Elementary Principles of Chemical Processes ch11

CHAPTER ELEVEN 11.1 a.

The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:

xp =

Mp M

Therefore, the leakage rate of hydrogen peroxide is m1 M p / M b. Balance on mass: Accumulation = input – output

E

dM = m0 − m1 dt t = 0, M = M 0 (mass in tank when leakage begins) Balance on H 2 O 2 : Accumulation = input – output – consumption

E dM p dt

= m0 x p 0 − m1

FG M IJ − kM HMK p

p

t = 0, M p = M p 0

11.2 a.

Balance on H3PO4: Accumulation = input . g / ml . Density of H3PO4: ρ = 1834 Molecular weight of H3PO4: M = 98.00 g / mol . dn p Accumulation = (kmol / min) dt 20.0 L 1000 ml 1.834 g mol 1 kmol = 0.3743 kmol / min Input = min L ml 98.00 g 1000 mol

E

dn p dt

= 0.3743

t = 0, n p0 = 150 × 0.05 = 7.5 kmol

z

z

np

b.

t

dn p = 0.3743 dt ⇒ n p = 7.5 + 0.3743t (kmol H 3PO 4 in tank )

7.5

xp =

c.

0

np

015 . =

n

=

np n0 + n p − n p 0

=

7.5 + 0.3743t 150 + 0.3743t

kmol H 3PO 4 kmol

7.5 + 0.3743t ⇒ t = 471 . min 150 + 0.3743t

11-1

11.3 a.

g b

b

g

b

g

bg

mw = a + bt t = 0, mw = 750 & t = 5, mw = 1000 ⇒ mw kg h = 750 + 50t h

Balance on methanol: Accumulation = Input – Output M = kg CH 3OH in tank dM = m f − mw = 1200 kg h − 750 + 50t kg h dt

b

g

E

b

dM = 450 − 50t kg h dt t = 0, M = 750 kg

z zb

M

b.

dM =

750

E

t

g

g

450 − 50t dt

0

M − 750 = 450t − 25t 2

E

M = 750 + 450t − 25t 2

Check the solution in two ways: (1) t = 0, M = 750 kg ⇒ satisfies the initial condition; dM (2) = 450 − 50t ⇒ reproduces the mass balance. dt c.

dM = 0 ⇒ t = 450 50 = 9 h ⇒ M = 750 + 450(9) − 25(9)2 = 2775 kg (maximum) dt M = 0 = 750 + 450t − 25t 2 t=

d.

−450 ±

b450g + 4b25gb750g ⇒ t = –1.54 h, 19.54 h 2b −25g 2

3.40 m 3 103 liter 0.792 kg = 2693 kg (capacity of tank) 1 m3 1 liter M = 2693 = 750 + 450t − 25t 2

t=

−450 ±

b450g + 4b25gb750 − 2693g ⇒ t = 719 . h,10.81 h 2b −25g 2

Expressions for M(t) are:

R|750 + 450t - 25t b0 ≤ t ≤ 719 . and 10.81 ≤ t ≤ 19.54g (tank is filling or draining) ( 719 . ≤ t ≤ 10.81) M(t) = S2693 (tank is overflowing) ||T0 (19.54 ≤ t ≤ 20.54) (tank is empty, draining 2

as fast as methanol is fed to it)

11-2

11.3 (cont’d) 3000 2500 M(kg)

2000 1500 1000 500 0 0

5

10

15

20

t(h)

11.4 a.

Air initially in tank: N 0 =

10.0 ft 3

492° R

1 lb - mole

b g = 0.0258 lb - mole

532° R 359 ft 3 STP

Air in tank after 15 s: Pf V P0V

=

N f RT N 0 RT

⇒ N f = N0

Rate of addition: n =

Pf P0

=

0.0258 lb - mole 114.7 psia = 0.2013 lb - mole 14.7 psia

b0.2013 − 0.0258g lb - mole air = 0.0117 lb - mole air s 15 s

b. Balance on air in tank: Accumulation = input

b

g

dN = 0.0117 lb - moles s ; t = 0, N = 0.0258 lb - mole dt

z z

N

c.

Integrate balance:

t

b

dN = n dt ⇒ N = 0.0258 + 0.0117t lb - mole air

0.0258

0

Check the solution in two ways: (1) t = 0, N = 0.0258 lb - mole ⇒ satisfies the initial condition ( 2)

d.

dN = 0.0117 lb - mole air / s ⇒ reproduces the mass balance dt

b

gb g

t = 120 s ⇒ N = 0.0258 + 0.0117 120 = 143 . lb - moles air

b g

O 2 in tank = 0.21 143 . = 0.30 lb - mole O 2

11-3

g

11.5 a.

Since the temperature and pressure of the gas are constant, a volume balance on the gas is equivalent to a mole balance (conversion factors cancel). Accumulation = Input – Output 1h dV 540 m 3 = − ν w m 3 min dt h 60 min

e

j

b

t = 0, V = 3.00 × 103 m 3 t = 0 corresponds to 8:00 AM

z zb

V

dV =

3.00×103

b.

t

g

0

0

b g

z

ν w dt ≅

0

z t

e j

9.00 − ν w dt ⇒ V m 3 = 3.00 × 103 + 9.00t − ν w dt t in minutes

Let ν w i = tabulated value of ν w at t = 10 i − 1 240

g

LM MN

i = 1, 2, … , 25

OP PQ

b

g b

24 24 10 10 . + 9.8 + 4 124.6 + 2 113.4 ν w1 + ν w 25 + 4 ∑ ν w i + 2 ∑ ν w i = 114 3 3 i = 3, 5, … i = 2, 4, …

= 2488 m 3

b g

V = 3.00 × 103 + 9.00 240 − 2488 = 2672 m 3

c.

