11- 1

CHAPTER ELEVEN

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

xM

Mpp=

Therefore, the leakage rate of hydrogen peroxide is & /m M Mp1 b. Balance on mass: Accumulation = input output

E= -

= =

dMdt

m m

t M M

& &

,

0 1

00 (mass in tank when leakage begins)

Balance on H O2 2: Accumulation = input output consumption

E= -

FHG

IKJ -

= =

dM

dtm x m

M

MkM

t M M

pp

pp

p p

& &

,

0 0 1

00

11.2 a. Balance on H3PO4: Accumulation = input Density of H3PO4: r = 1834. g / ml . Molecular weight of H3PO4: M = 9800. g / mol .

Accumulation =dndt

(kmol / min)

Input =20.0 L 1000 ml 1.834 g mol 1 kmol

min L ml 98.00 g 1000 mol kmol / min

dn

dt

n kmol

p

p

p0

=

E=

= = =

03743

0 3743

0 150 0 05 7 5

.

.

, . .t

b. dn dt n tp

n t

p

p

7 5 0

03743 75 0 3743.

. . . )= = + (kmol H PO in tank3 4

xn

n

n

n n nttp

p p

p p= =

+ -=

++0 0

7 5 03743150 03743

. ..

kmol H POkmol

3 4

c. 0157 5 03743150 03743

471.. .

..=

++

=tt

t min

11- 2

11.3 a. &m a btw = + t mw= =0 750, &b g & t m m tw w= = = +5 1000 750 50, & &b g b g b gkg h h Balance on methanol: Accumulation = Input Output

MdMdt

m m t

dMdt

t

t M

f w

=

= - = - +

E= -

= =

kg CH OH in tank

kg h kg h

kg h

kg

3

& &

,

1200 750 50

450 50

0 750

b g

b g

b. dM t dtM t

750 0

450 50= -b g

E- = -

E= + -

M t t

M t t

750 450 25

750 450 25

2

2

Check the solution in two ways:

( ) ,1 0 750

450 50

t M

t

= =

= -

kg satisfies the initial condition;

(2) dMdt

reproduces the mass balance.

c. dMdt

t M= = = = + - =0 450 50 9 750 450 9 25 9 27752 h kg (maximum)( ) ( )

M t t= = + -0 750 450 25 2

t =- +

-

450 450 4 25 750

2 25

2b g b gb gb g t = 1.54 h, 19.54 h

d. 3.40 m 10 liter kg

1 m 1 liter kg

3 3

30 792

2693.

= (capacity of tank)

M t t= = + -2693 750 450 25 2

t =- + -

-

450 450 4 25 750 2693

2 25

2b g b gb gb g t = 719 1081. , . h h

Expressions for M(t) are:

M(t) =

750 + 450t - 25t and (tank is filling or draining)

(tank is overflowing) (tank is empty, draining

as fast as methanol is fed to it)

2 0 719 1081 1954

2693 719 10 810 19 54 2054

RS|T||

t t

tt

. . .

( . . )( . . )

b g

11- 3

11.3 (contd)

11.4 a. Air initially in tank: N0492

0 0258=

=10.0 ft R 1 lb -mole

532 R 359 ft STP lb - mole

3

3b g . Air in tank after 15 s:

P V

P V

N RT

N RTN N

P

Pf f

ff

0 00

0

0 025802013= = = =

..

lb- mole 114.7 psia14.7 psia

lb- mole

Rate of addition: & . .n = - =02013 0 0258 0b g lb - mole air15 s

.0117 lb - mole air s

b. Balance on air in tank: Accumulation = input

dNdt

= 0 0117. lb - moles sb g ; t N= =0 00258, . lb - mole

c. Integrate balance: dN n dt N tN t

0 0258 0

0 0258 0 0117.

& . .= = + lb - mole airb g

Check the solution in two ways:

( ) = , = . lb - mole satisfies the initial condition

lb - moleair / s reproduces the mass balance

1 0 0 0258

2 0 0117

t NdNdt

= ( ) .

d. t N= = + =120 0 0258 0 0117 120 143 s lb - moles air. . .b gb g O in tank lb - mole O2 2= =0 21 143 030. . .b g

0

500

1000

1500

2000

2500

3000

0 5 10 15 20

t(h)

M(k

g)

11- 4

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).

( )( )

( ) ( )3

33

3 3

3 3

0 03.00 10

540 m 1 hAccumulation = input output m min

h 60 min

0, 3.00 10 m 0 corresponds to 8:00 AM

9.00 m 3.00 10 9.00 in minutes

w

V t t

w w

dVdt

t V t

dV dt V t dt t

n

n n

- = -

= = =

= - = + -

&

& &

b. Let &n wi = tabulated value of &n w at t i= -10 1b g i =1 2 25, , ,K

& & & & & . . . .

. .

, , , ,n n n n nw w w wi

iwi

idt

V

0

240

1 252 4

24

3 5

24

3

103

4 2103

114 98 4 1246 2 1134

2488

300 10 9 00 240 2488 2672

@ + + +LNMM

OQPP = + + +

=

= + - =

= =K Kb g b g

b g m

m

3

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 Simpsons 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 M Results: TIME (MIN) VOLUME (CUBIC M ETERS) 0.00 3000. 10.00 2974. 20.00 2944. M M 230.00 2683. 240.00 2674.

