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1. The gas-phase reaction
2
B
A  is taking place in a PFR. Pure A enters the reactor at
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/mol-
K. 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).
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 10 20 30 40 50 60 70 80 90 100
Volume (liters)
Conversion
(X)
0
0
0
0
0
0
50, 0.25, 350 27,000 0.25/30 575 K
100, 0.95, 350 27,000 0.95/30 1205 K
4 mol
0.1394
0.082 350 lit
(1 )
(1 )
0.1394 (1 0.25) 350
50, 0.25, 575 K,
(1 0.5 0.25) 575
Rx
pA
A
A A
A
H X
T T
C
V X T
V X T
P
C
RT
T
X
C C
X T
V X T C


 
     
     
  




 
    
 
mol
0.073
lit
0.1394 (1 0.95) 350 mol
100, 0.95, 1205 K, 0.00386
(1 0.5 0.95) 1205 lit
A
V X T C
 
    
 
2. The following liquid-phase consecutive reaction is taking place in a constant volume
batch reactor.
1 2
k k
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.
1
0 1
1 2 1 0 1 2
0 1 2
B
0
0 1 2
0
exp( )
exp( )
exp( ) ,
where, I is an integration constant
At t=0, C 0
(1 exp( ))
A
A
A A
B
A A
B A
A
B A
C A A B
dC
k C
dt
C C k t
dC
k C k k C k t k
dt
C C k t k t I
I C
C C k t k t
C C C C
 
 
    
    


   
  
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:
300
0 9 0 -1
40J/mol K; 40 J/mol K; 80 kJ/mol
(@27 ) 2 10 ; (@327 ) 1hr ; 30,000 J/mole
pA pB K
f
C C H
K C k C E
    
   
Determine the residence time in the reactor for 80% of the equilibrium
conversion.
2
1 1 2
2
2
1 1 2
-1
2
0
( ) 1 1 80,000 1 1
ln 18.3274
8.3144 300 700
21.956
1 1 30000 1 1
ln 0.8591
8.3144 700 600
2.361 hr
(1 )
Find equilibriu
Rx R
B
A f A f A
H T
K
K R T T
K
k E
k R T T
k
C X
r k C k C X
K K
 
   
     
   
 
 

 
   
    
   
 
 

   
     
   
   
1
m conversion; at equilibrium,
1 0.9564
21.956
80% of equilibrium conversion is =0.9852 0.8 0.7651
0.7651 0.9564
3.825
0.9564 0.7651
1.62 hr
e e
e e
A Ae
Ae A
X X
X X
K
X X
k
X X


    
 

  
 

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/mol-
K) 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.
 
 
i 0
0
i 0
0
0 0 0
0 0
Use equation 8-50 from the text
= ( )
( ( ) ) ˆ
( ) ( ) = ( )
ˆ
( ) ( )
ˆ
( )
ˆ
Rx pi i
A
a
Rx R p R pi i
A
A Rx R p R A A pA B pB a
A A pA B pB A p
Q W
X H C T T
F
U A T T W
X H T C T T C T T
F
F X H T C T F C C T UAT W
T
F C C UA F X C
C


 
 

  
   
     
       

   



 
220 180 200 150 50
2.0 0.8 30,000 50 300 2 (1 200 1.5 150) 310 300 0.75 293 12000
2 (1 200 1.5 150) 300 0.75 2 0.8 50
357.94 K
p
J
mol K
T
T
    

            

        

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.
0
0.2
0.4
0.6
0.8
1
1.2
300 350 400 450 500 550 600 650
Temperature
Equilibrium
Conversion
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?
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
-1
5000 cal/mole, 30 cal/mole K
7550
exp(15.32 ) s
Rx pA pB
H C C
k
T
     
 
The heat generated term given by 0
( )
Rx
H X
  is plotted in the figure shown.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
300 350 400 450 500 550 600 650 700
Series1
Series2
Series3
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?
0
Using CSTR Mole balance
1
1
1
Material balance for an adiabatic CSTR
( )
MB
Rx EB
pA
X
k
H X
T T
C




