l15-1 slides courtesy of prof m l kraft, chemical & biomolecular engr dept, university of...
TRANSCRIPT
L15-1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L15: Nonisothermal Reactor Example Problems
L15-2
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Multiple Steady States in CSTR
• 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!
0 100 200 300 400 500 6000
0.2
0.4
0.6
0.8
1
T (K)
XA
XA,EBXA,MB
L15-3
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
T
R(T)
Increase T0 T
R(T)
Increase
T0Ta
= 0 = ∞
For Ta < T0
Review: Heat Removal Term R(T) & T0
Ap0 C RX
A0
r VC 1 T T H
F
Heat removed: R(T) Heat generated G(T)
When T0 increases, slope stays same & line shifts to right
R(T) line has slope of CP0(1+)
p0 A0UA C F
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 ↑
a 0c
T TT
1
=0, then TC=T0 =∞, then TC=Ta
L15-4
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: CSTR StabilityG(T)
R(T)
T
1
2
3R(T) > G(T) → T falls to T=SS1
G(T) > R(T) → T rises to T=SS3
G(T) > R(T) → T rises to
T=SS1
R(T) > G(T) →T falls to T=SS3
• 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
Ap0 RXC
A0
r VC 1 T H
FT
Heat generated G(T)
p0 A0UA C F a 0c
T TT
1
Heat removed: R(T)
L15-5
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 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/molDHA°(273K)= -20 kcal/mol DHB°(273K)= -15 kcal/mol DHC°(273K)= -41 kcal/mol
3k 0.01 dm mol s at 300K
1. Mole balance2. Rate Law3. Stoichiometry4. Combine rate law & stoichiometry5. Energy balance6. Solve
Strategy:
6a. Solve CSTRi. 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 PFRi. 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 Tv. Use numeric technique to calculate V
L15-6
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 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?
Mole balance A0 ACSTR
A
F XV
r
CSTR PFR0.85
A0PFR A
A0
FV dX
r
Rate law3dm 10000cal mol 1 1
k 0.01 expmol s 1.987cal mol K 300K T
A A Br kC C
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/molDHA°(273K)= -20 kcal/mol DHB°(273K)= -15 kcal/mol DHC°(273K)= -41 kcal/mol
3k 0.01 dm mol s at 300K
Stoichiometry B A A0 AC C C 1 X
Combine: 22A A0 A
10000 1 1r 0.01exp C 1 X
1.987cal mol K 300K T
L15-7
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 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?
Combine MB, rate law & stoichiometry for CSTR:
A0 A
A
F XV
r
CSTR
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/molDHA°(273K)= -20 kcal/mol DHB°(273K)= -15 kcal/mol DHC°(273K)= -41 kcal/mol
3k 0.01 dm mol s at 300K
A0 0 ACSTR 3
22A0 A
C XV
dm 10000cal mol 1 10.01 exp C 1 X
mol s 1.987cal mol K 300K T
0 ACSTR 3
2A0 A
XV
dm 1 10.01 exp 5032.7K C 1 X
mol s 300K T
L15-8
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 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?
Energy balance
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/molDHA°(273K)= -20 kcal/mol DHB°(273K)= -15 kcal/mol DHC°(273K)= -41 kcal/mol
3k 0.01 dm mol s at 300K
n
s A0 i p,i i0 RX A0 Ai 1
0 Q W F C T T H (T)F X
0 0
n
A0 i p,i i0 RX A0 Ai 1
F C T T H (T)F X
RX RX R P RH H (T ) C T T
Solve for T:
Substitute:
n
i p,i i0 RX R P R Ai 1
C T T H (T ) C T T X
n
i p,i i0 RX R A P R Ai 1
n
i p,i P Ai 1
C T H (T )X C T XT
C C X
Multiply out quantities in brackets, bring T to 1 side of equation, factor out T, divide by quantity in bracket:
P pC pA pBC C C C Evaluate Cp:
Pcal
C 30 15 15 0mol K
L15-9
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 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?
