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L15-1
L15: Nonisothermal Reactor
Example Problems
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Multiple Steady States in
CSTR
1
L15-2
XA,EB
XA,MB
0.8
XA
0.6
0.4
0.2
0
0
100
200
300
400
500
600
T (K)
• Plot of XA,EB vs T and XA,MB vs T
• Intersections are the T and XA that satisfy both mass balance (MB) &
energy balance (EB) equations
• Each intersection is a steady state (temperature & conversion)
• Multiple sets of conditions are possible for the same reaction in the same
reactor with the same inlet conditions!
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-3
Review: Heat Removal Term R(T) & T0
Heat removed: R(T)
Heat generated G(T)
 r V 
Cp0 1     T  TC   HRX  A 
 FA0 
R(T) line has slope of CP0(1+)
  UA Cp0FA0 Tc   Ta  T0
1 
R(T)
=∞
=0
R(T)
Increase 
Increase T0
T
When T0 increases, slope stays
same & line shifts to right
For Ta < T0
Ta
T0
T
When  increases from lowering
FA0 or increasing heat exchange,
slope and x-intercept moves
Ta<T0: x-intercept shifts left as ↑
Ta>T0: x-intercept shifts right as ↑
=0, then TC=T0 =∞, then TC=Ta
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-4
Review: CSTR Stability
G(T)
Tc 
R(T)
 Ta  T0   UA Cp0FA0
1 
R(T) > G(T) → T
falls to T=SS1
G(T) > R(T)
→ T rises to
T=SS1
G(T) > R(T)
→ T rises to
T=SS3
2
Heat removed: R(T)
1
3
R(T) > G(T)
→T falls to T=SS3
Heat generated G(T)
 rA V 

 FA0 
Cp0 1     T  TC   HRX 
T
• Magnitude of G(T) to R(T) curve determines if reactor T will rise or fall
• G(T) = R(T) intersection, equal rate of heat generation & removal, no
change in T
• G(T) > R(T) (G(T) line above R(T) on graph): rate of heat generation > heat
removal, so reactor heats up until a steady state is reached
• R(T) > G(T) (R(T) line above G(T) on graph): rate of heat generation < heat
removal, so reactor cools off until a steady state is reached
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow L15-5
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Extra info: liquid phase rxn CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol
HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol
k  0.01 dm3 mol  s at 300K
Strategy:
1. Mole balance
2. Rate Law
3. Stoichiometry
4. Combine rate law & stoichiometry
5. Energy balance
6. Solve
6a. Solve CSTR
i.
Use EB to find T as a
function of XA
ii. Calculate V using the CSTR
design eq with k calculated
at that T
6b. Solve PFR
i.
Use EB to construct table of T as a
function of XA
ii. Use k = Ae-E/RT to construct table of k
as function of T & therefore XA
iii. Calculate -rA as a function of T & XA
iv. Calculate FA0/-rA for each T
v. Use numeric technique to calculate V
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-6
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol
HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol
k  0.01 dm3 mol  s at 300K
CSTR
 Mole balance
 Rate law
F X
VCSTR  A0 A
rA
PFR
0.85  F
A0

VPFR   
dX A

0  rA 
rA  kCACB
dm3
1 
 10000cal mol  1
k  0.01
exp 



mol  s
1.987cal
mol

K
300K
T



 Stoichiometry
 Combine:
CB  CA  CA0 1  X A 
10000
1 

 1
2
2
rA  0.01exp 

C
1

X



  A0
A
1.987cal mol  K  300K T  
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-7
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol
HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol
k  0.01 dm3 mol  s at 300K
 Combine MB, rate law & stoichiometry for CSTR:
CSTR
 VCSTR 
F X
V  A0 A
rA
CA0u0 X A
dm3
1 
 10000cal mol  1
2
2
0.01
exp 