Measure the height of the float roof (proportional to volume). The feed rate decreased, or the withdrawal rate increased between data points, or the storage tank has a leak, or Simpson’s rule introduced an error.

d.

REAL VW(25), T, V, V0, H INTEGER I DATA V0, H/3.0E3, 10./ READ (5, *) (VW(I), I = 1, 25) V= V0 T=0. WRITE (6, 1) WRITE (6, 2) T, V DO 10 I = 2, 25 T = H * (I – 1) V = V + 9.00 * H – 0.5 * H * (VW(I – 1) + VW(I)) WRITE (6, 2) T, V 10 CONTINUE 1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)') 2 FORMAT (F8.2, 7X, F6.0) END $DATA 11.4 11.9 12.1 11.8 11.5 11.3 Results: TIME (MIN) 0.00 10.00 20.00

VOLUME (CUBIC METERS) 3000. 2974. 2944.

230.00 240.00

2683. 2674.

Vtrapezoid = 2674 m3 ; VSimpson = 2672 m 3 ;

2674 − 2672 × 100% = 0.07% 2672

Simpson’s rule is more accurate.

11-4

g

11.6 a.

b

g

bg

ν out L min = kV L

⇒ ν out = 0.200V ν out = 20.0 L min ⇒ Vs = 100 L

V = 300 ν out = 60

b. Balance on water: Accumulation = input – output (L/min). (Balance volume directly since density is constant) dV = 20.0 − 0.200V dt t = 0, V = 300

c.

dV = 0 = 200 − 0.200Vs ⇒ Vs = 100 L dt

V

The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is 20.0 − 0.200(300) = −40.0. As t increases, V decreases. ⇒ dV / dt = 20.0 − 0.200V becomes less negative, approaches zero as t → ∞ . The curve is therefore concave up.

t

z

z

V

d.

t dV = dt 20.0 − 0.200V 0 300

⇒−

FG H

IJ K

1 20.0 − 0.200V ln =t −40.0 0.200

b

g

b

⇒ −0.5 + 0.005V = exp −0.200t ⇒ V = 100.0 + 200.0 exp −0.200t

b g

b

g

. 100 = 101 L 1% from steady state ⇒ V = 101

b

g

101 = 100 + 200 exp −0.200t ⇒ t =

b

ln 1 200 −0.200

g = 26.5 min

11-5

g

11.7 a.

A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points ( t = 1 week, D = 2385 kg week ) and ( t = 6 weeks, D = 755 kg week ). ln D = bt + ln a ⇔ D = ae bt b=

b

g

ln D2 D1 ln 755 2385 = = −0.230 6−1 t 2 − t1

b g b

gb g

ln a = ln D1 − bt1 = ln 2385 + 0.230 1 = 8.007 ⇒ a = e 8.007 = 3000

E

D = 3000e −0.230t

b. Inventory balance: Accumulation = –output

b

dI = −3000e −0.230t kg week dt t = 0, I = 18,000 kg

g

z z I

18,000

c.

11.8 a.

t

dI = −3000e −0.230t dt ⇒ I − 18,000 = 0

3000 −0.230t e 0.230

t 0

⇒ I = 4957 + 13,043e −0.230t

t = ∞ ⇒ I = 4957 kg

Total moles in room: N =

1100 m 3

Molar throughput rate: n =

273 K

103 mol

b g = 45,440 mol

295 K 22.4 m 3 STP 700 m 3 min

273 K

103 mol

b g = 28,920 mol min

295 K 22.4 m 3 STP

SO 2 balance ( t = 0 is the instant after the SO 2 is released into the room):

b gb

g

N mol x mol SO 2 mol = mol SO 2 in room

Accumulation = –output.

b g

d dx = −0.6364 x Nx = − nx ⇒ N = 45, 440 dt dt n = 28,920

t = 0, x =

15 . mol SO 2 = 330 . × 10 −5 mol SO 2 mol 45,440 mol

b. The plot of x vs. t begins at (t=0, x=3.30×10-5). When t=0, the slope (dx/dt) is −0.6364 × 330 . × 10 −5 = −210 . × 10 −5 . As t increases, x decreases. ⇒ dx dt = −0.6364 x becomes less negative, approaches zero as t → ∞ . The curve is therefore concave up.

11-6

x

11.8 (cont’d)

0 t

c.

Separate variables and integrate the balance equation:

z x

z t

3.30×10 −5

dx x = −0.6364dt ⇒ ln = −0.6364t ⇒ x = 330 . × 10 −5 e −0.6364t x 0 330 . × 10 −5

Check the solution in two ways: (1) t = 0, x = 3.30 × 10-5 mol SO 2 / mol ⇒ satisfies the initial condition; dx (2) = −0.6364 × 330 . × 10 −5 e −0.6364t = −0.6364 x ⇒ reproduces the mass balance. dt d.

e.

CSO 2 =

45,440 moles x mol SO 2 1100 m 3

mol

i)

t = 2 min ⇒ CSO 2 = 382 . × 10 −7

ii)

x = 10 −6 ⇒ t =

e

1 m3 103 L

mol SO 2 liter

ln 10 −6 3.30 × 10 −5 −0.6364

. × 10 −2 x = 13632 . = 4131 × 10 −6 e −0.6364t mol SO 2 / L

j = 55. min

The room air composition may not be uniform, so the actual concentration of the SO2 in parts of the room may still be higher than the safe level. Also, “safe” is on the average; someone would be particularly sensitive to SO2.

11-7

11.9 a.