Energy balance
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/molDHA°(273K)= -20 kcal/mol DHB°(273K)= -15 kcal/mol DHC°(273K)= -41 kcal/mol
3k 0.01 dm mol s at 300K
n
i p,i i0 RX R A R Ai 1
n
i p,i Ai 1
P
P
C T H (T )X C
C
T XT
C X
n
i pi pA pBi 1
cal calC C C 15 15 30
mol K mol K
Evaluate S iCp:
RX R Ai0 n
i p,ii 1
H (T )XT T
C
PC 0
A 1 B0B
A0
F1
F
Evaluate H°RX(TR):
RX R C A Bcal cal
H T H H H 41000 20000 15000 6000mol mol
AA
6000cal mol XT 300K T 300K 200K X
30cal mol K
Simplify EB:
L15-10
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 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 CSTR volume: use EB to find T when XA=0.85
Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/molDHA°(273K)= -20 kcal/mol DHB°(273K)= -15 kcal/mol DHC°(273K)= -41 kcal/mol
3k 0.01 dm mol s at 300K
AT 300K 200K X T 300K 200K 0.85 T 470K
0 ACSTR 3
2A0 A
XV
dm 1 10.01 exp 5032.7K C 1 X
mol s 300K T
3
CSTR 32
3
dm2 0.85
sV
dm 1 1 mol0.01 exp 5032.7K 0.1 1 0.85
mol s 300K 470K dm
CSTRV 175L
L15-11
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
0
0.2
0.4
0.6
0.8
0.85
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)
AT 300K 200K X T 300K 200K 0 300K
300
L15-12
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
0 300
0.2 340
0.4 380
0.6 420
0.8 460
0.85 470
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
AT 300K 200K X
ii. Calculate k(T) for each T in the table3dm 1 1
k 0.01 exp 5032.7Kmol s 300K T
L15-13
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 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
AT 300K 200K X
ii. Calculate k(T) for each T in the table3dm 1 1
k 0.01 exp 5032.7Kmol s 300K T
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
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
L15-14
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 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
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
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
iii. Calculate –rA each XA and k in the table
22A A0 Ar kC 1 X
232
A 3 3dm mol mol
r 0.01 0.1 1 0 0.0001mol s dm dm s
ii. Calculate k(T)
L15-15
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 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
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
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
ii. Calculate k(T)
iii. Calculate –rA each XA and k in the table
22A A0 Ar kC 1 X
232
A 3 3dm mol mol
r 0.01 0.1 1 0 0.0001mol s dm dm s
L15-16
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 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
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
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
ii. Calculate k(T)
iii. Calculate FA0/–rA for each XA in the table
3
A0 A0 0 3mol dm mol
F C 0.1 2 0.2s sdm
iii. Calculate –rA each XA and k in the table
L15-17
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 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
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
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
ii. Calculate k(T)
iii. Calculate FA0/–rA for each XA in the table
3
A0 A0 0 3mol dm mol
F C 0.1 2 0.2s sdm
iii. Calculate –rA each XA and k in the table
L15-18
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 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:
XA T (K) k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3)
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:
PFR0.2 0.05
V 2000 4 434.8 2 163.9 4 103.1 146 146 206.23 2
XX 54
1 20 1 2 3 4 4 5
X X0 4
h hf X dx f X dx f 4f 2f 4f f f f
3 2
2 5 4h X X
1 4 0h X X 4
PFRV 317L
L15-19
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
For these conditions, what is the max T0 that would keep T≤ 550K at complete conversion?
RX R Ai0 n
i p,ii 1
H (T )XT T
C
n
i pii 1
calC 30
mol K
RX R
calH T 6000
mol
A
i0
cal6000 X
molT Tcal
30mol K
i0 AT T 200K X
T≤ 550K at complete conversion, XA=1:
i0550K T 200K 1 i0350K T
Max T0 is 350K
L15-20
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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·KCpi =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
p C0R T C 1 T T
ARX
A0
r VG T H
F
p0 A0UA C F
a 0c
T TT
1
L15-21
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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·KCpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
Cp0R T 1 TC T p 00 AUA C F
a 0c
T TT
1
Put in terms of constants from the problem statement:
p 00 A
UAC F
n
p0 i pii 1
C C p0 A pA I pIC C C
A I B I I = inert
p0cal
C 20 30 50mol K
A I 1
cal mol
805
8000cal min K
0mol K min
2
Put R(T) in terms of constants in the problem statement, starting with CP0:
0ac
TT
T
1
Put TC in terms of constants in the problem statement:
c
300K2 450KT
1 2 cT 350K
L15-22
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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·KCpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