C
1

X



  A0
A
mol  s
1.987cal mol  K  300K T  
 VCSTR 
u0 XA
dm3
1 

2
 1
0.01
exp 5032.7K 
   CA0 1  X A 
mol  s
 300K T  

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-8
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol
HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol
k  0.01 dm3 mol  s at 300K
 Energy balance
n
0  Q  Ws  FA0  iCp,i  T  Ti0   HRX (T)FA0 X A
Solve for T:
i1
0 0
n
 FA0  iCp,i  T  Ti0   HRX (T)FA0 X A
i1
HRX
Substitute:
 H RX (TR )  CP  T  TR 
n
  iCp,i  T  Ti0     H RX (TR )  CP  T  TR   X A


i1
Multiply out quantities in brackets, bring T to 1 side of equation, factor
out T, divide by quantity in bracket:
n
Evaluate Cp:
 iCp,iTi0  H RX (TR )X A  CPTR X A C  C  C  C
P
pC
pA
pB
 T  i1
n
cal
CP   30  15  15 
0
 iCp,i  CP X A
mol  K
i1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-9
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol
HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol
k  0.01 dm3 mol  s at 300K
 Energy balance
n
 iCp,iTi0  H RX (TR )X A  CPTR XA
T  i1
n
 iCp,i  CP X A
i1
Evaluate SiCp:
FB0


1
A  1
B
FA0
CP  0
T  Ti0 
H RX (TR )X A
n
 iCp,i
i1
n
cal
cal
 30
 iCpi CpA  CpB  15  15 
mol  K
mol  K
i1
Evaluate H°RX(TR):
cal
cal
H RX  TR   HC  HA  HB   41000    20000    15000  
 6000
mol
mol
Simplify EB:
T  300K 
  6000 cal mol  X A
 T  300K   200K  X A
30 cal mol  K
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-10
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol
HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol
k  0.01 dm3 mol  s at 300K
Solve for CSTR volume: use EB to find T when XA=0.85
T  300K   200K  X A
VCSTR 
 T  300K   200K  0.85  T  470K
u0 XA
dm3
1 

2
 1
0.01
exp 5032.7K 
   CA0 1  X A 
mol  s
 300K T  

 dm3 
 2
 0.85
s 

 VCSTR 
dm3
1  
mol 

2
 1
0.01
exp 5032.7K 

0.1
1

0.85


 

mol  s
 300K 470K   

dm3 
 VCSTR  175L
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-11
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA (We’re
interested in range where XA = 0 to XA = 0.85)
T  300K   200K  X A
XA
T (K)
0
300
T  300K   200K  0  300K
k (dm3/mol·s) -rA (mol/dm3·s)
FA0/-rA (dm3)
0.2
0.4
0.6
0.8
0.85
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-12
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA -Temperature
range should cover XA = 0 to XA = 0.85
T  300K   200K  X A
ii.
XA
T (K)
0
300
0.2
340
0.4
380
0.6
420
0.8
460
0.85
470
k (dm3/mol·s) -rA (mol/dm3·s)
FA0/-rA (dm3)
Calculate k(T) for each T in the table
dm3
1 

 1
k  0.01
exp 5032.7K 
 
mol  s
 300K T  

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-13
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA -Temperature
range should cover XA = 0 to XA = 0.85
T  300K   200K  X A
ii.
k (dm3/mol·s) -rA (mol/dm3·s)
XA
T (K)
0
300
0.01
0.2
340
0.072
0.4
380
0.34
0.6
420
1.21
0.8
460
3.42
0.85
470
4.31
FA0/-rA (dm3)
Calculate k(T) for each T in the table
dm3
1 

 1
k  0.01
exp 5032.7K 
 
mol  s
 300K T  

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-14
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA -Temperature
range should cover XA = 0 to XA = 0.85
ii. Calculate k(T)
iii.
k (dm3/mol·s) -rA (mol/dm3·s)
XA
T (K)
0
300
0.01
0.2
340
0.072
0.4
380
0.34
0.6
420
1.21
0.8
460
3.42
0.85
470
4.31
FA0/-rA (dm3)
Calculate –rA each XA and k in the table
rA  kCA02 1  XA 
2
2
dm3 
mol 
mol
2
rA  0.01
0.1
1

0

0.0001


3
mol  s 
dm 
dm3  s
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-15
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA -Temperature
range should cover XA = 0 to XA = 0.85
ii. Calculate k(T)
iii.
k (dm3/mol·s) -rA (mol/dm3·s)
XA
T (K)
0
300
0.01
0.0001
0.2
340
0.072
0.00046
0.4
380
0.34
0.00122
0.6
420
1.21
0.00194
0.8
460
3.42
0.0014
0.85
470
4.31
0.00097
FA0/-rA (dm3)
Calculate –rA each XA and k in the table
rA  kCA02 1  XA 
2
2
dm3 
mol 
mol
2
rA  0.01
0.1
1

