 426.70 0.91

111.4

3203.65

509.46

ci ´ 106 – 574.71 – 7.03 1094.18 150.60

di ´ 109 137.20 – 1.69 – 412.34 – 51.84

663.04

– 328.67

Combustion Air Component

kmol

O2 N2 Ar H2O Total

C°mp equation constants ai

bi ´ 103

ci ´ 106

di ´ 109

0.21 0.78 0.01 0.0216

5.47 23.08 0.21 0.70

2.47 – 4.01 — —

– 0.49 10.28 — 0.29

– 0.12 – 3.87 — – 0.10

1.0216

29.46

– 1.54

10.08

– 4.09

Reference temperature, T0 = 298.15 K (25°C) Enthalpy of flue gas, entering RAH;

z

523.15

H1 =

(3203.65 + 509.46 ´ 10–3 T + 663.04 ´ 10–6 T 2 – 328.67 ´ 10–9 T 3) dT

298.15

= 788 175 kJ/100 kmol dry flue gas from boiler Enthalpy of flue gas, leaving RAH (excluding air being mixed),

z

433.15

H2 =

(3203.65 + 509.46 ´ 10–3 T + 663.04 ´ 10–6 T 2 – 328.67 ´ 10–9 T 3) dT

298.15

= 467 501 kJ/100 kmol dry flue gas from boiler 12 kmol dry air (with accompanying moisture) mixes with flue gas at 313.15 K and heat upto 433.15 K. Enthalpy of air at 313.15 K,

z

313.15

H3 =

(29.46 – 1.54 ´ 10–3 T + 10.08 ´ 10–6 T 2 – 4.09 ´ 10–9 T 3) dT

298.15

= 447.21 kJ/kmol dry air Enthalpy of air at 433.15 K,

z

433.15

H4 =

(29.46 – 1.54 ´ 10–3 T + 10.08 ´ 10–6 T 2 – 4.09 ´ 10–9 T 3) dT

298.15

= 4057.17 kJ/kmol dry air

278 Solutions Manual—Stoichiometry

Heat transferred in RAH to combustion air H5 = H1 – H2 – 12 (H4 – H3) = 788 175 – 467 501 – 12 (4057.17 – 447.21) = 277 354 kJ/100 kmol dry flue gas from boiler º 2773.5 kJ/kmol dry flue gas, entering RAH

Ans. (b)

100 1.045 = 95.694 kmol wet air º 93.67 kmol dry air Let T be the temperature of combustion air after RAH.

Combustion air required =

z T

(93.67)

(29.46 – 1.54 ´ 10–3 T + 10.08 ´ 10–6 T2 – 4.09 ´ 10–9 T3) dT = 277 354

298.15

Solving by Mathcad, T = 396.95 K or 123.8°C

Ans. (c)

12 ´ 100 93.67 = 12.81% Ans. (d) Instead of RAH, a tubular heater is used, heat transfer duty, H¢5 = 788 175 – 467 501 = 320 674 kJ/100 kmol dry flue gas

Increase in capacity of air blower =

(93.67)

T¢

–3

ò (29.46 – 1.54 ´ 10

T ¢ + 10.08 ´ 10–6 T¢2 – 4.09 ´ 10–9 T¢3) dT = 320 674

298.15

Solving by Mathcad, T ¢ = 412.25 K or 139.1°C

Ans.

EXERCISE 8.3 Basis: 4000 kg/h production rate of 1,3-butadiene Molar feed rate of n-butane = 1.818 ´ 3600/58 = 6545.5/58 = 112.85 kmol/h Butene-1 formed = 112.85 kmol/h H2 produced due to reaction (1) = 112.85 kmol/h Butadiene formed = 4000/54 = 74.07 kmol/h Butene-1 consumed to produce butadiene = 74.07 kmol/h H2 produced due to reaction (2) = 74.07 kmol/h Butene-1 consumed to produced ethylene = 112.85 – 74.07 = 38.78 kmol/h Butene-1 recycled = 6.642 kg/s = 23 911 kg/h = 426.96 kmol/h Ethylene produced = 38.78 ´ 2 = 77.56 kmol/h

Stoichiometry and Industrial Problems 279

Yield of butadiene = 74.07 ´ 100/112.85 = 65.64% Total H2 produced = 112.85 + 74.07 = 186.92 kmol/h

Offgas analysis:

. n i, kmol/h

mole %

H2 C2H4

186.92 77.56

70.67 29.33

Total

264.48

100.00

Component

Composition of hot gases, leaving reactor: . Component n i, kmol/h Butene-1 426.96 1, 3-Butadiene 74.07 Hydrogen 186.92 Ethylene 77.56 Total

mole % 55.77 9.68 24.42 10.13

765.51

100.00

Ans. (a) Enthalpy of reactants: Component

kmol/h . ni

Temp., K (°C)

n-Butane Butene-1

112.85 426.95

920 (647) 920 (647)

Total

539.80

—

(H° – H°o+ DH°f )i kJ/kmol

. n i × (H° – H°o + DH°f )i kW

24 494 131 000

767.82 15 536.24

—

16 304.06

Enthalpy of products: Component

kmol/h . ni

Temp., K (°C)

(H° – H°o+ DH°f )i kJ/kmol

. n i × (H° – H°o + DH°f )i kW

Butene-1 1, 3-Butadiene Hydrogen Ethylene

426.96 74.07 186.92 77.56

810 (537) 810 (537) 810 (537) 810 (537)

110 850 204 386 23 386 105 220

13 146.50 4 205.24 1 214.25 2 266.91

Total

765.50

20 832.90

Heat of reactions, DHr = 20 832.9 – 16 304.06 = 4528.84 kW (endothermic) Ans. (b) At T = 273.15 K: Antoine constants from Table .5.4, A = 3.997 98, B = 941.662 and C = – 32.753 p = 1.2047 bar or 120.5 kPa lv = 22 470

. I F 425 – 27315 H 425 – 268.7 K

0.38

280 Solutions Manual—Stoichiometry

= 22 223 kJ/kmol (H° – H°o + DH°f ) at 273.15 K of 1,3-butadiene = 139 270 kJ/kmol Total heat to be removed for condensation of 1,3-butadiene = (204 386 – 139 270) + 22 223 = 87 339 kJ/kmol Total heat removed = 87 339 ´ 74.07/3600 = 1797 kW

Ans. (d)

EXERCISE 8.4 It can be noticed that the concentration of a component in the effluents changes with respect to time. This is because when the acid is introduced, the bed is full of water which is replaced by acid during the course of time. Till the water is replaced by acid, no appreciable chlorides can be found in the effluents. After sometime (about 10 minutes in this case), chlorides start appearing in the effluents and its concentration reach to a peak value at the end of the acid introduction step. At the start of slow rinse, the bed is full with acidic water, and hence the concentration of chlorides taper out as the slow rinse proceeds. It can be further noticed that FMA is observed in the effluents after a longer period. This is because acid is completely consumed by cations in the beginning. Such an example can be solved by a geometric technique. A graph is plotted (refer Fig. E8.1) of the concentration of chloride and FMA vs. time. Area under the chloride curve from the start of acid introduction step to the end of slow rinse step = 191.67 square units. It is desired to segregate 85% of these chlorides. Chlorides to be segregated = 191.67 ´ 0.85 = 162.92 square units. Careful evaluation of the chloride curve indicates that area, corresponding to 163.60 square units, is represented by a period of 50 minutes; ranging from 20 minutes from the start of acid introduction step. Ans. (a) Area under the chloride curve 20 minutes to 52 minutes (i.e. 32 minutes) from the start of acid introduction step = 107.421 square units Area under the chloride curve from the start of slow rinse period to the end of segregation period = 163.60 – 107.421 = 56.179 square units Volume of concentrated effluent = (32 ´ 63/60) + (18 ´ 54/60) = 33.6 + 16.2 = 49.8 m3

Ans. (b – ii)

Volume of dilute effluents = (20 ´ 63/60) + (27 ´ 54/60) = 21 + 24.3 = 45.3 m3

Ans. (b – ii)

Stoichiometry and Industrial Problems 281 Slow Rinse

Acid Entry 60000 54000 48000

36000 30000

e urv

12000

FM AC

18000

e Curve

24000

Chlorid

Concentration in mg/L

42000

6000 0

10

20

30

40

50 60 70 Time in minutes Fig. E8.1 Solution of Exercise 8.3

80

90

100

Concentration of chlorides: Scales of the graph: x-axis: 2 units = 10 min y-axis: 2 unit = 6000 mg/L (i.e. ppm) Thus 4 square unit = 10 ´ 6000 = 60 000 (min × ppm) Chloride (Cl): Concentrated stream: Acid introduction step: Area = 60 000 ´ 107.421/4 = 1611 315 (min × ppm) Chlorides, eluted in 32 min = 1611 315/32 = 50 354 ppm or g/m3 Chlorides, eluted in acid introduction step, i.e. in 33.6 m3 = 50 354 ´ 33.6 = 1691 881 g

282 Solutions Manual—Stoichiometry

Slow rinse step: Area of chloride elution = 60 000 ´ 56.179/4 = 842 685 (min × ppm) Chlorides, eluted in 18 min = 842 685/18 = 46 816 ppm or g/m3 Chlorides, eluted in slow rinse step, i.e. in 16.2 m3 = 46 816 ´ 16.2 = 758 417 g Total chlorides in concentrated stream = 1691 881 + 758 417 = 2450 298 g º 2450.3 kg Average concentration of chlorides in concentrated stream = 2450 298/49.8 = 49 203 g/m3 or ppm Ans. (c–i) Weak stream: Acid introduction step: Area under the chloride curve upto first 20 min from start of acid introduction = 7.647 sq. units Area = 60 000 ´ 7.647/4 = 114 705 (min × ppm) 114 705 20 = 5732 ppm or g/m3 Chlorides, eluted in acid introduction step, i.e. in 21 m3 = 5732 ´ 21 = 120 377 g Slow rinse step: Area under the chloride curve after 72 min = 19.919 sq. units Chlorides, eluted in (52 + 45 – 70 =) 27 min = 19.919 ´ 60 000/4 = 298 785 (min × ppm) Chlorides, eluted = 298 785/27 = 11 066 ppm or g/m3 Chlorides eluted in 24.3 m3 = 11 066 ´ 24.3 = 268 907 g Total chlorides in weak effluent stream = 120 377 + 268 907 = 389 284 g º 389.3 kg Average concentration of chlorides in weak effluent stream = 389 284/45.3 = 8593 g/m3 or ppm Free Mineral Acidity (FMA): Concentrated stream: Acid introduction step: Area under FMA curve during acid introduction step = 36.232 sq. units

Chlorides, eluted in 20 min =

Stoichiometry and Industrial Problems 283

Area = 36.232 ´ 60 000/4 = 543 480 (min × ppm) FMA, eluted in 32 min = 543 480/32 = 16 984 ppm or g/m3 FMA, eluted in 33.6 m3 = 16 984 ´ 33.6 = 570 654 g Slow rinse step: Area under FMA curve, during slow rinse = 49.885 sq. units Area = 49.885 ´ 60 000/4 = 748 275 (min × ppm) FMA, eluted in 18 min = 748 275/18 = 41 571 ppm or g/m3 FMA, eluted in 16.2 m3 = 41 571 ´ 16.2 = 673 448 g Total FMA in concentrated stream = 570 654 + 673 448 = 1244 102 g º 1244.1 kg Average concentration of FMA in concentrated stream = 1244 102/49.8 = 24 982 g/m3 or ppm Ans. (c – ii) Weak stream There is no weak stream during acid introduction step. Slow rinse step: Area under FMA curve during slow rinse = 15.631 sq. units Area = 15.631 ´ 60 000/4 = 234 465 (min × ppm) FMA, eluted in 27 min = 234 465/27 = 8684 ppm or g/m3 FMA, eluted in 24.3 m3 = 8684 ´ 24.3 = 211 019 g º 211.0 kg Average concentration of FMA in weak effluent stream = 211 019/45.3 = 4658 g/m3 or ppm Chlorides curves give total acid introduced Total HCl (100%) consumed during regeneration = (2450 298 + 389 284) 36.5/(36.5 ´ 1000) = 2919.48 kg as 100% HCl º 9.268 t as 31.5% HCl

Ans.(d)

Total FMA = (673 448 + 211 019)/1000 = 884.47 kg Excess acid = 884.47 ´ 100/(2919.48 – 884.47) = 43.5% Ans. (e)

284 Solutions Manual—Stoichiometry

EXERCISE 8.5 Concentrations of all hydrocarbons, except ethane, are above the higher limit. By pressure purging, it is required to bring down the concentrations below the lower limits. It may further be noted that butane (total) concentration is 73.6% by volume (highest amongst all hydrocarbons) and its lower limit is 1.9% by volume (lowest). Hence, if butane concentration is brought down below 1.9%, bullet will be safe for hotwork. p1 = 0.25 bar g = 1.263 25 bar a p2 = 2.5 bar g = 3.513 25 bar a c0 = 73.6% by vol. c1 =

1.263 25 ´ 73.6 = 26.46% after first cycle of purging 3.513 25

Similarly, after repeated purgings c2 = 0.359 57 ´ 26.46 = 9.518% c3 = 0.359 57 ´ 9.518 = 3.422% c4 = 0.359 57 ´ 3.422 = 1.230% which is less than 1.9%. Thus 4 cycles of pressure purging are required. Similar calculations can be made for other hydrocarbons. Gas

Ans. (a)

Concentration after 4 cycles

Ethane Propane Butane

0.02 0.42 1.23

Total

1.67

Nitrogen in bullet after 4 cycles = 100 – 1.67 = 98.33% Ans.(b) Nitrogen requirement at 0.25 bar g and 308.15 K = (p2 – p1)/p1 (3.513 25 – 1.263 25) = 1.263 25 = 1.781 m3/(m3 vessel volume × cycle) Total nitrogen requirement = 1.781 ´ 4 ´ 22 = 156.728 m3 at 0.25 bar g and 308.15 K 1.263 25 27315 . ´ º 156.728 ´ 308.15 1.013 25 º 173.2 Nm3 Ans.(c)

Stoichiometry and Industrial Problems 285

EXERCISE 8.6 Solution Refer data of Exercise 8.5. Assume that bullet is at atmospheric pressure after nitrogen purgings. Specific volume of gas at 1.013 25 bar a and 308.15 K V m1 = Total gas inside the bullet =

0.083 14 ´ 308.15 = 25.284 57 m3/kmol 1.013 25

22 25.284 57

= 0.8701 kmol At the end of purging with nitrogen and at atmospheric pressure (i.e. at 101.325 kPa), hydrocarbons = 0.8701 ´ 0.0167 = 0.0145 kmol N2 in bullet = 0.8701 – 0.0145 = 0.8556 kmol Air pressurization is done upto 4 bar g or 5.013 25 bar a. 0.083 14 ´ 308.15 V m2 = = 5.1104 m3/kmol 5.013 25 Total gas in bullet after pressurization 22 = = 4.305 kmol 5.1104 Air introduced = 4.305 – 0.8701 = 3.4349 kmol O2 introduced = 3.4349 ´ 0.21 = 0.7213 kmol O2 in bullet after first pressurization cycle 0.7213 = 0.1675 or 16.75% (v/v) = 4.305 After depressurization to atmospheric pressure, again quantity of gas in bullet will be 0.8701 kmol. Oxygen in bullet = 0.8701 ´ 0.1675 = 0.1457 kmol After second cycle of pressurization, total oxygen in bullet = 0.1457 + 0.7213 = 0.867 kmol 0.867 O2 concentration in bullet = 4.305 = 0.2014 or 20.14% (v/v) Thus oxygen concentration will be well above 18% after 2 cycles of pressurization. Ans. (a) Compressed air requirement = 3.4349 ´ 2 = 6.8698 kmol Volumetric air requirement = 6.8698 ´ 22.414 = 154 Nm3 Ans. (b)

286 Solutions Manual—Stoichiometry

EXERCISE 8.7 Basis: 10 500 kg/h feed to bent tube evaporator Solids in feed = 10 500 ´ 0.3155 = 3312.75 kg/h Water in feed = 10 500 – 3312.75 = 7187.25 kg/h Bent tube evaporator : Amount of 60% liquor = 3312.75/0.6 = 5521.25 kg/h Evaporator’s load = 10 500 – 5521.25 – 4978.75 kg/h Vapours are produced at 383 K. Saturation pressure at 383 K for water = 143.3 kPa a From steam tables, h = 461.32 kJ/kg, H = 2691.3 kJ/kg and lv = 2230.0 kJ/kg (Table AIV.2) Take base temperature T0 = 298.15 K (25°C) Heat given to vapours = 2691.3 – 104.8 = 2586.5 kJ/kg Total heat in vapours = 2586.5 ´ 4978.75/3600 = 3577.09 kW Enthalpy to be supplied to feed = 5521.25 ´ 225.7/3600 = 346.15 kW Total heat load of the evaporator = 3577.09 + 346.15 = 3923.24 kW At p = 190 kPa g, Ts = 405.4 K (132.4°C) lv = 2166.6 kJ/kg Steam consumption in the evaporator = 3923.24 ´ 3600/2166.6 = 6518.8 kg/h Wiped film evaporator: Amount of 90% liquor = 3312.75/0.9 = 3680.83 kg/h Water evaporated = 5521.25 – 3680.83 = 1840.42 kg/h Enthalpy of liquid leaving the evaporator = 3680.83 ´ 202.2/3600 = 206.74 kW Enthalpy of liquid, entering the evaporator = 346.15 kW Temperature of water vapours = 410 K (137°C) At T = 410 K, ps = 3.32 bar a and H = 2729.2 kJ/kg Enthalpy of vapours = 1840.42 (2729.2 – 104.8)/3600 = 1341.67 kW Total heat load of evaporator = 1341.67 + 206.74 – 346.15 = 1202.26 kW For steam at p = 12 bar g, Ts = 464.6 K and lv = 1970.7 kJ/kg Steam consumption = 1202.26 ´ 3600/1970.7 = 2196.2 kg/h

Stoichiometry and Industrial Problems 287

Auger Agitated Calciner: Enthalpy of liquid, entering the calciner = 206.74 kW Enthalpy of dry solid mixture = 3312.75 ´ 581.5/3600 = 535.1 kW Water evaporated in calciner = 3680.83 – 3312.75 = 368.08 kg/h Temperature of vapours = 473 K(200°C) ps = 15.55 bar a of water However, the vapours are produced at 101.325 kPa a and 473 K (200°C). Total enthalpy of water vapours = 2875.4 kJ/kg Enthalpy of water vapours = [(2875.4 – 104.8) 368.08]/3600 = 283.28 kW Heat transfer in calciner = 535.1 + 283.28 – 206.74 = 611.64 kW Results are summarised in Table – 8.52 in the text.

EXERCISE 8.8 Basis: 1000 kg cell liquor NaOH content of liquor = 1000 ´ 0.109 = 109 kg NaCl content of liquor = 1000 ´ 0.1526 = 152.6 kg Water in the liquor = 1000 – 109 – 152.6 = 738.4 kg Reaction: 2 NaOH + SO2 = Na2SO3 + H2O Molar mass 2 ´ 40 64 126 18 Since reaction goes to completion, Na2SO3 produced = 126 ´ 109/80 = 171.67 kg Water produced = 18 ´ 109/80 = 24.53 kg Evaporation loss = 71 kg Final quantity of water = 738.4 + 24.53 – 71 = 691.93 kg Concentration of NaCl in sulphited liquor = 152.6 ´ 100/691.93 = 22 g/100 g water or 22 kg/100 kg water From Fig. 8.14, at 373.15 K (100°C), the solubility of Na2SO3 in the presence of 22 kg NaCl/100 kg water = 7.75 kg/100 kg water Dissolved Na2SO3 in solution = 7.75 ´ 691.93/100 = 53.63 kg Na2SO3 crystallized = 171.7 – 53.63 = 118.07 kg Yield of Na2SO3 = 118.07 ´ 100/171.7 = 68.8% Ans. (a)

288 Solutions Manual—Stoichiometry

In the second case, water is partially evaporated till the invariant composition is reached at 373.15 K (100°C). Solubility of NaCl at this point = 38 kg/100 kg water Solubility of Na2SO3 at this point = 3.0 kg/100 kg water To attain solubility of 38 kg NaCl/100 kg water, final water quantity after evapn. = 152.6 ´ 100/38 = 401.6 kg Water to be evaporated = 691.93 – 401.6 = 290.33 kg Na2SO3 in final solution = 3 ´ 4.016 = 12.05 kg Na2SO3 crystallised = 171.7 – 12.05 = 159.65 kg Ans.(b-i) Overall Yield = 159.65 ´ 100/171.7 = 93.0% Ans.(b-ii) In the third case, common salt is added to the extent that the invariant composition at 373.15 K (100°C) is attained. Final quantity of NaCl in solution = 38 ´ 691.93/100 = 262.93 kg Salt added = 262.93 – 152.6 = 110.33 kg Ans.(c-i) Na2SO3 in solution = 3 ´ 691.93/100 = 20.76 kg Yield of Na2SO3 = 171.7 – 20.76 = 150.94 kg Overall Yield = 150.94 ´ 100/171.7 = 87.9% Ans. (c – ii) Instead of concentrating the sulphited liquor, if cooling of solution is carried out to 273.15 K (0°C), yield of Na2SO3 can be improved. From Fig. 8.14 at invariant point, NaCl concn. = 33.6 kg/100 kg H2O at 273.15 K For NaCl concn. of 22.0 g/100 g H2O at 273.15 K(0°C), solubility of Na2SO3 = 5.0 kg/100 kg H2O These crystals are heptahydrate. Na2SO3 in solution = 5 ´ 691.93/100 = 34.6 kg Yield of crystals = 171.7 – 34.6 = 137.1 kg Water, associated with crystals = 18 ´ 7 ´ 137.1/126 = 137.1 kg Thus water, retained in the solution = 691.93 – 137.1 = 554.83 kg This change in water quantity will change the solubility of NaCl and Na2SO3. Due to this reason, a trial and error method is required. After a number of trial and error iterations, solubility of NaCl = 152.6 ´ 100/554.82 = 27.5 kg/100 kg H2O

Stoichiometry and Industrial Problems 289

This gives an indication that the final solubility if NaCl in the solution will be higher than that of 27.5 kg/100 kg H2O. Assume final solubility = 28 g NaCl/100 g H2O Final quantity of water = 152.6 ´ 100/28 = 545 kg From Fig. 8.14, solubility of Na2SO3 at 273.15 K (0°C) in presence of 28 kg NaCl/100 kg H2O = 4.25 kg/100 kg H2O. Na2SO3 in solution = 4.25 ´ 545/100 = 23.16 kg Na2SO3 crystallised = 171.7 – 23.16 = 148.54 kg Ans.(d-i) Water, associated with crystals = 148.54 ´ 18 ´ 7/126 = 148.54 kg Final water quantity = 691.93 – 148.54 = 543.39 kg This quantity closely tallies with 545 kg water. Overall Yield = 148.54 ´ 100/171.7 = 86.5% Ans. (d-ii)

EXERCISE 8.9 Basis: 5000 kg/h MgCl2 as product rate Amount of final product = 5000/0.9 = 5555.6 kg/h MgO present in final product = 5555.6 ´ 0.05 = 277.8 kg/h H2O in final product = 5555.6 – 5000 – 277.8 = 277.8 kg/h 1 mole MgO = 1 mole MgCl2 277.8 kg/h MgO corresponds to 95.3 ´ 277.8/40.3 = 656.9 kg/h MgCl2. HCl produced = 2 ´ 36.5 ´ 656.9/95.3 = 503.2 kg/h H2O consumed = 18 ´ 656.9/95.3 = 124.1 kg/h Total MgCl2 in the feed = 5000 + 656.9 = 5656.9 kg/h Feed contains 48% solids (by mass). Feed rate = 5656.9/0.48 = 11 785.2 kg/h Evaporation in the dryer = Total feed – (Amount of MgCl2 + Mass of water in product – Mass of water reacted) = 11 785.2 – (5656.9 + 277.8 – 124.1) = 5974.6 kg/h Ans. (a) Calculation of total heat load requires following considerations. (i) Latent heat of water vapours. (ii) Heat of crystallisation of MgCl2 from the solution. (iii) Heat of reaction between MgCl2 and H2O. (iv) Sensible heat of water vapour from the feed temperature to outlet gas temperature. (v) Sensible heat of solids from feed temperature to the outlet gas temperature. (vi) Heat loss from the system

290 Solutions Manual—Stoichiometry

lv of water at 101.325 kPa a = 2256.9 kJ/kg Heat supplied for evaporation, f1 = 5974.6 ´ 2256.9/3600 = 3745.6 kW In addition, sensible heat will be supplied to the water vapour to raise its temperature from 393 K to the outlet gas temperature, i.e. 573 K. Average heat capacity of water vapour between 393 K and 573 K = 1.985 kJ/(kg × K). Sensible heat, f2 = 5974.6 ´ 1.985 (573 – 393)/3600 f2 = 592.98 kW Reaction: MgCl2(s) + H2O(l) = MgO(s) + 2 HCl(g) DH°f – 601.7 – 285.83 – 601.7 – 2 ´ 92.31 DH°r = –601.7 – 2 ´ 92.31 – [– 641.32 – 285.83] = +140.83 kJ/mol MgCl2 at 298.15 K (25°C) (endothermic) º + 70.415 kJ/mol HCl at 298.15 K (25°C) (endothermic) º + 1931.25 kJ/kg HCl at 298.15 K (25°C) (endothermic) Change in DH°r with respect to the temperature is neglected. Heat absorbed during reaction, f3 = 1931.25 ´ 503.2/3600 = 269.95 kW Heat of crystallisation = – (heat of solution) = – [–801.15 – (–641.32)] = 159.83 kJ/mol MgCl2 = 1678.7 kJ/kg MgCl2 at 25°C (298.15 K) (endothermic) Heat absorbed during crystallisation, f4 = 5656.9 ´ 1678.7/3600 = 2637.84 kW Heat to be supplied to HCl, f5 = 503.2 ´ 0.816 (573 – 393)/3600 = 20.53 kW Heat, supplied to MgCl2, f6 = 5000 ´ 0.873 (573 – 393)/3600 = 218.25 kW Heat, supplied to MgO, f7 = 277.8 ´ 1.277 (573 – 393)/3600 = 17.74 kW Heat supplied to water present in final solid mixture, f8 = 277.8 ´ 4.1868 (573 – 393)/3600 = 58.15 kW 8

Total heat load, f = å fi 1

= 3745.6 + 592.98 + 269.95 + 2637.84 + 20.53 + 218.25 + 17.74 + 58.15 = 7561.04 kW Ans.(b)

Stoichiometry and Industrial Problems 291

. Enthalpy of dry flue gases, f9 = n ´ 1.006 (798 – 573)/3600 . = 0.0629 n kW . where n = dry flue gas flow, kg/h For enthalpy of water vapours, use total enthalpy of water vapour at 101.325 kPa and 798 K (525°C) and 573 K (300°C). Enthalpy of water at 798 K = 3541.7 kJ/kg over 273.15 K Enthalpy of water at 573 K = 3074.5 kJ/kg over 273.15 K Enthalpy of water vapours accompanying flue gas, . f10 = n ´ 0.03 (3541.7 – 3074.5)/3600 = 0.0039 n kW Total heat given up by flue gas = f9 + f10 . = (0.0629 + 0.0039)n . = 0.0668 n kW = 7561.04 kW . n = 113 189 kg/h of dry flue gas º 3903.07 kmol/h Water vapours = 113 189 ´ 0.03/18 = 188.65 kmol/h Total wet flue gas = 3903.07 + 188.65 = 4091.72 kmol/h Specific volume of gas at 101.325 kPa a and 798 K = (8.314 14 ´ 798.15)/101.325 = 65.492 m3/h Incoming flue gas rate = 4091.72 ´ 65.492 = 267 975 m3/h Ans.(c-i) Outgoing gas mixture = 4091.72 + (503.2/36.5) + (5726.4/18) – (124.1/18) = 4091.72 + 13.79 + 318.13 – 6.89 = 4416.75 kmol/h Specific volume of gas at 101.325 kPa a and 573 K = 8.314 14 ´ 573.15/101.325 = 47.029 m3/kmol Volumetric flow rate of outgoing gas mixture = 4416.75 ´ 47.029 = 207 715 m3/h Ans.(c-ii) Notes: (a) The reaction, mentioned in the above example, takes place in two steps. MgCl2 + 2 H2O = Mg(OH)2 + 2 HCl Mg(OH)2 = MgO + H2O

292 Solutions Manual—Stoichiometry

However, overall reaction is sufficient for mass and heat balances. (b) The system will be under slight positive pressure. However, in solving the problem, the pressure difference is neglected. (c) The product discharge temperature is assumed to be equal to the outlet flue gas temperature. This is a safe assumption.

