Review: SS Nonisothermal Reactors

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

L16-1

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


L16-2

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 HF

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


L16-3

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)


L16-4

L16: Unsteady State Nonisothermal Reactor Design

An open system (for example, CSTR)

Q

W

Fin

Hin

Fout

Hout

n nsysi i in i i out

i 1 i 1

dE Q W FE FE

dt

rate of heat flow from

surroundings to system

Rate of accumulation of energy in

system

=

Rate of work done by

system on surroundings

- +

Rate of energy added to

system by mass flow in

-

Rate of energy leaving system by mass flow

out

sysdE 0 steady state

dt sysdE

0 unsteady statedt

Goal: develop EB for unsteady state reactor


L16-5

Change in System Energy with Time

isysn

ii 1

EE N

Energy of system is the sum of products of each species specific energy Ei & the moles of each species:

i1

sysn

ii

iHE PVN

iii

sysi

n

1

d ddt dt

HN PVE

Differentiate wrt time

nin nisyi i

i 1 i 1i

s

i 1i

d d d ddt dt dt dtE

P VN NH

HN

Total V

n ini

ii 1i

s si

y

1P

d d d ddt dt dt d

NV

tN H

E H

n nS i i i i

i 1 i 1in o

nii

i 1

i

u

ni

i 1t

Hd dQ W FH FH H

Ndt dt

N

Total Energy Balance for unsteady state, constant PV

n ni i in i i o

syut

i 1 i 1

sdQ W FE FE

dtE

i i i i iH & so:U PVE U

For well-mixed reactor with constant PV- variation0


L16-6

Well-Mixed Reactors, Constant PV

Special case: well-mixed reactors (e.g., batch, CSTR or semibatch) with constant PV- variation in total P or V can be neglected

n n n ni i

S i i i i i ii 1 i 1 i 1 i 1in out

dH dN dQ W FH FH N H PVdt dt dt

Total Energy Balance for unsteady state, constant PV

= 0

n n n n iS i i i i i i

i 1 i 1 i 1 i 1in ou

i

t

dNQ W FH FH N H

dtdHdt

Total Energy Balance for unsteady state

T

i RX R pii i

TRpirecalling that H H

dHdHE dTC

dvaluate T C dT s o

td t

t d

n n n n iS i i i i i i

i 1 i 1 i 1 i 1in oupi

t

dNQ W FH FH N H

dtdTCdt

Need to put dNi/dt into terms that can be measured


L16-7

EB for Well-Mixed Reactors, PV=0n n

S i i i i pii 1 i 1in ou

n

ti

i 1

n ii

i 1

ddTQ W FH FH Cdt dt

N HN

From the mass balance:

i0 i ii

AdNdt

F F r V

In Out- + GenAccumulation =

i0 ii

i Ad

F Vt

NF

dr

n n

ii 1

nS i i i i pi

i 1 i 1ini

ni

itA

10 i i

ouN F F r VdTQ W FH FH C

dtH

n n

S i0 i0 i i pn

ii

n n ni i0 i i i A

i 1 i 11 i 1i i

i 1 i 1

dTQ W F H FH Cdt

HF HF H r VN

Substitute

Add SFiHi to both sides of equation:

n

i i0 RXn

ii

nS i0 i0 pi

i 1 1A

i 1

dTQ W F H Cdt

N HF H T r V

Substitute ΣiHi =H°RX(T):

n

S i0 i0 pin

in n

i i0 i i Ai 1 11 i ii 1

dTQ W F H Cdt

N HF H r V

H°RX(T)


L16-8

Simplified EB for Well-Mixed Reactors

n n nS i0 i0 i pi i i0 RX A

i 1 i 1 i 1

dTQ W F H NC HF H T r Vdt

Bring SFi0Hi and H°RX(T) terms to other side of equation:

n n n

S i0 i0 i i0 RX A i pii 1 i 1 i 1

dTQ W F H HF H T r V NCdt

Solve for dT/dt:

Factor SFi0Hi0 and SFi0Hi terms and divide by SNiCpi :

n

S i0 i i0 RX Ai 1

ni pi

i 1

Q W F H H H T r VdTdtNC

Energy balance for unsteady state reactor

with phase change:

n

S i0 pi i0 RX Ai 1

ni pi

i 1

Q W F C T T H T r VdTdtNC

Energy balance for

unsteady state reactor without phase change:


L16-9

Unsteady State EB, Liquid-Phase Rxns

For liquid-phase reactions, often Cp = SiCpi is so small it can be neglected

When Cp can be neglected, then:

S

ni pi A0 ps ps i pi

i 1 where is the heat capacity of the solNC N C C C ution

If the feed is well-mixed, it is convenient to use:S i0 pi A0 psF C F C

Plug these equations and Ti0 = T0 into the EB gives:

S i0 RA

A0

s X

p

0 Ap

s

Q W T T H TN C

VF C r dTdt

This equation for the EB is simultaneously solved with the mass balance (design eq) for unsteady state, nonisothermal reactor design

S i0 RX A

ni pi

i 1

ni0 pi

i 1Q W T T H T r V

NC

FT

dt

Cd


L16-10

Nonisothermal Batch Reactor Design

nS i0 pi i0 RX A

i 1n

i pii 1

Q W F C T T H T r VdTdtNC

No flow, so:

0

S RX A

in

pii 1

Q W H T r VdN

TtC

d

i A0 i i Ai0

iA

i0

i PpNN

where & CNN

X C

p

S RX An

i 1A0 i Api

Q W H T r V dT

CNt

Cd

X

Put the energy balance in terms of XA:

Solve with the batch reactor design equation using an ODE solver (Polymath)

AA0 A

dXN r Vdt


L16-11

Adiabatic Nonisothermal Batch Reactor Design

S RX An

A0 i pi p Ai 1

Q W H T r V dTdtN C C X

SW =0 & Q 0In the case of no stirring work and adiabatic operation,

RX A

A0 pn

i pii

A1

H T r V dTdt

N C XC

i pi psC CSubstitute:

ps

RX A

A0 p ACH T r V dT

dtN C X

A0 pR p AA SXdTN C C XT

tH r V

dRearrange:

AAR 0X RX R Ap R

dX-NH T H T anC T d T V

dt r Substitute:

RX R p R A0 pS pA

A A0dX

H T C T Ndt

dTN C Xt

T Cd

ARX R p R pS p A

dX dTH T C T T C C Xdt dt

00


L16-12 A

RX R p R pS p AdX dTH T C T T C C Xdt dt

A pS p ARX R p R dX CH C XT TCT T d

Get like terms together:

T

T RX R p R0

XA A

pS p AX 0A0

dTH T C T T

dXC C X

RXRX R p

p 0 p 0S SA

RA

C T T C

H TH T C T

T

T

TX X

Solving for T:

RX 0 A RX 0 A0 0 n

ps i pA p

A pii 1

H T X H T XT T T T

X C XC CC

Solve with the batch reactor design equation using an ODE solver (Polymath)XA A

A0A0

dXt Nr V

Integrate & solve for XA:

Heat capacity of soln (calculate Cps if not given)

Solve for how XA changes with T


L16-13

A 1st order, liquid-phase, exothermic reaction A→B is run in a batch reactor. The reactor is well-insulated, so no heat is lost to the surroundings. To control the temperature, an inert liquid C is added to the reaction. The flow rate of C is adjusted to keep T constant at 100 °F. What is the flow rate of C after 2h?

This is essentially a semi-batch reactor since only C is fed into the reactor

AAdN

Vdt

rDesign eq:

Rate eq: -rA = kCA

Note, using AA0 A

dXN V

dr

twould complicate

the calculation because V depends on t

Combine: AA

dNt

kCd

V AAdN

dtNk Rearrange and integrate for NA

N tA A

AN 0A0

dN kdtN

A

A0

Nln ktN

A A0N N exp kt

TC0 = 80 °F V0= 50 ft3 H°RX=-25000 Btu/lb molk(100 °F)= 1.2 x 10-4 s-1 Cp, (all components)= 0.5 Btu/lb mol °FCA0= 0.5 lb mol/ft3