0

0.0001


3
mol  s 
dm 
dm3  s
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-16
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA -Temperature
range should cover XA = 0 to XA = 0.85
ii. Calculate k(T)
iii. Calculate –rA each XA and k in the table
iii.
k (dm3/mol·s) -rA (mol/dm3·s)
XA
T (K)
0
300
0.01
0.0001
0.2
340
0.072
0.00046
0.4
380
0.34
0.00122
0.6
420
1.21
0.00194
0.8
460
3.42
0.0014
0.85
470
4.31
0.00097
FA0/-rA (dm3)
Calculate FA0/–rA for each XA in the table
mol  dm3 
mol
FA0  CA0u0  0.1 3  2

0.2


s
s
dm 

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-17
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
i.
Use EB to construct table of T as a function of XA -Temperature
range should cover XA = 0 to XA = 0.85
ii. Calculate k(T)
iii. Calculate –rA each XA and k in the table
iii.
k (dm3/mol·s) -rA (mol/dm3·s)
FA0/-rA (dm3)
XA
T (K)
0
300
0.01
0.0001
2000
0.2
340
0.072
0.00046
434.8
0.4
380
0.34
0.00122
163.9
0.6
420
1.21
0.00194
103.1
0.8
460
3.42
0.0014
142.9
0.85
470
4.31
0.00097
206.2
Calculate FA0/–rA for each XA in the table
mol  dm3 
mol
FA0  CA0u0  0.1 3  2

0.2


s
s
dm 

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-18
The elementary irreversible reaction A + B →C is carried out adiabatically in a flow
reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and
CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85?
Solve for PFR:
k (dm3/mol·s) -rA (mol/dm3·s)
FA0/-rA (dm3)
XA
T (K)
0
300
0.01
0.0001
2000
0.2
340
0.072
0.00046
434.8
0.4
380
0.34
0.00122
163.9
0.6
420
1.21
0.00194
103.1
0.8
460
3.42
0.00137
146
0.85
470
4.31
0.00097
206.2
Numeric evaluation by parts: 5-point rule for XA interval 0 to 0.8, 2-point rule
for XA interval from 0.8 to 0.85:
X5
X4
h1
h2
h1   X4  X0  4
f
X
dx

f
X
dx

f

4f

2f

4f

f

f

f










0
1
2
3
4
4
5
3
2
X
X
h X X
0
VPFR 
4
2
5
4
0.2
0.05
2000  4  434.8   2 163.9   4 103.1  146  
146  206.2
3
2
VPFR  317L
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-19
For these conditions, what is the max T0 that would keep T≤
550K at complete conversion?
T  Ti0 
H RX (TR )X A
n
 iCp,i
i1
n
cal
 iCpi 30
mol  K
i1
H RX  TR   6000
cal
mol
cal 

  6000
 XA
mol 
 T  Ti0  200K  X A 
 T  Ti0  
cal
30
mol  K
T≤ 550K at complete conversion, XA=1:
550K  Ti0  200K 1
 350K  Ti0
Max T0 is 350K
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-20
The first order irreversible reaction A(l) → B(l) is carried out in a jacketed
CSTR. The feed contains A and an inert liquid in equimolar amounts, where
FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K?
UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K
Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
Need to find where G(T)=R(T) for T0 = 450K
1. Put R(T) & G(T) in terms of constants in the problem
statement
2. Plot R(T) vs T and G(T) vs T on the same graph & find where
they intersect
R  T   Cp0 1     T  TC 
  UA Cp0FA0
Tc 
 Ta  T0
1 
 r V 
G  T   HRX  A 
 FA0 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-21
The first order irreversible reaction A(l) → B(l) is carried out in a jacketed
CSTR. The feed contains A and an inert liquid in equimolar amounts, where
FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K?
UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K
Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
I = inert
A I B I
Put R(T) in terms of constants in the problem statement, starting with CP0:
 T  T0
R  T   Cp0 1     T  TC 
  UA Cp0FA 0 Tc  a
1 
n
cal
Cp0   iCpi  Cp0   A CpA  ICpI  A  I  1 Cp0  20  30  50
mol  K
i1
Put  in terms of constants from the problem statement:
8000 cal min K
UA   
  2