EXERCISE 8.10 Basis: 186 minutes of operation Hydrogen liberated = Hydrogen from electrolysis of water – Hydrogen equivalent to SO4, migrated from catholyte + Hydrogen from electrolysis of H2SO4 = Hydrogen from electrolysis of water – Hydrogen equivalent to SO4, produced by decomposition of FeSO4 – Hydrogen equivalent to SO4 + Hydrogen from electrolysis of H2SO4 = Hydrogen from water + Hydrogen equivalent to SO4, produced by decomposition of FeSO4. O2 liberated = 3.331 Nm3 = 0.1486 kmol º 4.756 kg Equivalent hydrogen = 2 ´ 4.756/16 = 0.5945 kg Fe deposited = 5.313 kg Equivalent hydrogen = 2 ´ 5.313/56 = 0.1897 kg Hydrogen liberated = 0.5945 – 0.1897 = 0.4048 kg º 0.2024 kmol º 4.54 Nm3 Ans. (b) Let a = amount of final catholyte, b = amount of final anolyte and c = total amount of water evaporated; all in kg Fe balance at the cathode: Fe in final catholyte = Fe in initial catholyte – Fe deposited (0.018 ´ 56/152)a = (15.6 ´ 56/152) – 5.313 a = 65.6 kg Ans. (a – i) SO4 balance at the cathode: SO4 in final catholyte = SO4 in the initial catholyte – SO4 migrated to anode compartment H2SO4 in initial anotyte = 0.03 ´ 59 = 1.77 kg SO4 migrated = (0.177 b – 1.77) 96/98 kg SO4 in initial catholyte = (3.9 ´ 96/98) + (15.6 ´ 96/152) = 3.8204 + 9.8526 = 13.6730 kg SO4 in final catholyte = 65.6 [(0.018 ´ 96/152) + (0.0155 ´ 96/98)] = 1.7418 kg 1.7392 = 13.6730 – [(0.177 b – 1.77) 96/98] b = 78.81 kg Ans.(a-ii)

Stoichiometry and Industrial Problems 293

Overall material balance yields evaporation. Evaporation = 100 + 59 – 65.6 – 78.81 – 5.313 – 4.756 – 0.405 = 4.116 kg Ans.(c) 4 Current = 4000 ´ 186 ´ 60 = 4464 ´ 10 coulombs = 4464 ´ 104/96 580 = 462.208 Faradays Equivalent mass of Fe = 56/2 = 28 Theoretical Fe liberation = 462.208 ´ 28 = 12 942 g º 12.942 kg Current efficiency =

actual Fe liberation ´ 100 theoretical Fe liberation

= 5.313 ´ 100/12.942 = 41.05%

Ans. (d)

EXERCISE 8.11 Basis: 100 kmol NG (natural gas) input Let a, b, c and d be kmol of H2, CO, CO2 and H2O, respectively in the reactor outlet gases. Let e be kmol of CH4 and f be kmol of O2 supply (98% by vol.) Inerts (Ar + N2) in supply NG = 1.95 + 0.4 = 2.35 kmol Inerts (Ar) in O2 supply = 0.02f kmol Inerts in outgoing gas mixture = 2.35 + 0.02f kmol Total moles of outgoing gas mixture, n = a + b + c + d + e + 2.35 + 0.02f kmol CH4 balance : e = 0.0035 (1) n H2 balance: a + d + 2e = 93.25 ´ 2 + 3.32 ´ 3 + 0.88 ´ 4 + 0.2 ´ 5 = 186.5 + 9.96 + 3.52 + 1.00 = 200.98 (2) Carbon balance: b + c + e = 93.25 + 3.32 ´ 2 + 0.88 ´ 3 + 0.2 ´ 4 = 93.25 + 6.64 + 2.64 + 0.8 = 103.33 (3) O2 balance: 0.5 b + c + 0.5 d = 0.98 f b + 2 c + d = 1.96 f (4) Limiting reaction is steam-methane reaction. Its approach to equilibrium is 30 K. Since this reaction is endothermic, the composition of outgoing gas mixture will be determined by chemical equilibrium at 1395 – 30 = 1365 K.

294 Solutions Manual—Stoichiometry

Now as per Dalton’s law, pi = p × yi and yi = ni /n

æ (e / n) p × ( d / n ) p ö edn 2 = 3 2 = 2.5207 ´ 10–5 Therefore, Kp1 = ç 3÷ a bp è (b / n) p × [(a / n) p] ø System pressure, p = 2.0 MPa a = 19.7385 atm a Kp1 =

edn 2 = 2.5207 ´ 10–5 a b (19.7385) 2 3

edn 2 = 9.8209 ´ 10–3 (5) 3 a b Shift reaction is exothermic and hence it reaches almost to completion at such a high temperature. Thus chemical equilibrium of shift reaction at 1395 K is taken for calculations.

or

Kp2 =

( a / n) × (c / n) × p 2 = 0.4465 (b / n) × ( d / n) × p 2

ac = 0.4465 (6) bd Six simultaneous equations will yield six unknowns. Equations (5) and (6) are complicated. Therefore trial and error method is suggested. Iteration-1: Let n = 300 kmol. The increase in number of moles is evident as reforming reaction is a primary reaction. e = 0.0035 ´ 300 = 1.05 kmol a + d – 200.98 – 2 ´ 1.05 = 198.88 kmol b + c = 103.33 – 1.05 = 102.28 kmol Substitute values of n and e in Eq. (5),

or

or

1.05 ´ d (300) 2 = 9.8209 ´ 10–3 a 3b 1.05 ´ (300 ) 2 ´ 10 3 d d b= ´ 3 = 9.6223 ´ 106 3 9.8209 a a

Within this iteration, assume a value of d, obtain values of a, b and c. Calculate the ratio (ac/bd) which should be 0.4465 [Ref. Eq. (6)]. Let d1 = 30 kmol; a = 168.88 kmol b = 59.93 kmol,

c = 42.35 kmol

ac/ad = 168.88 ´ 42.35/(30 ´ 59.93) = 3.978 # 0.4465 Assume d 2 = 35 kmol; b = 77.22 kmol,

a = 163.38 kmol c = 25.06 kmol

ac/ad = 1.5149 # 0.4465

Stoichiometry and Industrial Problems 295

d, kmol (assumed) 30 35 37 39 39.5 39.1

a, kmol

b, kmol

c, kmol

168.88 163.88 161.88 159.88 159.88 159.78

59.93 77.22 83.94 91.83 93.88 92.23

42.35 25.06 18.34 10.45 8.40 10.05

ac /ad 3.978 1.5149 0.9559 0.4665 0.361 0.4453

> > > > < <

0.4465 0.4465 0.4465 0.4465 0.4465 0.4465

1.96 f = 92.23 + 2 ´ 10.05 + 39.1 = 151.43 f = 77.26 kmol n = 159.78 + 92.23 + 10.05 + 39.1 + 1.05 + 2.35 + 77.26 ´ 0.02 = 306.105 kmol Iteration-2: Let n = 306 kmol e = 0.0035 ´ 306 = 1.07 kmol a + d = 200.98 – 1.07 ´ 2 = 198.84 b + c = 103.33 – 1.07 = 102.26 1.07 ´ d ´ (306) 2 = 9.8209 ´ 10–2 3 a b or b = 10.2018 ´ 106 d/a3 Iteration: d, kmol (assumed) 39.1 38 37.8 37.85

a, kmol

b, kmol

c, kmol

159.74 160.84 161.04 160.99

97.862 93.17 92.335 92.544

4.398 9.09 9.925 9.716

ac / ad 0.1841 0.4130 0.4579 0.44655

< 0.4465 < 0.4465 > 0.4465 ; 0.4465

1.96 f = 92.544 + 2 ´ 9.716 + 37.85 = 149.826 or

f = 76.442 kmol n = 160.99 + 92.544 + 9.716 + 37.85 + 1.05 + 2.35 + 76.442 ´ 0.02 = 306.049 kmol ; 306 kmol (nearly same as assumed)

O2 supply = 0.7644 kmol/kmol natural gas Ans. (a) Mathcad can be of particular use for solving such simultaneous equations. Values of iterative calculations are used in following tables.

296 Solutions Manual—Stoichiometry

Analysis of exit gas mixture: Component H2 CO CO2 H2O CH4 N2 Ar

dry gas kmol 160.990 92.544 9.716 — 1.070 1.950 1.929

Total

268.199

mole % 60.03 34.50 3.62 — 0.40 0.73 0.72 100.00

kmol 160.990 92.544 9.716 37.850 1.070 1.950 1.929

wet gas mole % 52.60 30.24 3.17 12.37 0.35 0.64 0.63

306.049

100.00

Steam/dry exit gas = 37.850/268.199 = 0.1411 kmol/kmol

Ans.(b)

Enthalpy of reactants: Ref. Table – 5.22 Component CH4 C2H6 C3 H8 n-C4H10 N2 O2 Ar Total

kmol ni 93.25 3.32 0.88 0.20 1.95 74.91 1.929

Temperature K 422 422 422 422 422 400 400

176.439

—

Enthalpy, kJ/kmol (H° – H°o + DH°f )i – 52 247 – 49 736 – 56 164 – 64 005 – 12 187 – 12 277 — —

ni × (H° – H°o – DH°f )i kJ – 4 871 996 – 165 123 – 49 424 – 12 801 + 23 765 + 919 673 — – 4 155 906

Enthalpy of products: Ref. Table – 5.22 Temperature of product stream = 1395 K Component H2 CO CO2 H2O CH4 N2 Ar

kmol ni 160.99 92.544 9.716 37.85 1.07 1.95 1.929

Total

306.049

Temperature K 1395 1395 1395 1395 1395 1395 1395 —

Enthalpy, kJ/kmol (H° – H°o + DH°f )i + 41 315 – 70 120 – 328 410 – 186 030 + 12 386 + 43 372 —

ni × (H° – H°o – DH°f )i kJ + 6 651 282 – 6 489 197 – 3 190 833 – 7 041 238 + 13 253 + 84 576 —

—

Enthalpy, supplied to argon = 20.7723 (1395 – 400) ´ 1.929 = 39 869 kJ

– 9 972 157

Stoichiometry and Industrial Problems 297

Enthalpy change during reactions, DH = – 9972 157 – (4155 906) + (39 869) = – 5856 120 kJ (exothermic) per 100 kmol º – 5856 kJ/kmol feed gas Ans. (c) Relatively small exothermic enthalpy change can be attributed to heat loss from the system by radiation. As such the reaction proceeds autothermally. Note: For iterative calculations, a spreadsheet (such as Excel) or a mathematical software (such as Mathcad) can be very useful. Refer Example 9.14).

EXERCISE 8.12 Basis: 100 kg spent acid Let x kg be the fuel oil required for decomposition. Reactions: H2SO4 = H2O + SO3 SO3 = SO2 + 1/2 O2 2 NH4HSO4 + 1/2 O2 = N2 + 5 H2O + 2 SO2 C + O2 = CO2 H2 + 1/2 O2 = H2O Carbon of organics, present in spent acid = 5.1 ´ 5/6.1 = 4.18 kg Hydrogen of organics, present in spent acid = 5 – 4.18 = 0.82 kg Carbon content of fuel oil = 6.2 x/(7.2 ´ 12) = 0.0718 x kmol Hydrogen content of fuel oil = 1 ´ x/(7.2 ´ 2) = 0.0694 x kmol Component H2SO4 NH4HSO4 Cm Hn C H H2O Total

kg 20 45 4.18 0.82 30.0 100.00

Molar mass 98 115 12 2 18

kmol

(A) (B) (C) (D) (E)

O2 requirement, kmol

0.2041 0.3913 0.3483 0.4100 1.6667

– 0.1021 + 0.0978 + 0.3483 + 0.2050 —

3.0204

0.5490

Total SO2, formed by decomposition = 0.2041 + 0.3913 = 0.5954 kmol H2O formed by decomposition = 1.6667 + 0.41 + 2.5 ´ 0.3913 + 0.2041 = 3.2591 kmol CO2 formed = 0.3483 kmol N2 formed = 0.5 ´ 0.3913 = 0.1957 kmol Fuel oil combustion yields CO2 = 0.0718 x kmol H2O = 0.0694 x kmol O2 required in exit gas mixture = 1.2 ´ 0.5954 = 0.7145 kmol

298 Solutions Manual—Stoichiometry

Total dry gas mixture = SO2 content + CO2 content + O2 content + N2 content O2 required for combustion of fuel = 0.0718 x + (0.0694/2) x = 0.1065 x kmol Total oxygen supply = 0.1065 x + 0.5490 + 0.7145 = 1.2635 + 0.1065 x kmol Corresponding N2 supply from air = (79/21) (1.2635 + 0.1065 x) = 4.7532 + 0.4006 x kmol Total N2 in exit gas mixture = 0.1957 + 4.7532 + 0.4006 x = 4.9489 + 0.4006 x kmol Total dry gas mixture = 0.5954 + 0.3483 + 0.0718 x + 0.7145 + 0.4006 x + 4.9489 = 6.6071 + 0.4724 x kmol 0.5954 = 0.06 (6.6072 + 0.4724 x )

Solving the equation, x = 7.0219 kg fuel oil/100 kg spent acid Total dry air supply = 1.2635 + 0.1065 x + 4.7532 + 0.4006 x = 6.0167 + 0.5071 x = 6.0167 + 0.5071 ´ 7.0219 = 9.5775 kmol Humidity of air = 12.8 g/kg dry air @ DB = 308 K, and WB = 296 K = 0.0206 kmol/kmol dry air (Ref. Fig. 6.15) Moisture of air = 0.0206 ´ 9.5775 = 0.1975 kmol Total water in exit gas mixture = 3.2591 + 0.0694 x + 0.1975 = 3.9439 kmol Total exit gas mixture = 0.5954 + 0.3483 + 0.0718 ´ 7.0219 + 0.7145 + 0.4006 ´ 7.0219 + 4.9489 + 3.9439 = 13.8682 kmol (wet) New basis: 860 t/d spent acid Hourly feed rate = 860 000/24 = 35 833.3 kg/h Multiplying factor, based on original feed rate of 100 kg = 358.333 LHV of fuel oil = 43 000 kJ/kg Heat liberated by combustion = 43 000 ´ 7.0219 ´ 358.333/3600 = 30 054.35 kW Organics in spent acid = 5 ´ 358.333 = 1791.67 kg/h

Stoichiometry and Industrial Problems 299

LHV of organics = Heat liberated by combustion = = Total air requirements = = Combustion air requirements = = Enthalpy of dry air =

41 870 kJ/kg 1791.67 ´ 41 870/3600 20 838.12 kW (9.5775 + 0.1975) 9.775 kmol/100 kg spent acid 9.775 ´ 358.333 3502.7 kmol/h (moist) 1.006 ´ 9.5775 ´ 358.333 ´ 29 (723.15 – 298.15)/3600 = 11 820.11 kW Enthalpy of water vapour at 101.325 kPa and 723.15 K = 3382.4 kJ/kg Enthalpy of water vapour at 298.15 K = 2547.3 kJ/kg Enthalpy of water vapours at 723.15 K accompanying air = 0.1975 ´ 358.333 (3382.4 – 2547.3) ´ 18/3600 = 295.50 kW Total enthalpy of wet air = 11 820.11 + 295.50 = 12 115.61 kW Total heat supplied = enthalpy of air + combustion heat of fuel and organics = 12 115.61 + 30 054.25 + 20 838.12 = 63 007.98 kW Heat utilisation = 98% Heat utilised = 63 007.98 ´ 0.98 = 61 747.82 kW Heat of solution of NH4HSO4 = – 1019.85 – (–1026.96) = + 7.11 kJ/kmol NH4HSO4 (endothermic) º 61.768 kJ/kg NH4HSO4 Total NH4HSO4 dried = 45 ´ 358.333 = 16 125 kg/h Heat supplied during crystallization = 61.768 ´ 16 125/3600 f1 = 276.27 kW (exothermic) Acid concentration of spent acid (without NH4HSO4) = [20/(20 + 30)] ´ 100 = 40% This is to be dehydrated to 100%. SO3 in dilute acid = (40 ´ 80)/(98 ´ 80) = 0.408 kmol H2O in dilute acid = (0.408 ´ 18 + 60)/18 = 3.741 kmol (Ref. Table – 5.64) H 1 = – 2.093 kJ/mol H2O (Ref. Table – 5.64) H 2 = – 135.65 kJ/mol SO3 At 100% concentration, = H 1 = H 2 = – 44.34 kJ/mol Dehydration: Dilute acid + Heat of mixing = 100% acid + water

300 Solutions Manual—Stoichiometry

(– 2093 ´ 3.741) + (– 135 650 ´ 0.408) + DH = (– 0.408 ´ 44 340) + (– 0.408 ´ 44 340) + 0 Heat of dilution, DH = – 18 091 – 18 091 + 7830 + 55 345 = + 26 993 kJ/100 kg 40% acid (H2SO4) (endothermic) Original acid rate = (20/0.4) ´ 358.333 = 17 916.65 kg/h Heat to be supplied for dehydration, f2 = 17 916.65 ´ 26 993/(100 ´ 3600) = 1343.40 kW Enthalpy of vaporization of water at 291.15 K (18°C) = 2459 kJ/kg Water to be evaporated = 30 ´ 358.33 ´ 0.6 = 10 750 kg/h Heat, supplied for evaporation f3 = 10 750 ´ 2459/3600 = 7342.85 kW Reaction 1: H2SO4(l) = H2O(g) + SO2(g) + 1/2 O2(g) DH°r1 = – 241.82 – 296.83 – (– 813.99) = + 275.34 kJ/mol H2SO4 º 2807.3 kJ/kg H2SO4 (endothermic) Reaction 2: 4 NH4HSO4(s) + O2(g) = 2 N2(g) + 10 H2O(g) + 4 SO2(g) DH°r2 = [10 (– 241.82) + 4 (– 296.83)] – [4 (– 1026.96)] = + 502.32 kJ/4 kmol NH4HSO4 º + 125.58 kJ/mol NH4HSO4 (endothermic) º + 1091.0 kJ/kg NH4HSO4 Total heat of reaction = [(17 916.65 – 10 750) 2807.3 + 16 125 ´ 10 91]/3600 f4 = 10 475.36 kW (endothermic) Since calculations of heats of reactions involve water in the vapour phase, f3 need not be considered for product waters. 4

Total heat to be supplied = å fi i

= 1343.40 + 7342.85 + 104 75.36 – 276.27 = 18 885.34 kW Sensible heat of product gaseous stream over 298.15 K, f6 = 61 747.82 – 18 885.34 = 43 362.48 kW Assume mean specific heat of product gas stream to be 1.255 kJ/(kg × K). Flue gas flow rate = 13.8682 ´ 358.333 = 4969.43 kmol/h (wet) Average molar mass of product gas stream = 0.5954 ´ 64 + 0.8525 ´ 44 + 0.7145 ´ 32 + 7.7619 ´ 28 + 3.9439 ´ 18

Stoichiometry and Industrial Problems 301

= 386.8 kg/13.8682 kmol º 27.89 kg/kmol Let the final temperature of product gas stream be T K. 4969.43 ´ 27.89 ´ 1.255 (T – 298.15)/3600 = 43 362.48 T – 298.15 = 897.47 T = 1195.62 K (922.47°C) This temperature is lower than the desired decomposition temperature of about 1275 K (1002°C). 2nd iteration: Final desired temperature = 1275 K Additional heat to be supplied = 4969.43 ´ 27.89 (1275.15 – 1195.62) 1.255/3600 = 3842.62 kW Additional fuel, required to be fired = 3842.62 ´ 3600/(43 000 ´ 358.333) = 0.8978 kg/100 kg spent acid New x = 7.0219 + 0.8978 = 7.9197 kg/100 kg spent acid Total dry air supply = 6.0167 + 0.5071 ´ 7.9197 = 10.033 kmol/100 kg spent acid SO2 in exit gas mixture = 0.5954 ´ 100/(6.6071 + 0.4724 ´ 7.9197) = 5.75% Mositure, entering with air = 10.033 ´ 0.0206 = 0.2067 kmol/100 kg spent acid Moist air = 10.033 + 0.2067 = 10.2397 kmol/100 kg spent acid Total moisture in exit gas mixture = 3.2591 + 0.0694 ´ 7.9197 + 0.2067 = 4.0154 kmol/100 kg spent acid Flow rate of exit gas mixture = 6.6071 + 0.4724 ´ 7.9197 + 4.0154 = 14.3638 kmol/100 kg spent acid This means, exit gas mixture flow rate will increase by 14.3638/13.8682 = 1.0357 times or 3.57% increase. This increase in exit gas mixture will require additional fuel to raise the exit gas temperature to 1275 K. For a rise of (1275.15 – 1195.62) = 79.53 K, addition fuel required is 0.8978 kg/100 kg spent acid. Due to increase in exit gas flow rate, additional fuel required =

(14.3638 - 13.8682)1000 ´ 0.8978 13.8682 ´ 79.53

= 0.4034 kg/100 kg spent acid New x = (7.9197 + 0.4034) = 8.3231 kg/100 kg spent acid

302 Solutions Manual—Stoichiometry

3rd iteration: Assume x = 8.6 kg/100 kg spent acid SO3 in exit gas mixture = 0.5954 ´ 100/(6.6071 + 8.6 ´ 0.4724) = 5.57% Total dry air supply = 6.0167 + 0.5071 ´ 8.6 = 6.0167 + 4.3611 = 10.3778 kmol Moisture with air = 10.3778 ´ 0.0205 = 0.2138 kmol/100 kg spent acid Total moisture in exist gas mixture = 3.2591 + 0.0694 ´ 8.6 + 0.2138 = 4.0697 kmol/100 kg spent acid CO2 in exit gas mixture = 0.3483 + 0.0718 ´ 8.6 = 0.9658 kmol/100 kg spent acid N2 in exit gas mixture = 4.9489 + 0.4006 ´ 8.6 = 8.3941 kmol/100 kg spent acid Component

kmol ni

Heat capacity C°mp equation constants

SO2 CO2 O2 N2 H2O

0.5954 0.9658 0.7145 8.3941 4.0697

ai × ni 17.748 20.635 18.595 248.389 132.233

Total

14.7395

434.600

ni × bi ´ 103 37.479 62.086 8.399 – 43.154 0.324 65.134

ni × ci ´ 106 – 26.351 – 39.647 – 1.674 110.659 53.764 96.751

ni × di ´ 109 6.622 9.465 – 0.402 – 41.702 – 18.507 – 44.524

Additional heat liberated = (8.6 – 7.0219) 43 000 ´ 358.333/3600 = 6754.41 kW Total heat of exit gas mixture = 43 362.48 + 6754.41 = 50 116.89 kW º 180 420 804 kJ/h Additional heat available from air is neglected. Enthalpy of exit gas at T K over 298.15 K = 358.333 [434.6 (T – 298.15) + 65.134 ´ 10–3 (T2 – 298.152)/2 + 96.751 ´ 10–6 (T 3 – 298.153)/3 – 44.524 ´ 10–9 (T 4 – 298.154/4] = 180 420 804 = 155 732 T – 46 408 136 + 11.67 T2 – 1036 343 + 0.0116 T3 – 317 563 – 3.99 ´ 10–6 T4 + 31466 = 49 239.12 ´ 3600 = 177 260 832 – 3.99 ´ 10–6 T 4 + 0.0116 T3 + 11.67 T2 + 155 732 T = 224 991 408 By trial and error or by Mathcad, T = 1261.69 K or 988.54°C This temperature is lower than 1275 K.