Vtrapezoid3 m= 2674 ; VSimpson

3 m= 2672 ; 2674 2672

2672100% 0 07%

- = .

Simpsons rule is more accurate.

11- 5

11.6 a. & & .

&n n

n

outV

outkV V

out

L min Lb g b g= ==

=300

60

0 200 & .n out sV= =20 0 100 L min L

b. Balance on water: Accumulation = input output (L/min). (Balance volume directly since density is constant)

dVdt

V

t V

= -

= =

20 0 0200

0 300

. .

,

c. dVdt

V Vs s= = - =0 200 0200 100. L

The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is 200 0 200 300 40 0. . ( ) . .- = - As t increases, V decreases. = -dV dt V/ . .20 0 0 200 becomes less negative, approaches zero as t . The curve is therefore concave up.

d. dV

Vdt

V t

20 0 0 200300 0. .-=

- --

FHG

IKJ =

- + = - = + -

= =

= + - =-

=

10200

200 0 20040 0

0 5 0 005 0200 100 0 200 0 0 200

101 100 101

101 100 200 02001 2000 200

265

.ln . .

.. . exp . . . exp .

.

exp .ln

..

V t

V t V t

V

t t

b g b gb g b g

b g b g L 1% from steady state

min

t

V

11- 6

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 ln

ln ln.

ln ln ln . . .

.

D bt a D ae

bD D

t t

a D bt a e

D e

bt

t

= + =

=-

=-

= -

= - = + = = =

E= -

2 1

2 1

1 18 007

0 230

755 23856 1

0230

2385 0 230 1 8 007 3000

3000

b g

b g b gb g

b. Inventory balance: Accumulation = output

dIdt

e

t I

t= -

= =

-3000

0 18 000

0 230.

, ,

kg week

kg

b g

dI e dt I e I eI

tt

t t t

18 000

0 230

0

0 230

0

0 2303000 18 000 30000 230

4957 13 043,

. . .,.

,= - - = = +- - -

c. t I= = 4957 kg

11.8 a. Total moles in room: N = =1100 m K 10 mol

295 K 22.4 m STP mol

3 3

3273

45 440b g ,

Molar throughput rate: & ,n = =700 m K 10 molmin 295 K 22.4 m STP

mol min3 3

3273

28 920b g SO balance2 ( t = 0 is the instant after the SO 2 is released into the room):

N xmol mol SO mol mol SO in room2 2b g b g = Accumulation = output.

ddt

Nx nx dxdt

xNn

b g = - = -=

=

& .,

& ,45 44028 920

0 6364

t x= = = -01545 440

330 10 5,.

,.

mol SO mol

mol SO mol2 2

b. The plot of x vs. t begins at (t=0, x=3.3010-5). When t=0, the slope (dx/dt) is

- = - - -06364 330 10 210 105 5. . . . As t increases, x decreases. dx dt x= -0 6364. becomes less negative, approaches zero as t . The curve is therefore concave up.

11- 7

c. Separate variables and integrate the balance equation:

dxx

dt x t x ex t

t

3 30 10 05

5 0 6364

5

06364330 10

06364 330 10.

.. ln.

. .

-- -

-

= -

= - =

Check the solution in two ways:

( ) /

. . ..

1

0 6364 330 10 0 63645 0 6364

t = 0, x = 3.30 10 mol SO mol satisfies the initial condition;

(2) dxdt

reproduces the mass balance.

-52

= - = - - -e xt

d. C x x e tSO2

3

3 3 22 moles mol SO 1 m

1100 m mol 10 L mol SO L= = = - - -45 440 4131 10 13632 102 6 0 6364, . . /.

i) t C= = -2 382 10 7 min mol SOliterSO

22

.

ii) x t= =

-=-

- -

1010 3 30 10

0 6364556

6 5ln .

..

e j min

e. 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.

0

t

x

11.8 (contd)

11- 8

11.9 a. Balance on CO: Accumulation=-output

N x

nP

RT

n xP

RTx

d Nxdt

P

RTx

dxdt

PNRT

x

PV NRT

dxdt V

x

t x

p

p

pp

p

( ) (

& )&

& )&

( ) & &

&

, .

mol mol CO / mol) = total moles of CO in the laboratory

Molar flow rate of entering and leaving gas: (kmol

h

Rate at which CO leaves: (kmolh

kmol COkmol

=

CO balance: Accumulation = -output

kmol COkmol

=

FHG

IKJ

= - = -FHGIKJ

E == -

= =

n

n

nn

n

0 0 01

b. dxx V

dt tV

xx

pt

rp

r

0 01 0

100.

&& ln= - = -

n

nb g

c. V = 350 m3

tr = - =-350

700100 35 10 2836ln .e j hrs

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

dMdt

m m M= - = \ =& & kg min is a constant kgb g 0 200 b. Sodium nitrate balance: Accumulation = - output x = mass fraction of NaNO3

d xMdt

xm

dxdt

mM

xm

x

t x

b g b g= -E

= - = -

= = =

& min

& &

, .

kg

200

0 90 200 045

11- 9

dxdt

, x decreases when t increases

dxdt

becomes less negative until x reaches 0;

Each curve is concave up and approaches x = 0 as t ;

increases dxdt

becomes more negative x decreases faster.

= -