Balance on CO: Accumulation=-output N ( mol ) x ( mol CO / mol) = total moles of CO in the laboratory Pν p kmol )= h RT Pν p kmol CO kmol )x Rate at which CO leaves: n ( = x h kmol RT CO balance: Accumulation = -output Molar flow rate of entering and leaving gas: n (

FG H

FG H

IJ K

IJ K

Pν p d ( Nx ) dx P x⇒ =− =− ν px dt RT dt NRT

E PV = NRT

νp dx =− x dt V kmol CO kmol

t = 0, x = 0.01

z

z

νp r dx V =− ln 100 x dt ⇒ tr = − νp x V 0 0.01 t

x

b.

c.

b g

V = 350 m 3 350 tr = − ln 100 × 35 × 10 −6 = 2.83 hrs 700

e

j

d. The room air composition may not be uniform, so the actual concentration of CO in parts of the room may still be higher than the safe level. Also, “safe” is on the average; someone could be particularly sensitive to CO.

Precautionary steps: Purge the laboratory longer than the calculated purge time. Use a CO detector to measure the real concentration of CO in the laboratory and make sure it is lower than the safe level everywhere in the laboratory.

11.10 a.

Total mass balance: Accumulation = input – output

b

g

dM = m − m kg min = 0 ⇒∴ M is a constant = 200 kg dt

b. Sodium nitrate balance: Accumulation = - output x = mass fraction of NaNO 3

b g E

b

d xM = − xm kg min dt

g

dx m m =− x=− x 200 dt M t = 0, x = 90 200 = 0.45

11-8

11.10 (cont’d) c.

0.45

m = 50 kg / min m = 100 kg / min

x

m = 200 kg / min

0 t(min)

dx m =− x < 0 , x decreases when t increases dt 200 dx becomes less negative until x reaches 0; dt Each curve is concave up and approaches x = 0 as t → ∞; dx becomes more negative ⇒ x decreases faster. dt

m increases ⇒

z

z

x

d.

FG H

t

dx m x m mt dt ⇒ ln =− =− t ⇒ x = 0.45 exp − 0.45 200 200 x M 0.45 0

IJ K

Check the solution: (1) t = 0, x = 0.45 ⇒ satisfies the initial condition; (2)

dx m mt m = −0.45 × exp( − )=− x ⇒ satisfies the mass balance. dt 200 200 200

0.45 0.4

m = 50 kg / m in

0.35

m = 100 kg / m in

0.3

m = 200 kg / m in

x

0.25 0.2 0.15 0.1 0.05 0 0

5

10

15

20

t(m in)

e.

d

i

m = 100 kg min ⇒ t = −2 ln x f 0.45 90% ⇒ x f = 0.045 ⇒ t = 4.6 min 99% ⇒ x f = 0.0045 ⇒ t = 9.2 min

99.9% ⇒ x f = 0.00045 ⇒ t = 138 . min

11-9

25

11.11 a.

e je

Mass of tracer in tank: V m 3 C kg m 3

j

Tracer balance: Accumulation = –output. If perfectly mixed, Cout = C tank = C

b g = −ν C bkg ming

d VC

dC ν =− C dt V m t = 0, C = 0 V

V is constant

dt

b.

z

C

m0 V

c.

dC =− C

z

t

ν

0

V

dt ⇒ ln

FG C IJ = − νt ⇒ C = m V Hm VK V

0

FG νt IJ H VK

exp −

0

Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption

e

j e

j

of perfect mixing) through t = 1, C = 0.223 × 10 −3 & t = 2, C = 0.050 × 10 −3 . −

ν V

=

b

2 −1

E V = e30 m

11.12 a.

g = −1495 min .

ln 0.050 0.223

3

min

−1

min j = 201 . . m j e1495 −1

3

In tent at any time, P=14.7 psia, V=40.0 ft3, T=68°F=528°R 14.7 psia 40.0 ft 3 PV 3 ⇒N= = m(liquid) = = 01038 . lb - mole ft ⋅ psia RT 528 o R 10.73 o lb - mole ⋅ R

b. Molar throughout rate: 60 ft 3 492° R 16.0 psia 1 lb - mole lb - mole min = 01695 n in = n out = n = . min 528° R 14.7 psia 359 ft 3 STP Moles of O2 in tank= N (lb - mole) ×

FG lb - mole O IJ H lb - mole K

b g

2

Balance on O2: Accumulation = input – output dx b g = 0.35n − xn ⇒ 01038 . b0.35 − x g dx . = 01695 . b0.35 − xg ⇒ dt = 163

d Nx dt

c.

z

dt

z

t = 0, x = 0.21

b

g

t 0.35 − x dx = 163 . t = 163 . dt ⇒ − ln 0.21 0.35 − x 0 0.35 − 0.21 x

b

g

0.35 − x . e −1.63t ⇒ = e −1.63t ⇒ x = 0.35 − 014 014 . 1 0.35 − 0.27 x = 0.27 ⇒ t = − ln = 0.343 min (or 20.6 s) 163 . 0.35 − 0.21

LM FG N H

IJ OP KQ

11-10

11.13 a.

b gb

Mass of isotope at any time = V liters C mg isotope liter

g

Balance on isotope: Accumulation = –consumption

FG IJ b g H K

b g

dC = − kC dt t = 0, C = C0

Cancel V

mg d VC = − kC V L L⋅s dt

Separate variables and integrate

z z C

C0

dC = C

FG C IJ = − kt ⇒ t = − lnbC C g k HC K − lnb0.5g ln 2 = ⇒t =

t

0

0

C = 0.5C0 ⇒ t 1 2

b.