1. Solve design eq for comp as function of t2. Solve EB for FC0 using that info & T=100 °F


L16-14

Use EB to find how the flow rate of C depends on the rxn (solve EB for FC0)


nS i0 pi i0 RX A

i 1n

i pii 1

W F C T T H T r VdTdt NC

Q

n

i0 pi i0 RX Ai 1

0 F C T T H T r V

n

i0 pii 1

i0 RX AT T H VC T rF

C is the only species that flows, so: i0 AXpC RC0 T T H VF T rC

C0 pC i0 RXkt

A0F C T T H N ekT

rAV = -kCAV = -kNA A A0N N exwhere p kt

ktRX A0

C0pC i0

H T kN eF

C T T

0 0

0

Isolate FC0:


L16-15


TC0 = 80 °F V0= 50 ft3 H°RX=-25000 Btu/lb mol k(100 °F)= 1.2 x 10-4 s-1 Cp, (all components)= 0.5 Btu/lb mol °FCA0= 0.5 lb mol/ft3

AC0

ktX

i0

R 0

pC

H T

C

k e

TF

T

N

4 11.2 1 7200s

C

s4 1

0

0Btu25,000lb

Bt

m

u0.5lb m

25 lb mo1.2 1

10

0

0

e

F

ol

oll

FF

F

s

80

At 2h (7200s):

3A0 A0 0 3

lb molN C V 0.5 50ft 25 lb molft

C0lb molF 3.16

s


L16-16Instead of feeding coolant to the reactor, a solvent with a low boiling point is added (component D). The solvent has a heat of vaporization of 1000 Btu/lb mol, and initially 25 lb mol of A are placed in the tank. The reactor is well-insulated. What is the rate of solvent evaporation after 2 h if T is constant at 100 °F?

Still a semibatch reactor, where D is removed from the reactorUse EB that accounts for a phase change:

n

S i0 i0 i RX Ai 1

ni pi

i 1

Q W F H H H T r VdTdt NC

Q˙ =0ẆS=0

0 0

0

Additional info: k(100 °F)= 1.2 x 10-4 s-1 H°RX=-25000 Btu/lb mol

Clicker Question: Does dT/dt = 0?a) Yesb) No


L16-17Instead of feeding coolant to the reactor, a solvent with a low boiling point is added (component D). The solvent has a heat of vaporization of 1000 Btu/lb mol, and initially 25 lb mol of A are placed in the tank. What is the rate of solvent evaporation after 2 h?

Still a semibatch reactor, where D is removed from the reactorUse EB that accounts for a phase change:

n

S i0 i0 i RX Ai 1

ni pi

i 1

Q W F H H H T r VdTdt NC

Q˙ =0ẆS=0dT/dt = 0

n

i0 i0 i RXi 1

AF VH H T rH D is the only species that ‘flows’, and rAV = -kNA0(exp[-kt]), so:

ktD0 i0 i R 0X AF H H H N ekT

R

t

i0

XD0

i

A0kN ek

H

TF

H

H

0.00012 s

D0

7200s25 lb moBtu25,000F

0.00012

1000Bt

l

ulbmol

es

lb mol

Hi0-Hi = heat of vap

D0lb molF 0.0316

s

0 0

0

Additional info: k(100 °F)= 1.2 x 10-4 s-1 H°RX=-25000 Btu/lb mol


L16-18A liquid phase exothermic reaction A →B is carried out at 358K in a 0.2 m3 CSTR. The coolant temperature is 273K and the heat transfer coefficient (U) is 7200 J/min·m2·K. What is the heat exchange area required for steady state operation? Using this heat exchange area, plot T vs t for reactor start-up. CPA =CPS=20 J/g•K ẆS=0 CA0= 180 g/dm3 u0= 500 dm3/min T0= 313 Kr= 900 g/dm3 H°RX(T) = -2500 J/g E=94852 J/mol·K k(313K)= 1.1 min-1

a) Heat exchange area for steady state operation:

SS operation means that T is

constant, so:

n

S i0 pi i0 RX Ai 1

ni pi

i 1

Q W F C T T H T r VdT 0dt NC

S i0 RX An

i0 pii 1

W F C T T H r VQ T

Q˙ =UA(Ta-T), ẆS=0, and A is only species that flows A0 pA A0 RX Aa T TFUA H VT C T rT

