Assignment 10
Due April 12, 2006
Do not submit this assignment. This is for practice. Solutions will be posted on April 12.
B
is taking place in a PFR. Pure A enters the reactor at
2
a pressure of 4 atmospheres and 350 K. The heat of reaction which can be assumed
to be independent of temperature is -27 kJ/mol of A. Specific heat of A is 30 J/molK. The rate constant at 350 K is 0.2 s-1 and the activation energy is 18 kJ/mol. The
volumetric flow rate is 10 liters/s. When the conversion of A is plotted versus the
reactor volume, the following plot is obtained. Determine the temperature and
concentration of A in the reactor at the mid-point (V=50) and at the exit of the reactor
(V=100).
1. The gas-phase reaction A 
1
0.9
0.8
Conversion (X)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
Volume (liters)
60
70
80
90
100
T  T0 
H Rx X
C pA
V  50, X  0.25, T  350  27, 000  0.25 / 30  575 K
V  100, X  0.95, T  350  27, 000  0.95 / 30  1205 K
C A0 
P0
4
mol

 0.1394
RT0 0.082  350
lit
C A  C A0
(1  X ) T0
(1   X ) T
0.1394  (1  0.25) 350
mol
 0.073
(1  0.5  0.25) 575
lit
0.1394  (1  0.95) 350
mol
V  100, X  0.95, T  1205 K, C A 
 0.00386
(1  0.5  0.95) 1205
lit
V  50, X  0.25, T  575 K, C A 
2. The following liquid-phase consecutive reaction is taking place in a constant volume
batch reactor.
k1
k2
A 
 B 
C
The first reaction is first order and the second reaction is zero order. Derive expressions
for the concentrations of A, B and C as functions of time.
dC A
 k1C A
dt
C A  C A0 exp(k1t )
dCB
 k1C A  k2  k1C A0 exp(k1t )  k2
dt
CB  C A0 exp(k1t )  k2t  I ,
where, I is an integration constant
At t=0, C B  0
I  C A0
CB  C A0 (1  exp( k1t ))  k2t
CC  C A0  C A  CB
3. A first-order reversible liquid phase reaction is carried out in a CSTR at 4270C, with
the initial concentration of A being equal to 2 mol/liter.
A B
Following parameters are given:
C pA  40 J/mol K; C pB  40 J/mol K; H 300 K  80 kJ/mol
K (@ 27 0 C )  2 109 ; k f (@ 327 0 C )  1hr -1; E  30, 000 J/mole
Determine the residence time in the reactor for 80% of the equilibrium
conversion.
H Rx (TR )  1 1  80, 000  1
1 
K

ln 2 
  
  18.3274

R
K1
 T1 T2  8.3144  300 700 
K 2  21.956
1 
k2  E  1 1  30000  1


  
  0.8591

R  T1 T2  8.3144  700 600 
k1
k2  2.361 hr -1
ln
X
C 


rA  k f  C A  B   k f C A0  (1  X )  
K
K 


Find equilibrium conversion; at equilibrium,
Xe
X
 X e  0.9564
1 Xe  e 
21.956
K
80% of equilibrium conversion is =0.9852  0.8  0.7651
k1 
X A X Ae
0.7651 0.9564
 3.825

X Ae  X A 0.9564  0.7651
  1.62 hr
4. The following reaction goes to 80% completion in a CSTR. A and B are fed to the
reactor at rates of
2 mol/s and 3 mol/s respectively at a temperature of 370C. Specific heats (in J/molK) of A, B, C and D are 200, 150, 220 and 180 respectively.
A B C  D
The reactor is jacketed by water at a temperature of 200C. The overall heat transfer
coefficient has been estimated at 300 J/(m2.s.K), while the heat transfer area is 0.75
m2. Mixing is ensured through an agitator, which contributes a work of 12 KW to the
reactor. The heat of reaction is -30,000 J/mole of A at 300 K. Calculate the steady
state temperature in the reactor.
Use equation 8-50 from the text
Q W
 X H Rx =  iC pi (T  Ti 0 )
FA0
(U  A  (Ta  T )  W )
 X H Rx (TR )  Cˆ p (T  TR ) =  iC pi (T  Ti 0 )
FA0

T



FA0 X H Rx (TR )  Cˆ p  TR  FA0 ( AC pA   B C pB )T0  UATa  W
FA0 ( AC pA   B C pB )  UA  FA 0 X Cˆ p
J
mol  K
2.0  0.8  30, 000  50  300   2  (1  200  1.5 150)  310  300  0.75  293  12000
T
2  (1 200  1.5 150)  300  0.75  2  0.8  50
T  357.94 K
Cˆ p  220  180  200  150  50
5. A first-order, reversible reaction A  B is taking place adiabatically. The
equilibrium constant at 400 K is 3 and the heat of reaction, which can be assumed
constant is –50 kJ/mole. Pure A with a specific heat of 100 J/mole-K (which is the
same as that of B) enters the reactor at 300 K. The equilibrium conversion as a
function of temperature is plotted in the figure shown below.
1.2
Equilibrium Conversion
1
0.8
0.6
0.4
0.2
0
300
350
400
450
500
Temperature
550
600
650
a. Show the equations that were used for constructing the plot shown.
b. What is the maximum conversion that can be achieved, if the reactor is
operated adiabatically.
c. If two adiabatic reactors are used, with inter-stage cooling, and if the first
reactor exit stream is cooled to 350 K, what maximum conversion can be
achieved in the second reactor? What would be the exit temperature from this
second stage?
1.4
1.2
1
0.8
Series1
Series2
Series3
0.6
0.4
0.2
0
300
350
400
450
500
550
600
650
6.
A first-order irreversible liquid-phase reaction ( A  B ) is taking place in a CSTR.
Reactant A enters the reactor at 300 K at a concentration of 3 mol/liter. The reactor
volume is 18 liters and flow rate of pure A is 0.06 liters/s. Other parameters are:
0
H Rx
 5000 cal/mole, C pA  C pB  30 cal/mole  K
k  exp(15.32 
7550 -1
)s
T
0
 X ) is plotted in the figure shown.
The heat generated term given by (H Rx
700
6000
Heat Generation Function
5000
4000
3000
2000
1000
0
250
300
350
400
450
Temperature (K)
a. For the conditions given, determine the number of steady states, and the
conversions and temperatures at these steady states.
b. What is the minimum incoming temperature at which multiple steady states are
possible?
c. What is the maximum incoming temperature above which only one steady state
occurs?
Using CSTR Mole balance
1
X MB 
1
1
k
Material balance for an adiabatic CSTR
(H Rx X EB )
T  T0 
C pA
1.4
1.2
Xeb
Xmb
Xeb or Xmb
1
0.8
0.6
0.4
0.2
0
300
350
400
T
450
500