0

− kdt ⇒ ln

t 1 2 = 2.6 hr ⇒ k = C = 0.01C0

12

k

k

ln 2 = 0.267 hr −1 2.6 hr

t=-ln(C/C0)/k

t=

b g = 17.2 hr

− ln 0.01 0.267

11.14 A → products a.

Mole balance on A: Accumulation = –consumption

b g = − kC V

d C AV

A

dt

bV constant; cancelsg

t = 0, C A = C A0 ⇒

z

CA

CA0

dC A = CA

z

t

− kdt ⇒ ln

0

FG C IJ = − kt ⇒ C HC K A

A

b g

= C A0 exp − kt

A0

b. Plot C A (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies

b

g b

g

assumption of first-order) through t = 213 . , C A = 0.0262 & t = 120.0, C A = 0.0185 . ln C A = − kt + ln C A0 −k =

b

ln 0.0185 0.0262 120.0 − 213 .

g = −353 . × 10

−3

min −1 ⇒ k = 35 . × 10 −3 min −1

11.15 2 A → 2 B + C a.

Mole balance on A: Accumulation = –consumption

b g = − kC V

d C AV

2 A

dt

bV constant; cancelsg

t = 0, C A = C A0 ⇒

z

CA

dC A

CA0

C A2

z

LM N

t

1 1 1 = − kdt ⇒ − + = − kt ⇒ C A = + kt 0 C A C A0 C A0

11-11

OP Q

−1

11.15 (cont’d) b.

C A = 0.5C A0 ⇒ − n A = 0.5n A0

b = b0.5n

n P RT 1 1 1 + = − kt 1 2 ⇒ t 1 2 = ; but C A0 = A0 = 0 ⇒ t 1 2 = 0.5C A0 C A0 kC A0 kP0 V RT

gb g mol A react.gb1 mol C 2 mol A react.g = 0.25n

n B = 0.5n A0 mol A react. 2 mol B 2 mol A react. = 0.5n A0

nC

A0

total moles = 125 . n A0 ⇒ P1 2 = 125 . c.

A0

n A0 RT = 125 . P0 V

Plot t 1 2 vs. 1 P0 on rectangular paper. Data fall on straight line (verifying 2nd order

d

i d

i

. & t 1 2 = 209, 1 P0 = 1 0.683 decomposition) through t 1 2 = 1060, 1 P0 = 1 0135 RT 1060 − 209 = = 143.2 s ⋅ atm k 1 0135 . − 1 0.683

Slope:

⇒k =

d.

t1 2 =

b1015 Kgb0.08206 L ⋅ atm mol ⋅ Kg = 0.582 L mol ⋅ s 143.2 s ⋅ atm

I JK

F GH

FG IJ H K

t 1 2 P0 RT E 1 E 1 exp ⇒ ln = ln + k 0 P0 RT RT k0 R T

Plot t 1 2 P0 RT (log scale) vs. 1 T (rect. scale) on semilog paper.

bg

bg

t 1 2 s , P0 = 1 atm, R = 0.08206 L ⋅ atm / (mol ⋅ K), T K

d

i

Data fall on straight line through t 1 2 P0 RT = 74.0, 1 T = 1 900 &

dt

i

RT = 0.6383, 1 T = 1 1050

1 2 P0

b

g

E ln 0.6383 74.0 = = 29,940 K R 1 1050 − 1 900 ln

e.

b

R=8.314 J/ (mol ·K)

E = 2.49 × 10 5 J mol

g

1 29,940 = ln 0.6383 − = −28.96 ⇒ k 0 = 3.79 × 1012 L (mol ⋅ s) k0 1050

FG H

T = 980 K ⇒ k = k 0 exp − C A0 =

b

IJ K

E = 0.204 L (mol ⋅ s) RT

g

0.70 120 . atm

b0.08206 L ⋅ atm mol ⋅ Kgb980 Kg

= 1045 . × 10 −2 mol L

90% conversion

LM N

OP Q

LM N

1 1 1 1 1 1 − = − − 3 k C A C A0 0.204 1045 . × 10 . × 10 −2 1045 = 4222 s = 70.4 min

C A = 010 . C A0 ⇒ t =

11-12

OP Q

11.16 A → B a.

Mole balance on A: Accumulation = –consumption(V constant) dC A k C =− 1 A dt 1+ k2CA t = 0, C A = C A0

z

1+ k2CA dC A = C A0 k 1C A CA

b

z

t

− dt ⇒

0

g

b

g

b

g

k C C k 1 1 ln A + 2 C A − C A0 = − t ⇒ t = 2 C A0 − C A − ln A k1 k 1 C A0 k 1 C A0 k 1

b

b. Plot t C A − C A0 vs. ln C A / C A0

g bC

A0

g

− C A on rectangular paper:

x  y

  

  k t 1 ln C A C A0 =− + 2 k 1 C A0 − C A k1 C A0 − C A

b

g ;

b

g

slope

1

FG H

intercept y1

IJ FG K H

x1

y2

x2

IJ K

Data fall on straight line through 116.28, −0.2111 & 130.01, −0.2496 −

1 130.01 − 116.28 = = −356.62 ⇒ k 1 = 2.80 × 10 −3 L (mol ⋅ s) k 1 −0.2496 − −0.2111

b

g

b

g

k2 = 130.01 + 356.62 −0.2496 = 4100 . ⇒ k 2 = 0115 . L mol k1

11.17 CO + Cl 2 ⇒ COCl 2 a.

3.00 L

273 K

1 mol

. mol gas b g = 012035 . molg 3.00 L = 0.02407 mol L CO U bC g = 0.60b012035 |Vinitial concentrations . molg 3.00 L = 0.01605 mol L Cl | dC i = 0.40b012035 W C bt g = 0.02407 − C bt g U | Since 1 mol COCl formed requires 1 mol of each reactant C bt g = 0.01605 − C bt g V| W 303.8 K 22.4 L STP

CO i Cl 2

2

i

CO

p

Cl 2

p

2

b. Mole balance on Phosgene: Accumulation = generation

d i=

d VC p dt

c.

d1 + 58.6C

Cl 2

dC p

V=3.00 L

8.75CCO CCl 2 + 34.3C p

i

dt

2

=

d

id

2.92 0.02407 − C p 0.01605 − C p − 24.3C i . d1941 p

t = 0, C p = 0

b

g

Cl 2 limiting; 75% conversion ⇒ C p = 0.75 0.01605 = 0.01204 mol L

1 t= 2.92

z

. − 24.3C i d1941 dC d0.02407 − C id0.01605 − C i 2

0.01204

0

p

p

p

11-13

p

2

i

11.17 (cont’d) d.