cal  mol 
Cp0FA 0
50
 80

mol  K  min 
Put TC in terms of constants in the problem statement:
2  300K   450K
 T  T0
Tc  a
 Tc 
1 
1 2
 Tc  350K
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-22
The first order irreversible reaction A(l) → B(l) is carried out in a jacketed
CSTR. The feed contains A and an inert liquid in equimolar amounts, where
FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K?
UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K
Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
A I B I
I = inert
Find steady state temp [G(T)=R(T) ] for T0 = 450K
cal
R  T   Cp0 1     T  TC 
Tc  350K
Cp0  50
 2
mol  K
Plug  and Tc into R(T):
cal
R  T   Cp0 1     T  TC   R  T   50
1  2  T  350K 
mol  K
cal
cal
 R  T   150
T  52500
mol  K
mol
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-23
The first order irreversible reaction A(l) → B(l) is carried out in a jacketed
CSTR. The feed contains A and an inert liquid in equimolar amounts, where
FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K?
UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K
Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
A I B I
I = inert
Find steady state temp [G(T)=R(T) ] for T0 = 450K
cal
cal
R  T   150
T  52500
mol  K
mol
Now put G(T) in terms of constants from the problem statement:
 r V 
G  T   HRX  A 
 FA0 
 E  1 1 
k  k1 exp     
 R  T1 T  
Combine:
Rate law:
rA  kC A
Stoichiometry:
C A  CA0 1  X A 
 40000 cal

6.6  10 3
1
1


mol
k 
exp 
 

min
1.987 cal
 350 T  


mol  K
6.6  10 3
1 
 20130.85  1
 rA 
exp 


  C A0 1  X A 
min
K
 350 T  

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-24
The first order irreversible reaction A(l) → B(l) is carried out in a jacketed
CSTR. The feed contains A and an inert liquid in equimolar amounts, where
FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K?
UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K
Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
A I B I
I = inert
Find steady state temp [G(T)=R(T) ] for T0 = 450K
cal
cal
 rA V 
R  T   150
T  52500
G  T   HRX 

mol  K
mol
F
 A0 
Plug rate law into G(T) & simplify:
G  T   HRX
 6.6  10 3

1 
 20130.85  1
exp 
   C A0 1  X A  V 


min
K
350
T 






C A0u0


V


t
u0
 6.6  103

1 
 20130.85  1
 G  T   HRX 
exp 
   1  X A t 



min
K
350
T






Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-25
The first order irreversible reaction A(l) → B(l) is carried out in a jacketed
CSTR. The feed contains A and an inert liquid in equimolar amounts, where
FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K?
UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K
Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
A I B I
I = inert
Steady state temp [G(T)=R(T) ] for T0 = 450K
cal
cal
150
T  52500
 HRX
mol  K
mol
 6.6  103

1 
 20130.85  1
exp 
   1  X A t 



min
K
 350 T  



Rearrange so can be solved with Polymath nonlinear equation solver:
cal
cal
f  T   0  52500
 150
T  HRX
mol
mol  K
 6.6  10 3

1 
 20130.85  1
exp 
   1  X A t 



min
K
350
T 




Use design equation to get XA as an explicit equation:
VkC A0 1  X A 
V  rA 
FA0 X A