Stoichiometry and Industrial Problems 303

4th iteration: Let x = 8.6 + 0.3 = 8.9 kg/100 kg spent acid Dry air = 6.0167 + 0.5071 ´ 8.9 = 10.53 kmol Moisture with air = 10.53 ´ 0.0206 = 0.2169 kmol Total moisture in exit gas mixture = 3.2591 + 0.0694 ´ 8.9 + 0.2169 = 4.0937 kmol CO2 in exit gas mixture = 0.3483 + 0.0718 ´ 8.9 = 0.9873 kmol N2 in exit gas mixture = 4.9469 + 0.4006 ´ 8.9 = 8.5122 kmol Component

kmol ni

Dry gas mole %

Heat capacity C°mp equation constants ai × ni

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109 6.622 9.675 – 0.402 – 42.289 – 18.616

SO2 CO2 O2 N2 H2O

0.5954 0.9873 0.7145 8.5122 4.0937

5.51 9.13 6.61 78.75 —

14.748 21.094 18.595 251.884 133.013

37.479 63.468 8.399 – 43.761 0.326

– 26.351 – 40.529 – 1.674 112.215 54.081

Total

14.9031

100.00

439.334

65.911

97.742

– 45.01

Additional heat generted = 0.3 ´ 43 000 ´ 358.333/3600 = 1284.03 kW Total heat of exit gas mixture = 50 116.89 + 1284.03 = 51 400.92 kW º 185 043 312 kW Enthalpy of exit gas at T K over 298.15 K Ans. (i) –3 = 358.333 [439.334 (T – 298.15) + 65.911 – 10 (T – 298.15)/2 + 97.742 ´ 10–6 (T – 298.153)/3 – 45.01 ´ 10–9 (T – 298.154)/4] = 185 043 312 kJ By trial and error, T = 1274.28 K or 1001.13°C Ans.(i) This temperature is very close to 1275 K and hence acceptable. Total fuel oil fired = 8.9 ´ 358.333 = 3189.2 kg/h º 76.54 t/d Ans.(iii) Enthalpy of exit gas at 623.15 K over 298.15 K = 358.333 [439.334 (623.15 – 298.15) + 65.911 ´ 10–3 (623.15 – 298.152)/2 + 97.742 ´ 10–6 (623.153 – 298.153)/3

304 Solutions Manual—Stoichiometry

+ 45.01 ´ 10–9 (623.154 – 298.154)/4] = 358.333 ´ 158 063 = 56 639 189 kJ/h = 15 733.11 kW Heat liberated due to additional fuel firing = (8.9 – 7.0219) 43 000 ´ 358.333/3600 = 8038.43 kW Total moist air = (6.0167 + 0.5071 ´ 8.9) 1.0206 = 10.7469 kmol Additional heat due to air= (10.7469 – 9.775) 12 115.61/9.775 = 1204.62 kW Partial pressure of water vapours in exit gas stream, pH O = 4.0937 ´ 106.7/14.9031 2 = 29.309 kPa º 219.84 Torr A = ln (pH2O) = 3.3779 Water dew point = 341.6 K (68.6°C) (Ref. Table – 6.13) Ans. (ii) pSO = 0.5954 ´ 106.7/14.9031 = 4.263 kPa º 31.97 Torr 2

B = ln (pSO2) = 1.45 Sulphurous acid dew point (Ref. Exercise 7.28): 1000/TDP = 3.5752 – 0.1845 ´ 3.3779 – 9.333 ´ 10–4 ´ 1.45 –9.13 ´ 10–4 (3.3779 ´ 1.45) TDP(SO2) = 339.45 K (66.3°C) Therefore air must be introduced at about 345 K (72°C). a 5 K margine. Enthalpy of air 345 K (72°C) = 10.53 (1.006 + 1.84 ´ 0.0128) 358.33 (345 – 298.15)/3600 = 50.72 kW Enthalpy of air at 723 K (450°C) = 10.7469 ´ 12 115.61/9.775 = 13 320.23 kW Enthalpy supplied in air preheater = 13 320.23 – 50.72 = 13 269.51 kW º 12 404.73 kJ/kmol Total air required in furnace for decomposition = (10.7469) 358.333 = 3850.97 kmol/h (moist) Heat, given up by exit gas stream in air preheater = 51 400.92 – 15 733.11 = 35 667.81 kW Total air preheated = 35 667.81 ´ 3600 ´ 0.98/12 404.73 = 10 144.19 kmol/h

Stoichiometry and Industrial Problems 305

Portion of air, sent to furnace = 3850.97 ´ 100/10 144.19 = 38.0 % Air, used for steam generation = 10 144.19 – 3850.97 = 6293.22 kmol/h This air is utilised for saturated steam generation at 11 bar a. At p = 11 bar a, Ts = 457.1 K (184.1°C), h = 781.12 kJ/kg l v = 1998.5 kJ/kg and H = 2779.7 kJ/kg from Steam Tables (AIV.2) Since steam temperature is 457.1 K, it may not be economical to cool the air below 510 K (237 °C). Enthalpy of air at 510.15 K over 298.15 K = 6293.22 ´ 29 (1.006 + 1.84 ´ 0.0128) (510.15 – 298.15)/3600 = 11 065.54 kW Steam generation = 11 065.54 ´ 3600 ´ 0.98/(2779.7 – 104.8) = 14 594.7 kg/h say 14.59 t/h Ans. (iv)

EXERCISE 8.13 Basis: 1 kmol benzene formation in reactor (net) Profit a (conversion)0.5 Profit a (yield)3 Therefore porfit function a (conversion)0.5 (yield)3 Profit function is to be maximised. Since the values of conversion and corresponding yield are available in Table 8.56, a graph (Fig. E8.2) of conversion vs. profit [i.e. (conversion)0.5 (yield)3] is plotted. Another (second) graph is plotted for conversion vs. yield. Conversion, % Relative Profit Function

50 0.686

66 0.776

70 0.780

75 0.790

85 0.742

From the first graph, for a maximum profit, conversion = 76% Corresponding yield = 96.55% from the second curve Toluene charged to the reactor (by definition) = 1/(conversion ´ yield) = 1/(0.76 ´ 0.9685) = 1.3586 kmol Let x, y and z be the kmol of benzene, methane and diphenyl in the gross feed to the reactor, respectively. Total ‘oil’ in the gross feed = 1.3586 + x + z kmol H2 in the gross feed = 5(1.3585 + x + z) kmol Toluene reacted = 1.3586 ´ 0.76 = 1.0325 kmol (total) Total benzene produced = 1.0325 kmol H2 utilised in conversion to benzene = total benzene formed = 1.0325 kmol

100

0.80

99

98

0.76

97

0.72

96

n rsio nve Co V/S

tV

/S

Co

0.74

ld Yie

nv e

r si

on

96.55

95

0.70

0.68

94

50

60

70

76

80

Conversion Fig. E8.2 Solution of Exercise 8.11

90

100

93

Yield, %

0.78

Pr ofi

Relative Profit Function

Max. Profit

306 Solutions Manual—Stoichiometry

0.82

Stoichiometry and Industrial Problems 307

Hydrogen formed in side reaction = (1.0325 – 1)/2 = 0.016 25 kmol Toluene reacted = 1.0325 kmol (total) Methane formed = 1 + 0.0325 = 1.0325 kmol Amounts of the components in the exit of reaction: Hydrogen = 6.793 + 5 x + 5 z – 1.0325 + 0.016 25 = 5.776 75 + 5x + 5z kmol = 6.0933 Methane = 1.0325 + y kmol = 6.0877 Benzene = 1 + x kmol Toluene = 1.3586 – 1 – 0.0325 = 0.3261 kmol Diphenyl = z + 0.016 25 = 0.021 67 kmol The effluent gas mixture goes to the phase separator. Let x1 kmol benzene escape in the vapour phase while x2 kmol benzene are recycled to the reactor. This does not include the liquid product, used for quenching reactor exit stream. This small quantity is fixed on the basis of temperature control and is independent of liquid product, processed further for recovery of benzene.

Benzene in vapour phase = 0.005 Benzene in liquid phase x1/(1 + x2) = x2 Benzene column separates 95% benzene. 0.05(1 + x2) = x2 x2 = 0.052 63 kmol x1 = 0.005 (1.052 63) = 0.005 263 kmol Similarly, toluene column separates only 75% diphenyl (z + 0.016 25) 0.25 = z z = 0.005 42 kmol H2 in exit stream = 5.776 75 + 5 ´ 0.057 89 + 0.005 42 ´ 5 = 5.776 75 + 0.289 45 + 0.0271 = 6.0933 kmol CH4 in exit stream = 1.0325 + y kmol Both these components leave in total in vapour phase from the seperator as the concentration ratios in vapour phase to liquid phase is ¥ (infinite). Benzene in vapour phase, x1 = 0.005 263 kmol Toluene in vapour phase = 0.001 ´ (0.3261 – toluene in vapour phase) = 0.000 3258 kmol Diphenyl in vapour phase = 0 kmol

308 Solutions Manual—Stoichiometry

Compositions of vapour stream from the separator, recycle gas stream and purge gas stream are same.

H 2 in vapour phase = 0.5 Total gas in vapour phase 6.0933 = 0.5 (6.0933 + 1.0325 + y + 0.005 263 + 0.000 3258)

y = 5.0552 kmol Total vapour phase mixture = 6.0933/0.5 = 12.1866 kmol Methane in vapour phase = 1.0325 + 5.0552 = 6.0877 kmol Let A kmol are purged out and M kmol is the quantity of fresh feed. Methane balance: Methane in recycle stream + methane in make-up gas = Methane in gross fed [(12.1866 – A) 6.0877/12.1866] + 0.05 M = 5.0552 0.4995 A – 0.05 M = 1.0325 (1) Hydrogen balance: Hydrogen in recycle stream + Hydrogen in make-up gas = Hydrogen in gross feed (12.1866 – A)0.5 + 0.95 M = 6.793 + 5 ´ 0.057 89 + 5 ´ 0.005 42 = 7.109 55 – 0.5 A + 0.95 M = 1.016 25 (2) Solving the simultaneous equations, M = 2.2775 kmol and A = 2.2948 kmol Composition of steam from phase separator: Component H2 CH4 C6H6 C7H8 C12H10 Total

vapour phase kmol 6.093 3 6.087 7 0.005 363 0.000 3258 0

12.186 589

liquid phase

kg 12.186 6 79.062 4 0.410 5 0.029 97 0

mole % 50.00 49.95 0.043 0.007 0

kmol

kg

0 0 1.052 63 0.325 7742 0.021 67

82.105 14 29.971 23 3.337 18

71.14 25.97 2.89

91.689 47

100.00

1.400 0742

115.41355

100.00

Benzene column: Benzene recovered = 78 kg (final net product) Bottom product = 115.413 55 – 78 = 37.413 55 kg

mass %

Stoichiometry and Industrial Problems 309

Toluene column: Diphenyl in bottom product = 0.021 67 ´ 0.75 = 0.016 25 kmol º 2.502 89 kg Toluene in fresh feed = 1.3586 – 0.325 7742 = 1.032 858 kmol º 85.02 kg All these calculations can be prorated for 1000 kg/h benzene product from benzene column. These data are summarised in Table 8.55. Also these data are summarised in Fig. 3.6. Ans.

EXERCISE 8.14 Basis: Product gas stream from Reactor-II = 100 kmol/h HCN in the product gas stream = 5.9 kmol/h Reaction (ii): CH4 consumed = 5.9 kmol/h NO consumed = 5.9 kmol/h H2O produced = 5.9 kmol/h H2 produced =

5.9 = 2.95 kmol/h 2

Let a, b and c kmol/h of CH4, consumed as per reaction (iii), (iv) and (v), respectively. H2 produced = 4b + 3c as per reactions (iv) and (v) CO2 produced = a + b as per reactions (iii) and (iv) CO produced = c = 1.2 kmol/h Let d be the consumption (in kmol/h) of NH3 as per reaction (vi) H2O produced = 1.5 d Total H2 produced = 2.95 + 4b + 3c = 7.5 kmol/h Substituting value of c, b = 0.2375 kmol/h CO2 in product gas stream = 2.0 kmol/h = a + b a = 2 – 0.2375 = 1.7625 kmol/h O2 consumed as per reaction (iii) = 1.7625 ´ 2 = 3.525 kmol/h O2 consumed as per reaction (vi) =

3 d kmol/h 4

N2 produced by reaction (vi) = 0.5 d kmol/h N2 comes from air, introduced in Reactor I and that produced by reaction (vi). N2 from air = 56.7 – 0.5 d

310 Solutions Manual—Stoichiometry

100 (56.7 – 0.5d)= 79 H2O, entering with air = = NO production by reaction (i) = = H2O produced by reaction (i) = = Total H2O accounted = = = Substituting value of a, d = Total dry air, entering Reactor I = = H2O, entering Reactor I with air = =

Total dry air =

71.7722 – 0.6329 d

0.024 (71.7722 – 0.6329)d 1.7225 – 0.0152 d kmol/h NO consumed by reaction (ii) 5.9 kmol/h 1.5 ´ 5.9 8.85 kmol/h 5.9 + 2a + 1.5d + 8.85 + 1.7225 – 0.0152d 16.4725 + 2a + 1.4848 d 25.1 3.4365 kmol/h 71.7722 – 3.4365 ´ 0.6325 69.5972 kmol/h 1.7225 – 0.0152 ´ 3.4365 1.6699 kmol/h 21 (56.7 – 0.5d) O2, entering Reactor I in air = 79 21 (56.7 – 0.5 ´ 3.4365) = 79 = 14.6154 kmol/h N2, entering Reactor I = 56.7 – 0.5 ´ 3.4365 = 54.9818 kmol/h NH3, reacted as per reaction (vi) = d = 3.4365 kmol/h Total NH3, consumed in both reactors = 5.9 + 3.4365 + 1.6 = 10.9365 kmol/h Mixed feed entering Reactors I . . Component ni mole% Heat capacity (n i × Cop1) Equation constant . . . . kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 NH3 10.9365 13.30 280.52 366.16 3.85 –33.72 14.6154 17.78 380.38 171.81 –34.24 –8.22 O2 54.9818 66.89 1626.96 –282.66 724.82 –273.15 N2 1.6699 2.03 54.26 0.13 22.06 –7.59 H2O Total

82.2036

100.00

2342.12

Exit Gas Stream From Reactor I . Component ni mole % Heat . kmol/h ni × ai NH3 5.0365 6.02 129.19 7.2404 8.65 188.44 O2 54.9818 65.71 1626.96 N2

255.44

716.49

–322.68

. capacity (n i × Cop2) equation constant . . . ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 168.63 85.11 –282.66

1.77 –16.96 724.82

–15.53 –4.07 –273.15

Stoichiometry and Industrial Problems 311

NO H2O

5.9000 10.5199

7.05 12.57

173.97 341.81

–12.11 0.84

66.89 138.98

–28.44 –47.84

Total

83.6786

100.00

2460.37

–40.19

915.50

–369.03

Total CH4 consumed in Reactor II = a + b + c + 5.9 = 1.7625 + 0.2375 + 1.2 + 5.9 = 9.1 kmol/h Ans. (b) Gas Stream Entering Reactor II . . Component ni mole % Heat capacity (ni × C op3) equation Constant . . . . kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 NH3 5.0365 5.43 129.19 168.63 1.77 –15.53 7.2404 7.80 188.44 85.11 –16.96 –4.07 O2 54.9818 59.26 1626.96 –282.66 724.82 –273.15 N2 NO 5.9000 6.36 173.97 –12.11 66.89 –28.44 10.5199 11.34 341.81 0.84 138.98 –47.84 H2O CH4 9.1000 9.81 175.17 474.23 10.90 –102.99 Total

92.7786

100.00

2635.34

434.04

926.40

–472.02

Final Gas stream leaving Reactor II . . Component ni mole % Heat capacity (ni × C op4) equation Constant . . . . kmol/h ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 NH3 1.6 1.6 41.04 53.57 0.56 –4.93 56.7 56.7 1677.80 –291.49 747.47 –281.69 N2 HCN 5.9 5.9 192.89 133.29 –25.78 –2.41 25.1 25.1 815.55 2.00 331.59 –114.14 H2O 2.0 1.2 42.73 128.57 –82.10 19.60 CO2 CO 1.2 2.0 34.83 –3.38 13.97 –5.65 H2 7.5 7.5 214.58 7.65 –1.11 5.77 Total

100.00

100.0

3019.42

30.21

984.60

–383.45

Energy Balance: Enthalpy of Reactor I inlet stream 1 at t1/T1 = 900°C (1173.15 K) over 25°C (298.15 K), f1 =

1173.15

o ò (ni C p1 ) dT

298.15

= 244 0897 kJ/h º 678.027 kW Standard heat of reaction (i): DH or1 = 3(–241.82) + 2(90.25) – [2(–46.11) + 0] = – 452.74 kJ/2 mol NH3 º –226.37 kJ/mol NH3 Heat produced = 226.37 ´ 1000 ´ 5.9 f2 = 1335 583 kJ/h º 371.0 kW Enthalpy of Reactor I outlet stream over 25°C (298.15 K):

312 Solutions Manual—Stoichiometry

f3 = f1 + f2 = 2440 897 + 1335 583 = 3776 480 kJ/h º 1049.022 kW =

T2

o ò (ni × C p2 ) dT

298.15

Solving by Mathcad, T2 = 1595.4 K or t2 = 1322.25°C Ans. (c) Reactor II: Reaction (ii) DH or2 = –241.82 + 135.14 = – (–74.52 + 90.25) º –122.41 kJ/mol NO Reaction (iii) DHor3 = –393.51 + 2(–241.82) – (–74.52) = –802.63 kmol/CH4. It is NCV of CH4(g). Reaction (iv) DHor4 = –393.51 –[2(–241.82)] – (74.52)] = +164.65 kJ/mol CH4 Reaction (v): DHor5 = –110.53 –[–74.52 + 2 ´ (–241.82)] = +447.63 kJ/mol CH4 Reaction (vi): DHor6 = 3(–241.82) –2 (–46.11) = –633.24 kJ/2 mol NH3 º –316.52 kJ/mol NH3 Total heat of reaction in Reactor II. f4 = –122.41 ´ 1000 ´ 5.9 – 802.63 ´ 1000 ´ 1.7625 + 164.65 ´ 0.2375 ´ 1000 + 447.63 ´ 1000 ´ 1.2 – 316.52 ´ 1000 ´ 3.4365 = –722 219 – 1414 635 + 39 104 + 537 156 – 1087 721 = – 2648 315 kJ/h º –735.643 kW Enthalpy of exit stream from Reactor II at 1100°C (1373.15 K) over 25°C (298.15 K), f5 =

1373.15

o ò (ni × C p4 ) dT

298.15

= 3774 010 kJ/h º 1048.336 kW Enthalpy of gas stream, enteries Reactor II, f6 = f5 – f4 = 3774 010 – 2648 315 = 1125 695 kJ/h º 312.693 kW =

T3

o ò ( Sni × C p3 ) dT

298.15

Solving by Mathcad T3 = 671.80 K or t3 = 398.65°C

Stoichiometry and Industrial Problems 313

Methane is introduced at 35°C (308.15 K). Enthalpy of methane at 35°C (308.15 K) over 25°C (298.15 K) f7 = 175.17 (308.15 – 298.15) + 474.23 ´ 10–3 (308.152 – 298.152)/2 + 10.9 ´ 10–6 (308.153 – 298.153)/3 – 102.99 ´ 10–9 (308.154 – 298.154)/4 = 1751.7 + 1437.6 + 10.0 – 28.7 = 3170.6 kJ/h º 0.881 kW Enthalpy of exit stream of Reactor-I, required to be mixed to have a mixture temperature of 671.80 K. f8 = f6 – f7 = 1125 695 – 3170.6 = 1122 524 kJ/h º 311.812 kW T2¢

o ò ( Sni × C p2 ) dT

=

298.15

T ¢2 = 724.15 K or t¢2 = 451.0°C Ans. (d) Enthalpy to be extracted from exit gas stream from Reactor I (i.e. cooling from 1595.4 K to 724.15 K), f9 = 3776 480 – 1122 524 = 2653 956 kJ/h º 737.21 kW Enthalpy of mixed feed, entering Reactor I, at 35°C (308.15 K) over 25°C (298.15 K): 308.15

o ò (ni × C p1 ) dT

f10 =

298.15

= 247 642 kJ/h ´ 6.879 kW Enthalpy required to raise mixed feed, entering Reactor I, to 900°C (1173.15 K) from 35°C (308.15 K), f11 =

1173.15

o ò (ni × C p1 ) dT

298.15

= 2416 134 kJ/h º 671.148 kW Enthalpy amounting 671.148 kW is to be derived by exchanging heat with Reactor I exit gas stream. Enthalpy of exit gas stream from Reactor I, f12 = 3776 480 – 2416 134 = 1360 346 kJ/h º 377.874 kW In other words, enthalpy available from exit gas stream from Reactor I is more than required by mixed feed (to Reaactor I). Actual temperature of Reactor I exit gas stream will be higher than 451.0°C (724.15 K) T3

o ò (ni × C p2 ) dT = 1360 346

298.15

Solving by Mathacad, T3 = 809.05 K or t3 = 535.9°C Since exit gas stream from Reactor I is specified to be 1100°C (1373.15 K), there will be a heat loss from the system.

314 Solutions Manual—Stoichiometry

Heat loss = 1360 346 – 1122 524 = 237 822 kJ/h º 66.062 kW Ans. (f) Note: In normal run,heat input in the heater will not be required. It will be in use only during start-up. Air/Ammonia feed ratio in the mixed feed =

(69.5972 + 1.6699) = 6.519 kmol/kmol 10.9315

Ans. (a)

EXERCISE 8.15 Basis: Fresh feed rate F = 100 kmol/s H2 in F = 100 ´ 0.7 = 70 kmol/s Inerts in F = 100 ´ 0.01 = 1 kmol/s Let a = CO in mixed feed, kmol/s b = CO2 in mixed feed, kmol/s H2 in mixed feed = 1.05 (2a + 3b) = 2.1 a + 3.15 b Presence of CO, CO2 and H2 in MF = a + b + 2.1a + 3.15 b = 3.1a + 4.15 b kmol/s Methanol in MF = 0.45 mole % Inerts in MF = 10.0 mole % MF = (3.1a + 4.15 a)/(1.0000 – 0.1 – 0.0045) = 3.46175 a + 4.63 428b kmol/s Methanol in MF = (3.46175a + 4.634 28b) 0.0045 = 0.015 57 a + 0.020 85 b kmol/s Inerts in MF = (3.461 75 a + 4.634 28b)0.1 = 0.346 18 a + 0.463 43 b kmol/s Let x and y be kmol/s of CO and CO2 reacted, respectively. Composition of Converter Exit Gas Mixture Component CO CO2 H2 CH3OH H2O Inerts

kmol/s a–x b–y 2.1a + 3.15b – 2x – 3y 0.015 57 a + 0.020 85b + x + y y 0.346 18a + 0.463 43 b

Total 3.461 75a + 4.634 28 b – 2x – 2y Gas mixture, leaving the converter, contains 3 mole % methanol.

(0.015 57 a + 0.020 85b + x + y ) = 0.03 (3.461 75a + 4.634 28b - 2 x - 2 y )

Let c kmol/s of methanol are condensed alongwith y kmol/s of water.

(1)

Stoichiometry and Industrial Problems 315

Gas mixture after methanol condensation, T = 3.461 75 a + 4.634 28 b – 2x – 2y – y – c = 3.461 75a + 4.634 28 b – 2x – 3y – c kmol/s Methanol balance: Methanol in recycle gas stream = methanol in mixed feed (0.015 57a + 0.020 85 b + x + y – c) - (0.015 57 a + 0.020 85 b + x + y - c) P T = 0.015 57a + 0.020 85 b x + y – c = - (0.015 57 a + 0.020 85b + x + y - c) P (2) T Overall material balance: T – P + 100 = M = 3.461 75a + 4.634 28 b Simplifying, 2x + 3y + c = 100 – P (3) Hydrogen balance: 2x + 3y + (2.1a + 3.15 b - 2 x - 3 y ) P = 70 T

(4)

Inerts balance: (0.346 18a + 0.463 43b) P = 1 (5) T Reaction equilibrium of water gas reaction at 275°C (584.15 K) in converter exit gas mixture:

K=

(b - y ) (2.1a + 3.15 b - 2 x - 3 y ) (a - x) y

= 69 Solving six simultaneous equations by Mathcad, Mathcad output: a = 133.849 kmol/s b = 140.201 kmol/s c = 26.733 kmol/s x = 16.253 kmol/s y = 10.526 kmol/s P = 9.184 kmol/s Compositions of Gas streams Converter inlet (mixed feed – M) CO CO2

(6)

Converter outlet

kmol/s

mole %

kmol/s

mole %

133.849 140.201

12.02 12.60

117.596 129.675

11.10 12.24

316 Solutions Manual—Stoichiometry

H2 Inerts CH3OH H2O Total

722.716 111.308 5.009 Nil

64.93 10.00 0.45 Nil

658.632 111.308 31.788 10.526

62.16 10.51 3.00 0.99

1113.083

100.00

1059.525

100.00

Composition of Gas Streams Gas mixture after condensation CO CO2 H2 Inerts CH3OH Total

kmol/s

mole %

117.596 129.675 658.632 111.308 5.055 1022.266

Purge steam kmol/s

mole %

11.50 12.69 64.43 10.89 0.49

1.056 1.165 5.917 1.000 0.045

11.50 12.69 64.43 10.89 0.49

100.00

9.184

100.00

Compositions of Gas Streams Recycle (R) stream Component CO CO2 H2 Inerts CH3OH Total

kmol/s

Fresh feed (M-R)

mole %

kmol/s

mole %

116.540 128.510 652.715 110.308 5.010

11.50 12.69 64.43 10.89 0.49

17.309 11.691 70.000 1.000 Nil

17.31 11.69 70.00 1.00 Nil

1013.083

100.00

100.000

100.00

Aqueous methanol condensed = 26.733 + 10.526 = 37.259 kmol/s Mass flow rate of equeous methanol = 26.733 ´ 32.0419 + 10.526 ´ 18.0153 = 856.576 + 189.629 = 1046.205 kg/s Concentration of aqueous solution 856.576 ´ 100 1046.205 = 81.87 mass %

=

16.253 ´ 100 133.849 = 12.14% per pass

Conversion of CO =

Stoichiometry and Industrial Problems 317

10.526 ´ 100 140.201 = 7.51% per pass

Conversion of CO2 =

(16.253 ´ 2 + 10.526 ´ 3) 100 722.716 = 8.87% per pass

Conversion of H2 =

Recycle ratio =

1013.083 = 10.131 kmol/kmol fresh feed 100

Enthalpy balance: Reference temperature, T0 = 298.15 K Heat Capacity Equation Constants for Converter Inlet Gas Stream Component . Component ni Heat capacity equation constant . . . . kmol/s ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 CO 133.849 3885.3 –377.0 1558.5 –629.9 140.201 2995.5 9012.7 –5755.4 1374.0 CO2 722.716 20 677.3 736.7 –106.7 555.8 H2 Ar 33.392 693.6 – – – CH4 77.916 1499.8 4060.5 932.9 –881.8 5.009 124.6 254.8 293.7 –226.0 CH3OH Total

1113.083

29 876.1

13 687.7

–3077.0

192.1

Heat Capacity Equation Constants for Converter Exit Gas Stream . Component ni Heat capacity equation constant . . . . kmol/s ni × ai ni × bi ´ 103 ni × ci ´ 106 ni × di ´ 109 CO 117.596 3413.5 –331.2 1369.3 –553.4 129.675 2770.6 8336.0 –5323.2 1270.8 CO2 658.632 18843.8 671.4 –97.2 506.5 H2 Ar 33.392 693.6 – – – 77,916 1499.8 4060.5 932.9 –881.8 CH4 31.788 790.5 1617.2 1863.6 –1434.5 CH3OH H2O 10.526 342.0 0.8 139.1 –47.9 Total

1059.525

28 353.8

14 354.7

–1115.5

Standard heats of reaction: CO(g) + 2 H2(g) = CH2OH(g)

DH or1 = –200.94 – (–110.53)

= –90.41 kJ/mol CO reacted

CO2(g) + 3 H2(g) = CH3OH(g) + H2O(g)

DH or2 = –200.94 + (–241.82) – (–393.51)

–1140.3

318 Solutions Manual—Stoichiometry

= –49.25 kJ/mol CO2 reacted. Both are exothermic reactions. Total heat released f2 = 90.41 ´ 1000 ´ 16.253 + 49.25 ´ 1000 ´ 10.526 = 1469 433.7 + 518 405.5 = 1987 839.2 kJ/s or kW Enthalpy of converter exit gas stream at 275°C (548.15 K) over 298.15 K, f3 =

548.15

ò

298.15

(28.353.8 + 14 354.7 ´ 10–3 T – 1115.5 ´ 10–6

T2 – 1140.3 ´ 10–9 T3) dT = 8532 126.9 kJ/s or kW Enthalpy of converter inlet gas stream, f1 = 8532 126.9 – 1987 839.2 = 6544 287.7 kJ/s or kW =

T1

ò

298.15

(29876.1 + 13 687.7 ´ 10–3 T – 3077.0 ´ 10–6 T 2 + 192.1 ´ 10–9 T 3) dT

Solving by Mathacad Ans. T1 = 486.47 K or t1 = 213.32°C Thus ingoing gas stream is to be heated from 66°C (339.15 K) to 213.32°C (486.47 K). Heat exchange in the heat exchanger (HE-1), f4 =

486.47

ò

339.15

(29876.1 + 13 687.7 ´ 10–3 T – 3077.0 ´ 10–6 T2 + 192.1 ´ 10–9 T 3) dT

= 5157 753 kJ/h º 1432.709 kW If temperature of reactor exit gas stream after cooling in HE-1 is T 2, 248.15

ò

T2

(28 353.8 + 14 354.7 ´ 10 –3 T – 1115.5 ´ 10 –6 T2 – 1140.3 ´ 10–9 T 3) dT = 5157 753 kJ/h

Solving by Mathcad, T 2 = 399.86 K or t2 = 126.71°C

Ans.