11.18 a.

REAL F(51), SUM1, SUM2, SIMP INTEGER I, J, NPD(3), N, NM1, NM2 DATA NPD/5, 21, 51/ FN(C) = (1.441 – 24.3 * C) ** 2/(0.02407 – C)/(0.01605 – C) DO 10 I = 1, 3 N = NPD(I) NM1 = N – 1 NM2 = N – 2 DO 20 J = 1, N C = 0.01204 * FLOAT(J – 1)/FLOAT(NM1) F(J) = FN(C) 20 CONTINUE SUM1 = 0. DO 30 J = 2, NM1, 2 SUM = SUM1 + F(S) 30 CONTINUE SUM2 = 0. DO 40 J = 3, NM2, 2 SUM2 = SUM2 + F(J) 40 CONTINUE SIMP = 0.01204/FLOAT(NM1)/3.0 * (F(1) + F(N) + 4.0 * SUM1 + 2.0 * SUM2) T = SIMP/2.92 WRITE (6, 1) N, T 10 CONTINUE 1 FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES') END RESULTS 5 POINTS — 91.0 MINUTES 21 POINTS — 90.4 MINUTES 51 POINTS — 90.4 MINUTES t = 90.4 minutes

e j e

Moles of CO 2 in liquid phase at any time = V cm 3 C A mols cm 3

j

Balance on CO 2 in liquid phase: Accumulation = input

b g

e

d VC A = kS C *A − C A dt

FjG molsIJ ⇒ dCdt = kSV eC H s K t = 0, C = 0 A

* A

− CA

j

÷V

A

Separate variables and integrate. Since p A = y A P is constant, C *A = p A H is also a constant.

z

dC A

CA

0

C *A

⇒ ln

− CA

=

C *A − C A C *A

z

t

0

e

kS dt ⇒ − ln C *A − C A V

=−

j

CA CA =0

=

kS t V

C kS expb g t ⇒ 1 − *A = e − kSt V ⇒ C A = C *A 1 − e − kSt V V CA

e

1− C A C *A

11-14

j

11.18 (cont’d) b.

LM MN

C V ln 1 − *A kS CA

t=−

OP PQ

V = 5 L = 5000 cm , k = 0.020 cm s , S = 78.5 cm , C A = 0.62 × 10 3

2

−3

mol / cm

a fa f d9230 atm ⋅ cm moli = 0.65 × 10 mol cm e5000 cm j lnF1 − 0.62 × 10 I = 9800 s ⇒ 2.7 hr t=− b0.02 cm sge78.5 cm j GH 0.65 × 10 JK C *A = y A P H = 0.30 20 atm

−3

3

3

3

3

−3

−3

2

(We assume, in the absence of more information, that the gas-liquid interfacial surface area equals the cross sectional area of the tank. If the liquid is well agitated, S may in fact be much greater than this value, leading to a significantly lower t than that to be calculated) 11.19 A → B a.

Total Mass Balance: Accumulation = input dM d ( ρV ) = = ρv dt dt

E

dV = v dt t = 0, V = 0

A Balance: Accumulation = input – consumption dN A = C A0 v − ( kC A )V C =N /V A A dt

dN A = C Ao v − kN A dt t = 0, N A = 0

b. Steady State:

c.

z z z V

t

0

0

dV =

NA

0

⇒−

dN A C v = 0 ⇒ N A = A0 dt k

 ⇒ V = vt  vdt

dN A = C A0 v − kN A

z

t

dt

0

FG H

IJ K C v = 1 − expb− kt g k

C v − kN A C v − kN A 1 ln A0 = t ⇒ A0 = e − kt k C A0 v C A0 v

⇒ NA

CA =

A0

t →∞⇒ NA =

N A C A0 [1 − exp( − kt )] = V kt

11-15

C A0 v k

11.19 (cont’d)

When the feed rate of A equals the rate at which A reacts, NA reaches a steady value. NA would never reach the steady value in a real reactor. The reasons are:  ⇒ t → ∞, V → ∞. (1) In our calculation, V = vt But in a real reactor, the volume is limited by the reactor volume; (2) The steady value can only be reached at t → ∞. In a real reactor, the reaction time is finite. d.

C C A0 [1 − exp( − kt )] = lim A0 = 0 t →∞ t →∞ kt kt

lim C A = lim

t →∞

From part c, t → ∞, N A → a finite number, V → ∞ ⇒ C A = 11.20 a.

MCv

NA →0 V

dT = Q − W dt M = (3.00 L)(100 . kg / L) = 3.00 kg Cv = C p = (0.0754 kJ / mol ⋅ o C)(1 mol / 0.018 kg) = 4.184 kJ / kg ⋅ o C W = 0

dT = 0.0797Q (kJ / s) dt t = 0, T = 18 o C

b. c.

