Stoic-solution to Problems

Page 1 of 61 SOLUTION TO SOME PROBLEMS IN CHAPTER 1 1-8 1

2

3

piping dist. system leakage

4 5

Given: Reading of main meter: 250 m3; Consumptions in m3: tenant 1, 40; tenant 2, 35.5; tenant 3, 54.3; tenant 4, 4.36; and tenant 5, 55. Required: the amount of leakage. Solution: Arithmetic solution (algebraic solution also acceptable) Overall mass balance: Σmass inputs = Σmass outputs Total watery supplied = supply to tenant 1 + supply to tenant 2 + supply to tenant 3 + supply to tenant 4 + supply to tenant 5 + leakage (all in m3)

Leakage = 250 - (40 + 35.5 + 54.3 + 36 + 55) = 29.2 m3

1-10. As an irrigation system delivers water from the source to the point of usage, the following losses are encountered: 9% to evaporation, 7% to run-off, and 15% to infiltration. If 4,000 liters was finally used what was the volume of the water at the source? water source

DRAFT

evaporation water delivered

runoff

Infiltration

Given: Losses to evaporation, run-off, and infiltration with 4,000 liters finally used Required:

The volume of the water at the source

Solution: Let x = the volume of the water at the source 0.09x + 0.07x + 0.15x + 4000 = x x - 0.31x = 4000 x

=

4000/0.69

=

5,797

L

Page 2 of 61 SOLUTION TO SOME PROBLEMS IN CHAPTER 2 2-1

g kg p cm a. 8 cp% 100 cp % p $ s % 1000 g % 100mcm % kgN$ m % Pa = 0.8 Pa$s N 2 m s2 g g c % lb 12 in 2.54 cm kg m b. 150 ft$lbf % lbf % ft % % % % 9.8 m2 % Nm = 203.7 J 100 cm 2.2 lb in s kg 2 s g $ cm kg N s2 % m c. 1000 dynes % dyne =0.1 N % % 1000 g 100 cm kg $ m s2 3 3

gal qts (12 in) (2.54 cm) ft 3 m3 =0.016 % % % % % 3 3 (100 cm) 3 2 pints 4 qts 7.48 gal ft in 3 3 (2.54 cm) m = 0.344 m3 % 21, 000 in 3 % (100 cm) 3 in 3 gal 0.05 pint (12 in) 3 (2.54 cm) 3 ft 3 m3 = 0.021 m 3 % % % % % 900 fl. ounce% 3 3 (100 cm) 3 8 pints 7.48 gal fl. ounce ft in kg (100 cm) 3 ft 3 in 3 = 3.371x 104 kg/m3 2100 lbm 3 % 2.2 lbm % (12 in) 3 % (2.54 cm) 3 % m3 ft 252 cal 4.18 J 6 7150 Btu s % Btu % cal = 7.532 x 10 W

f. 34 pints% g. h. i.

DRAFT

j.

Page 3 of 61 2.4

Find out which of the following formulas is correct:

ØC v ØV

C, temperature, θ P, pressure, F/L2 V, specific volume, L3/M Cv, heat capacity, FL/MT Solution: Substituting the dimensions; Eq. 1:

FL Left side MT = F2 TL L3 M

Eq. 2

Left side

T

2 = T Ø P2 ØT

V

T

or

Right side T F/L T2

T FL/MT = F2 L 3 /M TL

2

T

=

2 = Ø P2 ØT

V

F TL 2

2 4 2 Right side F /L = F 2 4 2 T L T

The first equation is correct since it is dimensionally consistent.

DRAFT

ØC v ØV

Page 4 of 61 2-5 Find the value of the following dimensionless groups for the given data: Dv! a. The Reynolds No., N Re =  where D = diameter, L v = velocity L/θ ρ = density, M/L3 µ = viscosity, M/Lθ for a case where D = 16 in, v = 6 m/s, ρ = 4.0 kg/gal and

µ = 300

# ft $ h

Solution:

N Re

cm )(4 kg % 7.48 gal % ft 3 in 3 (16 in % 2.54 cm )(6 m % 100 % 3 3 3 cm 3 ) m s 3 2.54 Dv! gal in ft 12 in =  = kg in h 300 lb % % ft % % ft $ h 2.2 lb 12 in 2.54 cm 3600 s

DRAFT

N Re = = 2.073 % 10 4

 b. The Schmidt No., N Sc = !D v where

µ = viscosity, M/L ρ = density, M/L3 Dv = diffusivity, L2/θ

for a case where µ = 12 centipoise, ρ = 30 kg/cu ft, Dv = 745 cm2/h. Solution:

g p cm $s 12 cp % % p 100 cp ! N Sc = = 3 D v kg 1000 g cm 2 % in 3 h 30 3 % 745 % ft % % 3 3 3 2.54 cm 3600s h kg ft 12 3 in N Sc = 0.5473 2 c. The Froude No., N Fr = v Lg

v = velocity, L/θ L = length, L g = acceleration due to gravity for a case where L = 6 ft, v = 20 cm/s

where

N Fr = v = Lg

2 (20 cm s )

ft 2 in 2 2 % (12 in) 2 (2.54 cm) = 2.229 % 10 −3 ft 6 ft % 32.2 2 s

Page 5 of 61

2.7

The viscosity of a liquid at the critical point is given by formula  c = 61.6

MT c (V c ) 2/3

µ = critical viscosity in micropoise M = molecular weight Tc = critical temperature, K Vc = critical volume, cm3/mol For a similar formula using µc in Pa⋅s, and Vc in m3/kmol, and Tc in K, find the value of the constant corresponding to 61.6. The constant with the corresponding units is

where

3

( cm ) 2/3 mol 16 p K

DRAFT

Converting to the desired units 3 m3 ( cm % % 1000 mol ) 2/3 mol (1000 cm) 3 kmol P % % 0.1 Pa $ s new constant = 16 P P K 10 6 P 3 ( m ) 2/3 = Pa$s kmol K

Page 6 of 61

2-9

ft ( lb3 ) 0.4 ft 0.086 dyne (cP) 2 ( cm ) 3

is changed to

g 0.4 3) cm by conversion of units  dyne 3 lb 2 ( ) ( cm ) ft $ h cm (

454 g ft 3 % % 0.086 ft % 12 in % 2.54 cm ( lbm ) 0.4 ft in lbm (12 % 2.54 cm) 3 ft 3 = 1g/cm $ s poise % % lb % 12 % 2.54 cm % 3600 s ) 2 (cP % 454 g 100 cP poise ft h

1

dyne g $ cm/s kg cm 3 m n % % % % 100 ( cm % m ) 1000 g 100 cm kg $ m/s 2 dyne

DRAFT

%

= 8.59 % 10 7

c,(g/cm3 ) 0.4 N )3 ( lbm ) 2 % ( m ft $ h

Page 7 of 61 2-10. The temperature of a cold water being heated varies linearly with the distance from one end of the heat exchanger. From the following data find the suitable equation representing the variation graphically: Temperature, °C Distance, m

4.44 0

5.56 0.24

11.67 1.13

19.44 2.07

25.00 2.99

32.22 4.11

Solution: The equation is T = (slope) + mx T, temp x, distance Plot Temperature on the ordinate and x on the abcissa

Prob-2-10

30

T, temperature deg, Celsius

DRAFT

35

25 20 15 10 5 0 0

1

2

3

X , distance in m

Intercept: Slope:

4.2751052333 6.8955757685

Equation:

T = 4.275 + 6.896

4

5

Page 8 of 61

2.11

The following table gives the relationship between dissolved methane in a solution and the partial pressure of methane above the solution. (Chem. Eng. Prog. symp. series, Volume 65, 73 (1969).

Dissolved CH4, ppm 20 115 225 Partial pressure of CH4, kPa 138 689 1379 If the variation is linear, find the suitable relationship.

340 2068

450 2758

560 3447

Partial pressure of CH4, kPa

DRAFT

4000

3000

2000

1000

0

0

100

200

300

Dissolved CH4 , ppm

2-14

400

500

600

Page 9 of 61 y x

12.6 1.25

6.75 2.5

3.86 3.8

1.95 5.26

Solution: Plot log y vs x on a rectangular coordinate graph or plot y vs x on semilog graph. We use a semilog plot. intercept = 1.3437

Problem 2-14

Y Data

DRAFT

10

1 1

2

3

4

X Data Y vs X Plot 1 Regr

slope = - 0.2005

equation:

log y = -0.2005 x + 1.3437

y = 10 −0.2005x

+ 1.3437

5

6

Page 10 of 61 2-18 Convert the following temperature readings a. 180K to °C, °R, and °F Rdg in °C = 180 - 273 = -93°C Rdg in °R = 180× 1.8 = 324°R Rdg in °F = (180× 1.8) - 460 = 136°F b. 800°F to K, °C, and °R Rdg in K = (800 - 32)/1.8 + 273 = 699.7 K Rdg in °C = (800 -32)/1.8 = 426.7°C Rdg in °R = 800 + 460 = 1260°R c. 100,000°C to K, °R, and °F.

DRAFT

Rdg in K = 100,000 -273 = 99,723 K Rdg in °F = (100,000)1.8 + 32 = 180032°F Rdg in °R = (100,000)1.8 + 32 -460 = 179572°R

Page 11 of 61 2-26

Convert the following gage pressures to absolute values if the barometric pressure is 757 mm Hg:

a. b. c. d.

a.

200 psig 4 cm of Hg vacuum 29 mm of H20 head 15 mm of H20 draft Pabs = Pbar + P gage P abs = 757 mm Hg %

b.

Pabs = Pbar − P vac Pabs = 757 mm Hg %

c.

14.7 psi + 200 = 214.6 psi 760 mm Hg

76 cm Hg 4 cm Hg= 71.7 cm Hg 760 mm Hg −

Pabs = Pbar + head P abs = 757 mm Hg + 29 mm H 2 O%

d.