Find steady state temp [G(T)=R(T) ] for T0 = 450K
Cp0R T 1 TC T 2 cT 350K
A I B I I = inert
Plug and Tc into R(T):
Cp0R T 1 TC T
cal50
molR
KT 21 T 350K
p0
calC 50
mol K
cal calR T 150 T 52500
mol K mol
L15-23
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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·KCpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
Find steady state temp [G(T)=R(T) ] for T0 = 450K
A I B I I = inert
cal calR T 150 T 52500
mol K mol
Now put G(T) in terms of constants from the problem statement:
ARX
A0
r VG T H
F
AAr kCRate law: Stoichiometry:
A A0 AC C 1 X
1
1
E 1 1k k exp
R T T
3 cal400006.6 10 1 1molk expcalmin 350 T1.987 mol K
A0
3
A A6.6 10 20130.85 1 1
expmin K T
C50
13
XrCombine:
L15-24
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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·KCpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
Find steady state temp [G(T)=R(T) ] for T0 = 450K
A I B I I = inert
cal calR T 150 T 52500
mol K mol
Plug rate law into G(T) & simplify:
ARX
A0
r VG T H
F
RXA0 0
A
3
A06.6 10 20130.85 1 1
expmin K 350 T
C VT
1 XG H
C
3
RX A6.6 10 20130.85 1 1
G T H exp 1 Xmin K 350 T
t
0
Vt
L15-25
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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·KCpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K
Steady state temp [G(T)=R(T) ] for T0 = 450K
A I B I I = inert
3
RX Acal cal 6.6 10 20130.85 1 1
150 T 52500 H exp 1 Xmol K mol min K 350 T
t
Rearrange so can be solved with Polymath nonlinear equation solver:
t
3
RX Acal cal 6.6 10 20130.85 1 1
f T 0 52500 150 T H exp 1 Xmol mol K min K 350 T
Use design equation to get XA as an explicit equation:
A0 A
A
F X
rV
AA
A0
VX
F
r A0 A
AA0 0
C
C
1 XkVX
A Ak 1 X Xt
A Ak kX Xt t A Ak X kXt t Ak X 1 kt t
Ak 1 k Xt t Explicit equation for Polymath
L15-26
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Enter in Polymath:
Variable Value f(x) Initial Guess
1 T 399.9425 -4.547E-12 400. ( 300. < T < 500. )
Calculated values of NLE variables
If you want to see this graphically, click the Graph button above
Select graph
L15-27
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Format the graph
Create a table, save it as a text file, import into Excel, & make a graph with correct labels
L15-28
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Plot reactor temperature as a function of feed temperature, with T0 between 350 and 450K
CP0 and do not depend on T0, but TC does
Do we need to change G(T) or R(T) when T0 changes?
ARX
A0
r VG T H
F
G(T) does not depend on T0
Cp0R T 1 TC T p0 A0UA C F
a 0c
T TT
1
Need to re-run the Polymath program using various T0 between 350 and 450K to find the new steady state reactor temperatures
Enter equations into Polymath so that R(T) varies according to T0, & run program with varied values of T0 Run over and
over again, varying T0 from 350 K to 450 K
In Excel, create a table of T vs T0, and make a graph
L15-29
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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
350 370 390 410 430 450300
320
340
360
380
400
420
T0
T
Steady state temperature as a function of T0
L15-30
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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.)
H TRXG(T)V
=FA0
rA
RXRX P RRH T CH TT T
L15-31
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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.)
H TRXG(T)V
=FA0
rA
0
V
RXRX P RRH T CH TT T
pcal
C 15 15 0mol K
P PB PA
bC C C
a
X RX RR 0cal
H 10,000ol
H Tm
0 V
L15-32
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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.)
H TRXG(T)V
=FA0
rA
0
V
RXRX P RRH T CH TT T
pcal
C 15 15 0mol K
P PB PA
bC C C
a
X RX RR 0cal
H 10,000ol
H Tm
L300
minV 36,00120 n Lmi 0
0 V
G
mol
(T)=0.0015
L mimo
cal10,000
moln
l
36000
60min
L
calG(T) 9000
mol
L15-33
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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.)
Cp
UA
A00F
R T TCp0 1 T C
C n
C p0 ii 1
Pi
cal cal15 15
mol K mo K1 0
lCp0
calC 15p0 mol K
L15-34
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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.)
Cp
UA
A00F
R T TCp0 1 T C
C n
C p0 ii 1
Pi
cal cal15 15
mol K mo K1 0
lCp0
calC 15p0 mol K
cal15
momol
60mi
cal3
n
200min
l K
K
3.56
Ta Tc 1
0 T
L15-35
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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.)
Cp
UA
A00F
R T TCp0 1 T C
C n
C p0 ii 1
Pi
cal cal15 15
mol K mo K1 0
lCp0
calC 15p0 mol K
cal15
momol
60mi
cal3
n
200min
l K
K
3.56
Ta Tc 1
0 T
340K 313. 0KTc 1
56
3.56
3R . 333.45cal
15m
T 1 366o
0KK
Kl
T 33 4Kc 3.
calR T 1819
mol
L15-36
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
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 calG(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