11.21 a.

z z 100o C

240 s

18 C

0

dT = o

0.0797Q dt ⇒ Q =

100 − 18 kJ = 4.287 = 4.29 kW 240 × 0.0797 s

Stove output is much greater. Only a small fraction of energy goes to heat the water. Some energy heats the kettle. Some energy is lost to the surroundings (air). Energy balance: MCv

dT = Q − W dt M = 20.0 kg C v ≈ C p = ( 0.0754 kJ / mol ⋅ C)(1 mol / 0.0180 kg) = 4.184 kJ / (kg ⋅ C) Q = 0.97 ( 2.50) = 2.425 kJ s o

o

a f

W = 0

b g

dT = 0.0290 ° C s , t = 0, T = 25° C dt

The other 3% of the energy is used to heat the vessel or is lost to the surroundings.

z z T

b.

t

dT =

o

25 C

c.

bg

0.0290dt ⇒ T = 25° C + 0.0290t s

0

b

g

T = 100° C ⇒ t = 100 − 25 0.0290 = 2585 s ⇒ 43.1 min

No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal boiling point).

11-16

11.22 a. Energy balance on the bar MCv

b

dTb   = Q − W = −UA Tb − Tw dt

g

B

Table B.1

e

M = 60 cm

3

je7.7 g cm j = 462 g 3

Cv = 0.46 kJ (kg ⋅° C), Tw = 25° C 2

U = 0.050 J (min ⋅ cm ⋅° C)

a fa f a fa f a fa f

2

A = 2 2 3 + 2 10 + 3 10 cm = 112 cm

b

gb

dTb = −0.02635 Tb − 25 ° C min dt

2

g

t = 0, Tb = 95° C

d

i

dTb = 0 = −0.02635 Tbf − 25 ⇒ Tbf = 25° C dt

b.

95 85 75

Tb( oC)

65 55 45 35 25 15 5 0 t

z

z

Tb

c.

t dTb = −0.02635dt T − 25 0 95 b

FG T − 25 IJ = −0.02635t H 95 − 25K ⇒ T bt g = 25 + 70 expb−0.02635t g ⇒ ln

b

b

Check the solution in three ways: (1) t = 0, Tb = 25 + 70 = 95o C ⇒ satisfies the initial condition; dTb = −70 × 0.02635e −0.02635t = −0.02635(Tb − 25) ⇒ reproduces the mass balance; dt (3) t → ∞, Tb = 25o C ⇒ confirms the steady state condition.

(2)

Tb = 30° C ⇒ t = 100 min

11-17

11.23

12.0 kg/min 25oC

12.0 kg/min T (oC) Q (kJ/min) = UA (Tsteam-T)

a.

b

g

b

dT  p 25 − T + UA Tsteam − T = mC dt

Energy Balance: MCv

g

M = 760 kg m = 12.0 kg min

dT / dt = 150 . − 0.0224T ( o C min), t = 0, T = 25o C Cv ≈ C p = 2.30 kJ (min ⋅° C) UA = 115 . kJ (min ⋅° C)

a

f

Tsteam sat' d; 7.5bars = 167.8° C

b. Steady State:

dT = 0 = 150 . − 0.0224Ts ⇒ Ts = 67° C dt

T(oC)

67

25 0 t

z

c.

IJ K

FG H

z

Tf

dT 1 150 . − 0.0224T 150 . − 0.94 exp( −0.0224t ) ⇒T = = dt ⇒ t = − ln 150 . − 0.0224T 0 0.0224 0.94 0.0224 25 t

t = 40 min. ⇒ T = 49.8° C

d. U changed. Let x = (UA) new . The differential equation becomes: dT = 0.3947 + 0.096 x − ( 0.01579 + 5.721x )T dt

z

55

25

0.3947 + 0.096 x − (0.01579 + 5.721 × 10−4 x )T ⇒

−

1 0.01579 + 5.721 × 10−4

z

40

dT

=

dt

0

L 0.3947 + 0.096x − e0.01579 + 5.721 × 10 xj × 55 OP = 40 ln M x M 0.3947 + 0.096 x − e0.01579 + 5.721 × 10 x j × 25 P PQ NM −4

−4

⇒ x = 14.27 kJ / (min⋅o C)

ΔU Δ(UA) 14.27 − 115 . = = × 100% = 241% . U initial (UA)initial 115 .

11-18

11.24 a.

Energy balance: MCv

dT = Q − W dt . J g ⋅° C W = 0, Cv = 177 M = 350 g, Q = 40.2W = 40.2 J s

b gU|V ⇒ T = 20 + 0.0649tbsg |W T = 40° C ⇒ t = 308 s ⇒ 51. min

dT = 0.0649 ° C s dt t = 0, T = 20° C

b. The benzene temperature will continue to rise until it reaches Tb = 801 . ° C ; thereafter the heat input will serve to vaporize benzene isothermally. 801 . − 20 Time to reach Tb neglect evaporation : t = = 926 s 0.0649 Time remaining: 40 minutes 60 s min − 926 s = 1474 s Evaporation: ΔH = 30.765 kJ mol 1 mol 78.11 g 1000 J kJ = 393 J g

b

v

g

b

b

b

gb

g

gb

g

gb g Benzene remaining = 350 g − b0102 . g sgb1474 sg = 200 g

Evaporation rate = 40.2 J s / 393 J g = 0102 . g s

c.

11.25 a.

1. Used a dirty flask. Chemicals remaining in the flask could react with benzene. Use a clean flask. 2. Put an open flask on the burner. Benzene vaporizes⇒ toxicity, fire hazard. Use a covered container or work under a hood. 3. Left the burner unattended. 4. Looked down into the flask with the boiling chemicals. Damage eyes. Wear goggles. 5. Rubbed his eyes with his hand. Wash with water. 6. Picked up flask with bare hands. Use lab gloves. 7. Put hot flask on partner’s homework. Fire hazard. 60 m 3

273 K 1 kg - mole = 2.58 kg - moles 283 K 22.4 m 3 STP dT Energy balance on room air: nCv = Q − W dt Q = m ΔH H O, 3bars, sat' d − 30.0 T − T

Moles of air in room: n =

b g

W = 0 nCv

s

v

b

g

2

b

dT  s ΔH v − 30.0 T − T0 =m dt N = 2.58 kg - moles

b

0

g

g

Cv = 20.8 kJ (kg - mole⋅° C) ΔH v = 2163 kJ kg from Table B.6 T0 = 0° C

b

b

dT = 40.3m s − 0.559T ° C hr dt

g

g

t = 0, T = 10° C (Note: a real process of this type would involve air escaping from the room and a constant pressure being maintained. We simplify the analysis by assuming n is constant.)

