L15-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois,...

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


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


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


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


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



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


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



Combine MB, rate law & stoichiometry for CSTR:

A0 A

A

F XV

r

CSTR



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



Energy balance



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



Energy balance



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



Solve for CSTR volume: use EB to find T when XA=0.85



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


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


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



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





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


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






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






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)


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


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?




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


L15-17






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


L15-18



Solve for PFR:


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


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


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




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




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





A I B I I = inert


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





A I B I I = inert


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




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


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


Format the graph

Create a table, save it as a text file, import into Excel, & make a graph with correct labels

L15-28


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


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


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




H TRXG(T)V

=FA0

rA

0

V


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




H TRXG(T)V

=FA0

rA

0

V


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



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




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




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



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

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L15-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois,...

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