Stoichiometry and Industrial Problems 319

EXERCISE 8.16 Heat duty of chiller, fch = 1500 kW . Let mch be the mass flow rate of chilled water. . fch = mch × CL × Dt . mch =

1500 = 119.423 kg/s 4.1868 ´ (12 - 9)

» 430 m3/h In the chiller, water is used as a refrigerant. Latent heat of water at 6°C (279.15 K) = 2487.4 kJ/kg 1500 Evaporation rate of water = 2487.4 . m1 = 0.603 kg/s . Let m2 = Mass flow rate of SA solution, entering absorber, kg/h . m3 = Mass flow rate of WA solution, leaving absorber, kg/h . . m3 – m2 = Water evaporation in chiller + Water vapour formed due to flashing of water, coming from condenser = 0.603 ´ 3600 + x = 2170.8 + x Also considering overall material balance, . . m3 – m2 = Amount of water from condenser =y Enthalpy of water in condenser = 191.83 kJ/kg at 45.83°C (318.98 K) Enthalpy of water at 6°C = 25.31 kJ/kg Change in enthalpy will result in flashing. y (191.83 – 25.31) = x ´ 2487.4 y = 14.938 x = 2170.8 + x x = 155.75 kg/h . . m3 – m2 = 2170.8 + 155.75 = 2326.55 kg/h (1) LiBr balance: . . m2 ´ 0.633 = m3 ´ 0.595 . . m2 = 0.94 m3 . Substituting value of m2 in Eq. (1), . 0.06 m3 = 2326.55 . m3 = 38 775.83 kg/h . m2 = 36 449.28 kg/h

320 Solutions Manual—Stoichiometry

Heat of dilution will be assumed constant at 498 kJ/kg for determining heat duties of the generator and absorber. Heat duty of absorber fa = Heat of dilution of LiBr solution + Heat of condensation of water vapour + sensible heat change in solution = 2326.55 ´ 498 + 2326.55 ´ 2487.4 æ 38 775.83 + 36 449.28 ö + ç ÷ø 1.842 (48.9 – 40) è 2

Heat duty of generator,

= 1158 621.9 + 5787 060.5 + 616 612.7 = 7562 295.1 kJ/h º 2100.638 kW fg = Sensible heat change of WA from 75°C (348.15 K) to 88.9°C (362.05 K) + Sensible heat change from 88.9°C (362.05 K) to 101.6°C (374.75 K) + Heat of concentration of LiBr solution + Latent heat of vaporization of water at 45.83°C (318.98 K) = 38 775.83 ´ 1.842 (88.9 – 75)

é 38 775.83 + 36 449.28 ù + ê ú 1.842 (101.6 – 2 ë û 88.9) + 2326.55 ´ 498 + 2326.55 ´ 2392.97 = 992 808.6 + 879 885.5 + 1158 621.9 + 5567 364.4 = 8598 680.6 kJ/h º 2388.522 kW Saturated steam at 1.85 bar a: l v = 2208.5 kJ/kg Steam consumption in generator, . 8598 680.6 ms = = 3893.4 kg/h 2208.5 º 1.0815 kg/s Coefficient of performance: Effective refrigeration achieved COP = Total heat input to generator 1500 = = 0.628 kW/kW 2388.522 Heat duty of condenser, fc = Cooling superheated water vapour + condensation = 2326.55 (2690.6 – 2584.8) + 2326.55 ´ 2392.97 = 246 149 + 5567 364.4 = 58 13 513.4 kJ/h º 1614.865 kW

Stoichiometry and Industrial Problems 321

Cooling water enters at 32°C (305 K) into absorber coils and leave at 40°C (313 K) from the condenser. Total heat duty = fa + fc = 2100.638 + 1614.865 = 3715.503 kW

3715.503 . Required CW flow, mcw = 8 ´ 4.1868 = 110.929 kg/s º 399.3 m3/h Rise in CW temperature across absorber =

2100.638 = 4.523°C or K 110.929 ´ 4.1868

Rise in CW temperature across condenser = 8 – 4.523 = 3.477°C or K

EXERCISE 8.17 Before attempting the actual steam balance, enthalpies of steam at different pressure/temperature, of condensates and boiler feed water are to be tabulated. Steam or Pressure, Condensate or MPa a BFW Steam 4.6 Steam 0.95 Steam 0.95 Steam 0.1013 Steam 0.1013 Steam 0.014 Steam 0.014 Condensate 0.014 Condensate 0.014 Condendsate 0.014 BFW 4.9

Temperature, K (°C) 673 553 450.8 473 373 325.7 325.7 325.7 373 450.8 394

(400) (280) (177.7) (200) (100) (52.6) (52.6) (52.6) (100) (177.7) (121)

Condition

Enthalpy kJ/kg

Superheated Superheated Saturated Superheated Saturated 0.955 DF 0.867 DF Saturated Saturated Saturated Subcooled

3205.3 3010.5 2774.2 2875.3 2676.0 2489.7 2280.6 220.0 419.1 752.8 508.0

Turbine A: Heat transfer in turbine = 3205.3 – 3010.5 = 194.8 kJ/kg 1426 ´ 3600 = 27 740.2 kg/h Steam requirement in turbine = 0.95 ´ 194.8 Let x1 kg/h BFW is introduced in steam at 0.95 MPa a and 553 K (280°C). x1 ´ 508.0 + 27 740.2 ´ 3010.5 = (27 740.2 + x1) 2774.2 x1 = 2891.3 kg/h Total saturated steam at 0.95 MPa a = 27 740.2 + 2891.3 = 30 631.5 kg/h

322 Solutions Manual—Stoichiometry

Turbine B: Heat transfer in turbine = 3205.3 – 2489.7 = 715.6 kJ/kg 5974 ´ 3600 Steam requirement in turbine = = 31 635.4 kg/h 0.95 ´ 715.6 Turbine C: Total energy requirement = 2460 ´ 3600/0.95 = 9322 105.3 kJ/h or 2589.5 kW Saturated steam requirement for absorption refrigeration = 5940 kg/h Let x2 kg/h BFW is used for desuperheating steam @ 0.1013 MPa a and 473.15 K (200°C). 5940 ´ 2676.0 = x2 ´ 508 + (5940 – x2) 2875.3 x2 = 500.1 kg/h Superheated extraction steam = 5940 – 500.1 = 5439.9 kg/h from Turbine C Enthalpy removed in extraction part of Turbine C: = 5439.9 (3205.3 – 2875.3) = 1795 167 kJ/h º 498.7 kW Enthalpy removal in condensing part = 2589.5 – 498.7 = 2090.8 kW º 7526 938 kJ/h Steam requirement = 7526 938/(3205.3 – 2280.6) = 8139.9 kg/h Total steam input to Turbine C = 5439.9 + 8139.9 = 13 579.8 kg/h say 13 580 kg/h Total condensate to surface condenser = 8139.9 + 31 635.4 = 39 775.3 kg/h Enthalpy of total condensates, entering deaerator = 39 775.3 ´ 220 + 5940 ´ 419.1 + 30 631.5 ´ 752.8 = 34 299 413 kJ/h º 9527.6 kW Loss of enthalpy = 5% Let x3 kg/h be superheated steam, used in deaerator. Total condensate, entering deaerator = 39 775.3 + 30 631.5 + 5940.0 = 75 346.8 kg/h 34 299 413 ´ 0.95 + x3 ´ 3205.3 = (76 346.8 + x3) 508.0 x3 = 2298.5 kg/h BFW flow = 76 346.8 + 2298.5 = 78 645.3 kg/h

Stoichiometry and Industrial Problems 323

Total steam requirement @ 46 bar a and 673 K (400°C) = 2298.5 + 27 740.2 + 31 635.4 + 13 579.8 = 75 253.9 kg/h say 75 254 kg/h

Ans.

EXERCISE 8.18 Basis: Power requirement = 6832 kW Specific Steam Requirements of Turbines Initial Steam Conditions

Final steam conditions

Enthalpy available for energy conversion, kJ/kg

Actual energy conversion (97%), kJ/kg

Soevific steam consumption in turbine, kg/kWh

64 bar a 440°C (713 K) 3273.3 kJ/kg

39 bar a 385°C (658 K) 3182.1 kJ/kg

91.2

88.464

40.6945

303.804

11.8497

684.723

5.2576

596.259

6.0376

Final steam conditions

BFW

39 bar a 4.4 bar a 313.2 385°C 205°C (658 K) (478 K) 3182.1 kJ/kg 2868.9 kJ/kg 64 bar a 12 kPa a 705.9 440°C 50°C (323 K) (713 K) 99% dry 3273.4 kJ/kg 2567.4 kJ/kg 39 bar a 12 kPa a 614.7 385°C 50°C (323 K) (658 K) 99% dry 3182.1 kJ/kg 2567.4 kJ/kg BFW Requirements for Desuperheating Initial steam conditions Pressure bar a

Temp. °C(K)

Enthalpy kJ/kg

Pressure bar a

64 39 39 4.4

440 (713) 385 (658) 385 (658) 205 (478)

3273.3 3182.1 3182.1 2868.9

39 15 4.4 4.4

Temp. °C (K) 385 (658) Satd. Satd. Satd.

Enthalpy kJ/kg 3182.1 2789.9 2741.9 2741.9

requirement, kg/kg initial Steam 0.0333 0.1669 01912 0.0552

324 Solutions Manual—Stoichiometry

Nomenclature will be same as that followed in Example 8.7. Saturated 15 bar a steam requirement = 3 t/h Equivalent MP steam =

3 = 2.5709 t/h (1 + 0.1669)

Saturated LP steam header: Requirement = 13.5 + a t/h Exhaust from BFW and FD turbines = c + d Equivalent saturated LP steam Production = (1 + 0.0552) (c + d) + 1.1912 g 1.0552 (c + d) + 1.1912 g = 13.5 + a Deaerator: Equation remains unchanged a – 0.11 b – 0.11 e = 1.815 Power turbine: (b + c + d + g + 2,5709 - 1.0333 h) 1000 1000 e = 6832 + 40.6945 5.2476 or 5.2576 b + 5.2576 c + 5.2576 d + 5.2576 g + 40.6945 e – 5.4327 h = 1461.7433 FD fan turbine:

d=

(1) (2)

(3)

2.47 (b + c + d + e + g - 0.0333h + 2.5709) 11.8497 1000 b + c – 33.1661 d + e + g – 0.0333h = –2.5709 (4)

BFW pump turbine: Head developed = 70 bar = 742.907 m H2O at 105°C (378 K) Power requirement =

(a + b + e + 16.5) 742.907 ´ 0.7355 ´ 1000 3600 ´ 75 ´ 0.65

= 3.1134 (a + b + e + 16.5) 3.1134 (a + b + e + 16.5) 11.8497 =c 1000 a + b + e – 27.1055 c = –16.5 CW pump turbine: Enthalpy, given-up in SC = 2567.4 – 209.2 = 2358.2 kJ/kg steam

CW requirement in

SC =

2358.2 (b + e) 10 ´ 4.1868

= 56.3246 (b + e) t/h or m3/h Total CW requirement in the plant = 56.3246 (b + e) + 600 m3/h

(5)

Stoichiometry and Industrial Problems 325

Power requirement of CW pump =

[56.3246 (b + e) + 600] 51 ´ 0.7355 ´ 1000 3600 ´ 75 ´ 0.6

= 02315 [ 56.3246 (b + e) + 600] kW 0.2315[56.3246 (b + e) + 600] 6.0376 1000 656.1343 b – 56.3246 e = 600 All six equations are solved by Mathcad. Assume h = 0. HP steam generation, Z = b + c + d + e + g + 2.571 = 3.771 + 2.193 + 1.649 + 33.273 + 12.875 + 2.571 = 56.332 t/h Fig. 8.21 is the steam balance with all values. Increased in HP steam generation over that

b=

(6)

(56.332 - 53.618) 100 53.618 = 5.06 % Energy absorption in 64 bar a boiler = 56 332 (3273.3 – 440.17) = 159 595 879 kJ/h º 44 332.188 kW Energy absorption in 115 bar a boiler (Example 8.6) = 53 618 (3190.7 – 440.17) = 147 477 918 kJ/h º 40 966.088 kW Increase in energy absorption in boiler = 159 595 879 – 147 477 918 = 12 117 961 kJ/h º 3366.1 kW

in Example 8.7 =

Increase in energy consumption in boiler = Additional fuel requirement =

12 117 961 ´ 100 = 8.22% 147 477 918 12 117 961 0.75 ´ 40 000

= 403.93 kg/h

EXERCISE 8.19 Refer Example 8.7. Its basis is used for following calculations. LP steam header pressure = 4 bar a Enthalpy of superheated steam at 4 bar a and 438 K (165°C) = 2785.1 kJ/kg

326 Solutions Manual—Stoichiometry

At 4 bar a, Ts = 416.77 K (143.62 °C) h = 604.67 kJ/kg and H = 2737.6 kJ/kg For letdown from MP to LP header, let x1 be kg of BFW injected. 3020.4 + 440.17 x1 = (1 + x1) 2737.6 x1 = 0.123 kg/kg MP steam For production LP saturated steam from LP superheated steam, let x2 kg BFW is injected. 2785.1 + 440.17 x2 = (1 + x2) 2737.6 x2 = 0.0207 kg/kg LP superheated steam Steam ejectors for inter-after condenser and other LP ejectors will consume 10% more steam. LP steam consumption to inter-after condenser = 500 ´ 1.1 = 550 kg/h LP steam consumption to other ejectors = 2000 ´ 1.1 = 2200 kg/h BF make-up will increase from 2000 to 2200 kg/h. LP steam consumption in heaters will reduce by 1.75%. Steam consumption in brine heater = 2000 ´ 0.9825 = 1965 kg/h Steam consumption in first stage evaporator = 9000 ´ 0.9825 = 8842 kg/h Symbols of Example 8.6 are used. LP steam header: LP steam requirement = 8.842 + 1.965 + 2.2 + 0.55 + a = 13.557 + a t/h LP steam production = 1.0207 (c + d) + 1.123 g 1.0207 (c + d) + 1.123 g = 13.557 + a (1) Deaerator: a = 0.11 (b + e + 16.557) a – 0.11 b – 0.11 e = 1.821 (2) Power Turbine: No change in Eq. (3). FD fan: Enthalpy drop in turbine = 3020.4 – 2785.1 = 235.3 kJ/kg Specific steam consumption =

3600 235.3 ´ 0.97

= 15.773 kg/kWh

Stoichiometry and Industrial Problems 327

2.47 ´ 15.773(b + c + d + e + g + 2.74 ) 1000 b + c – 24.668 d + e + g = – 2.74 (4) BFW pump:

d=

5.493 ´ 15.773( a + b + e + 16.95) 1000 a + b + e – 11.542 c = – 16.95 CW pump turbine: No change in Eq. (6). Solving by Mathcad a = 4.674 t/h b = 3.778 t/h c = 4.776 t/h d = 2.069 t/h e = 29.718 t/h g = 10.014 t/h Z = 53.095 t/h Refer Fig. 8.22 in the text for the steam balance.

c=

(5)

Ans.

EXERCISE 8.20 Refer solution of Example 8.7. In this example, cooling water pump turbine is shown to be condensing. In this Exercise, this CW pump turbine is considered to be a back pressure type. Production of superheated LP steam = b + c + d t/h Corresponding saturated LP stream = 1.022 (b + c + d) + 1.121 g t/h 1.022 (b + c + d) + 1.121 g = 13.5 + a (1) Deaerator: Condensate from SC = e + 0.5 t/h a = 0.11 (e + 16.5) a = 0.11 e + 1.815 or a – 0.11 e = 1.815 (2) Equations, concerning power turbine and FD fan turbine remain unchanged. Thus equations (3) and (4) of Example 8.6 are unchanged. 5.16 b + 5.16 c + 5.16 d +21.79 e + 5.16 g – 5.50 h = 754.03 (3) b + c – 23.78 d + e + g – 0.066 h = – 2.74 (4) BFW pump: Total BFW = a + e + 16.5 t/h c = 5.493 (a + e + 16.5) 16.34/1000

328 Solutions Manual—Stoichiometry

a + e – 11.14 c = – 16.5

CW pump:

(5)

CW requirement = 54.05 e + 600 m3/h b = 0.2315 (54.05 e + 600) 16.34/1000 264.36 b – 54.05 e = 600

(6) Coefficient matrix of constraint equations: Eq. No.

a

6 2 5 1 3 4

0 1 1 –1 0 0

b

c

1 0 0 0 0 – 11.14 1.022 1.022 5.16 5.16 1 1

d

g

h

e

Constant

0 0 0 1.121 5.16 1

0 0 – 0.2045 2.2696 0 0 – 0.11 1.815 0 0 1 – 16.500 1.121 0 0 13.50 5.16 – 5.50 21.79 768.17 1 – 0.066 1 – 2.74

Mathcad solution Matrix

. O LM-1 1.022 1.022 1.022 0 1121 PP -0.11 0 0 0 MM 10 0516 . 5.16 5.16 21.79 5.16 P M: = M PP -23.78 1 1 1 MM 01 10 -1114 . 0 1 0 PP MN 0 264.36 0 -54.05 0 0 Q LM 13.5 OP MM 7541..815 P 03 P v: = M MM--162..574 PPP MN 600 PQ . OP LM 5153 474 PP MM 48..668 Solution = M × v Solution = M 061 P PP MM302..346 MN 2.78 PQ –1

Refer Fig. 8.23 in the text for steam balance.

Ans.

Stoichiometry and Industrial Problems 329

EXERCISE 8.21 Nomenclature will be same as that of Example 8.7. Initial steam conditions

Final steam conditions

Enthalpy available for energy conversion, kJ/kg 170.3

Actual energy Specific conversion steam (97%), consumption kJ/kg in turbine kg/kWh 165.2 21.79

115 bar a 713 K (440 °C) 3190.7 kJ/kg

39 bar a 593 K (320 °C) 3020.4 kJ/kg

36 bar a 673 K (400 °C) 3222.5 kJ/kg

4.4 bar a 458 K (185 °C) 2826.0 kJ/kg

396.5

384.6

9.36

115 bar a 713 K (440 °C) 3190.7 kJ/kg

12 kPa a 323 K (50 °C), 95% dry 2472.0 kJ/kg

718.7

697.1

5.16

36 bar a 673 K (400 °C) 3222.5 kJ/kg

12 kPa a 323 K (50 °C), 95% dry 2472.0 kJ/kg

750.5

728.0

4.945

BFW requirement for desuperheating LP steam: Let x2 kg BFW is to be injected in the desuperheater per kg of LP superheated steam at 458 K (185 °C) 2826.0 + 440.17 x2 = (1 + x2) 2741.9 x2 = 0.0365 kg/kg LP superheated steam Rest BFW requirements for desuperheating will remain unchanged (refer Table 8.17). Assume h = 0. Saturated LP Steam Header: 1.0365 (c + d ) + 1.121 g = 13.5 + a – a + 1.0365 c + 1.0365 d + 1.121 g = 13.5 (1) Dearerator: Balance is unchanged. a – 0.11 b – 0.11 e = 1.815 (2) Power Turbine: Balance is unchanged. 5.16 b + 5.16 c + 5.16 d + 21.79 e + 5.16 g = 754.03 (3) FD Fan Turbine: 2.47( b + c + d + e + g + 2.74)9.36 d= 1000 b + c – 42.254 d + e + g = – 2.74 (4) BFW Pump Turbine: 5.493( a + b + e + 16.5)9.36 c= 1000 a + b + e – 19.45 c = – 16.5 (5)

330 Solutions Manual—Stoichiometry

CW Pump Turbine: 0.2315(54.05b + 54.05e + 600 )4.945 1000 819.49 b – 54.05 e = 600 (6) Mathcad solution: a = 5.401 t/h b = 2.704 t/h c = 2.802 t/h d = 1.214 t/h e = 29.900 t/h g = 13.148 t/h Z = 52.508 t/h Ans. (53.624 - 52.508)100 Saving in HP steam generation = 53.624 = 2.08% (33.822 - 33104 . )100 Reduction in heat load of cooling tower = 33.822 = 2.12% Refer Fig. 8.24 in the text for steam balance. Ans.

b=

EXERCISE 8.22 Basis: 10 000 kg/h dry PP handling Inlet n-hexane content = 35% on wet basis Dry basis n-hexane content = 35 ´ 100/65 = 53.85% n-Hexane removal rate = 10 000 (0.5385 – 0.02) = 5185 kg/h Antoine constants for n-hexane (Table – 5.4) A = 4.002 66, B = 1171.53 and C = – 48.784 Nitrogen, ingoing to the spiral dryer has a dew point of T1 = 307.15 K (34°C). 1171.53 log10 p1 = 4.002 66 – (307.15 - 48.784) = 0.666 35 p1 = 0.294 bar º 29.4 kPa 29.4 86.1754 ´ 28.0134 (102 - 29.4) = 1.2457 kg n-hexane/kg dry N2 Nitrogen circulation rate = 10 000 ´ 0.18 = 1800 kg/h º 64.255 kmol/h

n-Hexane content =

Stoichiometry and Industrial Problems 331

n-Hexane circulation in N2 = 1800 ´ 1.2457 = 2242.26 kg/h º 26.02 kmol/h Nitrogen-n-Hexane mixture is heated from 343.15 K (70°C) to 403.15 K (130 °C). Heat input to nitrogen heater, f1 = 64.255 [29.5909 (403.15 – 343.15) – 5.141 ´ 10–3 (403.152 – 343.152)/2 + 13.1829 ´ 10–6 (403.153 – 343.153)/3 – 4.968 ´ 10–9 (403.154 – 343.154)/4] + 26.02 [– 4.4142 (403.15 – 343.15) + 581.9233 ´ 10–3 (403.152 – 343.152)/2 – 311.8584 ´ 10–6 (403.153 – 343.153)/3 + 64.9193 ´ 10–9 (403.154 – 343.154)/4] = 64.255 ´ 1755.1 + 26.02 ´ 10 591.47 = 112 774 + 275 590 = 388 364 kJ/h º 107.879 kW lv of saturated steam at 0.4 MPa a = 2133.0 kJ/kg (Table AIV.2) Steam requirement = 388 364/2133 = 182.1 kg/h Ans. (a) For T2 = 363.15 K (90°C), log10 p2 = 4.002 66 –

1171.53 (363.15 - 48.784)

= 0.276 02 p2 = 1.888 bar º 188.8 kPa Pressure of N2-n-Hexane mixture, leaving the separator (i.e. the mixture entering interchanger) = 110 kPa n-Hexane content of the mixture = 2242.3 + 5185 = 7427.3 kg/h Specific n-hexane content = 7427.3/1800 = 4.1263 kg/kg dry N2 If partial pressure of n-hexane is p3 after interchanger, p3 86.1754 ´ = 4.1263 110 - pa 28.0134

p3 = 63.02 kPa º 0.6302 bar log10 p3 = – 0.200 52 = 4.002 66 –

1171.53 (T3 - 48.784)

Saturation temperature T3 = 327.5 K (54.35°C) Reference temperature To = 298.15 K (25°C) Enthalpy of N2-n-hexane mixture, leaving the seperator H2= 64.255 [29.5909 (363.15 – 298.15) – 5.141 ´ 10–3 (363.152 – 298.152)/2

332 Solutions Manual—Stoichiometry

+ 13.1829 ´ 10–6 (363.153 – 298.153)/3 – 4.968 ´ 10–9 (363.154 – 298.154)/4] + (7411.6/86.1754) [– 4.4142 (363.15 – 298.15) + 581.9233 + 10–3 (363.152 – 298.152)/2 – 311.8584 ´ 10–6 (363.153 – 298.153)/3 + 64.9193 ´ 10–9 (363.154 – 298.154)/4 f2 = 64.255 ´ 1895 + 86.188 ´ 10 150.6 = 121 763 + 874 860 = 996 623 kJ/h º 276.840 kW In the interchanger, N2-n-hexane mixture from the blower at 307.15 K (34°C) is heated to 343.15 K (70°C). n-Hexane content = 25.837 kmol/h Heat added to the mixture, f3 = 64.255 [29.5909 (343.15 – 307) – 5.141 ´ 10–3 (343.152 – 307.152)/2 + 13.1829 ´ 10–6 (343.153 – 307.153)/3 – 4.968 ´ 10–9 (343.154 – 307.154)/4] + 26.02 [– 4.4142 (343.15 – 307.15) + 581.9233 ´ 10–3 (343.152 – 307.152)/2 – 311.8584 ´ 10–6 (343.153 – 307.153)/3 + 64.9193 ´ 10–9 (343.154 – 307.154)/4 = 64.255 ´ 1049.1 + 26.02 ´ 5543.0 = 67 409.9 + 144 229 = 211 638.9 kJ/h = 58.789 kW Average heat capacity of N2-n-hexane mixture, leaving the separator =

996 623 (64.255 + 86.188)(363.15 - 298.15)

= 101.92 kJ/(kmol × K) If it is assumed that there is no condensation of n-hexane from this mixture in the interchanger, drop in temperature = [211 638.9/(150.443 ´ 101.92)] = 13.8 K Temperature of the mixture, leaving the interchanger = 363.15 – 13.8 = 349.35 K or 76.2°C Ans. (c) This temperature is more than the calculated dew point = 327.5 K (54.35 °C) and hence no condensation will take place in the interchanger. Latent heat of evaporation (lv) of n-hexane at 298.15 K (25°C) is calculated using Watson equation [Ref. Eq. (5.25)].