Pabs = Pbar − draft

DRAFT

P abs = 757 mm Hg −15 mm H 2 O%

760 mm Hg 1 ft H 2 O = 778.3 mm Hg % 12 % 2.54 in 33.9 ft H 2 O

760 mm Hg 1 ft H 2 O = 746 mm Hg % 12 % 2.54 in 33.9 ft H 2 O

Page 12 of 61 2-27 solution Convert the following pressures to the indicated absolute pressures: a.a. 20.0 psig to atm abs with barometric pressure = 740 mm Hg

b. atm + 20 psi % 1 atm = atm abs c.P abs = 740 mm Hg % 7601 mm Hg 14.7 psi d. e.b. 28" Hg vacuum to mm Hg absolute with barometric pressure = 760 mm Hg f. 760 mm Hg g.P abs = 740 mm Hg − 28 in Hg % 29.92 in Hg = mm Hg h. i.c. 4" H2O draft to mm Hg abs with barometric pressure = 760 mm Hg j. 1 ft % 760 mm Hg = mm Hg k.P abs = 760 mm Hg − 4 in H 2 O % 12 in 33.9 ft H O 2

DRAFT

l. m.d. 5" H2O head to mm Hg abs with barometric pressure = 750 mm Hg. a. 760 mm Hg b. = mm Hg P abs = 750 mm Hg + 5 in H2 O% 1 ft % 12 in 33.9 ft H2 O

Page 13 of 61 2-28 A flue gas has the following composition on a molar basis: 12% CO2 77% N2 2% CO 2% H2O 7% O2 What is the composition in mass percent? What is the average molecular weight of the mixture? Solution: Basis: 100 mols mixture

DRAFT

Comp CO2 CO O2 N2 H2O

mol

MW 12 2 7 77 2 100

44 28 32 28 18

mass % mass 528 17.6 56 1.87 224 7.47 2,156 71.87 36 1.2 3,000

Page 14 of 61

2-29

Some ionic components of sea water are given in the following table: Cation mass % Anion mass % +l -1 Na 1.06 Cl 1.90 Mg+2 0.13 SO4-2 0.26 +2 -1 Ca 0.04 HCO3 0.01 K+1 0.04 Br-1 0.0065 +2 -1 Sr 0.01 F 10.0001 If the specific gravity of sea water is 1.03, how much NaC1 could be recovered from a cubic km of sea water? How much Mg(OH)2? MW Na = 23 MW Cl = 35.45 Solution: Basis:

1 km3 of sea water

3 3 kg kg Na + 57.45 kg NaCl 1 km 3 % 1000 3m % 1030 3 % 0.0106 % = 2.727 % 10 10 kg kg sea water 23 kg Na + m km

DRAFT

3 3 58.33 kg Mg(OH) 2 kg kg Mg +2 1 km3 % 1000 3m % 1030 3 % 0.0013 = 3.213 % 10 9 kg % kg sea water m km 24.31 kg Mg +2

Page 15 of 61 2-33 5.7 kg of sucrose is dissolved in a 20-liter can containing 18 liters of water. What is the sugar concentration of the solution formed in

a. b. c. d.

mass fraction? mole fraction? molality? molarity? Note: To find molarity, 1.383 at 25°C.

Solution: Basis: 5.7 kg of sucrose a. the mass fraction kg sucrose = 5.7 kg water = 18 L x 1 kg/L = 18 kg Total mass = 5.7 + 18 = 23.7 kg mass fraction sucrose = 5.7/23.7 = mass fraction water = 18/23.7 = b. the mol fraction (molecular formula of sucrose = C12H22O12 MW sucrose = 12(12) + 22(1) + 12(16) = 342 MW H2O = 18

DRAFT

kmol sucrose = 5.7/342 = 0.0167

kmol water = 18/18 = 1 kmol solution = 1.0167

mol fraction sucrose = 0.0167/1.0167 =

mol fraction water = 1/1.0167 =

c. molality kmol sucrose = 5.7/342 = 0.0167 g H2O = 18,000 g molality = gmol sucrose per 1000 g H20 gmol sucrose = 16.7 no. of 1000 g H2O = 18 therefore, per 1000 g H2O, the gmol of sucrose = 167/18 = and the molality is molal d. the molarity For this problem, we need the specific gravity of the sucrose solution, which we can obtain from handbooks or other references. Percent sucrose by mass = 76% the specific gravity = 1.383 from reference total mass = 57,000 + 18,000 = 75,000 g 1L Volume = 75, 000 g % ml = 54.23 % 1.383 g 1000 ml molarity = 143/54.23 = 2.64 M

Page 16 of 61 2.35 Ethane is flowing through a 1 m I.D. pipe at the rate of 6530 kg/hr. If the density of ethane is 1.3567 g/L what is

a. the flow velocity b. the mass velocity c. the volumetric flow rate? a. Flow velocity in m/s Q = w w = 6530 kg/h v= A !A

! 1.3567 g/L

A = (1) 2 /4 m 2

kg % 1h 3600 s h = 1.702 m/s v= g kg 1000 L % (1) 2 %  m 2 % 1.3567 % 3 L 1000 g 4 m 6530

b. mass velocity in kg/m2⋅s G= w = A

kg % 1h 3600 s = 2.31 kg/m 2 $s h (1) 2 %  m 2 4

6530

DRAFT

c. vol flow rate in m3/s Q= w ! =

kg % 1h 3600 s h = 1.337 m 3 /s g kg 1000 L % 1.3567 % L 1000 g m3 6530

Page 17 of 61 2-36 A lube oil (sp. gr. = 0.85) is pumped to a header at the rate of 4,000 liters/hr. At the header, the flow branches in two lines. One pipe has an inside diameter of 7 cm while the other, 15 cm. Assuming that the mass flow rate is directly proportional to the cross sectional area of flow, calculate for both pipes a. the flow velocity in m/s b. the mass velocity in kg/hr·m2 c. the volumetric flow rate in m3/hr Solution: Basis: 4,000 L/h cross sectional area of 7 cm dia (0.07 m dia) 2

A =  D =  0.07 4 4

2

= 0.003848

cross sectional area of 15 cm dia (0.15 m dia) 2

A =  D =  0.15 4 4

Let

2

= 0.01767

x = flow through 7 cm dia pipe 4000 - x = flow through the 15 dia m pipe D 27 4 = 0.07 2 = 49 x = 225 4000 − x 0.15 2 D2  15 4

DRAFT



225x = 49(4000 − x) x = 715 kg/h 4,000 - x = 3285 kg/h

a. flow velocity through 7 cm dia v = 715

3 kg h 1 = 0.061 m/s % % m % 3600 s 850 kg 0.003848 m 2 h

through 15 cm dia v = 3285

3 kg h 1 = 0.061 % % m % 3600 s 850 kg 0.01767 m 2 h

b. The mass velocity, G G= w A

715 for the 7 cm dia pipe G = w = = 1.846 % 10 6 kg/(s⋅m2) 0.003848 A 3285 = = 1.859 % 10 5 kg/(s⋅m2) for the 15 cm dia pipe G = w 0.017678 A

c. See above.

Page 18 of 61 2.39 Butane gas is stored in a cylinder at a pressure of 100 psig and at a temperature of 30°C. Some of the butane gas was used and after some time, the pressure had gone down to 50 psig. If the temperature is 25°C, what fraction of the butane had been used? Solution: The pressure is high. We must use the real gas law. Mass balance Butane used = Initial butane content - final butane content = P1V1/z1RT1 - P2V2/z2RT2

DRAFT

114.7 64.7 P1V1 P V P2 P1 − − 2 2 − z 1 (30 + 273) z 2 (25 + 273) z 1 RT 1 z 2 RT 2 z1T1 z2T 2 fraction used: = = = = P1V1 P1 114.7 z 1 (30 + 273) z 1 RT 1 z1T 1

Page 19 of 61 2.45 The flue gas from a furnace goes out with the following composition (dry basis): 12% CO2, 3% O2, 2% CO, and 83% N2. The partial pressure of water is 10 mm Hg. The total pressure is 4 cm H20 draft. The barometric pressure is 759 mm Hg. Calculate the composition on a wet basis (including H20). Solution: Basis: 100 mols of flue gas (dry basis It is best to use tabulation. PH2O = 10 mm Hg PT = 4 cm H2O draft Pbar = 759 mm Hg

Components CO2 O2 CO N2 H2O Total

mol (dry basis)

mol (wet basis) 12 3 2 83 100

12 3 2 83 1.34 101.34

mol % (wet basis 11.84 2.96 1.97 81.9 1.32 99.99

Patm = 759 mm Hg PH2O = 10 mm Hg PT = Patm - Pdraft = 759 − 4 cm H 2 O%

760 mm Hg ft H 2 O % =756.1 mm Hg (2.54 $ 12) cm H 2 O 33.9 ft H 2 O

DRAFT

Pdry gas + PH2O = 756.1 mm Hg Pdry gas = 756.1 - 10 = 746.1 mm Hg Mol water =100 mol dry gas%

10 mol H 2 O = 1.34 mols 746.1mol dry gas

Page 20 of 61 SOLUTION TO SOME PROBLEMS IN CHAPTER 3 3-7 A producer gas consisting of 8.4% CO2, 0.8% O2, 21.2% CO, 7.9% H2, and 61.7% N2 at a temperature of 30°C and 770 mm Hg total pressure, with Pwater = 25 mm Hg, flowing at a rate of 500 m3 per minute is mixed with a natural gas consisting of 81.11% CH4, 6.44% C2H6, 2.1% C3H8, 0.74% C4H10, 0.42% CO2, and 9.19% N2 at a total pressure of 2 atm and a temperature of 25°C and flowing at a rate of 100 m3 per minute. Its water vapor content has a partial pressure of 30 mm Hg. What is the composition of the resulting mixture on a dry basis? What is the water vapor content in grams per kmol of dry gas? Given: Producer gas: 500 m3 per minute at 30° C and 770 mm total Press PH2O = 25 mm Hg 8.4% CO2, 0.8% O2, 21.2% CO, 7.9% H2, and 61.7% N2 Natural gas: 100 m3/min at 25°C and total pres = 2 atm PH2O = 30 mm Hg 81.11% CH4, 6.44% C2H6, 2.1% C3H8, 0.74% C4H10, 0.42% CO2, and 9.19% N2 Required: Composition of the resulting mixture on a dry basis Water vapor content of the mixture, g/kmol dry gas Basis:

One minute operation Mixing of producer gas and natural gas

DRAFT

3 kmol producer gas = 500 m % 0 + 273 % 770 % kmol 3 = 20.376 kmol (with H 2 O) min 30 + 273 760 22.4 m

kmol H2O in prod. gas = 20.376 kmol wetb gas % 25 = 0.684 kmol 770 kmol dry producer gas = 20.376 - 0.684 = 19.629 3 kmol natural gas = 100 m % 0 + 273 % 2 % kmol 3 = 8.18 25 + 273 1 22.4 m min kmol H2O in nat. gas = 0.136 kmol dry gas %

30 = 0.165 kmol 2(760) − 30

kmol dry natural gas = 8.18 - 0.165 = 8.015 For the composition of the mixture Comp. Producer gas Nat. gas CO2 0.084(19.714) = 1.656 0.0042(8.018) = 0.034 N2 0.617 (19.714) = 12.572 0.0919(8.018) = 0.752 O2 0.008 (19.714) = 0.163 CO 0.212(19.714) = 4.32 H2 0.079 (19.714) = 1.61 CH4 0.8111 (8.018) = 6.635 C2H6 0.0644 (8.018) = 0.527 C3H8 0.021 (8.018) = 0.168 C4H10 0.0074 (8.018) = 0.061 Total kmol dry gas = 28.558 kmol H2O = 0.684 + 0.165 = 0.849

g H2O = (0.849)(18)(1000) = 1.528 x 104

g water vapor per kmol dry gas = (1.528 x 104)/28.558 = 535

kmol 1.69 12.9 0.158 4.179 1.557 6.503 0.516 0.172 0.059 27.734

mol % 6.093 46.66 0.571 15.13 5.64 23.23 1.84 0.602 0.212 99.978

Page 21 of 61 3-8 Moist air at 40°C and at a total pressure of 765 mm Hg is flowing through a pipeline. The partial pressure of water is 20 mm Hg. In order to measure the rate of flow, pure carbon dioxide gas is bled into the gas at rate of 20 kg/min. At a point downstream where complete mixing has occurred, it is found that the air contains 5% CO2 by volume. Find the volumetric flow rate of the moist air in m3/min. The % CO2 is on a dry basis. Given: mixed gases 5% CO2 vol moist air 40oC Pt = 765 mm Hg PH2O = 20 mm Hg

pure CO2 gas 20 kg/min

Required: Vol flow rate of moist air in m3/min Basis: 1 minute operation Solution: Vol flow rate of moist air = 3 kg CO 2 kmol 0.95 kmol dry air % % % 765 kmol moist air % 22.4 m % 40 + 273 % 760 0 + 273 765 0.05 kmol CO 2 min kmol 44 kg (765 − 20) kmol dry air

DRAFT

20

= 226.3 m3/min

Page 22 of 61 3-9 Basis:

1000 kg wet sand

Required: water evaporated A. Algebraic method Let x, water evap;

y, dried sand

OMB: 1000 = x + y

Eq. 1

BDS balance:

Eq. 2

(1-0.3)(1000) = (1- 0.09)y

Subst y = 1000 - x

in Eq. 2

700 = 0.91(1000 - x) x = 910 - 700 = 210/0.91 = 230.8 kg H2O evap Arithmetic solution: Basis: 1000 kg feed Required: water evaporated kg product =1000 kg feed %

(1 − .3) kg BDS kg product = % kg feed (1 − 0.09) kg BDS

DRAFT

769.2 kg prod OMB: kg H2O evap = 1000 - 769.2 = 230.8 kg

Page 23 of 61 3-10 Wood chips are to be dried in a rotary drier from 35 to 12% moisture. If 10 kg of wet wood is charged per minute, how much product can be obtained over an 8-hr period? Find the ratio kg H2O evaporated/kg wet wood charged Water evaporated 10 kg wet wood/ min 36 % moisture

Drier

Dried product 12% moisture

Required: Product for 8 hr period ratio kg water evap./kg wet wood charged. Solution: We can identify a key component, which is the bone-dry solids. Therefore, the arithmetic method is recommended. Basis: 1 min (10 kg wet wood charged)

DRAFT

kg dry prod/min =10 kg wet wood

Overall mass balance:

(1 − 0.35) kg BDS 1 kg product = 7.386 kg % 1 kg wet wood (1 − 0.12) kg BDS

For the 8 hr period: kg dry prod = 7.386 kg dry prod/min% 60 min/hr % 8 hr= 3545 kg kg water evap/kg wet wood charged = 2.614/10 = 0.261

Page 24 of 61 3-11 A drier is being used to reduce the moisture content of bagasse to be used for fiberboard manufacture. It contains 50% moisture. 5.75 kg of wet bagasse is charged into the drier. After 3 hours, it was found that the weight has gone down to 3.25 kg. What is the moisture content of the final product? Using the arithmetic method: Given:

5.75 kg wet bagasse with 50% moisture weight of product = 3.25 kg

Required: Moisture content of the final product Basis: 5.75 kg of wet bagasse BDS balance: BDS in feed = BDS in product BDS in feed = 0.5(5.75) = 2.875 kg = BDS in product = moisture in product = 3.25 - 2.875 = 0.375 kg

DRAFT

% moisture content = (0.375)(100)/3.25 = 11.54%

Page 25 of 61 3-14 It is desired to concentrate a 6% KN03 solution (in water) to 20% KN03 solution. 7000 kg of product liquor is to be needed per hour. How much solution should be charged? How much water is evaporated? Given:

input solution = 6% KNO2 7,000 kg/h product containing 20% KNO3

Required: solution charged; kg water evaporated. Basis:

1 hour operation

Solution: kg KNO3 = 0.2(7000) = 1400 kg KNO3 balance: KNO3 in product = KNO3 in feed = 1400 kg kg feed input = 1400 kg %

100 kg solution = 23,300 kg 6 kg KNO 3

OMB kg water evaporated

DRAFT

= kg input solution - kg product = 23300- 7000 = 16,300 kg

Page 26 of 61 3-19 An evaporator is concentrating solutions coming from two different sources. The solution from the first source containing 10% NaCl and 10% NaOH 8% NaC1 and 12% NaOH flows at the rate of 70 kg per minute. The two streams are fed directly to the evaporator. If 50% of the total water is to be evaporated, calculate the composition and the flow rate of the product.

Given: water evap = 50% of total H2O

50 kg/min 10% NaCl 10% NaOH Evaporator 70 kg/min 8% NaCl 12% NaOH

Product

Required: Composition and flow rate of product Basis: 1 min operation Solution: We can use the aritmetic solution.

DRAFT

Feed Stream 1: kg NaOH = 0.1(50) = 5 kg kg NaCl = 0.1(50) = 5 kg kg H2O = 0.8(50) = 40

Feed Stream 2 kg NaOH = 0.12(70) = 8.4 kg NaCl = 0.08(70) = 5.6 kg H2O = 0.8 (70) = 56

Product composition: kg NaOH = 5 + 84 = 13.4 kg NaCl = 5 + 5.6 = 10.6 kg H2O = 0.5(40+56) = 48 kg kg product = 13.4 + 10.6 + 48 = 72 kg

% NaOH = (13.4)(100)/72 = 18.6% % NaCl = (10.6)(100)/72 = 14.7% product flow rate = 72 kg/min

Page 27 of 61 3-20

A mixture containing 70% methanol and 30% water is to be distilled. If the distillate product is to contain 99.9% methanol and the bottoms product 0.004% methanol, how much distillate and bottoms product are obtained per 100 kg of feed distilled?

Given: Distillate , D

xm = 0.999 Feed = 100 kg 70% methanol 30% water

Bottoms, B

xM = .00004

DRAFT

Required: kg distillate and kg bottoms Basis: 100 kg feed Solution: We cannot identify a tie component. We have to use the algebraic solution. Let

D = kg distillate B = kg bottoms

Overall mass balance:

100 = D + B

Eq. 1

Methanol balance

0.7(100) = 0.999D + 0.00004B from Eq 1, B = 100 − D Substiting in Eq. 2,

70 = 0.999D + 0.00004(100 − D) Solving for D, D = 70.07 kg B = 29.93 kg

Eq. 2

Page 28 of 61 3-21 100 kmol of a benzene-toluene mixture in equimolal amounts is distilled. 96% of the benzene is obtained in the distillate product while 90% of the toluene is in the bottom. Calculate the composition of the product streams. Given: Distillate , D

96% of the benzene goes to the distillate

Feed, F = 100 kmol benzene-toluene solution, equimolar

90% of the toluene goes to the bottoms Bottoms, B

Required: Composition of the distillate and the bottoms Basis: 100 kmol feed Solution: While we cannot identify a tie component in this problem, we can analyze the problem and see that the given quantities make the problem amenable to arithmetic solution and we do not have to use the algebraic solution.

DRAFT

From the given quantities, benzene in the feed = 50 kmol 96% of the benzene in the feed goes to the distillate = 0.96(50) = 48 kmol the difference goes to the bottoms: 50 - 48 = 2 kmol 90% of the toluene in the feed goes to the bottoms = 0.9(50) = 45 kmol the difference goes to the bottoms: 50 - 45 = 5 kmol Therefore the distillate contains 48 kmol benzene and 5 kmol toluene The composition, mole fraction benzene = 48/(48 + 5) = 0.906 The bottoms contains 45 kmol toluene and 2 kmol toluene The composition, mole fraction benzene = 2/(45 + 2) = 0.043

While in general, problems without tie components are usually solved by algebraic methods, some problems are so structured that they can be solved by arith metic method.

