CP303 – Assignment for self study (January – May 2012)
Solutions:
1) Calculate the following:
a) Conversion and heat duty for an isothermal reactor operating at 298 K.
Calculation of conversion for an isothermal reactor operating at 298 K:
Mass balance for A over CSTR gives FA0  FAf  V ( rA ) exit
Since ( rA ) exit  kCAf , we get v0C A0  v f C Af  VkCAf , where v0 is the volumetric flow rate at
the inlet and v f is the volumetric flow rate at the exit.
FAf  FA0 v f C Af  v0C A0

FA0
v0C A0
For a liquid-phase reaction, we can assume v0 = v f . Therefore, we get C Af  C A0 (1  x Af ) .
Thus, the mass balance for A over the reactor becomes v0 x Af  Vk (1  x Af ) , which gives
Conversion is defined as x Af 
1  x Af
x Af
v
1
where   V / v0 is the space time.
 0 
Vk k
k
1

1  k 1  1
k
Since k(298) = 1.7 x 10-4 s-1 and  = (0.5/0.86) x 103 s = 0.581 x 103 s,
Therefore, we get x Af 
(1.1)
We get k = 0.0988
Thus, x Af at 298 K = 0.0988 / 1.0988 = 0.09
Calculation of heat duty for an isothermal reactor operating at 298 K:
For the data given, we could start with equation (8.18) from the Lecture Note Set 8, which is
Q  W
 m C (T  T )  F x H (@ T )
in


shaft in
T0 p
f
0
A0 Af
R
f
Neglecting shaft work and using m T 0  v0  , the above equation can be reduced to
Q  v C (T  T )  v C x H (@ T )
in
0
p
f
0
0 A0 Af
R
f
For an isothermal operation, (1.2) reduces to
Q in  v0C A0 x Af H R (@ 298)
 (0.86 x 10- 3 m 3/s)  (2.0 kmol/m 3 )  (0.09)  ( 167,500 kJ/kmol of A reacted )
 25.9 kJ/s  26 kW
That is, heat must be removed at a rate of 26 kW to maintain isothermal condition.
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(1.2)
b) Conversion and reactor temperature for an adiabatic reactor with inlet temperature of
298 K.
(1.1) gives the mass balance where k is a function of temperature described by the Arrhenius Law as
 E 
k (T )  k 0 exp  
.
 RT 
Since k(@298) is given as 1.7 x 10-4 s-1, we can determine k0 as follows:
E 
k (298)
1.7  10 4

k ( 298)  k 0 exp  
which
gives
k


s-1
0

E 
41,870 


 298 R 
exp 
exp 

 298R 
 298  8.314 
Therefore,
1 
1  -1
 E1
 41,870  1
4
(1.3)
k (T )  k (298) exp   
  1.7  10 exp 
 
 s
 R  T 298 
 8.314  T 298 
Heat balance given by (1.2) for an adiabatic case which negligible shaft work reduces to
v0 C p (T f  T0 )  v0C A0 x Af H R (@ T f )  0 .
Since an average specific heat capacity is assumed, from Lecture Note Set 8, we could see that the
heat of reaction can be taken to remain a constant. Therefore, we get
C p (T f  T0 )
x Af 
C A0 H R (@ T f )

(1050 kg/m 3 )  (4.19 kJ/kg.K) (T f  T0 ) K
( 2.0 kmol/m 3 )  ( 167,500 kJ/kmol of A reacted )

(1050)  (4.19) (T f  T0 )
( 2.0)  ( 167,500)
x Af  0.0131(T f  T0 )
(1.4)
Plotting (1.1) [with k given by (1.3)] and (1.4) [with T0 = 298 K] in the same graph we get
0.25
(x_Af)_mass balance
0.2
(x_Af)_heat balance
0.15
0.1
0.05
0
295
300
305
Temperature (K)
2
310
315
Numerical data related to the plot is
T
298
300
302
304
306
308
310
310.5
311
311.5
312
k
0.00017
0.00019
0.000213
0.000237
0.000264
0.000294
0.000327
0.000336
0.000345
0.000354
0.000363
(x_Af)_mass balance
0.08994709
0.09960537
0.110028968
0.12124038
0.13325719
0.146091301
0.159748213
0.163290984
0.166885022
0.170530201
0.174226361
(x_Af)_heat balance
0
0.0262
0.0524
0.0786
0.1048
0.131
0.1572
0.16375
0.1703
0.17685
0.1834
The temperature is 310.5 K and the conversion is 0.163 for the adiabatic case.
c) Conversion and preheating temperature for an adiabatic reactor with a reactor
temperature of 363 K.
Mass balance would remain the same as (1.1), in which k must be calculated at the reactor
temperature of 363 K as
1  -1
 41,870  1
-1
k (363)  1.7  10  4 exp 


 s  0.003505 s
8.314
363
298



k
1
(1.1) therefore gives x Af 

 0.67
1
1  k 1 
0.581 x 103 s  0.003505 s-1



The inlet temperature can be calculated using the heat balance given by (1.4) as follows:
T0  T f 
x Af
0.0131
 363 
0.67
 311.86  312 K
0.0131
Conversion is 0.67 and preheating temperature is 312 K for an adiabatic reactor with a
reactor temperature of 363 K.
d) Conversion and heat duty if the reactor is operated non-adiabatically without preheating
and at a temperature of 363 K.
Conversion is as the same as in case (c) since it is influenced by the reactor temperature which is the
same as in (c). Heat duty shall be calculated using (1.2) as follows:


 1050 kg/m 3  4.19 kJ/kg.K   (363  298) K
Q in  0.86 x 10- 3 m 3/s 
 2.0 kmol/m 3  0.67    167,500 kJ/kmol of A reacted





 0.86 x 10- 3 1050   4.19   (363  298)  2.0 0.67  167,500 kW  53 kW
Conversion is 0.67 and heat duty is 53 kW if the reactor is operated non-adiabatically without
preheating and at a temperature of 363 K.
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Tabulating the results:
Part
Feed
Temperature,
T0 K
Reactor
Temperature,
Tf K
Conversion
of A attained,
Heat duty of
the reactor,
x Af
Q in kW
Observations
a)
298
298
0.09
26
b)
298
310.5
0.163
0
c)
312
363
0.67
0
d)
298
363
0.67
53
For isothermal operation of CSTR
at 298 K, we reach only 9%
conversion of A, which is very low.
For adiabatic operation with feed
stream at 298 K, the reactor
temperature stabilizes at 310.5 and
conversion of A stabilizes at
16.3%, which is still low.
For adiabatic operation with
preheating feed stream to 312 K so
as to maintain the reactor
temperature at 363, conversion of
A stabilizes at 67%.
Heat required to heat the feed
stream from 298 K to 312 K is
given below.*
Without preheating the feed
stream, we could heat the reacting
mixture to 363K in a non-adiabatic
operation of the CSTR.
Conversion of A stabilizes at 67%
as in part (c).
Heat required to heat the reacting
mixture is the same as the heat
required to heat the feed stream
from 298 K to 312 K.
* Heat required to heat the feed stream from 298 K to 312 K
= v0 C p (312  298)
 (0.86 x 10-3 m3/s)  (1050 kg/m 3 )  (4.19 kJ/kg.K)(312  298)K
 (0.86 x 10-3 m3/s)  (1050 kg/m 3 )  (4.19 kJ/kg.K)(312  298)K
= 53 kW
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2) This problem will be worked out using the notation used in Lecture Note Set 10.
k
Adsorption reaction can be written as A( g )  S  A  S , which is said to be at rapid equilibrium.
k
Therefore, forward reaction rate equals the backward reaction rate. Hence we write the following:
(2.1)
kpACv  k C A S
kS
Surface reaction A  S  A( g )  A2 ( g )  S is rate limiting, and therefore rate of production of A2 is
given as
(2.2)
rA2  rAS  rA  k S C AS p A
Since C AS in (2.2) is a variable, it must be converted to a constant. To do that, first we write the
site balance as
Ct  Cv  C AS
(2.3)
Eliminating Cv from (2.1) and (2.3), we get
 k

 1

k C
Ct   A S  C A S     1C A S  
 1C A S
kpA
 kpA

 K A pA

where K A  k / k  is the adsorption equilibrium constant.
Eliminating C A S from (2.2) and (2.4), we get
k S K AC t p 2A
rA2  k S
pA 
1  K A pA
 1