Stoichiometry and Industrial Problems 333 0.98

é 507.6 - 298.15 ù lv = ê ´ 28 850 ú ë 507.6 - 341.9 û = 31 537 kJ/kmol n-hexane Heat balance across spiral dryer: Reference temperature To = 298.15 K (25 °C) Enthalpy of ingoing N2-n-hexane gas mixture, f4 = 64.255 [29.5909 (403.15 – 298.15) – 5.141 ´ 10–3 (403.152 – 298.152)/2 + 13.1829 ´ 10–6 (403.153 – 298.153)/3 – 4.968 ´ 10–9 (403.154 – 298.154)/4] + 26.02 [– 4.4142 (403.15 – 298.15) + 581.9233 ´ 10–3 (403.152 – 298.152)/2 – 311.8584 ´ 10–6 (403.153 – 298.153)/3 + 64.9193 ´ 10–9 (403.154 – 298.154)/4 = 64.255 ´ 3066.2 + 26.02 ´ 17 000.0 = 197 019 + 442 340 = 639 359 kJ/h º 177.6 kW Enthalpy of outcoming N2-n-hexane gas mixture, f2 = 996 623 kJ/h = 276.840 kW Enthalpy of feed (wet PP), f5 = 10 000 ´ 1.926 (338.15 – 298.15) + 5385 (338.15 – 298.15) ´ 2.512 = 770 400 + 541 085 = 1311 485 kJ/h º 364.301 kW Enthalpy of outgoing PP-n-hexane mixture, f6 = 10 000 ´ 1.926 (353.15 – 298.15) + 200 ´ 2.512 (353.15 – 298.15) = 1059 300 + 27 632 = 1086 932 kJ/h º 301.926 kW Heat required for evaporation, f7 = 31 537 ´ 5185/86.1754 = 1897 518 kJ/h º 527.088 kW Net heat input to jacket = f2 + f7 + f6 – f4 – f5 = 276.840 + 527.088 + 301.926 – 177.600 – 364.301 = 563.953 kW

334 Solutions Manual—Stoichiometry

Steam requirement in jacket = 563.953 ´ 3600/2133 = 951.8 kg/h

Ans. (b)

EXERCISE 8.23 Basis: Dry nitrogen flow to AFD = 100 Nm3/h Molar flow rate =

100 = 4.4615 kmol/h 22.414

Solvent laden nitrogen comes out from the dryer at 105 kPa and 90°C (363.15 K) with a DP = 10°C (283.15 K). Refer Table 5.4 for Antoine constants for CH2Cl2. At 10°C, log10 pv1 = 4.53691 +

1327.016 T - 20.474

T = 283.15 K log pv1 = 4.536 91 – 5.50 191 = – 0.515 pv1 = 0.3055 bar or 30.55 kPa CH2Cl2 content of N2 from ldryer, 30.55 = 0.4103 kmol/kmol dry N2 (105 - 30.55) . CH2Cl2 in N2 stream from dryer, n1 = 4.4615 ´ 0.4103 = 1.8306 kmol/h N2 stream from dryer is cooled from 90°C to 40°C in a cooler with cooling water. Since DP = 10°C, no condensation will take place.

y1 =

f1 = 4.4615

363.15

363.15

313.15

313.15

o ò C pN2 dT + 1.8306 ò C pmc dT

= 4.4615 [29.5908 (363.15 – 313.15) – 5.141 ´ 10–3 (363.152 – 313.152)/2 + 13.1829 ´ 10–6 (363.153 – 313.153)/3 – 4.968 ´ 10–9 (363.154 – 313.154)/4] + 1.8306 [16.9092 (363.15 – 313.15) + 140.4533 ´ 10–3 (363.152 – 313.152)/2 – 94.028 ´ 10–6(363.153 – 313.153)/3 + 23.7905 ´ 10–9(363.154 – 313.154/4] f1 = 4.4615 [1479.54 – 86.92 + 75.51 – 9.66] + 1.8306 [845.46 + 2374.71 – 538.56 + 46.25] = 4.4615 ´ 1458.47 + 1.8306 ´ 2727.86 = 6507 + 4994 = 11 501 kJ/h º 3.195 kW

Stoichiometry and Industrial Problems 335

Solvent laden N2 enters cryo condenser at 40°C and leaves at –31°C (242.15 K). log10pv2 = 4.536 91 +

1327.016 (242.15 - 20.474)

= – 1.449 38 pv2 = 0.0355 bar or 3.55 kPa CH2Cl2 content of saturated N2 stream, leaving cryo condenser, y2 =

3.55 102 - 3.55

= 0.0361 kmol/kmol dry N2 . CH2Cl2, leaving with N2, n2 = 4.4615 ´ 0.0361 = 0.1611 kmol/h CH2Cl2 condensed in cryo condenser . . . n3 = n1 – n2 = 1.8306 – 0.1611 = 1.6695 kmol/h º 141.8 kg/h Ideal Gas Heat Capacity Equation Constants for Incoming GAs Mixture . Component ni Equation constants . . . . kmol/h ai × n i bi × n i ´ 103 ci × n i ´ 106 di × n i ´ 10–9 N2 4.4615 132.02 –22.94 58.82 –22.16 1.8306 21.73 315.36 –273.22 95.71 CH2Cl2 Total

6.2921

153.75

292.42

-214.40

73.55

Reference temperature, T0 = 298.15 K, T1 = 313.15 K Enthalpy of N2-stream, entering cryo condenser, f2 = 153.75 (313.15 – 298.15) + 292.42 ´ 10–3 (313.152 – 298.152)/2 – 214.40 ´ 10–6 (313.153 – 298.153)/3 + 73.55 ´ 10–9 (313.154 – 298.154)/4 = 2306.25 + 893.78 – 300.51 + 31.52 = 2931.04 kJ/h º 0.814 kW Ideal Gas Heat Equation Constants for Outgoing Gas Mixture . Component ni Equation constants . . . . kmol/h ai × n i bi × n i ´ 103 ci × n i ´ 106 di × n i ´ 10–9 N2 4.4615 132.02 –22.94 58.82 –22.16 CH2Cl2 0.1611 1.91 27.75 –24.04 8.42 Total

4.6226

133.93

3.81

34.78

–13.74

336 Solutions Manual—Stoichiometry

T2 = 242.15 K enthalpy of N2-stream, leaving cryo condenser, f3 = 133.93 (242.15 – 298.15) + 3.81 ´ 10–3 (242.152 – 298.152)/2 + 34.78 ´ 10–6 (242.153 – 298.153)/3 – 13.74 ´ 10–9 (242.154 – 198.154)/4 = –7500.08 – 57.64 – 142.65 + 15.33 = – 7685.04 kJ/h º – 2.135 kW Sensible heat load of cryo condenser f ¢4 = 2931.04 – (–7685.04) = 10 616.08 kJ/h º 2.949 kW Average temperature of condensation =

10 + (-31) = – 10.5°C 2

lv1 at TB (= 313 K) = 28.06 kJ/mol Tc = 508.K Using Watson equation,

or 262.65 K

(Ref. Table 5.5)

é 508 - 262.65 ù lv2 = 28.06 ê ú ë 508 - 313 û

0.38

= 30.619 kJ/mol CH2Cl2 Condensation heat duty, f¢¢4 = 30 619 ´ 1.6695 = 51 118.42 kJ/h º 14.20 kW Total heat duty of cryo condenser f4 = f¢4 + f¢¢4 = 61 734.5 kJ/h º 17.149 kW (a) Conventional cryo condenser: N2 vapour (as refrigerant) enters the coil at 5 bar a and –170°C (103.15 K) and leaves at 4.8 bar a and 4°C (277.15 K). Enthalpy available for refreigeration = 408.41 – 220.23 = 185.16 kJ/kg N2 51 118.42 = 276.08 kg/h 185.16 (b) Modified cryo condenser: Refrigeration in one coil = 408.53 – 220.23 = 188.03 kJ/kg

LN requirement =

Stoichiometry and Industrial Problems 337

Enthalpy requirement in vaporizer for LN = 220.23 – 35.21 = 185.02 kJ/kg Enthalpy required for cooling gaseous N2 = 408.53 – 223.40 = 185.13 kJ/kg Both values match closely and hence vaporization heat duty is satisfied by cooling of N2 gas stream from the first coil on a mass to mass basis. Refrigeration in 2nd coil = 408.53 – 223.40 = 185.13 kJ/kg Total refrigeration available = 188.18 + 185.3 = 373.48 kJ/kg LN LN requirement for the modified system = Reduction in LN requirement =

51 118.42 = 136.87 kg/h 373.48 (276.08 - 136.87) 100 276.08

= 50.42 %

Ans.

EXERCISE 8.24 (a) Basis: 900 kg liquid chlorine at 303.15 K (30°C) At 303.15 K, rl1 = 1378.83 kg/m3 lv1 = 245.5 kJ/kg Volume of liquid chlorine V1 = 900/1378.83 = 0.652 73 m3 Length of container = 1.8 m Cross-sectional area of chlorine filled circular surface = A1 = 0.652 73/1.8 = 0.362 63 m2 A1 = where

1 2 r a1 2

a1 = angle in radians 2 p radians = 360° 6.283 18 radians = 360° 1 radian = 57.296° r = 0.76/2 = 0.38 m

338 Solutions Manual—Stoichiometry

1 (0.38)2 a1 = 0.362 63 2

a1 = 5.022 58 radian º 287.77° Arc length of cylinder in contact with liquid chlorine, S = ra S = 0.38 ´ 5.022 58 = 1.9086 m Area in contact with liquid chlorine, A¢1 = 1.9086 ´ 1.8 = 3.435 48 m2 Energy balance: . Chlorine withdrawl rate, m3 = 5.56 g/s = 20.016 kg/h . m3 lv1 = UA¢1(T0 – Ts) T0 = 303.15 K U = 11.4 W/(m2 × K) = 41.04 kJ/(h × m2 × K) 41.04 ´ 3.435 48 (303.15 – Ts) = 20.016 ´ 245.5 Ts = 268.3 K Iteraction 2: At this Ts = 268.3 K, r1 = 1483.68 kg/m3 lv1 = 270.82 kJ/kg Average lv =

(245.5 + 270.82) 2

= 258.16 kJ/kg V 1 = 900/1483.68 = 0.6066 m3 A 1 = 0.6066/1.8 = 0.337 m2 a1 = 0.337 ´ 2/(0.38)2 = 4.66759 radian S = ra = 0.38 ´ 4.667 59 = 1.773 68 m A¢1 = 1.773 68 ´ 1.8 = 3.192 62 m2 41.04 ´ 3.192 62 (303.15 – Ts) = 20.016 ´ 258.16 Ts = 263.71 K

Stoichiometry and Industrial Problems 339

Iteration 3: At Ts = 263.71 K, r1 = 1496.31 kg/m3 l 1 = 273.73 kJ/kg Average lv = 272.27 kJ/kg V 1 = 900/1496.31 = 0.601 48 m3 A 1 = 0.601 48/1.8 = 0.334 16 m2 a1 = 0.334 16 ´ 2/(0.38)2 = 4.628 25 radian S = r a1 = 0.38 ´ 4.628 25 = 1.758 74 m A¢ = 1.758 74 ´ 1.8 = 3.165 73 m2 41.04 ´ 3.165 73 (303.15 – Ts) = 20.016 ´ 272.27 Ts = 261.20 K Iteration 4: At Ts = 261.20 K, r1 = 1503.1 kg/m3 lv1 = 275.28 kJ/kg Average lv = 274.5 kJ/kg V 1 = 900/1503.1 = 0.598 76 m3 A 1 = 0.596 76/1.8 = 0.332 65 m2 a1 = 0.332 65 ´ 2/(0.38)2 = 4.607 28 radian S = 4.607 28 ´ 0.38 = 1.750 77 m A¢1 = 1.750 77 ´ 1.8 = 3.151 38 m2 41.04 ´ 3.151 38 (303.15 – Ts) = 20.016 ´ 274.5 Ts = 260.67 K Iteration 5: At Ts = 260.67, r1 = 1504.53 kg/m3 lv1 = 275.6 kJ/kg Average lv = 275.44 kJ/kg V 1 = 900/1504.53 = 0.598 19 m3 A 1 = 0.598.19/1.8 = 0.332 33 m2

340 Solutions Manual—Stoichiometry

a1 = 0.332 33 ´ 2/(0.38)2 = 4.6029 radian S = 4.6029 ´ 0.38 = 1.749 10 m A¢1 = 1.749 10 ´ 1.8 = 3.148 39 m2 41.04 ´ 3.148 39 (303.15 – Ts) = 20.016 ´ 275.44 Ts = 260.48 K or 012.67°C This Ts value is close enough to the value of iteration 4. Steady state temperature, t/T = – 12.67°C/260.48 K Average latent heat between 303.15 K and 260.48 K l=

Ans.

(245.5 + 275.44) = 260.47 kJ/kg 2

Average heat transfer rate, f = 20.016 ´ 260.47 = 5213.6 kJ/h º 1.448 kW Initially the container has 900 kg liquid chlorine at 303.15 K. Its enthalpy = 62.64 kJ/kg (with reference enthalpy of saturated liquid chlorine at atmospheric (1.013 25 bar) and –33.98°C (239.17 K) = 0 kJ/kg). Enthalpy of liquid chlorine at 260.47 K is 20.68 kJ/kg. Thus for cooling down the mass of 900 kg liquid chlorine, enthalpy equivalent to h = 900 (62.64 – 20.68) = 37 764 kJ needs to be removed. This will take q = 37 764/5213.6 = 7.24 h for cooling. Thus it will take approximately 7 h 14 min to attain the steady state conditions to achieve 260.48 K. In 7 h 14 min, chlorine consumption will be 144.92 kg which is substantial. Thus surface area in contact with liquid chlorine (S) would be reduced to a definite extent. This will mean heat pick-up from atmosphere will be lower. Thus in actual practice, time required to achieve 260.45 K will be more than 7.24 h. (b) Basis: 100 kg liquid chlorine at 303.15 K (i) Dq1 = 0.2 h = 720 s At 303.15 K, r1 = 1378.83 kJ/kg lv1 = 245.5 kJ/kg V 1 = 100/1378.83 = 0.072 53 m3 A 1 = 0.072 53/1.8 = 0.040 29 m2 a1 = 0.040 29 ´ 2/(0.38)2 = 0.55803 radian Arc length, S1 = 0.38 ´ 0.558 03 = 0.212 05 m

Stoichiometry and Industrial Problems 341

A¢1 = 0.212 05 ´ 1.8 = 0.381 69 m2 Ts = 303.15 – [20.016 ´ 245.5/(41.04 ´ 0.381 69)] = –10.55 K - irrational answer Therefore, transient material and energy balance is required. Energy balance: Heat input – Heat output = Accumulation . . UA¢ (Ta – T) – mlv = (M0 – mq) Cl (dT/dq) (1) where M 0 = initial mass of liquid chlorine = 100 kg at q = 0 h Cl = Heat capacity of liquid chlorine = 0.985 kJ/(kg ×K) q = time, h T = Temperature of liquid chlorine in the container after time q, K Integrating above equation T2 Cl dT dq = ò  lv ]  T1 [UA ¢ (Ta - T ) - m 0 ( M 0 - mq ) . a = UA¢ (Ta – T) – mlv da = – UA¢ dT . T = T0, a = UA¢ (Ta – T0) – mlv = a0 . b = Cl (M0 – mq) . db = –Cl m dq q = 0, b = ClM0 = b0

q

ò

Let

When Let

when Rewriting Eq. (2)

a

ò

a0

(2)

b da db = ò  -UA¢ a b0 - Cl mb

a = 1 ln b 1 ln Cl m b0 UA¢ a0

UA¢ / C m

l é C [ M 0 - m q ] ù UA ¢ (Ta - T ) - m lv ln = ln ê l ú UA ¢ (Ta - T0 ) - m lv Cl M 0 ë û Simplifying, . . . . UA¢ (Ta – T) – mlv = (UA (Ta – T0) – mlv) (1 – mq/M0)UA/Clm Rearranging,

T=

1  (UA ¢ Ta - m lv ) + [m lv = UA ¢ (Ta - T0 )] (1 - m q / M 0 )UA¢ / Cl m (3) UA ¢

342 Solutions Manual—Stoichiometry

Equation (3) can be utilized to know T after time q (is elapsed). Since surface (A¢ ) in contact with liquid chlorine (of the container) varies as soon as some mass is vaporized, it is advisable to adopt numerical integration. Period-I : q1 = 0.2 h Initially liquid in the container is at the ambient temperature, i.e., T1 = 303.15 K Ta – T1 = 0 Liquid density at T1, rl = 1378.83 kg/m3 Latent heat of vaporization at T1, l v = 245.50 kJ/kg Inital mass M0 = 100 kg Volume, V 1 = 100/1378.83 = 0.072 53 m3 Cross-section area of liquid of circular surface, A1 = 0.072 53/1.8 = 0.040 29 m2 Central angle, a1 = 0.040 29 ´ 2/(0.38)2 = 0.558 06 radian Arc length of curved surfaqce in contact with liquid, S1 = 0.558 06 ´ 0.38 = 0.212 06 m Area in contact with liquid, A¢1 = 0.212 06 ´ 1.8 = 0.381 71 m2 . mq/M0 = 20.016 ´ 0.2/100 = 0.040 03 . 1 – mq/M0 = 1 – 0.040 03 = 0.959 97 . UA¢1/mCl = 41.04 ´ 0.381 71/(20.016 ´ 0.985) = 0.794 56 . . (UA¢1/mCl) [1 – mq/M0] = (0.959 97)0.794 56 = 0.968 06 T2 =

1 [ (41.04 ´ 0.38171 ´ 303.15 – 20.016 ´ 245.5) (41.04 ´ 0.381 71)

+ {20.016 ´ 245.5 – 0} 0.968 06] = 293.13 K T2 is substantially lower (10.02 K) than T1. Hence, liquid chlorine properties will change. Also, reduction in mass will be 4.0032 kg in 0.2 h which will change the surface area (A¢ ) in contact with liquid chlorine. Mass of chlorine after 0.2 h = 100 – 4.0032 = 95.9968 kg Because of these variations, iterations will have to be performed till calculated T2 matches closely with assumed T1. Density of chlorine at 293.13 K = 1410.59 kg/m3

Stoichiometry and Industrial Problems 343

lv at 293.13 K = 253.13 kJ/kg Average mass of liquid chlorine for 0.2 h period, M0 =

(100 + 95.9968) = 97.9984 kg 2

Using these values, T2 is recalculated as T2 = 287.74 K simlar iterations are performed 10 times. Final T2 = 282.70 K after 0.2 h During second period of 0.2 h, M 0 = 95.9968 kg (initial value) Nine iterations are performed and T3 = 260.90 K. For the third period nine iterations are performed. T3 = 238.40 K (–34.75°C) Spreadsheet Excel was used for calculations as attached.

Ans.

EXERCISE 8.25 Basis: F = 100 kmol/h Let D and B be kmol/h of distilled and bottom products. D + B = 100 Benzene balance: 0.228 D = 2.2 or D = 9.63 kmol/h B = 100 – 9.63 = 90.37 kmol/h A check can be made for other components.

Ans. (a)

Heat load of overhead condenser: For finding the heat load of the condenser, bubble and dew points of the distillate products are required to be found. Component Benzene (1) Toluene (2) Ethylebenzene (3) Styrene (4)

mole fraction 0.228 0.722 0.050 .–

Antoine constants (Ref. Table – 5.4) A 4.018 14 4.078 27 4.074 88 4.219 48

B 1203.835 1343.943 1419.315 1525.059

Total pressure in the column, p = 21.3 kPa a = 0.213 bar a Using, Antoine equation, i.e. Eq. (5.24): T1b = saturation temp. of benzene at 21.3 kPa = 309.92 K or 36.77°C

C – 53.226 –53.773 – 60.539 – 56.379

Initial value of T Tin,K 303.15 Liquid density at T ri, kg/m3 1378.83 Liquid latent lv, kJ/kg 245.50 heat of vap. Average liquid density ri, kg/m3 1378.83 Average latent lV, kJ/kg 245.50 heat of vap. Average Mo Mo, kg 100 0.07253 Volume of liquid V1, m3 Cross-sectional area A1, m2 0.04029 Central angle a1, radian 0.55806 Arc length Si, m 0.21206 0.38171 Area in contact A¢1, m2 with liquid m*q/Mo 0.04003 1–m*q/Mo 0.95997 UA ¢1 15.6654 303.15 To UA¢1/mC1 0.79456 (1-m*q/Mo)(UA 1¢ /mC1) 0.96806 UATa 4748.968 4913.851 mlv (avg) UA (Ta – To) 0.000 4756.899 Calculated T Tcal,K 293.13

293.13 1410.59 253.36

287.91 1426.64 257.28

285.29 1434.53 259.19

283.98 1438.45 260.13

283.33 1440.40 260.60

283.00 1441.37 260.83

282.84 1441.85 260.95

282.76 1442.10 261.01

282.72 1442.22 261.04

1394.71 249.43

1402.74 251.39

1406.68 252.34

1408.64 252.81

1409.62 253.05

1410.10 253.16

1410.34 253.22

1410.47 253.25

1410.53 253.27

97.9984 0.07026 0.03904 0.54066 0.20545 0.36981

97.9984 0.06986 0.03881 0.53757 0.20428 0.36770

97.9984 0.06967 0.03870 0.53606 0.20370 0.36666

97.9984 0.06957 0.03865 0.53531 0.20342 0.36615

97.9984 0.06952 0.03862 0.53494 0.20328 0.36590

97.9984 0.06950 0.03861 0.53476 0.20321 0.36578

97.9984 0.06949 0.03860 0.53467 0.20317 0.36571

97.9984 0.06948 0.03860 0.53462 0.20316 0.36568

97.9984 0.06948 0.03860 0.53460 0.20315 0.36567

0.04085 0.95915 15.1771 298.14 0.76979 0.96840 4600.926 4992.569 76.030 4761.195 287.91

0.04085 0.95915 15.0902 295.53 0.76539 0.96858 4574.608 5031.750 115.025 4762.250 285.29

0.04085 0.95915 15.0479 294.22 0.76324 0.96867 4561.772 5050.868 134.372 4762.454 283.98

0.04085 0.95915 15.0270 293.57 0.76218 0.96871 4555.434 5060.306 144.006 4762.475 283.33

0.04085 0.95915 15.0166 293.24 0.76166 0.96873 4552.287 5064.994 148.813 4762.465 283.00

0.04085 0.95915 15.0114 293.08 0.76139 0.96874 4550.718 5067.329 151.212 4762.454 282.84

0.04085 0.95915 15.0089 293.00 0.76126 0.96875 4549.936 5068.494 152.411 4762.448 282.76

0.04085 0.95915 15.0076 292.95 0.76120 0.96875 4549.545 5069.076 153.010 4762.444 282.72

0.04085 0.95915 15.0069 292.93 0.76116 0.96875 4549.350 5069.366 153.309 4762.442 282.70

344 Solutions Manual—Stoichiometry

Excel outputs for Exercise 8.24 Basis: 100 kg liquid chlorine in the container Chlorination rate, m = 20.016 kg/h Interval q = 0.2 h Ambient temperature, ta/Ta = 30°C/303.15 K First Interval q1, = 0.2 h Mo = 100 kg (initial value) Overall heat transfer coefficient, U = 11.4 W/(m*-K) = 41.04 kJ/(h-m*K)

Second interval q2 = 0.2 h

Calculated T

Tin.K ri, kg/m3 lv, kJ/kg ri, kg/m3 lv, kJ/kg Mo, kg V1, m3 A1, m2 a1, radian S1, m A¢1 m2

282.70

Tcal,K

282.70 1442.27 261.05 1442.27 261.05 95.9968 0.06795 0.03775 0.52283 0.19868 0.35762 0.04085 0.95915 14.6766 277.55 0.74441 0.96943 4449.216 5225.141 300.137 4774.445 272.44

272.44 1472.70 268.26 1457.49 264.65 93.9952 0.06383 0.03546 0.49111 0.18662 0.33592 0.04259 0.95741 13.7862 274.64 0.69925 0.97003 4179.288 5369.526 352.985 4866.171 266.64