 
0
1000
2000
3000
4000
5000
6000
250 300 350 400 450
Temperature (K)
Heat
Generation
Function
0
0.2
0.4
0.6
0.8
1
1.2
1.4
300 350 400 450 500
T
Xeb
or
Xmb
Xeb
Xmb

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Hw10sol

  • 1. 1. The gas-phase reaction 2 B A  is taking place in a PFR. Pure A enters the reactor at 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/mol- K. 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). 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 10 20 30 40 50 60 70 80 90 100 Volume (liters) Conversion (X)
  • 2. 0 0 0 0 0 0 50, 0.25, 350 27,000 0.25/30 575 K 100, 0.95, 350 27,000 0.95/30 1205 K 4 mol 0.1394 0.082 350 lit (1 ) (1 ) 0.1394 (1 0.25) 350 50, 0.25, 575 K, (1 0.5 0.25) 575 Rx pA A A A A H X T T C V X T V X T P C RT T X C C X T V X T C                                 mol 0.073 lit 0.1394 (1 0.95) 350 mol 100, 0.95, 1205 K, 0.00386 (1 0.5 0.95) 1205 lit A V X T C          2. The following liquid-phase consecutive reaction is taking place in a constant volume batch reactor. 1 2 k k 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. 1 0 1 1 2 1 0 1 2 0 1 2 B 0 0 1 2 0 exp( ) exp( ) exp( ) , where, I is an integration constant At t=0, C 0 (1 exp( )) A A A A B A A B A A B A C A A B dC k C dt C C k t dC k C k k C k t k dt C C k t k t I I C C C k t k t C C C C                        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: 300 0 9 0 -1 40J/mol K; 40 J/mol K; 80 kJ/mol (@27 ) 2 10 ; (@327 ) 1hr ; 30,000 J/mole pA pB K f C C H K C k C E         
  • 3. Determine the residence time in the reactor for 80% of the equilibrium conversion. 2 1 1 2 2 2 1 1 2 -1 2 0 ( ) 1 1 80,000 1 1 ln 18.3274 8.3144 300 700 21.956 1 1 30000 1 1 ln 0.8591 8.3144 700 600 2.361 hr (1 ) Find equilibriu Rx R B A f A f A H T K K R T T K k E k R T T k C X r k C k C X K K                                                            1 m conversion; at equilibrium, 1 0.9564 21.956 80% of equilibrium conversion is =0.9852 0.8 0.7651 0.7651 0.9564 3.825 0.9564 0.7651 1.62 hr e e e e A Ae Ae A X X X X K X X k X X                 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/mol- K) 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.
  • 4.     i 0 0 i 0 0 0 0 0 0 0 Use equation 8-50 from the text = ( ) ( ( ) ) ˆ ( ) ( ) = ( ) ˆ ( ) ( ) ˆ ( ) ˆ Rx pi i A a Rx R p R pi i A A Rx R p R A A pA B pB a A A pA B pB A p Q W X H C T T F U A T T W X H T C T T C T T F F X H T C T F C C T UAT W T F C C UA F X C C                                       220 180 200 150 50 2.0 0.8 30,000 50 300 2 (1 200 1.5 150) 310 300 0.75 293 12000 2 (1 200 1.5 150) 300 0.75 2 0.8 50 357.94 K p J mol K T T                               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. 0 0.2 0.4 0.6 0.8 1 1.2 300 350 400 450 500 550 600 650 Temperature Equilibrium Conversion
  • 5. 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? 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 -1 5000 cal/mole, 30 cal/mole K 7550 exp(15.32 ) s Rx pA pB H C C k T         The heat generated term given by 0 ( ) Rx H X   is plotted in the figure shown. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 300 350 400 450 500 550 600 650 700 Series1 Series2 Series3
  • 6. 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? 0 Using CSTR Mole balance 1 1 1 Material balance for an adiabatic CSTR ( ) MB Rx EB pA X k H X T T C       0 1000 2000 3000 4000 5000 6000 250 300 350 400 450 Temperature (K) Heat Generation Function
  • 7. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 300 350 400 450 500 T Xeb or Xmb Xeb Xmb