Page 29 of 61 3-23 A mixture consisting of isobutane, n-butane, and n-pentane is to be distilled in batch mode. The following tabulation gives the composition of the streams in mole fraction: Component n-butane isopentane n-pentane Per 100 moles of feed, find the the composition.

feed distillate bottoms 0.23 x 0.129 0.32 y 0.3 0.45 0.024 z amount of the distillate and the bottom streams. Complete the tabulation of

Solution: Given: tabulation of the composition with unknowns Basis: 100 moles of feed Examining the table, we see that in the bottoms column, z can be solved from the equation, z + 0.129 + 0.3 = 1 and

z = 0.571

The unknown compositions are x and y Let D = mol distillate

and B mol bottoms

We have four unknowns and we need four equations

DRAFT

O.M. Balance: D + B = 100

(1)

n-butane balance: 0.23(100) = xD + 0.129B

(2)

isopentane balance: 0.32(100) = yD + 0.3B

(3)

n-pentane balance: 0.45(100) = 0.024D + 0.571B

(4)

Substituting the value of D (from Eq. 1) in Eq. 4 45 = 0.024(100 - B) + 0.571B B = 77.88 mol

D = 22.12 mol

Substituting the value of B and D in (2) 0.23(100) = x(22.12) + 0.129(77.88) x = (23 - 0.129(77.88))/22.12 = 0.586 Substituting the value of B and D in (3) 0.32(100) = y(22.12) + 0.3(77.88) y = (32 - 0.3(77.88))/22.12 = 0.390 As a check,

the sum of the composition of the distillate = 0.390 + 0.586 + 0.024 = 1

Page 30 of 61 3-24 Ethane is to be removed in an absorber from an ethane-hydrogen mixture by using a mixture of hexane containing 2% ethane. The feed contains 40% ethane and 60% hydrogen. If 99% of the ethane is removed, what is the composition of the effluent at the top and at the bottom? The liquid to gas molar ratio is 2. Given: Hexane 2% ethane

hydrogen and ethane 2

Required: Composition of outlet gas and liquid Basis: 100 kmol gas 200 kmol of liquid Hexane in:

0.98(200) = 196 kmol

Ethane in with hexane = 0.02(200) = 4 kmol 99% of the ethane is removed

Ethane carried by hydrogen in: 0.4(100) = 40 kmol

Absorber

Ethane absorbed = 0.99(40) = 39.6 kmol

DRAFT

Hexane + absorbed ethane

40% ethane 60% hydrogen

1

Hydrogen in = 0.6(100) = 60 kmol Liquid out = 196 kmol hexane + 39.6 kmols ethane + 4

Composition of outgoing liquid =

43.6 % 100 =18.2 % 43.6 + 196

Gas out = 60 + 0.4 = 60.4 Composition of outgoing gas =

0.4 % 100 = 0.662% 0.4 + 196

Page 31 of 61 3-26 In order to recover the benzene from a benzene-air mixture, the mixture is scrubbed with kerosene in a packed tower. 8000 m3/h of a benzene-air mixture containing 2 mole % benzene (T = 25°C, P = 765 mm Hg) is treated with 10,000 kg/h of kerosene. If 98% of the benzene is recovered, find the composition of the gas stream and liquid stream leaving the tower. What is the volumetric flow rate of the outgoing gas stream (T = 25°C, P = 759 mm Hg)? Solution:

Kerosene

Outlet gas 2

10,000 kg/h

DRAFT

Absorber

kerosene + absorbed benzene

25oC 759 mm Hg

98% of the benzene is recovered 8000 m 3/h Air-benzene mixture 2% mol benzene 25oC 765 mm Hg

1

Basis: 1 hour operation 3 kmol air-benzene in = 8000 m % 0 + 273 % 765 % kmol 3 = 325 kmol 25 + 273 760 22.4 m h

kmol air = (0.98)(325) = 318.5 kmol kmol benzene = 0.98(325) = 6.5 kmol kmol benzene in gas out = 0.02(6.5) = 0.13 kmol kmol benzene in liquid out = 0.98(6.5) = 6.37 kmol = 6.37(80) = 509.6 kg composition of air-benzene out =

0.13(100) = 0.041% mol benzene 318.5 + 0.13

composition of benzene-kerosene mixture =

509.6(100) = 4.85% 10, 000 + 509.6

Page 32 of 61

3-34 Eight hundred kg/h of halibut livers containing 23% oil is extracted with 570 kg/h of pure ether. The extracted livers is analyzed and is found to contain 1.12% oil, 32.96% ether, and 65.92% oil-free livers. Find the composition and the weight of the extract, and the percentage recovery of the oil. Given: 570 kg/hr ether

Halibut livers

Extract Extractor

800 kg/hr 23% oil

Extracted livers 1.12 % oil 32.96% ether 65.92% oil-free livers

Required: Composition and mass of the extract Basis: 800 kg of halibut livers Solution: The oil-free livers is a tie component. We can use the arithmetic method.

DRAFT

kg livers = 800 kg kg oil = 0.23(800) = 184 kg kg oil-free livers = 800 - 184 = 616 kg Oil-free livers balance: Oil-free livers in feed = oil-free livers in extracted livers = 616 kg

1.12 kg oil = 10.466 kg 65.92 kg oil-free livers 32.96 kg ether Ether in extracted livers = 616 kg oil-free livers % = 308 kg 65.92 kg oil-free livers Oil in extracted livers = 616 kg oil-free livers %

Oil balance: Oil in extract = oil in charge - oil in extracted livers = 184 - 10.466 = 173.53 kg Ether balance: Ether in extract = ether input - ether in extracted livers = 570 - 308 = 262 kg kg extract = ether + extracted oil = 173.53 + 262 = 435.53 kg % oil = 173.53(100)/435.53 = 39.84% % oil recovery = 173.53(100)/184 = 94.31%

Page 33 of 61 3-40 A calcium carbonate slurry is to be filtered in a plate-and-frame filter press with 15 frames measuring 90 cm × 120 cm × 2.5 cm. The solids in the slurry is 5 mass percent. The filter cake is 32% porous. Assuming that the filter cake is incompressible, how much volume of the slurry is required to fill the filter press? The specific gravity of the solids in the slurry is 2.63. The density of the liquid is 1.0. Assume that the specific gravity of the slurry is almost equal to 1. Given: filter press with 15 frames 90 cm x 120 cm x 2.5 cm (assume the inside volume) material to be filtered: 5% (mass) calcium carbonate slurry filter cake: 32% porous; incompressible Sp. g of solids 2.63 sp. g. of liquid and slurry = 1 Required: Volume of slurry required to fill the filter press Solution:

arithmetic method

Total volume of the frames: 0.9 m % 1.2 m % 0.025 m = 4.05 m3 = volume of the cake Volume of the solids in the filter cake = (1 - 0.32)(4.05) = 2.754 m3 density of the solid = 2630 kg/m3

DRAFT

kg solids in filter cake = 2.754 m3 % 2630 kg/m3 = 7245 kg 1.0 kg slurry = 1.449 % 10 5kg slurry 0.05 kg solids 3 volume of the slurry = 144900 kg slurry % 1 m = 144.9 m3 1000 kg

liquid in the slurry = 7243 kg solids %

Page 34 of 61 3-47 A double-effect evaporator is to concentrate 1,000,000 kg/day of a liquor containing 5% solids to 40% solids. Assuming equal evaporations are obtained from each effect, calculate the composition of the solution from the first effect (if forward feeding is used) and the flow rate of the product in kg/h. How much evaporation from each effect is obtained? Given:

S

F = 106 kg/day

E2 .

E1

P1

Effect 1

xF = 0.05

xP1

P2

Effect 2

C1

xP2 = 0.4 C2

E1 = E2 Required:

DRAFT

a. Composition of the product from the first effect b. The flow rate of the product in kg/h c. Evaporation from each effect Solution: Basis: One-day operation Use combination of arithmetic and algebraic methods Let

E1 equals the evaporation from Effect No. 1 E2 equals the evaporation from Effect No. 2

Overall Mass Balance around the whole system, solution side

1, 000, 000 = P2 + E2 + C2

Eq. 1

C2 = E1

Eq. 2

E1 = E2

Eq. 3

Solids balance around the whole system, solution side

0.05(1, 000, 000) = 0.4P2 P2 =

Eq. 4

kg day 0.05(1, 000, 000) = 125, 000 = 5208 kg/h % 0.4 day 24 h

Subst Eq 2 and Eq 3 in Eq1

1, 000, 000 = 125, 000 + E1 + E1 E1 = (1, 000, 000 − 125, 000)/2 = 875,000 kg/day = 36,460 kg/h E2 = E1 = 36,460 kg/h Overall Mass Balance around effect 1

1, 000, 000 = E1 + P1

Page 35 of 61

DRAFT

P1 = 1, 000, 000 − 437, 500 = 562, 500 kg/day = 23,440 kg/h 0.05(1, 000, 000) x P1 = = 0.089 562, 500

Page 36 of 61 3-48. It is desired to produce 7% NaNO3 solution continuously. The water line (NaNO3-free) is split into two. 500 kg per hour is sent to a tank where NaNO3 is added. The mixture is stirred well to form a saturated solution of Bypass stream NaNO3 (47.9%). The other line bypasses the tank and is mixed with the 47.9% solution. What is the flow rate of the bypass stream and the final NaNO3 product?