266.64 1488.13 271.85 1465.20 266.45 93.9952 0.06349 0.03527 0.48855 0.18565 0.33417 0.04259 0.95741 13.7144 273.20 0.69560 0.97018 4157.508 5405.415 390.938 4864.940 263.74

263.74 1495.70 273.59 1468.98 267.32 93.9952 0.06333 0.03518 0.48731 0.18518 0.33332 0.04259 0.95741 13.6794 272.47 0.69383 0.97025 4146.910 5422.821 409.763 4863.939 262.29

262.29 1499.44 274.44 1470.86 267.75 93.9952 0.06325 0.03514 0.48669 0.18494 0.33290 0.04259 0.95741 13.6622 272.11 0.69296 0.97029 4141.686 5431.384 419.126 4863.347 261.57

261.57 1501.30 274.87 1471.79 267.96 93.9952 0.06321 0.03512 0.48639 0.18483 0.33269 0.04259 0.95741 13.6536 271.93 0.69252 0.97031 4139.095 5435.627 423.790 4863.031 261.21

261.21 1502.23 275.08 1472.25 268.06 93.9952 0.06319 0.03511 0.48624 0.18477 0.33259 0.04259 0.95741 13.6494 271.84 0.69231 0.97032 4137.806 5437.738 426.115 4862.869 261.03

261.03 1502.69 275.18 1472.48 268.12 93.9952 0.06318 0.03510 0.48616 0.18474 0.33254 0.04259 0.95741 13.6472 271.80 0.69220 0.97032 4137.163 5438.790 427.274 4862.787 260.94

260.94 1502.92 275.23 1472.60 268.14 93.9952 0.06318 0.03510 0.48612 0.18473 0.33251 0.04259 0.95741 13.6462 0.69215 0.97032 4136.842 5439.314 427.852 4862.747 260.90

Stoichiometry and Industrial Problems 345

Initial value of T Liquid density at T Liquid latent heat of vap. Average liquid density Average latent heat of vap. Average Mo Volume of liquid Cross-sectional area Central angle Arc length Area in contact with liquid m*q/Mo 1–m*q/Mo UA ¢1 To UA¢1/mC1 (1-m*q/Mo)(UA¢1/mC1) UATa mlv (avg) UA (Ta – To)

Initial value of T Liquid density at T Liquid latent heat of vap. Average liquid density Average latent heat of vap. Average Mo Volume of liquid Cross-sectional area Central angle Arc length Area in contact with liquid m*q/Mo 1–m*q/Mo UA ¢1 To UA¢1/mC1 (1-m*q/Mo)(UA¢1/mC1) UATa mlv (avg) UA (Ta – To)

Tin,K ri, kg/m3 lv, kJ/kg ri, kg/m3 lv,kJ/kg Mo, kg V1,m3 A1 m2 a1 radian S1 m A¢1 m2

260.92 250.22 244.44 241.55 1502.99 1531.84 1546.36 1553.45 275.25 281.70 284.86 286.38 1502.99 1517.42 1524.68 1528.22 275.25 278.48 280.06 280.81 91.9936 89.9920 89.9920 89.9920 0.06121 0.05875 0.05847 0.05834 0.03400 0.03264 0.03248 0.03241 0.47097 0.45204 0.44991 0.44888 0.17897 0.17178 0.17097 0.17057 0.32214 0.30920 0.30774 0.30703 0.04352 0.04448 0.04448 0.04448 0.95648 0.95552 0.95552 0.95552 13.2207 12.6895 12.6296 12.6006 260.92 255.73 252.76 251.27 250.53 0.67056 0.64362 0.64058 0.63911 0.97061 0.97114 0.97127 0.97134 4007.851 3846.812 3828.667 3819.875 5509.428 5638.517 5670.126 5685.314 554.079 601.717 636.437 653.672 4809.693 4891.426 4889.079 4887.419 Calculated T Tcal, K 250.22 244.28 241.31 239.83 Final temperature of liquid chlorine in container after 0.6 h (2160 s) = 238.40 K (–34.75°C)

240.11 1556.94 287.12 1529.97 281.18 89.9920 0.05827 0.03237 0.44837 0.17038 0.30669 0.04448 0.95552 12.5864 250.17 0.63839 0.97137 3815.554 5692.743 662.244 4886.470 239.09

239.39 1558.67 287.49 1530.83 281.37 89.9920 0.05824 0.03235 0.44812 0.17029 0.30651 0.04448 0.95552 12.5793 249.98 0.63803 0.97138 3813.416 5696.411 666.510 4885.968 238.72

239.03 1559.53 287.67 1531.26 281.46 89.9920 0.05822 0.03235 0.44799 0.17024 0.30643 0.04448 0.95552 12.5758 249.89 0.63786 0.97139 3812.355 5698.230 668.635 4885.711 238.54

238.85 1559.96 287.76 1531.48 281.50 89.9920 0.05821 0.03234 0.44793 0.17021 0.30639 0.04448 0.95552 12.5741 249.84 0.63777 0.97140 3811.826 5699.135 669.693 4885.581 238.45

238.76 1560.18 287.80 1531.58 281.53 89.9920 0.05821 0.03234 0.44790 0.17020 0.30636 0.04448 0.95552 12.5732 0.63772 0.97140 3811.563 5699.585 670.220 4885.516 238.40

346 Solutions Manual—Stoichiometry

Third interval q2 = 0.2 h

Stoichiometry and Industrial Problems 347

T1t = saturation temp. of toluene at 21.3 kPa = 336.71 K or 63.56°C T1e = saturation temp. of ethylbenzene at 21.3 kPa = 359.56 K or 86.41°C T1 = 309.92 ´ 0.228 + 336.71 ´ 0.722 + 359.56 ´ 0.05 = 331.7 K (58.55°C) —first assumption for the bubble point and dew point calculations Bubble point: All pressures are in kPa. Component

xi

Benzene(1) 0.228 Toluene (2) 0.722 Ethyl 0.050 benzene (3) Total

Temperature, K T1 = 331.7 K p si 49.606 17.497 6.932

1.000

T2 = 328.7 K

yi 0.531 0.582 0.016

p si 44.512 15.496 6.059

T3 = 328.3 K

yi 0.477 0.525 0.014

1.129

p si 43.826 15.233 5.946

1.016

0.999

Bubble point of distillate product TBB = 328.3 K (55.15 °C) Dew Point: All pressures are in kPa. Component

yi

yi 0.469 0.516 0.014

Ans. (b – i)

Temperature, K T1 = 33.17 K p si xi

T2 = 336 K p si xi

T3 = 335 K p si xi

T4 = 334.7 K p si xi

Benzene(1) 0.228 49.606 0.098 58.011 Toluene(2) 0.722 17.497 0.879 20.852 Ethyl 0.050 6.932 0.154 8.420 benzene(3)

0.084 0.738 0.126

56.030 0.087 55.112 20.050 0.767 19.693 8.065 0.132 7.906

0.088 0.781 0.135

Total

0.948

0.986

1.004

1.000

1.131

Dew point of distillate product, TDP = 334.7 K (61.55°C) Ans. (b – ii) Average temperature of condensation = (TBB + TDP)/2 = (328.3 + 334.7)/2 = 331.5 K (58.35°C) Consider reference temperature, T0 = 331.5 K Use Watson equation [Eq. (5.26)] and data from Table–5.5 for calculation of enthalpy of vaporisation. é 562.05 – 331.5 ù lv1 = 30 720 ê ú ë 562.05 - 353.3 û

0.98

= 31 902 kJ/kmol of benzene

348 Solutions Manual—Stoichiometry

é 591.75 - 331.5 ù lv2 = 33 180 ê ú ë 591.75 - 383.8 û

0.98

= 36 133 kJ/kmol of toluene 0.98

é 617.15 - 331.5 ù = 40 138 kJ/kmol of ethylebenzene lv3 = 35 570 ê ú ë 617.15 - 409.3 û Since distillate product is leaving in saturated conditions, its enthalpy = 0 kJ/kmol. Heat load of condenser = 9.63 [0.228 ´ 31 902 + 0.722 ´ 36 133 + 0.05 ´ 40 138] ´ (1 + 6) = 2384 196 kJ/h = 662.277 kW Ans.(c) Enthalpy of bottom product: Tower is considered to operate at near constant pressure T1s = saturation temperature of styrene at 21.3 kPa = 368.18 K or 95.03°C T1 = 336.7 ´ 0.005 ´ 359.6 ´ 0.475 + 368 ´ 0.52 = 364 K (90.85°C)

Bubble Point: All pressures are in kPa Component

Temperature, K

xi

T1 = 364.0 K p si yi

Toluene(2) 0.005 Ethyl 0.475 benzene(3) Styrene(4) 0.520 Total

T2 = 363.0 K p si yi

T3 = 363.55 K p si yi

52.350 23.304

0.012 0.520

54.078 24.154

0.013 0.539

54.935 24.594

0.013 0.548

17.018

0.415

17.668

0.431

17.973

0.439

1.000

0.947

0.983

1.000

T BB = 363.55 K (90.4°C) Data on heat capacity constants: Component

xi

liquid heat capacity (C°mp) equation constants bi xi ´ 103 4.061 427.583

ci xi ´ 106 – 7.563 – 688.774

di xi ´ 109 8.150 680.960

Toluene(2) Ethyl benzene(3) Styrene(4)

0.005 0.475

ai xi 0.009 2.049

0.520

– 19.770

622.549

– 1441.738

1005.222

Total

1.000

– 17.712

1054.193

– 1838.075

1694.332

Enthalpy of bottom product = 90.37 [– 17.712 (363.55 – 331.5) + 1054.193 ´ 10–3 (363.552 – 331.52)/2

Stoichiometry and Industrial Problems 349

– 1838.075 ´ 10–6 (363.553 – 331.53)/3 + 1694.332 ´ 10–9 (363.554 – 331.54)/4] = 572 794 kJ/h º 159.11 kW (liquid) Enthalpy of feed: T1 = 0.022 ´ 309.9 + 0.074 ´ 336.7 + 0.434 ´ 359.6 + 0.47 ´ 368 = 360.7 K (87.7°C) Bubble Point: All pressures are in kPa. Component

Temperature, K

xi

T1 = 360.7 K p si yi

T2 = 358.0 K p si yi

T3 = 357.75 K p si yi

Benzene(1) Toluene(2) Ethyl benzene(3) Styrene(4)

0.022 0.074 0.434

124.80 49.19 21.757

0.129 0.171 0.443

117.27 45.88 20.145

0.121 0.159 0.410

116.137 45.402 19.927

0.120 0.158 0.406

0.470

15.838

0.349

14.614

0.322

14.424

0.323

Total

1.000

1.092

1.012

1.007

TBB = 357.75 K (84.75 °C) Heat capacity data: Component

xi

Lliquid heat capacity C°mp equation constants ai xi

bi xi ´ 103

ci xi ´ 106

di xi ´ 109

Benzene(1) Toluene(2) Ethyl benzene(3) Styrene (4)

0.022 0.074 0.434

– 0.160 0.134 1.872

16.952 60.105 390.676

– 36.26 – 111.93 – 629.322

41.755 120.621 622.182

0.470

– 17.869

562.689

– 1031.956

908.566

Total

1.000

– 16.023

1030.422

– 1809.476

1693.124

Enthalpy of feed = 100.0[– 16.023 (357.75 – 331.5) + 1030.422 ´ 10–3 (357.752 – 331.52)/2 – 1809.475 ´ 10–6 (357.753 – 331.53)/3 + 1693.124 ´ 10–9 (357.754 – 331.54)/4] = 506 369 kJ/h º 140.66 kW Reboiler duty = Heat load of condenser + Enthalpy of distillate

350 Solutions Manual—Stoichiometry

+ Enthalpy of bottoms – Enthalpy of feed = 662.277 + 0 + 159.11 – 140.66 = 680.727 kW

Ans. (d – i)

Latent heat (lv) of steam at 0.2 MPa a = 2201.6 kJ/kg Steam consumption in reboiler = 680.727 ´ 3600/2201.6 = 1113.1 kg/h

Ans. (d – ii)

EXERCISE 8.26 Basis: 40 000 Nm3/h offgas mixture at 0.25 bar g and 523.15 K (250°C) Offgas mixture flow rate =

40 000 = 1784.6 kmol/h 22.414

New basis: 100 kmol dry offgas mixture Case I: Element Balance of Offgas Mixture Component

kmol

C, kmol

H2, kmol

O2, kmol

H2 N2 Ar CH4 CO CO2 C2H2

15.3 61.9 0.9 0.4 17.2 3.9 0.4

— — — 0.4 17.2 3.9 0.8

15.3 — — 0.8 — — 0.4

— — — — 8.6 3.9 —

Total

100.0

22.3

16.5

12.5

Moisture in offgas mixture = 0.15 ´ 100 = 15 kmol 16.5 – 12.5 2 18.05 kmol 10% 18.05 ´ 1.1 = 19.855 kmol 79 ´ 19.855/21 = 74.693 kmol 19.855 + 74.693 = 94.548 kmol 2726.76 kg 94.548 ´ 0.017 1.607 kmol

Theoretical O2 requirement = 22.3 + = = = = = º Moisture, entering with air = = Excess air Actual O2 supply N2, entering with O2 Total dry air

Stoichiometry and Industrial Problems 351

Boiler Flue Gas Analysis Component N2 + Ar O2 CO2 H2O

kmol

Dry analysis, % (Orsat)

Wet analysis %

85.08 1.12 13.80 —

70.62 0.93 11.45 17.00

194.705 (wet) 100.00 161.598 (dry)

100.00

61.9 + 0.9 + 74.693 = 137.693 19.855 – 18.05 = 1.805 22.300 16.5 + 15.0 + 1.607 = 33.107

Total

Heat Input to Boiler: (For nomenclature, refer Table – 7.13). Gross Calorific value of fuel: H1 = 15.3 ´ 285.83 ´ 1000 + 17.2 ´ 282.98 ´ 1000 + 0.4 ´ 890.65 ´ 1000 + 0.4 ´ 1301.0 ´ 1000 = 10 117 115 kJ/100 kmol offgas mixture Heat of Offgas mixture at 523.15 K (250°C) T0 = 298.15 K (25 °C) Component

kmol, ni

Cºmp equation constants ni × ai

ni × bi ´ 103

H2 N2 Ar CH4 CO CO2 C2 H 2 H2O

15.3 61.9 0.9 0.4 17.2 3.9 0.4 15.0

437.741 1831.676 18.695 7.700 499.276 144.979 9.002 487.382

15.600 – 318.230 — 20.845 – 48.444 90.625 36.050 1.194

Total

115.0

3436.451

– 202.360 –3

ni × ci ´ 106

ni × di ´ 109

– 2.258 816.022 — 4.789 200.272 – 28.777 – 25.494 198.161

11.628 – 307.519 — – 4.527 – 80.948 3.203 7.146 – 68.211

1162.716

– 439.228

2

H2 = 3436.451 (523.15 – 298.15) – 202.36 ´ 10 (523.15 – 298.152)/2 + 1162.716 ´ 10–6 (523.153 – 298.153)/3 – 439.228 (523.154 – 298.154)/4 = 791 829 kJ Heat of input (combustion) air at 313.15 K (40°C): Cs = 1.006 + 1.84 ´ 0.017 = 1.037 kJ/(kg dry air × K) H3 = 2726.76 ´ 1.037 (313.15 – 298.15) = 42 415 kJ Total heat input HI = H1 + H2 + H3 = 10 951 359 kJ/100 kmol offgas mixture

352 Solutions Manual—Stoichiometry

Heat Output: Heat loss in flue gases at 448.15 K (175ºC): Component

kmol, ni

Cºmp equation constants ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

1812.556 – 4.228 – 915.428 437.367

– 683.065 – 1.015 218.538 – 150.5511

N2 O2 CO2 H2O

137.493 1.805 22.300 33.107

4068.542 46.976 476.451 1075.716

– 706.852 21.218 1433.535 2.635

Total

194.705

5667.685

750.536 –3

1330.267 2

– 616.093 2

H7 = 5667.685 (448 – 298.15) + 750.536 ´ 10 (448 – 298.15 )/2 + 1330.267 ´ 10–6 (4483 – 298.153)/3 – 616.093 ´ 10–9 (4484 – 298.154)/4 = 914 392.4 kJ Heat loss due to evaporation: Humidity of flue gas =

kmol 33107 . = 0.2047 161598 . kmol dry gas

p = 745 Torr = 99.3 kPa a Let pw = saturation pressure of water in flue gas pw1 = 0.2049 p - pw1 pw1 = 16.887 kPa From Table 6.13, Ts = 329.55 K (56.4ºC) From steam tables (Appendix AIV.1), lv = 2367 kJ/kg Heat loss due to evaporation of moisture, H8 = 16.5 ´ 18.0153 ´ 2367 = 703 597 kJ Unaccounted heat loss, H13 = 10 278 006 ´ 0.015 = 119 833 kJ Total heat loss = 914 392.4 + 703 597 + 119 833 = 1772 159 kJ Heat utilised for production of superheated steam H14 = 10 951 359 – 1772 159 = 9179 200 kJ/115 kmol wet incoming gas Total heat available for steam production for original basis of 1784.6 kmol/h offgas mixture 9179 200 ´ 1784.6 f14 = 115 = 142 445 220 kJ/h

Stoichiometry and Industrial Problems 353

º 39 568.117 kW From steam tables: Enthalpy of superheated steam at 43 bar a and 698.15 K (425ºC) = 3269 kJ/kg Enthalpy of BFW at 303.15 K (30ºC) = 121.41 kJ/kg Heat required for production of superheated steam = 3269 – 121.41 = 3147.59 kJ/kg 97 005 216.7 Production of superheated steam = 3147.59 = 45 255.3 kg/h Ans. Case II: Loss of offgas in CO2 processing plant = 1% CO2 recovery in CO2 processing plant = 97% CO2 recovery = 3.9 ´ 0.99 ´ 0.97 = 3.745 kmol/100 kmol dry offgas mixture CO2 in offgas mixture to boiler = 3.9 ´ 0.99 – 3.745 = 0.116 kmol Pressure and temperature of offgas mixture at the outlet of CO2 absorber = 0.1 bar g and 328 K p = 0.1 bar g = 1.113 25 bar a T = 328 K (55ºC), pw2 = 15.741 kPa kmol 15.741 Humidity of offgas = = 0.1647 111325 . - 15.741 kmol dry gas Element Balance of Offgas Mixture Component H2 N2 Ar CH4 CO CO2 C2H2

kmol 15.147 61.281 0.891 0.396 17.028 0.116 0.396

C, kmol — — — 0.396 17.028 0.116 0.792

H2, kmol 15.147 — — 0.792 — — 0.396

O2, kmol — — — — 8.514 0.116 —

Total 95.255 (dry) 18.332 16.335 Water, entering with gas mixture = 95.255 ´ 0.1647 = 15.688 kmol 16.335 Theoretical O2 requirement = 18.332 + – 8.63 2 = 17.87 kmol Actual O2 supply = 17.87 ´ 1.1 = 19.657 kmol N2, entering with O2 = 79 ´ 19.657/21 = 73.948 kmol Total dry air = 19.657 + 73.948 = 93.605 kmol

8.630

354 Solutions Manual—Stoichiometry

º 2699.568 kg Moisture entering with air = 93.605 ´ 0.017 = 1.591 kmol Boiler Flue Gas Component N2 + Ar O2 CO2 H2O

kmol 61.281 + 0.891 + 73.948 = 19.657 – 17.87 = 0.116 + 18.332 = 15.688 + 16.335 + 1.591 =

Total

Dry analysis, % (Orsat) 136.12 87.06 1.787 1.14 18.448 11.80 33.614 —

wet189.969 dry 156.355

100.00

Wet analysis % 71.65 0.94 9.71 17.70 100.00

Heat Input to Boiler: Heat of offgas mixture to boiler at 328.15 K (55ºC): Component

kmol, ni

C ºmp equation constants ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

H2 N2 Ar CH4 CO CO2 C2H2 H2O

15.147 61.281 0.891 0.396 17.028 0.116 0.396 15.688

433.363 1813.360 18.508 7.623 494.284 2.478 8.912 509.736

15.441 – 315.046 — 20.637 – 47.959 7.457 35.689 1.249

– 2.236 807.861 — 4.741 198.270 – 4.762 – 25.239 207.249

11.648 – 304.444 — – 4.482 – 80.139 1.137 7.074 – 71.340

Total

110.943

3288.264

– 282.532

1185.884

– 440.548

H2 = 3288.264 (328 – 298.15) – 282.532 ´ 10–3 (3282 – 298.152)/2 + 1185.884 ´ 10–6 (3283 – 298.153)/3 – 440.548 ´ 10–9 (3284 – 298.154)/4 = 101 870.5 kJ Heat of input air at 313 K (40ºC): H3 = 2699.6 ´ 1.037 ´ 15 = 41 992.3 kJ GCV of offgas mixture, H1 = 15.147 ´ 285 830 + 17.028 ´ 282.98 ´ 1000 + 0.396 ´ 890 650 + 0.396 ´ 1301 000 = 10 015 660 kJ Total heat input, HI = 10 015.660 + 101 870.5 + 41 992.3 = 10 159 523 kJ

Stoichiometry and Industrial Problems 355

Heat output: Unaccounted heat loss, H13 = 10 159 523 ´ 0.015 = 152 393 kJ Heat loss in flue gases at 448.15 K (175ºC): Component

kmol, ni

C mp º equation constants ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

1794.456 – 4.186 – 757.301 444.064

– 676.244 – 1.005 180.789 – 152.856

1477.033

– 649.316

N2 O2 CO2 H2O

136.12 1.787 18.448 33.614

4027.913 46.508 394.151 1092.189

– 699.793 21.006 1185.913 2.676

Total

189.969

5560.761

509.802 –3

2

H7 = 5560.761 (448.15 – 298.15) + 509.802 ´ 10 (448.15 – 298.152)/2 + 1477.033 ´ 10–6 (4483 – 298.153)/3 – 649.316 ´ 10–9 (448.154 – 298.154)/4 = 887 744.8 kJ pw1 kmol 33.614 = = 0.215 p - pw 3 kmol dry gas 156.355

pw3 = 17.572 kPa From Table – 6.13, From Steam Tables,

Ts = 330.45 K (57.3ºC). l v = 2365.2 kJ/kg H8 = 16.335 ´ 18.0153 ´ 2365.2 = 696 019.3 kJ Total heat loss = 887 744.8 + 696 019.3 + 152 393 = 1735 157 kJ Heat utilised for steam generation = 10 159 523 – 1735 137 = 8423 366 kJ Total heat available for steam production for 40 000 Nm2/h offgas mixture, 8423 366 ´ 1784.6 ¢ = B14 115 = 130 715 991 kJ/h º 36 310.0 kW 130 715 991 Steam generation = = 41 528.9 kg/h Ans. (i) 3147.59 1784.6 ´ 44.0098 ´ 0.95 CO2 production = 3.745 ´ 115 = 2429.79 kg/h º 51 315 kg/d º 51.315 t/d Ans. (ii)

356 Solutions Manual—Stoichiometry

Case III: Additional loss of gas in cryogenic separation plant = 1% Pressure of gas at the outlet of cold box = 1 bar g Temperature = 298.15 K (25ºC) Moisture = Nil Recovery of H2 in cryogenic plant = 90% H2 recovery = 15.147 ´ 0.99 ´ 0.9 = 13.496 kmol H2 in reject gas stream = 15.147 ´ 0.99 – 13.496 = 1.5 kmol 13.496 CO requirement for methanol production = 2 = 6.748 kmol CO in reject gas stream = 17.028 ´ 0.99 – 6.748 = 10.11 kmol Reject Gas to Boiler at 298.15 K (25ºC): Element Balance of Reject Gas Stream: Component H2 N2 Ar CH4 CO CO2 C2H2 H2O Total

kmol 1.5 60.668 0.882 0.392 10.110 0.115 0.392 Nil

C, kmol — — — 0.392 10.110 0.115 0.784 —

H2, kmol 1.5 — — 0.784 — — 0.392 —

O2, kmol — — — — 5.055 0.115 — —

74.059

11.401

2.676

5.170

Theoretical O2 requirement = 11.401 + = Actual O2 supply = Excess O2 = N2, entering with O2 = Total dry air = º Moisture, entering with air = Boiler Flue Gas

2.676 – 5.17 2

7.569 kmol 7.569 ´ 1.1 = 8.326 kmol 8.326 – 7.569 = 0.757 kmol 79 ´ 8.326/21 = 31.322 kmol 8.326 + 31.322 = 39.648 kmol 1143.448 kg 39.648 ´ 0.017 = 0.674 kmol

Stoichiometry and Industrial Problems 357

Component N2 + Ar O2 CO2 H2O

kmol

Dry analysis, % (Orsat) 60.668 + 0.882 + 31.322 = 92.872 88.42 0.757 0.72 11.401 10.86 2.676 + 0.674 = 3.350 —

Total

wet 108.380 dry 105.030

Wet analysis % 85.69 0.70 10.52 3.09

100.00

100.00

Heat Input to Boiler: Since reject gas stream enters at 298.15 K, H2 = 0 H1 = 1.5 ´ 285 830 + 10.11 ´ 282 980 + 0.392 ´ 890 650 + 0.392 ´ 1301 000 = 4148 800 kJ Enthalpy of input air, H3 = 1143.448 ´ 1.037 ´ 15 = 17 786.3 kJ Total heat input, H1 = 4148 800 + 0 + 17 786.3 = 2426 683.6 kJ Heat Output from Boiler: Unaccounted heat loss, H13 = 4166 586 ´ 0.015 = 62 49.9 kJ Heat loss in flue gases at 448.15 K (175ºC): Component

kmol, ni

C mp º equation constants ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

N2 O2 CO2 H2O

92.872 0.757 11.401 3.350

2748.166 19.701 243.588 108.848

– 477.455 8.899 732.903 0.267

1224.322 – 1.773 – 468.016 44.256

– 461.388 – 0.426 111.729 – 15.234

Total

105.030

3120.303

264.614

798.787

– 465.319

H7 = 3120.303 (448.15 – 298.15) + 264.614 ´ 10–3 (448.152 – 298.152)/2 + 798.787 ´ 10–6 (448.15 3 – 298.153)/3 – 465.319 ´ 10–9 (448.15 4 – 298.154)/4 = 495 488.1 kJ Moisture in fine gas =

3.35 = 0.0329 kmol/kmol dry gas (105.03 - 3.35)

