CHFEN 3553 Chemical Reaction Engineering
April 14, 2003; 12:55 PM – 1:45 PM
Answer all questions
1. The value of the overall heat transfer coefficient U is 20
Btu
Joules
. Find the value of U in 2
.
0
ft  hour  F
m s  K
2
15 points
2
0
Btu
J
1 h
F
2 ft

1055.6

3.28


1.8
2
0
2
ft  hr  F
Btu
m 3600 s
K
Joules
 113.6 2
m s  K
U  20
No partial credit
2. The following multiple reactions are taking place in a reactor system.
2 A  B  C , r1 A  k1C A2
A  C  D, r2 A  k2C ACC
A  D  E , r3 A  k3C ACD
Write down the composite rates for species A, B, C, D and E.
15 points
rA  k1C A2  k2C ACB  k3C ACD 4 points
rB  k1C A2 / 2 2 points
rC  k1C A2 / 2  k2C ACC 4 points
rD  k2C ACC  k3C ACD 3 points
rE  k3C ACD 2 points
3. The following liquid-phase consecutive reaction is taking place in a PFR.
k1
k2
A 
 B 
C
The reactor volume is 100 liters, volumetric flow rate is 20 liters/second and CA0=0.2 mol/liter. The rate
constants k1 and k2 are 0.5 s-1 and 0.3 s-1 respectively. Determine the concentrations of A, B and C at the exit of
the reactor.
20 points
dC A
 k1C A ; C A  C A0e  k1
2 points
d
dCB
 k1C A  k2CB  k1C A0 e  k1  k2CB
4 points
d
dCB
 k2CB  k1C A0 e  k1 Integrating factor is e k2 - Multiply both sides
d
dCB
e k2
 e k2 k2CB  k1C A0 e( k2 k1 )
d
d
CB e k2   k1C A0 e( k2  k1 )

d
k1
CB e k2 
C A0e( k2  k1 )  C1 where C1 is the integration constant
k2  k1
CB  0 at  =0, C1  
CB 
k1
k2  k1
k1
C A0  e  k1  e  k2  5 points Solution was given in class and is also on page 293 of the text.
k2  k1
CC  C A0  C A  CB
C A  0.01642
3 points
mol
mol
mol
; CB  0.0705
; CC  0.1131
lit
lit
lit
6 points
4. This first order reversible reaction takes place in a flow reactor.
A B
The equilibrium constant at 300 K is 35 and the heat of reaction is –5,000 cal/mol. A conversion of 80% is
desired. The engineer designing the reactor contends that you need to operate below 1250 C to achieve this
conversion. Verify is this is correct or incorrect.
20 points
 H Rx  1 1  
K e (T )  K e (T1 ) exp 
  
 R  T1 T  
 5000  1
1 
K e (T )  35  exp 
 

 1.987  300 T  
 4.437 at 398 K
8 points

C 
First-order rate law:  rA  k  C A  B 
Ke 

At equilibrium,  rA  0  C Ae  CBe / K e
C A0 (1  X e ) 
C A0 X e
Ke
 Xe 
Ke
1  Ke
X e  0.816
4 points
8 points
Above this temperature, X will decrease and fall below 0.8. Hence, the contention is correct.
5. The following reaction goes to 75% completion in a 300 liter CSTR. A and B are fed to the reactor at
identical rates of 1.5 mol/s at a temperature of 270C. Specific heats (in J/mol-K) of A, B and C are 200, 180
and 190, respectively.
A  B  2C
The reactor is jacketed by water at a temperature of 200C. The overall heat transfer coefficient has been
estimated at 225 J/(m2.s.K), while the heat transfer area is 1 m2. Mixing is ensured through an agitator, which
contributes a work of 15 KW to the reactor. The heat of reaction is -35,000 J/mole of A. Calculate the steady
state temperature in the reactor by assuming that the heat of reaction is independent of temperature.
30 points
Use equation 8-50 from the text
Q W
 X H Rx =   iC pi (T  Ti 0 )
FA0
T
T
FA0 X (H Rx )  F0 (C pA  C pB )T0  UATa  W
F0 (C pA  C pB )  UA
12 points
1.5  0.75  (35000)  1.5  (200  180)  300  225 1  293  15000
1.5  (200  180)  225 1
=366.4 K
Recognition of parameters and solution – 18 points