Water

500kg/hr

Saturator

47.9% NaNO3

DRAFT

Solution: Basis: 1 hr operation Required: Flow rate of bypass stream and the final product Balance around saturator inside the bypass loop Water balance water in 47.9% NaNO3 solution = 500 kg NaNO3 balance: 47.9 kg NaNO 3 = 459.693 kg NaNO3 input = 500 kg water % (100 − 47.9) kg H 2 O Balance around saturator outside the bypass loop NaNO3 balance: NaNO3 input = 459.693 kg = NaNO3 in 7% NaNO3 solution Water balance: 93 kg H 2 O Water in 7% NaNO3 solution = 459.693 kg NaNO 3 % = 6107.35 kg 7 kg NaNO 3

Balance around point of splitting of bypass stream bypass stream = 6107.35 - 500 = 5607.36 kg/h product stream

= 459.693 + 6107.35 = 6567.04 kg/h

7% NaNO3

Page 37 of 61 3-49 A process needs an air supply which should contain 0.12 mol H2O/ mol dry air exactly. 1500 m3/min of air at 25°C and 101.3 kPa is to be treated. Part of this air goes to a spray chamber where the air picks up water and goes out with 0.3 mol H2O/mol dry air. The other part of the air feed bypasses the water spray and is mixed with the humidified air to produce a mixture containing 0.12 mol H2O per mol dry air. What is the water consumption? What is the ratio of the flowmeter readings of the bypass stream and the stream to the water spray?

Basis: 1 minute operation Bypass stream Dry Air 1500 m3 25oC 101.5 kPa

Net Feed

0.12 kmol H2O/ kmol dry air

Spray chamber 0.3 kmol H2O/ kmol dry air water

Using the arithmetic method m % 0 + 273 % kmol = 63.35 kmol BDA kmol bone-dry air = 1500 min 25 + 273 22.4 m 3 3

DRAFT

BDA balance around spray chamber outside the bypass loop BDA in feed = BDA in process air = 61.35 kmol BDA 0.12 kmol H 2 O kmol H2O in process air = 61.35 kmol BDA % = 7.36 kmol H 2 O kmol BDA H2O balance around spray chamber outside the recycle loop water spray = H2O in process air = 7.36 kmol = 7.36 x 18 = 132.14 kmol H2O/ min

H2O balance around the spray chamber within the bypass loop water in stream of air going out of the chamber = 7.36 kmol

BDA balance around the spray chamber within the bypass loop BDA in net feed = BDA in air out or chamber BDA = 7.36 kmol H 2 O %

kmol BDA =24.53 kmol 0.3 kmol H 2 O

bypass stream = 61.35 =2.52 stream to spray chamber 24.53

Page 38 of 61 3-50

An air conditioning system supplies 1000 m3/min of air containing 0.01 mol H2O/mol dry air. It is at 20°C and 1 atm. To conserve energy, part of the exhaust air containing 0.08 mol H2O/mol dry air is recycled and mixed with the fresh air from the air conditioner to produce a gross air feed to the room containing 0.035 mol H20/mol dry air. How many kg of water is picked up by the air per minute? What is the volumetric flow rate of the recycle stream? (27°C and 99 kPa) Given:

Recycle stream 0.01 kmol H 2O/ kmol dry air

0.035 mol H 20 per mol dry air

0.08 mol H 20 per mol dry air Room

1000 m 3 air 20oC 1 atm

27oC and 99 kPa

Required: a. kg water picked up by the air per minute

DRAFT

b. volumetric flow rate of the recycle stream Basis: 1 minute operation Bone-dry air balance around whole system 3 bone-dry in = bone-dry out = 1000 m % 0 + 273 % kmol 3 = 41.6 kmol min 20 + 273 22.4 m kmol H 2 O Water in air in = 41.6 kmol % 0.01 = 0.416 kmol kmol bda min kmol H 2 O Water in air out = 41.6 kmol % 0.08 = 3.328 kmol kmol bda min

Water picked up by the air per minute Using water balance, Water picked up by the air = kg (3.328 kmol water out − 0.416 kmol water in ) % 18 = 52.42 kg kmol kmol bda kmol bda Balance around point of mixing Let R = kmol recycle (dry basis) water balance

41.6(0.01) + 0.08R = 0.035(R + 41.6) Solving for R, R=

41.6(0.035 − 0.01) = 23.11 kmol 0.08 − 0.035

3 Volumetric flow rate of recycle = 23.11 kmol %22.4 m % 27 + 273 % 101.325 = 582.2 m 3 0 + 273 99 kmol

room inside recycle loop:

Page 39 of 61 Water picked up by the air = 52.42 kg Let x = kmol dry air entering the room = kmol dry air leaving the room Water balance:

0.08x − 0.035x = 52.42/18 x=

52.42 = 64.72 kmol (18)(0.025)

Balance around point of splitting, Kmol recycle = 64.72 - 41.6 = 23.12 kmol

DRAFT

3 Volumetric flow rate of recycle = 23.12 kmol %22.4 m % 27 + 273 % 101.325 = 582.5 m 3 0 + 273 99 kmol

Page 40 of 61 SOLUTION TO SOME PROBLEMS IN CHAPTER 4

4-3 For each of the following reactions, calculate the amount of the compounds asked for: (Note: The equations are not balanced.) a) Benzene + Oxygen → Carbon dioxide + water per 120 kg of benzene burned, how many kilograms CO2 and kilograms H2O are produced? How much oxygen is needed? b) NaCl + H2O + SO3 → Na2SO4 + HC1 Per 300 kg NaC1, how much Na2SO4 and HC1 are formed? Solution: a) C6H6 + 7.5O2 → 6CO2 + 3H2O Basis: 120 kg benzene 6(44.01) 6MWCO 2 = 120 % = 405.7 kg kg CO 2 = 120 kg C 6 H 6 % MWC 6 H 6 78.114 kg H 2 O = 120 kg C 6 H 6 % kg O 2 = 120 kg C 6 H 6 %

DRAFT

b) 2NaCl + H2O + SO3 Basis: 300 kg NaCl

3(18.015) 3MWH 2 O = 120 % = 83.03 kg MWC 6 H 6 78.114

7.5(31.999) 7.5MWO 2 = 120 % = 368.7 kg MWC 6 H 6 78.114

→ Na2SO4 + 2HC1

kg Na 2 SO 4 = 300 kg NaCl %

MWNa 2 SO 4 = 300 % 142.04 = 364.6 kg 2(58.44) 2MWNaCl

2(46.46) kg HCl = 300 kg NaCl % 2MWHCl = 300 % = 187.2 kg 2(58.44) 2MWNaCl

c) Ca3(PO4)2 + 3SiO2 + 5 C → 3CaSiO3 + 2P + 5 CO2 2 2 Per 100 tons Ca3(PO4)2 how much of each of the products are formed? Basis: 100 tons Ca3(PO4)2 3(116.16) 3MWCaSiO 3 tons CaSiO 2 = 100 tons Ca 3 (PO 4 ) 2 % = 100 % = 112.4 tons MWCa 3 (PO 4 ) 2 (310.18) 2(30.97) 2MWP tons P = 100 tons Ca 3 (PO 4 ) 2 % = 100 % = 19.97 tons MWCa 3 (PO 4 ) 2 (310.18) tons CO 2 = 100 tons Ca 3 (PO 4 ) 2 %

(5/2)(44.01) (5/2)MWCO 2 = 100 % = 35.47 tons (310.18) MWCa 3 (PO 4 ) 2

d) C2H6 + (7/2)O2 → 2CO2 + 3H2O How many kg CO2 and H2O can be produced from 13 kg of ethane? 2(44.01) 2MWCO 2 = 13 % = 38.05 kg MWC 2 H 6 32.07 3(18.015) 3MWH 2 O kg CO 2 = 13 kg C 2 H 6 % = 13 % = 32.35 kg MWC 2 H 6 32.07 kg CO 2 = 13 kg C 2 H 6 %

e)

C12H22O11 (sucrose) + H2O → C6H12O6 C6H12O6

→ C2H5OH (ethyl alcohol) + CO2

DRAFT

Page 41 of 61

Page 42 of 61 4-6 Basis: 1,000 kg of fertilizer kg N2 = 0.02(1000) = 20 kg kg P2O5 = 0.12(1000) = 120 kg kg K2O = 0.06(1000) = 60 kg

MW (NaNO3)= 85.0 MW Ca3(PO4)2 =310.2 MW KCl = 74.55 MW P2O5 = 141.94

MW K2O = 94.18

kg 98% NaNO3 needed = 20 %

2(85) 2(MW NaNO 3 ) 1 kg 98% NaNO 3 = 20 % % % 1 = 123.8 kg MW N 2 0.98 28 0.98 kg NaNO 3

kg Ca3(PO4)2 = 120 %

MW Ca 3 (PO 4 ) 2 = 120 % 310.2 = 262.5 kg 141.94 MW(P 2 O 5 )

kg 97% KCl = 60 %

2(74.55) 2(MW KCl) 1 kg 97% KCl = 60 % % % 1 = 97.93 kg 0.97 94.18 MW K 2 O 0.97 kg KCl

DRAFT

Overall Mass Balance: kg inerts = kg mixed fertilizer - (kg 98% NaNO3 + kg 97% KCl + kg Ca3(PO4)2) kg inerts = 1,000 - (123.8 + 262.5 + 97.93) = 515.8 kg

Page 43 of 61 4-7 Basis: 1,000 kg of fertilizer Calcium nitrate, pure Phosphoric acid KNO3 Inert fillers.