358 Solutions Manual—Stoichiometry

p = 1 bar g = 2.013 25 bar a º 201.325 kPa pw3

= 0.0329 p - pw3 p w = 6.413 kPa DP of flue gas = 37.4°C or 310.55 K lv of water at DP = 2413.14 kJ/kg Heat loss due to evaporation = 2.676 ´ 18.0153 ´ 2413.15 = 116 335 kJ Total heat loss = 66 499 + 116 335 + 495 488.1 = 674 322 kJ Heat utilised for steam generation = 4166 586 – 674 322 = 3492 264 kJ Total heat available for steam generation for 40 000 Nm3/h offgas mixture,

¢¢ = 3492 264 ´ 1784.6/115 B14 = 54 193 864 kJ/h º 15 053.85 kW Steam generation = 54 193 864/3147.59 = 17 217.6 kg/h Ans. (i) CO2 production = 51 315 kg/d – same as Case II Ans.(ii) 13.496 1784.6 Methanol (100%) production = ´ 0.9 ´ 2 115 = 94.246 kmol/h º 3019.8 kg/h º 72 476 kg/d º 72.476 t/d Ans. (iii)

EXERCISE 8.27

Basis: 50 kmol/h of dry offgas mixture to first bed Vapour pressure of water, fw = 7.375 kPa at 313.15 K (40ºC) Total pressure, p = 2 bar g = 301.325 kPa a 7.375 Moisture content of offgases = 301325 . - 7.375 = 0.025 kmol/kmol dry gas Offgas Mixture from Adipic Acid Plant Component N2O O2 CO2 N2 H2O

kmol 30 8 3 59 2.5

Total

102.5

Molar mass 44 32 44 28 18

(Table 6.13)

kg 1320 256 132 1652 45 3405

Stoichiometry and Industrial Problems 359

Average molar mass of wet offgas =

3405 = 33.22 102.5

Average molar mass of dry offgas =

3405 - 45 = 33.60 100 3405 ´ 50 100

Flow of wet offgas mixture of first bed =

= 1702.5 kg/h Reaction:

N2O =

N2

+

1 2

O2

DHºf 82.05 0 0 kJ/mol at 298.15 K Considering 100% conversion, following will be the composition of treated gas mixture from the first bed. Component

kmol/100 kmol offgas (dry)

N2O

20.00

32

736

CO2 N2 H2O

Nil 0 30 8+ = 23 2 3 59 + 30 = 89 2.5

2.61 77.39 —

44 28 18

132 2492 45

Total

117.5

100.00

O2

mole % (dry)

Molar mass

44

kg

0

3405

In 117.5 kmol wet treated gas mixture, 115 kmol dry gas mixture is present. Thus for 1 kmol dry offgas mixture, production of treated dry gas mixture will be 1.15 kmol. Compositions of recycle gas mixture and treated gas mixtures will be same as 100% conversion. Offgas mixture: Component

kmol ni

Cºmp equation constant ni × ai

ni × bi ´ 103

ni × ci ´ 106

ni × di ´ 109

N2O O2 CO2 N2 H2O

0.30 0.08 0.03 0.59 0.025

6.962 2.082 0.641 17.459 0.812

19.101 0.940 1.929 – 3.033 0.002

– 12.702 – 0.187 – 1.232 7.778 0.330

3.121 – 0.045 0.294 – 2.931 – 0.114

Total

1.025

27.956

18.939

– 6.013

0.325

360 Solutions Manual—Stoichiometry

Treated gas mixture: Component

kmol

Cºmp equation constant

O2 CO2 N2 H2O

ni 0.2000 0.0261 0.7739 0.0217

ni × ai 5.205 0.558 22.900 0.705

Total

1.0217

29.368

ni × bi ´ 103 2.351 1.678 – 3.979 0.002 0.052

ni × ci ´ 106 – 0.469 – 1.071 10.202 0.287 8.949

ni × di ´ 109 – 0.112 0.256 – 3.845 0.099 – 3.602

Base temperature, To = 298.15 K (25°C) Based on Cºmp equation, following enthalpies are calculated. Temperature, T K (°C) 313.15 323.15 353.15 373.15 443.15 483.15 723.15 913.15 973.15

(40) (50) (80) (100) (170) (210) (450) (640) (700)

Enthalpy at T over To, kJ/kmol dry gas Offgas mixture 497.9 831.7 1 842.3 2 523.4 4 952.7 6 371.4 15 308.3 22 830.1 25 274.5

Treated gas mixture 451.8 753.5 1 661.6 2 269.5 4 414.1 5 652.3 13 302.6 19 654.0 21 715.3

Catalyst Bed I: Let R kmol/h of dry treated gas mixture mixed (i.e. recycled) with 50 kmol/h of dry offgas mixture at 373.15 K (100 ºC). Total enthalpy of mixed gas at 723.15 K = 15 308.3 ´ 50 + 13 302.6 R = 765 415 + 13 302.6 R kJ/h Heat of reaction at 298.15 K = – 82.05 kJ/mol N2O (exothermic) Heat generated in 1st bed = 50 ´ 0.3 ´ 82 050 = 1230 750 kJ/h Enthalpy of total treated gas mixture = 1230 750 + 765 415 + 13 302.6 R = 1996 165 + 13 302.6 R kJ/h at 973.15 K Total flow of dry treated gas mixture from 1st bed = 50 ´ 1.15 + R = 57.5 + R kmol/h Since treated gas mixture leaves at 973.15 K from 1st bed, 21 715.3 (57.5 + R) = 1996 165 + 13 302.6 R R = 88.86 kmol/h

Stoichiometry and Industrial Problems 361

Catalyst Bed II: Flow of dry gas mixture, leaving 1st bed = 57.5 + 88.86 = 146.36 kmol/h Enthalpy of gas mixture, leaving 1st bed = 21 715.3 ´ 146.36 = 3178 251 kJ/h º 882.848 kW Let the flow of dry offgas mixture (i.e. quenching) to 2nd bed be F1 kmol/h at 373.15 K. Mixed gas temperature is 723.15 K. Heat balance of mixing: 3178 251 + 2523.4 F1 = 13 302.6 ´ 146.36 + 15 308.3 F1 F 1 = 96.31 kmol/h at 723.15 K Enthalpy of total gas mixture, entering 2nd bed = 3178 251 + 2523.4 ´ 96.31 = 3421 280 kJ/h Heat generated in 2nd bed = 96.31 ´ 0.3 ´ 82 050 = 2370 671 kJ/h Enthalpy of gas mixture, leaving 2nd bed = 3421 280 + 2370 671 = 5791 951 kJ/h º 1608.875 kW Flow of dry gas mixture, leaving 2nd bed = 146.361 + (96.31 ´ 1.15) = 257.12 kmol/h If T1 is the temperature of gas mixture, leaving 2nd bed,

z

T1

257.12

C mp º dT = 5791 951

298.15

Solving by Mathcad, T1 = 996.55 K or 723.4 ºC Catalyst Bed III: Let the flow of dry offgas mixture (i.e. quenching) to 3rd bed be F2 kmol/h at 373.15 K. Again mixed gas temperature is 723.15 K. Heat balance of mixing: 5791 951 + 2523.4 F2 = 13 302.6 ´ 257.12 + 15 308.3 F2 F 2 = 185.50 kmol/h Enthalpy of total gas mixture, entering 3rd bed = 5791 951 + 2523.4 ´ 185.50 = 6260 042 kJ/h at 723.15 K Heat generated in 3rd bed = 185.50 ´ 0.3 ´ 82 050 = 4566 083 kJ/h

362 Solutions Manual—Stoichiometry

Enthalpy off gas mixture, leaving 3rd bed = 6260 042 + 4566 083 = 10 826 125 kJ/h º 3007.257 kW Flow of dry gas mixture, leaving 3rd bed = 257.12 + (185.50 ´ 1.15) = 470.45 kmol/h If T2 is the temperature off gas mixture, leaving 3rd bed

z

T2

470.45

Cºmp dT = 10 826 124

298.15

Solving by Matched T2 = 1010.52 K or 737.37 ºC Total dry offgas flow treated = 50 + 96.31 + 185.50 = 331.81 kmol/h Offgas mixture bypassed to 2nd bed = Offgas bypassed to 3rd bed = Recycle =

96.31 ´ 100 = 29.03% 331.81 185.50 ´ 100 = 55.91% 331.81

88.86 kmol = 0.233 (470.45 - 88.86) kmol gas emitted

Heat exchange E2: Enthalpy of mixed feed gas, entering E2 = 50 ´ 2523.4 + 88.86 ´ 1661.6 = 273 820 kJ/h Enthalpy of mixed feed gas, leaving E2 = 50 ´ 15 308.3 + 13 302.6 ´ 88.86 = 1947 484 kJ/h Heat duty of E2, f1 = 1947 484 – 273 820 = 1673 664 kJ/h º 464.91 kW Enthalpy of reactor effluent stream, leaving E2 = 10 826 124 – 1673 664 = 9152 460 kJ/h If T3 is the temperature of reactor effluent stream, leaving E2,

z

T3

470.45

C°mp dT = 9152 460

298.15

Stoichiometry and Industrial Problems 363

Solving by Mathcad, T3 = 907.3 K or 634.15 K Heat exchanger E3: Heat duty of E3, f2 = 9152 460 – (470.45 ´ 5652.3) = 6493 335 kJ/h º 1803.7 kW Enthalpy of saturated steam at 4.5 bar a = 2742.6 kJ/kg Enthalpy of feed water at 303.15 K (30°C) = 126 kJ/kg Steam generated in steam drum =

6493 335 (2742.6 - 126)

= 2481.6 kg/h (saturated) Heat exchanger E1: Offgas (total) enters E1 at 313.15 K and leaves at 373.15 K Heat duty of E1, f3 = 331.81 (2523.4 – 497.9) = 672 081 kJ/h º 186.69 kW Enthalpy of reactor effluent stream, leaving E1 = 470.45 ´ 5652.3 – 672 081 = 1987 043 kJ/h or 4223.71 kJ/kmol dry gas mixture If T4 is the temperature of reactor effluent stream, leaving E1,

z

T4

470.45

Cºmp dT = 1987 043

298.15

Solving by Mathcad, T4 = 436.97 K or 163.82 ºC Heat exchanger E4: In this heat exchanger, 88.86 kmol/h (= R) dry gas mixture is cooled from 436.97 K to 323.15 K. Heat duty of E4, f4 = 88.86 (4223.71 – 753.5) = 308 363 kJ/h º 85.66 kW Assuming 5 K rise in cooling water temperature, cooling water flow =

308 363 = 14 730 kg/h 5 ´ 4.1868

º 14.7 m3/h

364 Solutions Manual—Stoichiometry

Heat exchanger E5: Flow of dry gas, being emitted to atmosphere = 470.45 – 88.86 = 381.59 kmol/h Recycle dry gas (88.86 kmol/h) is heated in E5 from 341.15 K to 353.15 K.

z

353.15

Heat duty of E5, f5 = 88.86

Cºmp dT

341.15

= 32 327 kJ/h º 8.98 kW Enthalpy of emitted gas mixture to stack = 381.59 ´ 4223.71 – 32 327 = 1579 398 kJ/h º 438.722 kW If T5 is the gas mixture to stack,

z

T5

381.59

Cºmp dT = 1579 398

298.15

Solving by Mathcad, T5 = 434.2 K or 161.05ºC.

Ans.

EXERCISE 8.28 Basis: 1 kmol dry acid gas fed to the burner Reactions taking place in burner: (1) H2S (g) + 3/2 O2 (g) = SO2 (g) + H2O (g) (2) H2 (g) + ½ O2 (g) = H2O (g) mö æ CnHm(g) + çè n + ÷ø O2(g) = n (CO2(g) + m H2O(g) (3) 2 In the burner, combustion reactions go to 100% completion. H2S burnt = 0.3 kmol H2 burnt = 0.01 kmol O2 requirement for acid gas = 0.3 ´ 1.5 + 0.01 ´ 0.5 = 0.455 kmol Let a kmol NG are required for firing in the burner. From Example 7.4, 2.064 = 1.032 kmol/kmol NG O2 required = 2 N2 entering in NG = 0.02 kmol/kmol NG Total O2 required = 0.455 + 1.032a kmol Excess O2 supply = 5% = 0.022 75 + 0.0516 a kmol Actual O2 supply = 1.05 (0.455 + 1.032 a) = 0.477 75 + 1.0836 a kmol 78 N2 entering with O2 = (0.477 75 + 1.0836 a) = 1.7745 + 4.0248 a kmol 21

Stoichiometry and Industrial Problems 365

1 (0.477 75 + 1.0836 a) 21 = 0.022 75 + 0.0516 a kmol Total dry air = 2.275 + 5.16 a kmol Moisture entering with air = 0.02 (2.275 + 5.16 a) = 0.0455 + 0.1032 a kmol Acid gas enters saturated at 0.2 bar g and 50°C (323.15 K). pw at 50°C = 12.335 kPa p = 0.2 bar g = 1.21325 bar a = 121.325 kPa a 12.335 Moisture content of acid gas = (121.325 - 12.335) = 0.1132 kmol/kmol dry gas Total moisture in the burner exit gas mixture = 0.3 + 0.01 + 0.1132 + 0.0455 + 0.1032 a + 2.064 a = 0.4687 + 2.1672 a kmol CO2 produced by NG burning = 1.098 kmol/kmol NG Composition of Gas Exit of Burner

Ar entering with O2 =

Component

kmol

SO2 CO2 Ar N2 O2 H2O

0.03 0.69 + 1.098 a 0.022 75 + 0.0516 a 1.7745 + 4.0248a + 0.02a = 1.7745 + 4.0448a 0.02275 + 0.0516a 0.4687 + 2.1672a

Total

3.2787 + 7.4132 a wet 2.81 + 5.246 a dry

Heat effect:

D H ro1 = –296.83 – 241.82 – (–20.63) = –518.02 kJ/mol H2S at 298.15 K (exothermic) D H ro2 = –241.82 kJ/mol H2 at 298.15 K (exothermic) Similarly, D H ro3 = –855.32 kJ/mol NG at 298.15 K (exothermic) Enthalpy of feed (acid) gas:

Heat Capacity Equation Constants Component

kmol

Heat capacity equation constants

n

ai × ni

bi × ni ´ 103

ci × ni ´ 106

di × ni ´ 109

H2S CO2 H2 H2O

0.30 0.69 0.01 0.1132

10.357 14.472 0.286 3.678

–5.294 44.356 0.010 0.009

20.300 –28.325 – 0.001 1.495

–15.974 6.762 0.008 –0.515

Total

1.1132

29.063

39.081

– 6.531

– 9.719

366 Solutions Manual—Stoichiometry

Reference temperature, To = 298.15 K Feed (acid) gas temperature, T1 = 323.15 K H1 = 29.063 (323.15 – 298.15) +

39.081 ´ 10 -3 (323.152 – 298.152) 2

-6.531 ´ 10 -9 9.719 ´ 10 -9 (323.153 – 298.153) – (323.154 – 298.154) 3 4 = 726.6 + 303.5 – 15.8 – 7.3 = 1007.0 kJ/kmol dry acid gas –

Enthalpy of Combustion air: Heat Capacity Equation Constants Component

kmol ni

ai × ni

Heat capacity equation constants bi × ni ´ 103 ci × ni ´ 106 di × ni ´ 109

O2 N2 Ar H2O

0.21 0.78 0.01 0.02

4.314 23.081 0.208 0.650

16.820 – 4.100 — 0.002

–13.112 10.283 — 0.264

3.564 –3.875 — –0.091

Total

1.02

28.253

12.722

–2.565

–0.402

T2 = 313.15 K H2 = 28.253(313.15 – 298.15) +

12.722 ´ 10-3 2

(313.15 2 – 298.15 2 )

2.565 ´ 10-6 0.402 ´ 10-9 (313.153 – 298.153) – (313.154 – 298.154) 3 4 = 423.8 + 58.3 – 3.6 – 0.2 = 478.3 kJ/mol dry air –

Enthalpy (Sensible heat) of NG: Heat Capacity Equation Constants Component

kmol ni

Heat capacity equation constants bi × ni ´ 103 ci × ni ´ 106 di × ni ´ 109

ni × ai

CH4 C2H6 C3H8 i-C4H10 n-C4H10 CO2 N2

0.894 0.050 0.019 0.004 0.006 0.007 0.020

17.209 0.271 –0.080 –0.036 –0.017 0.150 0.592

46.589 8.904 5.819 1.678 2.743 0.450 –0.103

10.704 –3.369 –3.014 0.935 –0.147 –0.287 0.264

–10.118 0.436 0.611 0.204 0.286 0.069 0.099

Total

1.000

18.089

66.080

5.086

–8.413

Stoichiometry and Industrial Problems 367

T3 = 308.15 H3 = 18.089 (308.15 – 298.15) +

66.080 ´ 10-3 (308.15 2 – 298.15 2 ) 2

5.086 ´ 10-6 8.413 ´ 10-9 (308.153 – 298.153) – (308.154 – 298.154) 3 4 = 180.9 + 209.3 + 4.7 – 2.3 = 383.6 kJ/kmol NG Total enthalpy of product gas mixture, exit of burner: H4 = 518.02 ´ 0.3 ´ 1000 + 241.82 ´ 0.01 ´ 1000 + 855.32 ´ 1000 a + 1007.0 + 478.3 (2.275 + 5.16 a) + 383.6 a = 155 406 + 2418 + 855 320 a + 1007.0 + 1088.1 + 2468.0 a + 383.6 a = 159 919.1 + 858 171.6 a kJ Burner exit gas mixture enter reaction furnace where 60% of H2S is converted to elemental sulphur. (4) 2 H2S(g) + SO2(g) = 3 S(g) + 2 H2O(g)

+

1 ´ 0.6 0.4 = 1.5 mol 1.5 ´ 0.3 = 0.45 kmol 0.45 ´ 0.6 = 0.27 kmol 0.27 = 0.135 kmol 2 3 ´ 0.27 = 0.405 kmol 2 1.5 ´ 0.01 = 0.015 kmol 0.015 + 0.27 = 0.285 kmol 1.5 ´ 0.1132 = 0.1698 kmol

Acid gas bypassed and introduced to reaction furnace = H2S entering furnace = H2S reacted by Eq. (2) = SO2 reacted = S produced = H2 reacted = H2O produced = H2O entering with acid gas =

Composition of Gas Mixture Exit of Reaction Furnace Component

kmol

H2S SO2 CO2 Ar N2 O2 S H2O

0.45 – 0.27 = 0.180 0.3 – 0.135 = 0.185 0.69 + 0.69 ´ 1.5 + 1.098 a = 1.725 + 1.098 a 0.022 75 + 0.0516 a 1.7745 + 4.0448 a 0.022 75 + 0.0516 a – (0.015/2) = 0.015 25 + 0.0516 a 0.405 0.4687 + 2.1672 a + 0.285 + 0.1698 = 0.9235 + 2.1672 a

Total

5.211 + 7.4132 a wet 4.2875 + 5.246 a dry

368 Solutions Manual—Stoichiometry

It is specified that gas mixture, leaving reaction furnace should have 1100°C (1373.15 K). Heat effect: D H ro4 = 2(–241.82) + 3(278.81) – [2(–20.63) – 296.83] = +690.88 kJ/2 mol H2S (endothermic) = +345.44 kJ/mol H2S Enthalpy of product gas mixture, leaving reaction furnace, H5 = 159 919.1 + 858 171.6 a – 345.44 ´ 0.27 ´ 1000 + 241.82 ´ 0.015 ´ 1000 + 1.5 ´ 1007.0 = 159 919.1 + 858 171.6 a – 932 688 + 3627.3 + 1510.5 = –767 631.1 + 858 171.6 a kJ Enthalpy, H5 is the enthalpy of product gas at 1100°C (1373.15 K) over 25°C (298.15 K). Enthalpy (H ¢5) of the gas mixture, exit of reaction furnace, can be calculated using heat capacity equation constants of Table 5.2. H ¢5 and H5 should match for a given value of a. Thus it can be a trial and error method in which iterations are to be performed by varying value of a. Such calculations can be done with ease on Excel worksheet, attached herewith. Trial-1: Assume a = 0.1 kmol NG/kmol dry acid gas fed to burner For the exit gas mixture (from reaction furnace),

Cmo pmix (1) = 163.1492 + 110.2838 ´ 10 –3 T – 23.8516 ´ 10 –6 T 2 – 6.5103 ´ 10–9 T 3 1373.15

H 5¢ =

ò

298.15

o Cmp (1) dT mix

= 175 385.3 + 99 070.6 – 20 374.3 – 5773.6 = 248 308.0 kJ/kmol mixture H5 = –767 631.1 + (858 171.6 ´ 0.1) = –621 814 kJ/kmol mixture H 5¢ ¹ H5 After several trails, for a = 1.6395 kmol NG/kmol dry acid gas

Cmo pmix (2) = 495.2154 + 193.5631 ´ 10–3 T + 27.9635 ´ 10–6 T 2 – 34.7041 ´ 10–9 T 3 1373.15

H 5¢ =

ò

298.15

Cmo pmix (2) dT

= 532 356.5 + 173 882.3 + 23 886.7 – 30777.0 = 699 348.5 kJ/kmol mixture H5 = –767 631.1 + (858 171.6 ´ 1.6395) = 699 341.2 kJ/kmol mixture H 5¢ = H5

Iterative Calculations for Claus Process Case (a): Combustion Air Feed Temperature: 40°C (313.15 K) Gas mixture at the exit of the reaction furnace Trial 1: a = 0.1 kmol/kmol dry acid gas fed to burner Component a kmol ai ‘< a* n, SO2 H2S CO2 Ar N2 02 S H20

0.165 0.18 1.725 0.02275 1.7745 0.01525 0.405 0.9235 5.2110

0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1

0 0 0.1098 0.00516 0.40448 0.00516 0 0.21672

0.1650 0.1800 1.8348 0.0279 2.1790 0.0204 0.4050 1.1402 5.9523

24.7706 34.5234 21.3655 20.7723 29.5909 20.5451 27.4597 32.4921

4.0871 6.2142 39.2014 0.5798 64.4780 0.4193 11.1212 37.0481 163.1492

bi

b*n*E3

ci

ci;*ni*E6

1.6395 1.6395 1.6395 1.6395 1.6395 1.6395 1.6395 1.6395

0 0 1.800171 0.084598 6.63145 0.084598 0 3.553124

0.1650 0.1800 3.5252 0.1073 8.4059 0.0998 0.4050 4.4766 17.3649

24.7706 4.0871 34.5234 6.2142 21.3655 75.3170 20.7723 2.2299 29.5909 248.7396 20.5451 2.0514 27.4597 11.1212 32.4921 145.4549 495.2154

di

di*n*E9

62.9481 10.3864 –44.2582 –7.3026 11.1220 1.8351 –17.6481 –3.1767 67.6664 12.1800 –53.2454 –9.5842 64.2841 226.6124 –41.0506 –144.7104 9.7999 34.5463 0.0000 0.0000 0.0000 –5.1410 –43.2150 13.1829 110.8148 –4.9680 –41.7608 80.0947 7.9973 –62.4369 –6.2342 16.9722 1.6946 –13.3278 –5.3978 10.0657 4.0766 –2.6624 –1.0783 0.0796 0.3563 13.2107 59.1393 –4.5474 –20.3570 193.5631 27.9635 –34.7041

Stoichiometry and Industrial Problems 369

0.165 0.18 1.725 0.02275 1.7745 0.01525 0.405 0.9235 5.2110

di*n*E9

62.9481 10.3864 –44.2582 –7.3026 11.1220 1.8351 –17.6481 –3.1767 67.6664 12.1800 –53.2454 –9.5842 64.2841 117.9485 –41.0506 –75.3196 9.7999 17.9809 0.0000 0.0000 0.0000 –5.1410 –11.2021 13.1829 28.7253 –4.9680 –10.8252 80.0947 1.6347 –62.4369 –1.2743 16.9722 0.3464 –13.3278 –5.3978 10.0657 4.0766 –2.6624 –1.0783 0.0796 0.0908 13.2107 15.0631 –4.5474 –5.1850 110.2838 –23.8516 –6.5103

Enthalpy at 1373.15 Kover 298.15 K Total H¢5 175385.3 99070.58–20374.27 –5773.56 248308.1kJ/kmol acidgas to burnerby integration –621813.9kJ/kmol acidgas to burner Difference –870122 kJ H5 Final trial 1: a = 1.6395 kmol/kmol dryacid gas fed to burner Component a kmol ai ‘< a* n, bi b*n*E3 ci ci;*ni*E6 SO2 H2S CO2 Ar N2 02 S H2O

di

SO2 CO2 Ar N2 O2 H2O

0.3 2.49017 0.10735 8.40595 0.10735 4.02182 15.4326

0 0 0 0 0 0

0 0 0 0 0 0

.5 173882.3 23886.68 –30777 699348.5kJ/kmol acidI gas to burnerby integration .2kJ/kmol acidgas to burner Difference –7.308321 kJ

0.3 2.49017 0.10735 8.40595 0.10735 4.02182 15.43264

ai 24.7706 21.3655 20.7723 29.5909 20.5451 32.4921

‘< a* n, 7.43118 53.20373 2.229906 248.7396 2.205516 130.6774 444.4873

bi

b*n*E3

ci

ci;*ni*E6

62.9481 18.88443 –44.2582 –13.27746 64.2841 160.0783 –41.0506 –102.223 0 0 –5.141 –43.21499 13.1829 110.8148 80.0947 8.598166 –62.4369 –6.702601 0.0796 0.320137 13.2107 53.131057 144.6661 41.742822

di

di*n*E9

11.122 3.3366 9.7999 24.40342 0 –4.968 –41.76076 16.9722 1.821966 –4.5474 –18.28882 –30.4876

Enthalpy at 2918.26 K over 298.15 K Total 1164606 609575.7 345437.2–552728.5 1566890 kJ/kmol acid gas to burner Case (b): Combustion Air Feed Temperature: 250°C (523.15 K) Component a kmol ai ‘< a* n, bi b*n*E3 ci ci;*ni*E6 di di*n*E9 SO2 H2S CO2 Ar N2 02 S H2O

0.165 0.18 1.725 0.02275 1.7745 0.01525 0.405 0.9235 5.2110

1.4106 1.4106 1.4106 1.4106 1.4106 1.4106 1.4106 1.4106

0 0 1.548839 0.072787 5.705595 0.072787 0 0.145574

0.1650 0.1800 3.2738 0.0955 7.4801 0.0880 0.4050 1.0691 12.7566

24.7706 4.0871 34.5234 6.2142 21.3655 69.9472 20.7723 1.9845 29.5909 221.3427 20.5451 1.8087 27.4597 11.1212 32.4921 34.7365 351.2422

62.9481 10.3864 –44.2582 –7.3026 11.1220 1.8351 –17.6481 –3.1767 67.6664 12.1800 –53.2454 –9.5842 64.2841 210.4558 –41.0506 –134.3930 9.7999 32.0833 0.0000 0.0000 0.0000 –5.1410 –38.4552 13.1829 98.6093 –4.9680 –37.1611 80.0947 7.0513 –62.4369 –5.4968^ 16.9722 1.4942 –13.3278 –5.3978 10.0657 4.0766 –2.6624 –1.0783 0.0796 0.0851 13.2107 14.1232 –4.5474 –4.8615 180.9490 –18.2033 –17.2725

Enthalpy at 1373.15 K over 298.15 KTotal 377585.3 162550.8 –15549.4 –15317.91 509268.8 kJ/kmol acid gas to burner H5' 509281 kJ/kmol acid gas to burnerDifference 12.17642 kJ H5

370 Solutions Manual—Stoichiometry

Enthalpy at 1373.15 K over 298.15 K Total 532356 H5' 699341 H5 Gas mixtureat the exitof the burner Component a kmol

Stoichiometry and Industrial Problems 371

Hence, a = 1.6395 kmol NG/kmol dry acid gas fed to burner Substitute a = 1.6395. Gas Mixture, Exit of Gas Burner Component SO2 CO2 Ar N2 O2 H2O Total

Ans. (a–i)

kmol

mole % (wet)

mole % (dry)

0.3 2.49017 0.10735 8.40595 0.10735 4.02182

1.94 16.14 0.70 54.46 0.70

2.63 21.82 0.94 73.67 0.94

100.00

100.00

15.43264 (wet) 11.41082 (dry)

Using heat capacity equation constants, C°mp (3) = 444.4873 + 144.6661 ´ 10–3 T + 41.7428 ´ 10–6 T2 – 30.4876 mix ´ 10–9 T 3 Enthalpy of the gas mixture, H4 = 159 919.1 + (858 171.6 ´ 1.6395) = 159 919.1 + 1406 972.3 = 1566 891.4 kJ T

=

ò

298.15

Cmo pmix (3) dT

Solving by Mathcad, T = 2918.25 K or 2645.1°C (b) Combustion air is preheated to 250°C (523.15 K). Enthalpy of combustion air at 523.15 K H 2¢ = 6256.9 + 1175.5 – 99.8 – 6.7 = 7425.9 kJ/kmol dry air Enthalpy of product gas mixture, leaving reaction furnace, H5 = –767 631.1 + 858 171.6a + (7425.9 – 478.3) (2.275 + 5.16a) = –751 825.3 + 894 021.2 a kJ For a = 1.4106

Cmo pmix (4) = 351.2422 + 180.9490 ´ 10–3 T – 18.2033 ´ 10–6 T2 – 17.2725 ´ 10–9 T3 1373.15

H ¢5 =

ò

298.15

Cmo pmix (4) dT

= 509 269/kmol mixture H5 = 4\509 281 kJ/kmol mixture H 5¢ = H5 Hence, a = 1.4106 kmol NG/kmol dry acid to burner Ans.(b) A reduction of 13.96% NG can be realized by preheating combustion air.