 X A  t k 1  X A   X A


X
V
A
C A0u0
FA0
rA
 t k  t kXA  XA  t k  XA  t kXA  t k  X A 1  t k 
t k 1  t k   X A Explicit equation for Polymath
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-26
Enter in Polymath:
Select graph
If you want to see this graphically,
click the Graph button above
Calculated values of NLE variables
1
Variable
Value
T
399.9425
Initial
Guess
400. ( 300.
-4.547E-12
< T < 500. )
f(x)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-27
Format the graph
Create a table, save it as
a text file, import into
Excel, & make a graph
with correct labels
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-28
Plot reactor temperature as a function of feed temperature, with T0
between 350 and 450K
Need to re-run the Polymath program using various T0 between 350 and 450K
to find the new steady state reactor temperatures
Do we need to change G(T) or R(T) when T0 changes?
 rA V 
G  T   HRX 
 G(T) does not depend on T0
 FA0 
 T  T0
R  T   Cp0 1     T  TC    UA Cp0FA0 Tc  a
1 
CP0 and  do not depend on T0, but TC does
Enter equations into Polymath
so that R(T) varies according to
T0, & run program with varied
values of T0
In Excel, create a table of T vs
T0, and make a graph
Run over and
over again,
varying T0 from
350 K to 450 K
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-29
Steady state temperature as a function of T0
T0 (K)
T1 (K)
T2 (K)
T3 (K)
350
316.7
360
320.15
370
323.6
357.16
370.3
380
327.3
353.4
375.1
390
331.2
350.1
379.1
400
336.3
346
382.8
410
386
420
389.8
430
393.2
440
396.6
450
399.9
420
400
380
T 360
340
320
300
350
370
390
410
430
450
T0
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-30
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
What is the value of the heat generated term (with units) when a disturbance causes the
temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The
reactor is NOT at the steady state.)
 rA V 
G(T)=  HRX  T  

F
 A0 
HRX  T   H RX  TR   CP  T  TR 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-31
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
What is the value of the heat generated term (with units) when a disturbance causes the
temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The
reactor is NOT at the steady state.)
 rA V 
G(T)=  HRX  T  

F
 A0 
HRX  T   H RX  TR   CP  T  TR 
b
cal
CPB  CPA  Cp  15  15 
0
mol  K
a
cal
 H RX  H RX  TR   0  10,000
mol
CP 
t
V
u0
 tu0  V
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-32
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
What is the value of the heat generated term (with units) when a disturbance causes the
temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The
reactor is NOT at the steady state.)
 rA V 
G(T)=  HRX  T  

F
 A0 
HRX  T   H RX  TR   CP  T  TR 
b
cal
CPB  CPA  Cp  15  15 
0
mol  K
a
cal
 H RX  H RX  TR   0  10,000
mol
CP 
t
V
u0
L 

 tu0  V  V  120 min  300
  36,000L
min 

mol 


0.0015
36000
L


cal   
cal
L min 

G(T)=   10,000



G(T)

9000

mol
mol  
mol


60


min


Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-33
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
What is the value of the heat removal term (with units) when a disturbance causes the
temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The
reactor is NOT at the steady state.)
R  T   Cp0 1    T  TC 
n
cal 
cal 
cal

Cp0   iCP  Cp0  1 15

0
15

C

15



p0
mol

K
mo
l

K
mol  K
i




i 1

UA
Cp0FA0
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-34
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
What is the value of the heat removal term (with units) when a disturbance causes the
temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The
reactor is NOT at the steady state.)
R  T   Cp0 1    T  TC 
n
cal 
cal 
cal

Cp0   iCP  Cp0  1 15

0
15

C

15



p0
mol

K
mo
l

K
mol  K
i




i 1
cal
3200
UA
min K

 
   3.56
Cp0FA0
cal  mol 
15
 60

mol  K  min 
 Ta  T0
Tc 
1 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-35
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
What is the value of the heat removal term (with units) when a disturbance causes the
temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The
reactor is NOT at the steady state.)
R  T   Cp0 1    T  TC 
n
cal 
cal 
cal

Cp0   iCP  Cp0  1 15

0
15

C

15



p0
mol

K
mo
l

K
mol  K
i




i 1
cal
3200
UA
min K

 
   3.56
Cp0FA0
cal  mol 
15
 60

mol  K  min 
 Ta  T0
3.56  340K   310K
Tc 
 Tc  333.4K
 Tc 
1 
1  3.56
R  T   15
cal
cal
1  3.56  360K  333.4K   R  T   1819
mol
m ol  K
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15-36
The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The
feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 =
300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol
CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1
When a disturbance causes the temperature in the reactor to drift to 360 K if –rA =
0.0015 mol/L•min at 360K,
cal
cal
G(T)  9000
and
R  T   1819
mol
mol
Will the reactor temperature heat up, cool down, or stay at 360 K?
G(T) > R(T) so the reactor will heat up
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.