2% N2, 12% P2O5, 6% K2O

kg N2 = 0.02(1000) = 20 kg MW Ca(NO3)2 = 226.1 MW H3PO4 = 98 kg P2O5 = 0.12(1000) = 120 kg 0.06(1000) = 60 kg MW KNO3 = 101.1 kg K2O = MW P2O5 = 141.94 kg KNO3 needed = 60 %

MW K2O = 94.18

2(101.1) 2(MW KNO 3 ) = 60 % = 128.8 kg MW K 2 O 94.18

N2 from KNO3 = 128.8 kg KNO 3 %

MWN 2 28 =17.84 %= 128.8 % 2MW KNO 3 2(101.1)

N2 balance

DRAFT

N2 from Ca(NO3)3 = total N2 - N2 from KNO3 = 20 - 17.84 = 2.16

g Ca(NO3)3 = 2.16 kg N 2 %

2(226.1) 2MW Ca(NO 3 ) 3 = 2.16 % = 11.63 kg 3MW(N 2 ) 3(28)

kg 97% H3PO4 = 120 %

2(98) 2(MW H 3 PO 4 ) = 120 = 165.8 kg 141.9 MW P 2 O 5

Overall Mass Balance: kg inerts = kg mixed fertilizer - ( kg KNO3 + kg Ca (NO3)3 KCl + kg H3PO4)2) kg inerts = 1,000 - (128.8 + 11.63 + 165.8) = 693.8 kg

Page 44 of 61 4-8 To make 5,000 kg of mixed fertilizer containing 6% N2, 12% P2O5 and 12% K2O, find the amount of the following ingredients necessary: (NH4)3PO4 Ca3(PO4)2 KNO3 Inerts fillers Solution: Basis: 5,000 kg mixed fertilizer kg N2 = 0.06(5000) = 300 kg kg P2O5 = 0.12(5000) = 600 kg kg K2O = 0.12(5000) = 600 kg kg KNO3 needed = 600 %

MW (NH4)3PO4 = 149 MW Ca3(PO4)2 = 214.24 MW KNO3 = 101.09 MW P2O5 = 141.94

MW K2O = 94.18

2(101.09) 2(MW KNO 3 ) = 600 % = 1288 kg MW K 2 O 94.18

N2 balance: N2 in mixed fertilizer = N2 from KNO3 + N2 from (NH4)3PO4 2(14) N2 from KNO3 = 1288 % = 173.376 101.09 N2 from (NH4)3PO4 = 300 - 173.376 = 121.624

DRAFT

kg (NH4)3PO4 = 600 %

2(MW (NH 4 ) 3 PO 4 ) 2(149) = 600 % = 431.176 kg 3(MW N 2 ) 3(28)

kg P2O5 from (NH4)3PO4 = 431.476 %

MW P 2 O 5 = 431.476 % 141.94 = 205.516 2(149) 2(MW (NH 4 ) 3 PO 4 )

P2O5 balance: P2O5 in mixed fertilizer = P2O5 from (NH4)3PO4 + P2O5 from Ca3(PO4)2 P2O5 from Ca3(PO4)2 = 600 - 205.516 = 394.484 kg

kg Ca3(PO4)2 = 394.484 %

MW Ca 3 (PO 4 ) 2 = 394.484 % 212.24 = 589.864 kg MW(P 2 O 5 141.94

Overall Mass Balance: kg inerts = kg mixed fertilizer - (kg KNO3 + kg (NH4)3PO4 + kg Ca3(PO4)2) kg inerts = 5,000 - (1288 + 431.1746 + 589.864) = 2691 kg Answer: kg inerts = 2691 kg kg KNO3 = 1288 kg kg (NH4)3PO4 = 431 kg kg Ca3(PO4)2 = 590 kg

Page 45 of 61 4-10 Solution: Basis: 1,000 kg of solution at 30°C We have no tie component and we have to use the algebraic method Let x = Na2SO4⋅10H2O produced 1,000 - x = kg solution at 15°C (Using an overall mass balance) Na2SO4 balance: Na2SO4 in feed = Na2SO4 in Na2SO4⋅10H2O

1000 %

+ Na2SO4 in final solution

13.5 kg Na 2 SO 4 40 kg Na 2 SO 4 MW Na 2 SO 4 =x + (1000 − x) % MW Na 2 SO 4 $ 10H 2 O (100 + 40)kg solution (100 + 13.5) kg solution 1000 % 40 = x 142 + (1000 − x) 13.5 140 322.2 113.5

Solving for x,

DRAFT

x=

40000 − 13500 140 113.5 = 518.3 kg Na SO ⋅10H O 2 4 2 142 − 13.5 322.2 113.5

Page 46 of 61 4-13 For the following reactions, find the composition the final product. If the second reactant used is 10% in excess of the theoretical amount or complete reaction. (The first is the limiting reactant) The reaction is 100% complete and the reactants are pure. a. 2KCl + H2SO4 → K2SO4 + 2HCl b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2 Solution: a. 2KCl (s) + H2SO4(l) → K2SO4(s) + 2HCl(g) MW Cl: 35.45 H: 1 S: 32 O: 16 K: 39 basis: 100 kg KCl MWH 2 SO 4 98 kg theoretical H2SO4 needed = 100 kg KCl % = = 65.82 kg 2 % 74.45 2MWKCl excess H2SO4 needed = 0.1( 65.82 ) = 6.58 kg kg K2SO4 produced = 100 kg KCl %

MWK 2 SO 4 174 = = 116.9 kg 2 % 74.45 2MWKCl

kg HCL produced = 100 kg KCl % MWHCl = 36.45 = 48.96 kg 74.45 MWKCl

DRAFT

product composition: HCl is a gas and normally will not mix with the other substances. The product mixture consist of H2SO4 and K2SO4.

% H 2 SO 4 =

6.58 % 100 = 5.33% 6.58 + 116.9

% K 2 SO 4 =

116.9 % 100 = 94.67% 6.58 + 116.9

b. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2 Basis: 100 kg Na2CO3 kg theoretical Br2 needed = 100 kg Na 2 CO 3 % excess Br2 needed = 0.1(150.8) = 15.08 kg kg NaBr produced = 100 kg Na 2 CO 3 %

MWBr 2 = 159.8 = 150.8 kg 106 MWNa 2 CO 3

5MWNaBr = 5(102.8) = 32.33 kg 3MWNa 2 CO 3 3(106)

kg NaBrO3 produced = 100 kg Na 2 CO 3 %

MWNaBrO 3 = 342.6 = 107.7 kg 3MWNa 2 CO 3 3(106)

kg product = 15.08 + 32.33 + 107.7 = 155.1 kg % Br2 = 15.08 % 100 = 9.722% 155.1 % NaBr = 32.33 % 100 = 20.84% 155.1 % NaBrO3 = 107.7 % 100 = 69.44% 155.1 CO2 is a gas and will not be in the solid and liquid substances.

Page 47 of 61 4.14a For the following, the reaction goes only to 95% completion. Pure reactants and theoretical amounts are used. For every 500 kg of the impure products, how much of each of the reactants are used? a. KCl +NaNO3 → KNO3 + NaCl Basis: 100 kg KCl MWNaNO

kg NaNO3 needed = 100 kg KCl % MWKCl 3 =

(100)(84.995) = 114.009 74.551

MWKNO

kg KNO3 produced = 100 kg KCl % MWKCl3 % 0.95 = kg CaCl2 produced = 100 kg KCl % MWNaCl % 0.95 = MWKCl

(100)(101.103)(0.95 = 128.835 74.551

(100)(58.442)(0.95) = 74.473 74.551

The product consists of KNO3, NaCl2, and unreacted KCl and NaNO3. unreacted KCl = 5 kg unreacted NaNO3 = 0.05(114.009) = 5.7 kg kg product = 5 + 5.7 + 128.835 + 74.473 = 214.008 kg By ratio and proportion, we can obtain the reactants needed to form 500 kg product.

DRAFT

kg KCl : 500 = 100 : 214.008 100 = 233.636 kg kg KCl = 500 % 214.008

kg NaNO3 : 500 = 114.009 : 214.008

kg NaNO3 = 500 % 114.009 = 266.366 kg 214.008 4.14b For the following, the reactions go only to 95% completion. Pure reactants and theoretical amounts are used. For every 500 kg of the impure products, how much of each of the reactants are used? MgO + CaCl2 + CO2 → MgCl2 +CaCO3

Basis: 100 kg MgO MWCO

2 = kg CO2 needed = 100 kg MgO % MWMgO

(100)(44.01) = 109.2 kg (40.3)

MWCaCl

kg CaCl2 needed = 100 kg MgO % MWMgO2 = MWMgCl

(100)(111) = 275.4 kg (40.3)

kg MgCl2 produced = 100 kg MgO% MWMgO2 % 0.95 = MWCaCO

(100)(95.21)(0.95) = 224.4 kg 40.3

kg CaCO3 produced = 100 kg MgO % MWMgO 3 % 0.95 =

(100)(100.1)(0.95) = 236 kg 40.3

The product consists of MgCl2, CaCO3, and unreacted MgO and CaCl2 unreacted MgO = 5 kg unreacted CaCl2 = 0.05(275.4) = 13.77kg kg product = 5 + 13.77 + 224.4 + 236 = 479.2 kg By ratio and proportion, we can obtain the reactants needed to form 500 kg product.