 1
 K A pA

Ct
Relating partial pressure of A to the concentration of A, we get pA = CART.
Therefore the rate expression becomes,
2
k S K ACt ( RT ) 2 C A
rA2 
1  K A RTC A
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(2.4)
3) It is desired to make a product X-Y via the following reaction:
X-OH + Y-H → X-Y + H2O
An equimolar feed of liquid X-OH and Y-H at 25oC are fed to a CSTR.
At 25oC, where all four material species are liquids.
At 25oC, the heat of reaction ∆Hrxn = -200 kJ/mole.
At 25oC, the heat capacity of each liquid-phase species is 4 kJ/(kg oC).
The molecular weight of X-OH is 150 g/mole, and the molecular weight of Y-H is 100 g/mole.
The temperature inside the reactor (T) is controlled by putting the reactor in thermal contact with a
fluid flowing over the outside of the reactor at temperature Ta.
To a good approximation, the heat transfer rate (Q, in watts) from the fluid flowing over the outside
the reactor to the contents of the reactor is given by the linear expression: Q = UA(Ta-T)
a) If the reaction is carried out with the reactor at steady-state at the inlet temperature of
25oC, is T greater than, less than, or equal to Ta?
For an exothermic reaction, to maintain the products at the same temperature as the reactants one
must remove heat. So T must be greater than Ta, i.e. Ta must be below room temperature (25oC).
b) When running the reactor at T = 25oC to 50% conversion, the productivity is
unacceptably low. To try to accelerate the reaction, it is decided to increase the steadystate reactor temperature to T = 105oC. At this temperature, all of the H2O formed
evaporates, but the other species are still liquids. The heat of vaporization of H2O at 105oC
is +40 kJ/mole. When T = 105oC, the reaction runs to 50% conversion 10x faster than it
did at 25oC, so we increase the flow rates until the reactor is making 10x as much product
as it did at 25oC (still at 50% conversion). When we achieve the new steady-state highproductivity operation at 105oC, will the magnitude of Q (i.e. |Q|) be larger, smaller, or the
same as it was when we were operating at 25oC? At this steady-state condition, is T greater
than, less than, or equal to Ta?
When running the reactor at T = 25oC (i.e., isothermal conditions) to 50% conversion, the heat
requirment can be calculated as follows:
Q in isothermal  Heat released by the reaction
 FA0 x Af H R (@ 298)  ( FA0 mole/s)  (0.5)  ( 200 kJ/mole)  -100 FA0 kW
Note that so much heat will be removed from the reactor.
When running the reactor at T = 105oC (i.e., non-isothermal conditions) to 50% conversion, the heat
required can be calculated as follows:
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Q in nonisothermal
 Heat required to heat up X - OH stream from 25o C to 105o C
 Heat required to heat up Y - H stream from 25o C to 105o C
 Heat released by the reaction  Heat required to vapourise the water produced in the reaction
Data provided states that the feed flow rate is 10 times more than the isothermal case and the
conversion remains 50%.
Q in nonisothermal
 (10 FA0 mole/s)  ( 4 kJ/kg  o C)  (105  25) o C  (10 FB 0 mole/s)  ( 4 kJ/kg  o C)  (105  25) o C
 (10 FA0 mole/s)  ( x Af )  H R (@ 298)  (10 FA0 mole/s)  x Af  ( heat of vapouriza tion of water)
Since the feed is equimolar in X-OH and Y-H,
Q in nonisothermal
 (10 FA0  0.25 kg/s)  ( 4 kJ/kg)  (80)  (10 FA0  0.10 kg/s)  ( 4 kJ/kg)  (80)
 (10 FA0 mole/s)  (0.5)  ( 200 kJ/mole)  (10 FA0 mole/s)  (0.5)  ( 40 kJ/mole)
 10  (0.15  0.1)  4  80  1000  200FA0 kW
 800  1000  200FA0 kW
0
Note that the heat released by the reaction balances the heat required to warm up the mixture and
evaporate the water. Therefore, T should approximately equal Ta.
In contrast, when we ran the reactor at 298 K, we had to remove heat at a rate of 200 kJ/kg.
c) Since operating hot improved our productivity, but conversion is still pretty low, the
operator tries to improve things by cranking up the temperature, preheating the inlet
streams to 185oC and increasing Ta. For good measure the operator simultaneously cranks
up the reactor pressure from 1 bar to 100 bar; at this high pressure all the species remain
as liquids. (The reactor is safe at this condition, and even up T = 300oC.) Curiously, the
conversion and productivity of the reactor do not increase under these severe conditions,
instead they decrease. Propose an explanation for this experimental observation.
It is possible that we are running into equilibrium limitations on the reaction. For exothermic
reactions, Keq decreases with increasing temperature. So if we are equilibrium limited at high T, we
would expect the conversion to decrease as T increases. (Note that by increasing the pressure so
dramatically, the operator prevented most of the H2O from evaporating, hence the concentration of
H2O in the liquid phase is probably much higher now than it was in the 105oC case where the water
evaporated; this contributes to the equilibrium limitation).
d) Your manager gives the operator who turned up the temperature (without doing any
calculations first) a formal reprimand, saying the operator is probably lucky that the
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conversion went down instead of increasing. Why do you think the manager was happy
that conversion was low instead of increasing a lot?
Exothermic reactions can “run away” if T goes high enough, i.e. past a certain point, the steadystate conversion will suddenly jump from a low number to a very high conversion, releasing
essentially the whole exothermicity, and jumping the T to a temperature above the safety limits of
the reaction vessel. In the present case, if the reverse reaction were negligible the reaction would
release 800 kJ/kg, enough to increase the temperature inside the reactor by up to 200 degrees, so it
could have exceeded the T = 300oC safety limit. The results could have been fatal to the operators
or anyone else nearby. But fortunately in this case it was lucky that high conversion was not
achievable due to the small value of the equilibrium constant.
Acknowledgement: MIT open courseware
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