11-19

11.25 (cont’d) 0.559T 40.3

b. At steady-state, dT dt = 0 ⇒ 40.3m s − 0.559T = 0 ⇒ m s = T = 24° C ⇒ m s = 0.333 kg hr

c.

Separate variables and integrate the balance equation:

z

Tf

10

dT = 40.3m s − 0.559T

z

t

0

dt

z

m s = 0.333

E

T f = 23°C

t=−

11.26 a.

b

Integral energy balance t = 0 to t = 20 min Q = ΔU = MCv ΔT =

dT

23

10 13.4 − 0.559T

=t

LM MN

b g OP = 4.8 hr b g PQ

13.4 − 0.559 23 1 ln 0.559 13.4 − 0.559 10

g

b60 − 20g° C = 4.00 × 10

250 kg 4.00 kJ kg⋅° C

4

kJ

4.00 × 104 kJ 1 min 1 kW Required power input: Q = = 333 . kW 20 min 60 s 1 kJ s b. Differential energy balance: MCv

dT = Q dt

bg

dT = 0.001Q t dt

M = 250 kg Cv = 4.00 kJ kg⋅°C

z z

T

Integrate:

t = 0, T = 20° C

z

t

t

 dT = 0.001 Q dT ⇒ T = 20 o C + Qdt

20o C

0

0

Evaluate the integral by Simpson's Rule (Appendix A.3) 600 s  = 30 33 + 4 33 + 35 + 39 + 44 + 50 + 58 + 66 + 75 + 85 + 95 Qdt 3 0

z

b

g

b

g

+2 34 + 37 + 41 + 47 + 54 + 62 + 70 + 80 + 90 + 100 = 34830 kJ

b g

jb

e

g

⇒ T 600 s = 20o C + 0.001 oC / kJ 34830 kJ = 54.8° C

c.

b

g

10 kW Past 600 s, Q = 100 + t − 600 s = t 6 60 s

LM  = 20 + 0.001M Qdt T = 20 + 0.001 Qdt MM  + MN

0.001 F t 600 I ⇒ T = 54.8 + − ⇒ t bsg = G 6 H6 2 JK

z

z z

600

t

0

0 34830

2

2

t

OP t P dt 6 P PP Q

600

b

12000 T − 24.8

g

T = 85° C ⇒ t = 850 s = 14 min, 10 s ⇒ explosion at 10:14 + 10 s

11-20

11.27 a. Total Mass Balance:

Accumulation=Input– Output

E dM tot d( ρV) i − m o ⇒ =m = 8.00ρ − 4.00ρ dt dt

ρ=constant

dV = 4.00 L / s dt t = 0, V0 = 400 L

KCl Balance: dM KCl d( CV )  i, KCl − m  o, KCl ⇒ . × 8.00 − 4.00C =m = 100 dt dt dC 8 − 8C = dV dt = 4 dV dC dt V ⇒V +C = 8 − 4C dt dt t = 0, C 0 = 0 g / L

Accumulation=Input-Output ⇒

b. (i)The plot of V vs. t begins at (t=0, V=400). The slope (=dV/dt) is 4 (a positive constant). V increases linearly with increasing t until V reaches 2000. Then the tank begins to overflow and V stays constant at 2000.

V

2000

400 0 t

(ii) The plot of C vs. t begins at (t=0, C=0). When t=0, the slope (=dC/dt) is (8-0)/400=0.02. As t increases, C increases and V increases (or stays constant)⇒ dC/dt=(8-8C)/V becomes less positive, approaches zero as t→ ∞. The curve is therefore concave down.

C

1

0 t

c.

dV =4⇒ dt

z

V

z

t

dV = 4 dt ⇒ V = 400 + 4t

400

0

11-21

11.27 (cont’d) dC 8 − 8C dC 1− C = = V = 400 + 4 t dt V dt 50 + 0.5t C dC t dt C t = ⇒ − ln(1 − C ) 0 = 2 ln(50 + 0.5t ) 0 0 1− C 0 50 + 0.5t 50 + 0.5t ⇒ ln(1- C)-1 = 2 ln = ln(1 + 0.01t )2 50 1 1 ⇒ = (1 + 0.01t )2 ⇒ C = 1 − 1- C (1 + 0.01t )2

z

z

When the tank overflows, V = 400 + 4t = 2000 ⇒ t = 400 s 1 = 0.96 g / L C = 12 1+ 0.01 × 400

b

g

11.28 a. Salt Balance on the 1st tank: Accumulation=-Output

E

v d(CS1V1 ) dC = − CS1v ⇒ S1 = − CS1 = −0.08CS1 V1 dt dt CS1 ( 0) = 1500 500 = 3 g / L

Salt Balance on the 2nd tank: Accumulation=Input-Output

E

v d(CS2V2 ) dC = CS1v − CS 2 v ⇒ S2 = ( CS1 − CS 2 ) = 0.08( CS1 − CS 2 ) dt V2 dt CS 2 ( 0) = 0 g / L

Salt Balance on the 3rd tank: Accumulation=Input-Output

E

v d(CS3V3 ) dC = CS 2 v − CS 3v ⇒ S3 = ( CS 2 − CS 3 ) = 0.04( CS 2 − CS 3 ) dt V3 dt CS 3 ( 0) = 0 g / L

b.

