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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
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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
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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)
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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.
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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
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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.
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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)
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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
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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.
=