Elementary Principles of Chemical Processes ch11

7/27/2019 Elementary Principles of Chemical Processes ch11

1/26

11-1

CHAPTER ELEVEN

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

xM

Mp

p=

Therefore, the leakage rate of hydrogen peroxide is /m M Mp1

b. Balance on mass: Accumulation = input output

E

=

= =

dM

dtm 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

= =

dMdt

m x m MM

kM

t M M

pp

pp

p p

,

0 0 1

00

11.2 a. Balance on H3PO4: Accumulation = inputDensity of H3PO4: = 1834. g / ml.

Molecular weight of H3PO4: M = 9800. g / mol .

Accumulation =dn

dt(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

03743

0 150 0 05 7 5

.

.

, . .t

b. dn dt n t p

n t

p

p

7 5 0

03743 7 5 03743.

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

xn

n

n

n n n

t

tp

p p

p p

= =+

=+

+0 0

7 5 0 3743

150 0 3743

. .

.

kmol H PO

kmol3 4

c. 0157 5 0 3743

150 0 3743471.

. .

..=

+

+ =

t

tt min


2/26

11-2

11.3 a. m a bt w = + t mw= =0 750, b g & t m m t w w= = = +5 1000 750 50, b g b g b gkg h hBalance on methanol: Accumulation = Input OutputM

dM

dt

m m t

dM

dtt

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 dt M t

750 0

450 50z z= 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)dM

dtreproduces the mass balance.

c.dM

dtt M= = = = + =0 450 50 9 750 450 9 25 9 27752h 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

3

07922693

.= (capacity of tank)

M t t= = + 2693 750 450 25 2

t =

+

450 450 4 25 750 2693

2 25

2b g b gb g

b 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 1081

0 1954 20 54

RS|

T||

t t

t

t

. . .

( . . )

( . . )

b g


3/26

11-3

11.3 (contd)

11.4 a. Air initially in tank: N0492

00258=

=10.0 ft R 1 lb - mole

532 R 359 ft STP

lb- mole3

3

b g

.

Air in tank after 15 s:

P V

P V

N RT

N RTN N

P

P

f f

f

f

0 00

0

0 025802013= = = =

..

lb- mole 114.7 psia

14.7 psialb- mole

Rate of addition: . .

n =

=02013 00258

0b g lb- mole air

15 s.0117 lb- mole air s

b. Balance on air in tank: Accumulation = input

dNdt

= 00117. lb - moles sb g ; t N= =0 0 0258, . lb - mole

c. Integrate balance: dN n dt N t N t

0 0258 0

0 0258 0 0117

.

. .z z= = + lb- mole airb gCheck 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 N

dN

dt

= ( ) .

d. t N= = + =120 0 0258 0 0117 120 143s lb - moles air . . .b gb g O in tank lb - mole O2 2= =0 21 143 0 30. . .b g

0

500

1000

1500

2000

2500

3000

0 5 10 15 20

t(h)

M(kg)


4/26

11-4

11.5 a. Since the temperature and pressure of the gas are constant, a volume balance on the gasis equivalent to a mole balance (conversion factors cancel).

Accumulation = Input Output

dV

dt

t V t

dV dt V t dt t

w

V

w

t

w

t

=

= = =

= = +

z z z

540 1

0 300 10 0

9 00 300 10 9 00

3

3 00 10 0

3

03

m h

h 60 minm min

m corresponds to 8:00 AM

m in minutes

33

3

3

, .

. . .

.

e j

b gb g e j

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

. . . .

. .

, , , ,

w w w wii

wii

dt

V

0

240

1 252 4

24

3 5

24

3

10

34 2

10

3114 9 8 4 124 6 2 1134

2488

300 10 9 00 240 2488 2672

z + + +L

NMM

O

QPP

= + + +

=

= + =

= =

b g b g

b gm

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, HINTEGER IDATA V0, H/3.0E3, 10./READ (5, *) (VW(I), I = 1, 25)V= V0T=0.WRITE (6, 1)WRITE (6, 2) T, VDO 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 CONTINUE1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)')2 FORMAT (F8.2, 7X, F6.0)

END$DATA11.4 11.9 12.1 11.8 11.5 11.3

Results:TIME (MIN) VOLUME (CUBIC METERS)0.00 3000.

10.00 2974.20.00 2944.

230.00 2683.240.00 2674.