Page 48 of 61 500 % 100 = 104.3 kg kg MgO = 479.2

kg CaCl2 : 500 = 275.4 : 479.2

500 % 275.4 = 287.4 kg kg CaCl2 = 479.2

kg CO2 : 500 = 109.2 : 479.2

500 % 109.2 = 113.9 kg kg CO2 = 479.2

DRAFT

kg MgO : 500 = 100 : 479.2

Page 49 of 61 4-18 100 kg of pure carbon is burned. For each of the following cases, calculate the composition of the combustion gases: a. Theoretical amount of pure O2 used; complete combustion b. Theoretical amount of air used; complete combustion c. 30% excess air is used: 90% of the C burns to CO2 and 10% to CO d. 20% excess air is used: 90% of the C burns to CO2 while the 10% goes out as unburned solid. Basis: 100 kg pure carbon a. Theoretical pure O2 used; complete combustion Reaction: C + O2 → CO2 katom C = 100/12 = 8.333 katom kmol CO2 formed = 8.333 kmol O2 needed = 8.333 kmol combustion gas = 100% CO2

DRAFT

b. Theoretical air used; complete combustion kmol CO2 formed = 8.333 kmol Theo O2 needed = 8.333 kmol N2 from air = 8.333(79/21) = 31.348 kmol Total mols combustion gas = 8.333 + 31.348 = 39.681 kmol % CO2 = 8.333(100)/39.681 = 21% % N2 = 31.348(100)/39.681 = 79%

c. 30% excess air is used: 90% of the C burns to CO2 and 10% to CO Theo O2 needed = 8.333 excess O2 = 0.3(8.333) = 2.5 kmol total O2 needed = 1.3(8.333) = 10.833 kmol ) = 40.753kmol N2 from air = 10.833( 79 21 CO2 formed = 0.9(8.333) = 7.5 kmol CO formed = 0.1(8.333) = 0.833 kmol O2 left due to incomplete reaction to CO = 0.833/2 = 0.416 kmol O2 in combustion gas = 2.5 + 0.416 = 2.92 kmol Component CO2 O2 CO N2 Total

kmol 7.5 2.92 0.83 40.75 52

% 14.42 5.62 1.6 78.36 100

d. 20% excess air is used: 90% of the C burns to CO2 while the 10% goes out as unburned solid. Theo O2 needed = 8.333 excess O2 = 0.2(8.333) = 1.667 kmol total O2 needed = 1.2(8.333) = 10 kmol ) = 37.619kmol N2 from air = 10( 79 21 CO2 formed = 0.9(8.333) = 7.5 kmol O2 left due to incomplete reaction = 0.833 kmol O2 in combustion gas = 1.667 + 0.833 = 2.5 kmol

Page 50 of 61

DRAFT

Comp CO2 O2 N2 Total

kmol 7.5 2.5 37.62 47.62

% 15.75 5.25 79 100

Page 51 of 61 4.19e The following pure compound (CO) is burned with 25% excess of air required over the theoretical. Basis: 100 kmol CO Reaction: CO + 1/2 O2 → CO2 Theo O2 required = 100/2 = 50 kmol Excess O2 required = 0.25(50) = 12.5 kmol Total O2 required = 50+12.5 = 62.5 kmol N2 from air = 62.5(79/21) = 235.1 CO2 formed = 100 kmol

DRAFT

Component CO2 O2 N2 total

kmol 100 12.5 235.1 347.6

% 28.77 3.6 67.64 100.01

Page 52 of 61 4-34 500 m3 per min of a fuel gas at 30°C and 800 mm Hg consisting of 15% CH4, 40% C2H6, 15% C3H8, 10% C6H6, and 20% N2 is burned with 20% excess air. All the C burns to CO2 and H2 to H2O. The air enters at a temperature of 60°C and a total pressure of 760 mm Hg wing PH2O = 12 mm Hg. Calculate: a. the flue gas analysis (dry basis) b. the partial pressure of water in the flue gas c. m3 humid air used/m3 fuel at the given conditions PT of flue gas = 755 mm Hg. Given: Required: a. flue gas analysis b. partial pressure of water in the flue gas c. m3 humid air used/m3 Basis:

1 minute operation

3 fuel input = 500 m % 0 + 273 % 800 % 1 kmol3 = 21.17 kmol min 30 + 273 760 22.4 m

DRAFT

Comp CH4 C2H6 C3H8 C6H6 N2 total

%

kmol 15 40 15 10 20 100

katoms C 3.18 8.47 3.18 2.12 4.23 21.18

Theo O2 required = 42.38 + 50.85 = 67.805 kmol 2 Excess O2 = 0.02(67.805) = 13.56 kmol total O2 required = 81.37 kmol N2 from air = 81.37(79/21) = 306.09 kmol Air required = 81.37 + 306.09 = 387.46 kmol H2O in air = 387.46 %

12 760 − 12

(a) Products of combustion

Comp CO2 O2 N2

kmol 42.38 13.56 306.09

% 11.83 3.74 84.43

(b) Partial pressure of the flue gas

total H2O in flue gas = 50.85 + 6.22 = 57.07 kmol 57.07 % 760 = 102.68 mm Hg 363.03 + 57.07 (387.46 + 6.22)(22.4) % 50 + 273 3 0 + 273 = 21.50 m humid air = (c) 500 m 3 fuel P H2O =

kmol H2 3.18 16.94 9.54 12.72 42.38

6.36 25.41 12.72 6.36 50.85 50.85

Page 53 of 61 SOLUTION TO SOME PROBLEMS IN CHAPTER 5 5-2 One kg of iron is initially at 35°C. It absorbs 500 joules of heat? What will be its final temperature? Solution: Basis: 1 kg of iron at 35°C that absorbs 500 joules of heat Required: Final temperature of the iron Let Tf as the final temperature of the iron The sensible heat gain of the iron is 500 J. m ¶ T C p dT = 500 Tf i

m = 1 kg Ti = 273 + 35 = 315 K Cp = 4.13 + 0.00635T Cp, cal/(mol)(K)

T, K

(1000) ¶ 315 (4.13 + 0.00635T)dT = 500 Tf

DRAFT

2

(1000) 4.13T + 0.00635 T 2

Tf 315

= (1000) 4.13(T f − 315) + 0.00635

4.13Tf − (4.13)(315) + (0.00635/2)T2f − (0.00635/2)(315)2 = 4.18/2

Simplifying, Tf = 315.3 K = 42.3 °C

T 2f − 315 2 = (500)(4.18) 2

Page 54 of 61 SOME SOLUTION TO PROBLEMS IN CHAPTER 7 7-1

Find the degrees of freedom for the following systems in equilibrium: Formula: F = C - P + 2 a. salt and sugar (with undissolved quantities) in water. F=2-2+2=2 b. hexane, heptane and octane in equilibrium with the vapor. F=3-2+2=3 c. oil and water in a closed vessel at 100°C. F = 2 - 2 + 2 - 1= 1 d. carbon, iron and silicon. F=3-1+2=4 e. coffee, milk, and sugar. F=3-1+2=4

DRAFT

7-2

Calculate the boiling point of a. benzene at 30 in Hg. 760 mm Hg = 762.0 mm Hg 30 in Hg % 29.92 in Hg From Cox Chart, BP at 762.0 mm Hg = 81°C From Antoine Equation,

t=

1211.033 − 220.790 = 80.2°C 6.90565 − log 762.0

b. water at 10 mm Hg. From Cox Chart, BP at 10 mm Hg = 12°C c. toluene at 10 atm. 10 atm %

760 mm Hg = 7600 mm Hg 1 atm

From Cox Chart, BP at 7600 mm Hg = 220°C From Antoine equation, 1344.800 t= − 219.482 = 218°C 6.95464 − log 7600 d. propane at 50 psig. 760 mm Hg + 760 mm Hg = 3345 mm Hg 50 psig % 14.7 psi From Cox Chart, BP at 3345 mm Hg = −4°C From Antoine equation, 813.20 t= − 248.00 = −2.0°C 6.82973 − log 3345

Page 55 of 61

7-3 What is the vapor pressure of a. water at 80°F? (80-32)/1.8 = 26.7°C From Cox Chart, Pv at 26.7°C = 27 mm Hg b. n-octane at 250°C? From Cox Chart, Pv at 250°C = 9000 mm Hg c. n-pentane at 125°C? From Cox Chart, Pv at 125°C = 7400 mm Hg From Antoine equation, 1064.63 = 7414 mm Hg Pv = anti log 6.85221 − 232.000 + 125

DRAFT

d. mercury at 200°C? From Cox Chart, Pv at 200°C = 16 mm Hg

7-4 Given that glycerol at mm Hg boils at 167.2°C and at 760 mm Hg, 290°C, calculate the constants A and B in the equation Log P V = A - B T+ C where PV is vapor pressure in mm Hg T, temperature in K, and C = 230 What is the vapor pressure of glycerol at 200°C? B 167.2 + 273 + 230 B log 760 = A − 290 + 273 + 230

log 10 = A −

(1) (2)

Solving (1) and (2) simultaneously, A = 13.1456 B = 8140.01

Pv = anti log 13.1456 − 7-5

8140.01 = 36.9 mm Hg 200 + 273 + 230

A flue gas on a wet basis contains 10% CO2 , 1% CO, 5% O2, 5% H2O and 79% N2. It is at 200°C and 758 mm Hg. To what temperature should it be cooled to make it saturated with water vapor, keeping the total pressure constant at 758 mm Hg? Given: Flue gas composition on a wet basis: 10% CO2, 1% CO, 5% O2, 79% N2, and 5%H2O T = 200°C P = 758 mm Hg Required: Temperature to which it should be cooled to make it saturated with water vapor PH2O = 0.05(758) = 37.9 mm Hg

Page 56 of 61 The equilibrium temperature corresponding to the an equilibrium vapor pressure of 37.9 mm Hg is what we want. We can find this from Appendix 9 to Appendix 7. From Appendix 9 (Cox Chart) the temperature corresponding to P = 37.9 mm Hg for water is 33°C. From the Appendix 7 (Steam Table), the pressure is 5.053 kPa. By interpolation, the corresponding equilibrium temperature is T = 32 + 2 (5.053 - 4.759)/(5.324 - 4.759) = 32 + 0.98 = 33°C 7-6 In a room measuring 4m × 5m × 4m was left an open tank containing ethyl alcohol. Considering that the room is sealed from the outside, calculate the amount of alcohol that has evaporated into the room if the temperature is 30°C and the atmospheric pressure is 755 mm Hg. Assume that the room becomes completely saturated with alcohol.