372 Solutions Manual—Stoichiometry

EXERCISE 8.29 Basis: qr1 = 120 m3/h waste water feed to stripper Benzene content = 200 mg/L = 200 g/m3 waste water . Benzene entering in waste water, m1 = 120 ´ 200/100 = 24 kg/h . Molar flow rate of benzene, n1 = 24/78.1118 = 0.3073 kmol/h Since waste water, leaving the stripper, has negligible benzene content, all benzene, entering in feed, will be stripped. Nitrogen flow rate, qv2 = 3400 m3/h Nitrogen enters stripper at 0.2 bar g and 50°C (323.15 K). 0.08314 ´ 323.15 Specific volume of gas = (1.013 25 + 0.20)

= 22.144 m3/kmol . Molar flow rate of dry nitrogen, n 2 = 3400/22.144 = 153.54 kmol/h Benzene in ingoing nitrogen = 20 ´ 10–6 ´ 3400 = 0.068 m3/h º 0.0031 kmol/h Benzene in outcoming nitrogen from stripper = 0.0031 + 0.3073 = 0.3104 kmol/h Specific benzene content of nitrogen stream, leaving stripper 0.3104 153.54 = 0.002 022 kmol/kmol º 2022 ppm (v/v) Nitrogen stream leaves stripper at 0.14 bar g, saturated with water at 50°C (323.15 K) Nitrogen leaves heater at 0.12 bar g and 45°C (318.15 K) and 40% RH. Vapour pressure of water at 45°C, pv1 = 9.582 kPa Partial pressure of water, p1 = 9.582 ´ 0.4 = 3.8328 kPa Moisture content of nitrogen at heater outlet,

=

3.8328 H2 = (113.325 - 3.8328)

= 0.035 kmol/kmol dry nitrogen Vapour pressure of water at 50°C = 12.335 kPa Moisture content of nitrogen at stripper outlet,

Stoichiometry and Industrial Problems 373

12.335 H1 = (115.325 - 12.335) = 0.1198 kmol/kmol dry nitrogen Moisture, required to be condensed = H1 – H2 = 0.1198 – 0.035 = 0.0848 kmol/kmol Total moisture removal = 0.0848 ´ 153.54 = 13.02 kmol/h º 234.56 kg/h Since maximum moisture content of 0.035 kmol/kmol is desired of the nitrogen stream from chiller, the stream should be cooled to a saturation level at a system pressure of 0.13 bar g. pv2 114.325 - p = 0.035 v2

pv2 = 3.866 kPa From Table 6.13, saturation temperature t2 = 28.4°C or T2 = 301.55 K Thus nitrogen stream will leave stripper at 0.14 bar g and 50°C (323.15 K), will be cooled to 301.55 K and 0.13 bars in chiller and then rehated to 45°C (318.15 K) at 0.12 bar g. Chiller: Average condensation temperature in 50 + 28.4 chiller = 2 = 39.3°C or 312.35 K lv at 39.2°C = 2408.8 kJ/kg Gas stream entering chiller . Component ni Heat capacity equation constants . . . . kmol/h ai ×n i bi ×n i ´ 103 ci × n i ´ 106 di × n i ´ 109 N2 153.54 4548.387 –789.349 2024.102 –762.787 18.39 597.530 1.464 242.945 –83.628 H2O 0.3104 –11.791 152.227 –99.759 24.635 C6H6 Total

172.2402

5134.126

-635.658

2167.288

-821.779

Gas stream leaving chiller . Component ni Heat capacity equation constants . . . . kmol/h ai × n i bi × n i ´ 103 ci × n i ´ 106 di × n i ´ 109 N2 153.54 4548.387 –789.349 2024.102 –762.787 5.37 174.483 0.427 70.941 –24.420 H2O 0.3104 – 11.791 152.227 –99.759 24.635 C6H6 Total

159.2204

4711.079

–636.695

1995.284

–762.572

374 Solutions Manual—Stoichiometry

Reference temperature, T0 = 298.15 K Enthalpy of gas mixture, entering chiller 323.15

f1 =

ò

298.15

5134.126 – 635.658 ´ 10–3 T + 2167.288 ´ 10–6 T2 – 821.779

´ 10–9 T3) dT = 128 034 kJ/h º 35.565 Enthalpy of gas mixture, leaving chiller 301.55

f2 =

ò

298.15

(4711.079 – 639.695 ´ 10–3 T + 1995.284 ´ 10–6 T2 – 762.572

´ 10–9 T3) dT = 15 909 kJ/h º 4.419 kW Heat load of chiller, fc = f1 – f2 + 2409.8 ´ 234.56 = 128 034 – 15 909 + 565 008 = 677 133 kJ/h º 188.093 kW º 53.52 TR Heat load of heater, 318.15

fh =

ò

301.55

(4711.079 – 636.695 ´ 10–3 T + 1995.284 ´ 10–6 T 2 – 762.572

´ 10–9 T 3) dT = 77.733 kJ/h º 21.593 kW Saturated steam at 2 bar a is used for heating lv at 2 bar a = 2201.6 kJ/kg 77 733 2201.6 = 35.3 kJ/h Regeneration of adsorber is carried out in 4 hours at an uniform rate after heating the bed for 3 hours. Flow rate of vapours to condenser

Steam consumption in heater =

24 ´ 8 4 = 748 kg/h During regeneration, pressure of about 0.1 bar g is maintained in the adsorber. ts/Ts = 102.32°C/375.47 K lv1 at 1.113 25 bar a = 2251.4 kJ/kg

= 700 +

0.38

é 562.05 - 375.47 ù lv2 of benzene at 375.47 K, lv2 = 30720 ê ú ë 562.05 - 353.3 û = 29 437 kJ/kmol º 376.86 kJ/kg

Stoichiometry and Industrial Problems 375

Heat load of condensation = 700 ´ 2251.4 + 48 ´ 376.86 = 1575 980 + 18 089 = 1594 069 kJ/h º 442.797 kW Both liquids are cooled from 375.47 K to 318.15 K. Cooling load of benzene 375.47

= 0.3073

ò

318.15

(–484.365 + 5056.235 ´ 10–3 T – 14 292.204 ´ 10–6 T2

+ 14 419.754 ´ 10–9 T3) dT = 0.3073 ´ 8704 = 2675 kJ/h º 0.743 kW Cooling load of water = 700 (428.8 – 188.4) = 168 280 kJ/h º 46.744 kW Total heat load of condenser fcd = 1594 069 + 2675 + 168 280 = 1765 024 kJ/h º 490.284 kW

Ans.

EXERCISE 8.30 Reaction pressure,

p = 50 atm = 50.6625 bar Since the reaction take place at very high pressure, use of partial pressures may lead to wrong answes. It is therefore necessary to consider gas-vapour mixture to be non-ideal. Rewriting Eq. (8.1). yAA = and

yMe =

pv AA x AA g AA f sat AA f AA p sat pvMe xMe g Me fMe

fMe p where gi = activity coefficient of i-th component in solution (assumed for similar compoinds) fi = fugacity coefficient of i-th component in pure state sat fi = fugacity coefficient of i-th component in pure state at saturated conditions From Ref 13, and Eq. (8.2) é pri o ù ( Bi + w i × Bi1 ) ú fi = exp ê ë Tri û

and

fi

sat

é prisat o ù 1 = exp ê T ( Bi + w i × Bi ) ú ëê ri ûú

376 Solutions Manual—Stoichiometry

where

pri = reduced pressure p = p ci Tri = reduced temperature T = Tei

and

pvi prisat = p ci and w i = acentric factor for ith component Boi and Bi1 are Pitzer correlation coefficients and are functions of temperature only 0.422 Boi = 0.083 – T 1.6 ri

p cisat =

pvi pci

0.172 Bi1 = 0.139 – T 4.2 ri

and Component

Tci, K

pci, bar

wi

pvi, bar

Tri

Acetic acid Methanol

590.7 512.64

57.8 80.92

0467 0.564

5 27

Component

psat ri

B 0i

B1i

fi

Acetic acid Methanol

0.0864 0.3337

–0.5655 –0.431

–03923 –0.1497

0.4242 0.6942

0.7645 0.884

p ri 0.8756 0.626 fsat i 0.9189 0.8232

5 ´ 0.15 ´ 0.9189 yAA = 50.6625 ´ 0.4242 = 0.032 27 ´ 0.85 ´ 0.8232 yMe = 50.6625 ´ 0.6842 = 0.5372

For gas-vapour mixture, yCO + yH2 + yAA + yMe = 1 yCO + yH2 = 1 – 0.032 – 0.5372 = 0.4308 =

nCO + nH 2 nt

=

65 + 35 nt

100 100 = = 232.126 kmol/h 0.4308 0.4308 . nMe = 0.5372 ´ 232.126 = 124.7 kmol/h . nAA = 0.032 ´ 232.126 = 7.43 kmol/h

. nt =

Stoichiometry and Industrial Problems 377

If the gas-vapour mixture is considered ideal, vapour-liquid equilibrium will follow Raoult's law. y¢AA =

x AA × pv AA p

=

0.15 ´ 5 50 × 6625

= 0.0148 0.85 ´ 27 50.6625 = 0.453 + yH2 = 1 – 0.0148 – 0.453 = 0.5322

y¢Me =

yCO2

100 = 187.899 kmol/h 0.5322 . nMe = 0.453 ´ 187.899 = 85.12 kmol/h . nAA = 0.0148 ´ 187.899 = 2.78 kmol/h Molar flow rates of acetic acid and methanol are significantly high with nonideality. Ans.

. nt =

9

Stoichiometry and Digital Computation EXERCISE 9.1 a. by spreadsheet Component

y

H2 C2H6 C2H4 C3H8 C3H6 N2 NH3

0.81 0.02 0.04 0.01 0.02 0.08 0.02

Mm 2.0159 30.0696 28.0538 44.0965 42.0806 28.0135 17.0306

y.Mm

Tc, K

1.6329 0.6014 1.1222 0.441 0.8416 2.2411 0.3406

32.200 305.420 282.340 369.820 364.850 126.090 405.500

7.2207 p T pc Tc R Mm e dV a B

2.00E+06 333.15 1.98E+06 72.68 8314 7.2207 0.00100 0.10 77682.8 0.038093

y.Tc, K 26.082 6.108 11.294 3.698 7.297 10.087 8.110

pc, Pa 1.297E+06 4.880E+06 5.039E+06 4.247E+06 4.601E+06 3.394E+06 1.135E+07

72.676

y.pc, Pa 1.051E+06 9.760E+04 2.016E+05 4.247E+04 9.202E+04 2.715E+05 2.270E+05 1.983E+06

Pa K Pa K kJ/(kmol × K)

Trial

V

F(V)

dV

new dV

Test

sp. vol.

0 1 2 3 4 5 6

0.1 0.20000 0.30000 0.40000 0.50000 0.60000 0.70000

–0.22164 –0.54072 –0.70550 –0.75247 –0.73677 –0.68813 –0.62128

0.10000 0.10000 0.10000 0.10000 0.10000 0.10000 0.10000

0.10000 0.10000 0.10000 0.10000 0.10000 0.10000 0.10000

FALSE FALSE FALSE FALSE FALSE FALSE FALSE

# N/A # N/A # N/A # N/A # N/A # N/A # N/A

Stoichiometry and Digital Computation 377

7 8 9 10 11 12 13 14 15 16 17

0.80000 0.90000 1.00000 1.10000 1.20000 1.30000 1.35000 1.37500 1.38750 1.39375 1.39688

–0.54376 –0.45963 –0.37122 –0.27992 –0.18662 –0.09188 –0.04410 –0.02012 –0.00811 –0.00210 0.00091

b. by Mathcad

FG H

a (R, Tc, pc) := 27 × R2 b (R, Tc, pc) := R × f (V, p, R, T, Tc, pc) :=

0.10000 0.10000 0.10000 0.10000 0.10000 0.10000 0.05000 0.02500 0.01250 0.00625 0.00313

Tc2 64 × pc

Tc 8× pc

0.10000 0.10000 0.10000 0.10000 0.10000 0.05000 0.02500 0.01250 0.00625 0.00313 0.00156

FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE

# N/A # N/A # N/A # N/A # N/A # N/A # N/A # N/A # N/A # N/A 0.19345

IJ K

LMF p + a( R, T , p ) I ×(V - b( R, T , p ))OP – (R × T) K V NH Q c 2

c

c

c

Guess value of v V := 1.358 V := root ( f (V, 2 × 106, 8314, 333.15, 72.676, 1.983 × 106),V) V = 1.39593 m3/kmol V 7.2207 spvol = 0.19332 m3/kg

spvol :=

EXERCISE 9.2

F aI H vK b B(b0, b, v) := b0 × F1 - I H vK A(a0, a, v) := a0 × 1 -

e(c, T, v) :=

c v×T 3

f(a0, b0, a, b, c, T, R, p, v) := p –

LM R× T ×(1 - A(c, T, v) ×(v + B(b0, b, v)OP v N Q 2

+

A( a0, a, v ) v2

378 Solutions Manual—Stoichiometry

For Guess value of T use Ideal Gas Law. T := T1 soln := root (f(a0, b0, a, b, c, T, R, p, v), T) soln = 1 Solution of Exercise 2.37 T := 840 soln := root (f (0.588, 0.094, 0.05861, 0.01915, 90 × 104, T, 0.008314, 4.2, 1.6628), T) soln = 835.183 K

EXERCISE 9.3 (a) Exercise 4.36 Case (a). F := 100 x := 0.24 y := 0.1 z := 0.65 yf := 0.01 af := 24.75 M := 10 P := 8 a := 2 For equations refer the solution of Exercise 4.36. Given y × M × P = yf × F × (M – 2 × x × a × (1 + z) – M × z × (1 – y) + 4 × a × z) 100 + (M – 2 × x × a × (1 + z) – M × z × (1 – y) + 4 × a × z) – P = M (1 – x) × a ×

LM OP + a = a P 2 ( 1 ) ( 1 ) 4 M x a z M z y a z × × × + × + × × N Q f

vec := Find (M, P, a)

LM455.293 OP vec = MMN 967..9789 P 4188PQ

M := vec0 M = 455.293 R := M – F

P := vec1 P = 7.9789 R = 355.293

a := vec2 a = 96.4188

R rr = 3.5539 F nh3 := (2 × x × a + M × (1 – y) – 4 × a) × z nh3 = 45.7401 (b) Exercise 4.36 Case (b). F := 100 x := 0.25 y := 0.1 z := 0.7 yf := 0.01 af := 24.75 M := 10 P := 8 a := 2 Given y × M × P = yf × F × (M – 2 × x × a × (1 + z) – M × z × (1 – y) + 4 × a × z) 100 + (M – 2 × x × a × (1 + z) – M × z × (1 – y) + 4 × a × z) – P = M

rr :=

LM N

(1 – x) × a × 1 -

OP Q

P + af = a M - 2 × x × a × (1 + z ) - M × z × (1 - y ) + 4 × a × z

Stoichiometry and Digital Computation 379

vec := Find (M, P, a)

LM432.801 OP vec := MMN 927..8713 P 5826 PQ M := vec0 M = 432.801 R := M – F

P := vec1 P = 7.8713 R = 332.801

a := vec2 a = 92.5826

R rr = 3.328 F nh3 := (2 × x × a + M × (1 – y) – 4 × a) × z

rr :=

nh3 = 45.8374

EXERCISE 9.4 HP steam enthalpy = MP superheated steam enthalpy = LP superheated steam enthalpy = LP saturated steam enthalpy = 15 bar saturated steam enthalpy = Surface condenser steam enthalpy = BFW enthalpy = Surface condenser condensate enthalpy =

H1 H2 H3 H4 H5 H6 H7 H8

kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg kJ/kg

Let x1 kg BFW required to quench 1 kg of 39 bar MP superheated steam to 15 bar saturated steam. H - H5 x1(H2, H5, H7) := 2 H5 - H7 Q1(H2, H5, H7) :=

3 t MP steam required/h 1 + x1 ( H2 , H5 , H7 )

Let x2 kg BFW required to quench 1 kg of 39 bar MP superheated steam to 4.4 bar saturated steam. H - H4 x2(H2, H4, H7) := 2 H 4 - H7 Let x3 kg BFW required to quench 1kg of 4.4 bar superheated steam to 4.4 bar saturated steam. H - H4 x3(H3, H4, H7) := 3 H 4 - H7 Power Turbine: Specific Steam consumption in Power Turbine from HP to MP steam: 3600 S1(H1, H2) := ( H1 - H2 ) × 0.97

380 Solutions Manual—Stoichiometry

Steam consumption in turbine from MP to LP: 3600 S2(H2, H3) := ( H2 - H3 ) × 0.97 Specific Steam consumption in Power turbine from HP to SC steam: 3600 S3(H1, H6) := ( H1 - H6 ) × 0.97 Steam turbine from MP to surface condenser: 3600 S4(H2, H6) := ( H2 - H6 ) × 0.97 Cooling water requirement in surface condenser: H 6 - H8 Qcw(H6, H8) := 10 × 4.1868 Guess a := 1 b := 1 c := 1 d := 1 e := 1 g := 1 Given 13.5 + a = (1 + x3(H3, H4, H7)) × (b + c + d) + (1+ x2(H2, H4, H7)) × g a = 0.11 × (e + 16.5)

(b + c + d + g + Q1 ( H2 , H5 , H7 )) e 6832 + = S3 ( H1 , H6 ) 1000 S1 ( H1 , H2 ) 2.47 × (b + c + d + e + g + Q1(H2, H5, H7)) × S2(H2, H3) = 1000d 5.493 (a + e + 16.5) × S2(H2, H3) = 1000c 0.2315 [(Qcw(H6, H8) × e) + (600)] × S2(H2, H3) = 1000b vec(H1, H2, H3, H4, H5, H6, H7, H8) := find(a, b, c, d, e, g) Case – I vec(3190.7, 3020.4, 2793.3, 2741.9, 2800.8, 2472.0, 440.17, 209.2)

a b c d e g Z

LM 5.1556 OP 4795 MM 48..6702 PP = MM 2.0625PP MM302..3692 PP 7687 N Q

= Saturated LP steam input to deaerator = 5.156 t/h = MP steam input to CW pump turbine = 8.480 t/h = MP steam input to BFW pump turbine = 4.670 t/h = MP steam input to FD fan turbine = 2.063 t/h = Exhaust (under vacuum) from power turbine = 30.369 t/h = Letdown from MP steam to LP steam = 2.769 t/h = Total HP steam production = 51.095 t/h

Stoichiometry and Digital Computation 381

By substituting appropriate values of variables in vec(H1, H2, H3, H4, H5, H6, H7,H8) for Cases – II to V, values of a, b, c, d, e and g can be obtained. These are tabulated in Table 9.7 in the text.

EXERCISE 9.5 E1 is Ethylene required for EO production: 28.0538 ×E E 1(E) := 44.0532 Ethylene reacted: E ( E) E2(E, yE) := 1 yE Ethylene converted to CO2 and H2O: Eb(E, yE) := E2(E, yE) – E1(E) Total ethylene mixed feed: E3(E, yE, xE) := Total mixed feed in the reactor: E4(E, yE, xE, yI1) :=

E2 ( E, yE ) x E × 28.0538 E3 ( E, yE , x E ) 0.1

Inerts in mixed feed: Imf (E, yE, xE, yI1) := E4(E, yE, xE, yI1) × yI1 Oxygen reacted by reaction 1: O21(E, yE, xE) := Oxygen reacted by reaction 2: O22(E, yE, xE) :=

E3 ( E, yE , x E ) × x E × yE 2 Eb ( E, yE ) × 3 28.0538

Total oxygen fed: O23(E, yE, xE) := O21(E, yE, xE) + O22(E, yE, xE) Inerts fed along with oxygen: I1(E, yE, xE, yI2) :=

O23 ( E, yE , x E ) × yI 2 (1 - yI 2 )

CO2 and H2O in mixed feed: E5(E, yE, xE, yI1) := E4(E, yE, xE, yI1) – E3(E, yE, xE) – O23(E, yE, xE) – Im(E, yE, xE, yI1) CO2 and H2O produced: E6(E, yE, xE) :=

O22 ( E, yE , x E ) × 4 3

Reactor outlet stream: E7(E, yE, xE, yI1, yI2, x, P) := E3(E, yE, xE) × (1– xE) + (I1(E, yE, xE, yI2)

382 Solutions Manual—Stoichiometry

LM N Recycle stream 1 contains:

+ O23(E, yE, xE) + E3 ( E, yE , x E ) × (1 - xE ) ×

P R1 ( E, yE , xE , yI1 , x )

OP Q

R1(E, yE , xE, yI1, x) := E3(E, yE, xE) (1 – xE) + Imf (E, yE, xE, yI1) + x Recycle stream 2 contains: R1(E, yE , xE, yI1, x, P) := R1(E, yE , xE , yI1, x) – P Given

x × R2 ( E, y E , x E , yI1 , x, P) = E5(E, yE, xE, yI1) R1 ( E, y E , x E , yI1 , x )

Imf ( E, yE , x E , yI1 ) × P R1 ( E, yE , x E , yI1, x )

=

yI 2 × O23 ( E, yE , x E ) (1 - yI 2 )

soln(E, yE, xE, yI1, yI2, x, P) := find (x, P) x := 1600

P := 44

soln(3500, 0.7, 0.5, 0.1, 0.03, x, P) =

LM 1.707 × 10 OP N39.58 Q 3

x := soln(3500, 0.7, 0.5, 0.1, 0.03, x, P)0 P := soln(3500, 0.7, 0.5, 0.1, 0.03, x, P)1 x = 1.707 × 103

E := 3500

P = 39.58 kmol/h yE := 0.7

xE := 0.5

yI1 := 0.1

yI2 := 0.03

Recycle Stream R2 to Reactor:

R2(E, yE, xE, yI1, yI2, x, P) = 2.008.103 kmol/h

Fresh Ethylene feed: FE := E3(E, yE, xE)

LM N

– E3 ( E, yE , xE ) × (1 - xE ) - E3 ( E, yE , xE ) × FE = 115.693

FE × 28.0538 = 3.246.103 kg/h

(1 - x E ) × P R1 ( E, yE , xE , yI1 , x )

Inerts entering with oxygen : I1(E, yE, xE, yI2) = 4.388 Oxygen fed : O23(E, yE, xE) = 141.874 Total Fresh Feed = Fresh Ethylene Feed + Inerts with oxygen + oxygen fed : F E := FE + I1(E, yE, xE, yI2) + O23(E, yE, xE) F E = 261.955 kmol/h

OP Q

Stoichiometry and Digital Computation 383

Recycle ratio:

R2 ( E, yE , x E , yI1 , x, P) FF ratio := 7.666 kmol/kmol fresh feed By substituting appropriate values of yI1, various parameters can be calculated. Please refer Table 9.8 in the text for the answers.

ratio :=

EXERCISE 9.6 Double recycle system of ammonia synthesis loop

384 Solutions Manual—Stoichiometry

Stoichiometry and Digital Computation 385

386 Solutions Manual—Stoichiometry

By varying values of inerts in mixed feed (i), ratio of H2 to carbon oxides (r) and K, Tables 9.10, 9.11 and 9.12 are prepared.