CS1, CS2, CS3

3

CS1

CS2 CS3

0 t

11-22

11.28 (cont’d)

The plot of CS1 vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is −0.08 × 3 = −0.24 . As t increases, CS1 decreases ⇒ dCS1/dt=-0.08CS1 becomes less negative, approaches zero as t→ ∞. The curve is therefore concave up. The plot of CS2 vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 0.08(3 − 0) = 0.24 . As t increases, CS2 increases, CS1 decreases (CS2 < CS1)⇒ dCS2/dt =0.08(CS1-CS2) becomes less positive until dCS2/dt changes to negative (CS2 > CS1). Then CS2 decreases with increasing t as well as CS1. Finally dCS2/dt approaches zero as t→∞. Therefore, CS2 increases until it reaches a maximum value, then it decreases. The plot of CS3 vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 0.04(0 − 0) = 0 . As t increases, CS2 increases (CS3 < CS2)⇒ dCS3/dt =0.04(CS2-CS3) becomes positive ⇒ CS2 increases with increasing t until dCS3/dt changes to negative (CS3 > CS1). Finally dCS3/dt approaches zero as t→∞. Therefore, CS3 increases until it reaches a maximum value then it decreases. c. 3

CS1, CS2, CS3 (g/L)

2.5 2 CS1 1.5 CS2

1

CS3

0.5 0 0

20

40

60

80

100

120

140

160

t (s)

11.29 a. (i) Rate of generation of B in the 1st reaction: rB1 = 2r1 = 0.2C A

(ii) Rate of consumption of B in the 2nd reaction: − rB 2 = r2 = 0.2CB2 b. Mole Balance on A: Accumulation=-Consumption

E

d ( C AV ) dC A . C AV ⇒ . CA = −01 = −01 dt dt . mol / L t = 0, C A0 = 100

Mole Balance on B: Accumulation= Generation-Consumption

E

d ( CBV ) dCB = 0.2C AV − 0.2CB2V ⇒ = 0.2C A − 0.2CB2 dt dt t = 0, CB 0 = 0 mol / L

11-23

11.29 (cont’d) c. 2

CA, CB, CC

CC

1 CB CA 0 t

The plot of CA vs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is −01 . × 1 = −01 . . As t increases, CA decreases ⇒ dCA/dt=-0.1CA becomes less negative, approaches zero as t→∞. CA→0 as t→∞. The curve is therefore concave up. The plot of CB vs. t begins at (t=0, CB=0). When t=0, the slope (=dCB/dt) is 0.2(1 − 0) = 0.2 . As t increases, CB increases, CA decreases ( C B2 < CA)⇒ dCB/dt =0.2(CA- C B2 ) becomes less positive until dCB/dt changes to negative ( CB2 > CA). Then CB decreases with increasing t as well as CA. Finally dCB/dt approaches zero as t→∞. Therefore, CB increases first until it reaches a maximum value, then it decreases. CB→0 as t→∞. The plot of CC vs. t begins at (t=0, CC=0). When t=0, the slope (=dCC/dt) is 0.2(0) = 0 . As t increases, CB increases ⇒ dCc/dt =0.2 C B2 becomes positive also increases with increasing t ⇒ CC increases faster until CB decreases with increasing t ⇒ dCc/dt =0.2 CB2 becomes less positive, approaches zero as t→∞ so CC increases more slowly. Finally CC→2 as t→∞. The curve is therefore S-shaped.

CA, CB, CC (mol/L)

d. 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0

CC

CB CA

0

10

20

30 t (s)

11-24

40

50

11.30 a. When x = 1, y = 1 . y=

a ax x =1, y =1 ⇒ 1= ⇒ a = 1+ b 1+ b x+b pC5 H12 = yP = xp *C5 H12 ( 46o C ) ⇒ y =

b. Raoult’s Law:

p *C5 H12 (46o C) = 10

Antoine Equation:

(6.84471−

1060.793 ) 46 + 231.541

o

xp *C5 H12 ( 46 C )

⇒y=

P

=

xp *C5 H12 ( 46o C ) P

= 1053 mm Hg

0.7 × 1053 = 0.970 760

0.70a R| y = ax . 0.970 = ""(1) |Ra = 1078 ⇒S 0.70 + b S| x + b TFrom part (a), a = 1+ b"""""""""""(2) |Tb = 0.078 x=0.70, y=0.970

c. Mole Balance on Residual Liquid: Accumulation=-Output

E

dN L = − nV dt t = 0, N L = 100 mol Balance on Pentane: Accumulation=-Output

E

ax dx dN L d(NLx) = − nV y ⇒ x + NL = − nV x+b dt dt dt dN L / dt = − nV

E

FG H

n ax dx =− V −x NL x + b dt t = 0, x = 0.70

IJ K

d. Energy Balance: Consumption=Input

E

 nV ΔH vap = Q

 ΔH vap =27.0 kJ/ mol

t = 0, N L = 100 mol

From part (c),

dN L = − nV dt nV Q 27.0 =  NL Qt 100 27.0

nV =

b

Q 27.0 kJ / mol

N L = 100 − nV t = 100 −

Substitute this expression into the equation for dx/dt from part (c):

11-25

g

 Qt 27.0

11.30 (cont’d)

IJ K

FG H

FG H

dx n ax Q 27.0 ax =− V −x =− −x  dt NL x + b x+b Qt 100 27.0 x(0) = 0.70

IJ K

e. 1 0.9 0.8 y (Q=1.5 kJ/s)

0.7 x, y

0.6 x (Q=1.5 kJ/s)

0.5 0.4

x (Q=3 kJ/s)

0.3

y (Q=3 kJ/s)

0.2 0.1 0 0

200

400

600

800

1000 1200 1400 1600 1800

t(s)

f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The higher the heating rate, the faster x and y decrease.

11-26