1623.22 = 78.552 mm Hg 228.98 + 30 78.522 = = 0.11612 kmol EtOH 755 − 78.552 kmol dry air

Pv EtOH, 30o C = anti log 8.16290 − Y EtOH,sat. =

Pv EtOH P T − Pv EtOH

DRAFT

Vol. of room = 4 m % 5 m % 4 m = 80 m 3 = vol. of dry air 80 m 3 dry air %

(755 − 78.552) mm Hg 46.07 kg % 273 K % kmol 3 % 0.11612 kmol EtOH % 760 mm Hg 30 + 273 K 22.4 m kmol dry air kmol

= 15.3 kg EtOH

7-7 The atmospheric pressure varies with elevation according to the equation P = 29.92 − 0.00111h where P is pressure in in Hg and h is the altitude in ft. What is the elevation where water boils at 95°C? At 95°C, from Cox chart, Pvwater = 650 mm Hg = 25.59 in. Hg h = 29.92 − 25.59 = 3900 ft. 0.00111 7-8 A liquid mixture at 0°C consists of 30% ethane, 40% propane, and 30% n-butane. Find the composition of the vapor mixture in equilibrium with this liquid. What is the total equilibrium pressure? From Antoine equation, Pv ethane = anti log 6.80266 − 656.40 = 17322 mm Hg 256.00 Pv propane = anti log 6.82973 − 813.20 = 3553.8 mm Hg 248.00 Pv bu tan e = anti log 6.83029 − 945.90 = 774.53 mm Hg 240.00 PT = 0.30(17322) + 0.40(3553.8) + 0.30(774.53) = 6850 mm Hg 0.30(17322) = 0.7586 6850 0.40(3553.8) = = 0.2075 6850

y ethane = y propane

Page 57 of 61 y bu tan e = 7-9

0.30(774.53) = 0.0339 6850

A gaseous mixture consists of 61% n-hexane, 26% n-octane and 13% n-decane at a temperature of 50°C. Find the equilibrium pressure and the composition of the liquid in equilibrium with this mixture.

DRAFT

At 50°C, from Cox chart, Pvhexane = 390 mm Hg Pvoctane = 50 mm Hg Pvdecane = 8 mm Hg P P x hexane = Pv T $ y hexane = T (0.61) 390 hexane P P x oc tan e = Pv T $ y oc tan e = T (0.26) 50 oc tan e P P x decane = Pv T $ y decane = T (0.13) 8 decane x hexane + x oc tan e + x decane = 1 P P PT 0.61 390 + 0.26 50T + 0.13 8T = 1 1 PT = = 43.45 mm Hg 0.61 + 0.26 + 0.13 390 50 8 xhexane = 0.6796 xoctane = 0.2259 xdecane = 0.7061

7-10 Find the dew point of a gaseous mixture of 40% ethyl alcohol and 60% water at a total pressure of 2 atm. What is the composition of the first liquid that appears? P T = 2 atm = 1520 mm Hg Pv EtOH $ x EtOH PT Pv water $ x EtOH = 0.60 = PT

1520(0.4) = 608 Pv EtOH Pv EtOH 1520(0.6) = = 912 Pv water Pv water

y EtOH = 0.40 =

x EtOH =

y water

x water

At the correct T:

x EtOH + x water = 1 608 912 Pv EtOH + Pv water = 1

Assume T = 125°C PvEtOH = 3750 mm Hg

Pvwater = 1700 mm Hg

608 + 912 =? 1 3750 1700

0.162 + .536 = 0.698 ! 1

Page 58 of 61 T = 117°C PvEtOH = 3000 mm Hg

Pvwater = 1300 mm Hg

912 ? 1 608 3000 + 1300 =

0.203 + 0.702 = 0.905 ! 1

T = 114°C PvEtOH = 2200 mm Hg 912 ? 1 608 2200 + 1000 =

Pvwater = 1000 mm Hg 0.276 + 0.912 = 1.188 ! 1

T = 115°C Pvwater = 1150 mm Hg PvEtOH = 2500 mm Hg 912 0.243 + 0.793 = 1.036 ! 1 608 2500 + 1150 = T = 116°C

DRAFT

PvEtOH = 2530 mm Hg

Pvwater = 1200 mm Hg

912 0.24 + 0.76 = 1 608 2530 + 1200 =

7-11 Find the bubble point of a mixture of 50% n-pentane and 50% n-butane at a pressure of 4 atm. What is the composition of the first vapors formed? PT = 4 atm = 3040 mm Hg

Pv pentane 0.50 P T $ x pentane = 3040 Pv pentane Pv bu tan e $ x bu tan e = 0.50 Pv bu tan e = PT 3040

xpentane = 0.50

y pen tan e =

xbutane = 0.50

y bu tan e

At the correct T:

ypentane + ybutane = 1

Assume T = 60°C Pvpentane = 1700 mm Hg

Pvbutane = 4750 mm Hg

0.50(1700) 0.50(4750) ? + =1 3040 3040

0.2796 + 0.7813 = 1.0609 l 1

â Bubble point = 60°C Using Antoine equation: 1064.63 232.000 + T Pv bu tan e = anti log 6.83029 − 945.90 240.00 + T 0.50 0.50 Pv + =1 Pv 3040 pen tan e 3040 bu tan e Pv pen tan e = anti log 6.85221 −

(1) (2) (3)

Solving simultaneously (1), (2) and (3), bubble point = 58.2°C

Page 59 of 61 7-12 The composition of a liquefied petroleum gas (LPG) is 0.5% ethane, 0.1% ethylene, 16.4% propane, 2.1% propylene, 74.0% n-butane, and 6.9% n-butene. If the room temperature is 30°C, what is the pressure inside the tank and the composition of the gas that first issues from the mixture? At T = 30°C 656.40 = 32177.7 mm Hg 256.00 + 30 = 49536.9 mm Hg Pv ethylene = anti log 6.74756 − 585.00 255.00 + 30 Pv propane = anti log 6.82973 − 813.20 = 8026.94 mm Hg 248.00 + 30 Pv propylene = anti log 6.81960 − 785.00 = 9675.31 mm Hg 247.00 + 30 = 2123.03 mm Hg Pv n−bu tan e = anti log 6.83029 − 945.90 240.00 + 30 = 2587.62 mm Hg Pv n−butene = anti log 6.84290 − 926.10 240.00 + 30 Pv ethane = anti log 6.80266 −

DRAFT

P T = 0.005(32177.7) + 0.001(49536.9) + 0.164(8026.94) + 0.021(9675.31) + 0.74(2123.03) +0.069(2587.62) = 3479.62 0.005(32177.7) = 0.04624 3479.62 0.001(49536.9) = 0.01424 y ethylene = 3479.62 0.164(8026.94) = 0.37832 y propane = 3479.62 0.021(9675.31) = 0.05839 y propylene = 3479.62 0.74(2123.03) y n−bu tan e = = 0.45150 3479.62 0.069(2587.62) y n−butene = = 0.05131 3479.62

y ethane =

7-16 A helium-toluene mixture contains 30% mole toluene at 100°C and 760 mm Hg. Calculate the percentage saturation, relative saturation, dew point and the molar humidity of the mixture. T = 100°C PT = 760 mm Hg Ptoluene = 0.30(760) = 228 mm Hg = 556.3 mm Hg Pv toluene = anti log 6.95464 − 1344.800 219.482 + 100 P toluene 228 P T − P toluene 760 − 228 % 100 = 15.69 % a) % saturation = % 100 = Pv toluene 556.3 760 − 556.3 P T − Pv toluene

Page 60 of 61

b) c) d)

P toluene % 100 = 228 % 100 = 40.99 % Pv toluene 556.3 1344.800 − 219.482 = 73.1°C dew point: t = 6.95464 − log 228 molar humidity, Y = 0.30 = 0.429 1 − 0.30 rel. saturation =

7-17 200 grams of benzene is allowed to evaporate in a 10 m3 tank containing CO2 gas at 40°C and 750 mm Hg. If the mixture is brought to 60°C and 1.2 atm, what is the percentage saturation and the relative saturation of the mixture? At 60°C

Pv benzene = anti log 6.90565 −

DRAFT

P T = 1.2 atm %

1211.033 = 391.5 mm Hg 220.790 + 60

760 mm Hg = 912 mm Hg 1 atm

10 m 3 CO 2 % kmol 3 % 750 % 273 K = 0.384 kmol CO 2 22.4 m 760 273 + 40 K kmol = 0.00256 kmol benzene 0.2 kg benzene % 78.12 kg 0.00256 (912) = 6.04 P benzene = y benzene $P T = 0.00256 + 0.38425 P benzene 6.04 P T − P benzene 912 − 6.04 % 100 = 0.886 % % 100 = percent saturation = Pv benzene 391.5 912 − 391.5 P T − Pv benzene P relative saturation = benzene % 100 = 6.04 % 100 = 1.54 % Pv benzene 391.5

7-18 A nitrogen-acetone mixture containing 20% acetone and 80% nitrogen is at a temperature of 30°C and a pressure of 755 mm Hg. To what temperature should the mixture be chilled so that when heated back to 30°C at the same pressure, the percentage saturation will be 40%? At 30°C,

1277.03 = 283.7 mm Hg 237.23 + 30 P acetone 755 − P acetone % 100 = % 100 = 40 283.7 755 − 283.7

Pv acetone = anti log 7.23157 −

P acetone P T − P acetone percent saturation = Pv acetone P T − Pv acetone â P acetone = 146.5 mm Hg

At chilling temperature, Pacetone = Pvacetone t=

1277.03 − 237.23 = 14.86°C 7.23157 − log 146.5

Page 61 of 61 7-19 A mixture of benzene-nitrogen contains 15% benzene and is at 40°C and 780 mm Hg. What will happen if the mixture is compressed until the final pressure becomes 1500 mm Hg with temperature kept at 40°C? What is the final percentage saturation and concentration, (kg benzene)/(kg benzene-free nitrogen)?

1211.033 = 182.8 mm Hg 220.790 + 40 0.15(780) P benzene 1500 − 0.15(780) P T − P benzene % 100 = % 100 = percent saturation = Pv benzene 182.8 1500 − 182.8 P T − Pv benzene

DRAFT

Pv benzene = anti log 6.90565 −