Vtrapezoid3m= 2674 ; VSimpson

3m= 2672 ;2674 2672

2672100% 0 07%

= .

Simpsons rule is more accurate.


5/26

11-5

11.6 a. .

outV

outkV V

out

L min Lb g b g= ==

=300

60

0200 .out sV= =20 0 100L min L

b. Balance on water: Accumulation = input output (L/min).

(Balance volume directly since density is constant)dV

dtV

t V

=

= =

20 0 0 200

0 300

. .

,

c.dV

dtV Vs s= = =0 200 0 200 100. L

The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is20 0 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

. .=z z

FHG

IKJ=

+ = = +

= =

= + = =

1

0200

20 0 0 200

40 0

0 5 0 005 0 200 100 0 200 0 0 200

101 100 101

101 100 200 0 200

1 200

0200 265

.ln

. .

.

. . exp . . . exp .

.

exp .

ln

. .

Vt

V t V t

V

t t

b g b g

b g b g

b gb g

L 1% from steady state

min

t

V


6/26

11-6

11.7 a. A plot ofD (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

lnln

.

ln ln ln . . .

.

D bt a D ae

b D Dt t

a D bt a e

D e

bt

t

= + =

=

=

=

= = + = = =

E=

2 1

2 1

1 18 007

0 230

755 2385

6 10230

2385 0 230 1 8 007 3000

3000

b gb g b gb g

b. Inventory balance: Accumulation = output

dI

dte

t I

t=

= =

3000

0 18 000

0 230.

, ,

kg week

kg

b g

dI e dt I e I e

It

tt

tt

18 000

0 230

0

0 230

0

0 2303000 18 0003000

02304957 13043

,

. . .,.

,z z= = = +

c. t I= = 4957 kg

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

295 K 22.4 m STPmol

3 3

3

27345 440

b g,

Molar throughput rate: ,n = =700 m K 10 mol

min 295 K 22.4 m STP

mol min3 3

3

27328 920

b g

SO balance2 ( t = 0 is the instant after the SO2 is released into the room):

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

d

dtNx nx

dx

dtx

Nn

b g = = =

=

.,

,45 440

28 920

06364

t x= = = 0

15

45 440330 10 5,

.

,.

mol SO

molmol 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

= 0 6364 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.


7/26

11-7

c. Separate variables and integrate the balance equation:dx

xdt

xt x e

x tt

3 30 10 05

5 0 6364

5

06364330 10

0 6364 330 10

.

.. ln.

. .

z z=

= =

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)dx

dtreproduces the mass balance.

-52

= =

e xt

d. Cx

x et

SO2

3

3 3 22

moles mol SO 1 m

1100 m mol 10 Lmol SO L= = =

45 4404131 10 13632 102 6 0 6364

,. . /.

i) t C= = 2 382 10 7minmol SO

literSO

22

.

ii) x t= =

=

1010 330 10

06364556

6 5ln .

..

e jmin

e. The room air composition may not be uniform, so the actual concentration of the SO2in 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)


8/26

11-8

11.9 a. Balance on CO: Accumulation=-output

N x

nP

RT

n xP

RTx

d Nx

dt

P

RTx

dx

dt

P

NRTx

PV NRT

dx

dt Vx

t x

p

p

p

p

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: (kmol

h

kmol CO

kmol=

CO balance: Accumulation = -output

kmol CO

kmol

=

FHG IKJ

= = FHG

IKJ

E =

=

= =

0 0 01

b.dx

x Vdt t

Vx

xp

t

rp

r

0 01 0

100.

lnz z= =

b g

c. V = 350 m3

tr = =350

700100 35 10 2 836ln .e j hrs

d. The room air composition may not be uniform, so the actual concentration of COin parts of the room may still be higher than the safe level. Also, safe is on theaverage; 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

dM

dtm m M= = = kg min is a constant kgb g 0 200

b. Sodium nitrate balance: Accumulation = - output

x = mass fraction ofNaNO3

d xM

dtxm

dx

dt

m

Mx

mx

t x

b g b g=

E

= =

= = =

min

, .

kg

200

0 90 200 0 45


9/26

11-9

dx

dt, x decreases when t increases

dx

dt becomes less negative until x reaches 0;

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

increasesdx

dtbecomes more negative x decreases faster.

=

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Elementary Principles of Chemical Processes ch11