Plug in rA = -kCA and solve for A

a A0 pA A0 ARXU T T F C T T H T VkCA

RX A A0 pA A0

a

H T kC V F C T TU T T

A



A0 A ArF F V 0 A0 0 A 0 AkC 0CC Vu u A0 0 A 0 AC C kC Vu u

Use material balance to determine steady state value of CA

A0 0A

0

Ck

CV

uu

1.1 94852J mol 1 1k 358K expmin 8.314J mol K 313 358

= 107.4k 358Kmin

1

3

3A

3

3

500dm min

500d 107.4min

180

m 2

g dm

mC

n 00dmi 3

AC 4.1g dm

R pA A0X

a

A0Ak FSolve EB for A:

H TU

T TC CVT T

A

3

A0 33 3

3

3F 90000g 1000dmdm500

mV =0.2g180

dm 200dm

mm n

n

mi

i



Solve for heat exchange area at SS:

33

2

g90000m

313200dmA

J7

g4.1d

200min m

10 358

K

J2500g

7.4m

Ki

2

J20g

73K

m Kin nK

358K

RX AA 0AA p

a

0FH T

VCk TTTU T

CA

A=227.4 m2

FA0=90,000 g/min V=200 dm3

H°RX(T) = -2500 J/g k= 107.4 min-1 CA= 4.1 g/dm3 CPA =20 J/g•K U= 7200 J/min·m2·K TA0= 313 K T= 358K Ta=273K


L16-21Will use Polymath to plot T vs t for CSTR start-up (unsteady-state). Need equations for dCA/dt, dT/dt, and k.

AA0 A A

dN F F r Vdt

a A0 pA i0 RX An

i0 psi 1

UA T T F C T T H r VdTdt N C

1.1 94852J mol 1 1k= expmin 8.314J mol K 313 T

1.1 11408.7 1 1k= expmin K 313 T

U= 7200 J/min·m2·K A=227.4m2 CA0=180g/dm3 t=V/u0 Ta=273KHRX=-2500 J/g Ni0=mi0 V=200 dm3 CPs=20 J/g˙K u0=500 dm3 T0=313

Mass balance:

CPs is in terms of mass (J/g·K), so FA0 & Ni0 must also be in terms of mass

rA=-kCA

FA0=90,000 g/minSubstitute ṁi0 for Ni0, & use r for the solution to calculate:

i0 33 g9m 200d 0m 0

dmi0 Vm r i0m 180,000g

Amount of gas leaving reactor (L7)

A0A AA

CdC Cr

dt

t t


L16-22Will use Polymath to plot T vs t for CSTR start-up (unsteady-state). Need equations for dCA/dt, dT/dt, and k.

A0A AA

CdC Cr

dt

t t

a A0 pA i0 RX An

i0 psi 1

UA T T F C T T H r VdTdt N C

1.1 11408.7 1 1k= expmin K 313 T

U= 7200 J/min·m2·K A=227.4m2 CA0=180g/dm3 t=V/u0 Ta=273KHRX=-2500 J/g Ni0=mi0 V=200 dm3 CPs=20 J/g˙K u0=500 dm3 T0=313

rA=-kCA

FA0=90,000 g/mini0m 180,000g


L16-23

T (K

)

t (min)

Reaches steady state at ~12 minutes


L16-24The elementary, liquid phase, exothermic reaction A →B is carried out in a 2 m3

CSTR that is equipped with a heat jacket. Pure A enters the reactor at 60 mol/min and a temperature of 310K. The coolant in the heat jacket is kept at 280 K. Provide all equations, all constants, the initial time, and the final time that must be entered into Polymath in order to plot temperature vs time for the first 20 min of reactor start-up. ΔHRX(TR) = -10,000 cal/mol CPA = CPS =15 cal/mol·K CpB=15 cal/mol·K ẆS=0 E = 20,000 cal/mol k = 1 min-1 at 400 K UA= 3200 cal/min•K u0= 300 L/minNeed equations for how T changes with time, CA changes with time, & k changes with T.

n

pi RiS i0 A

ni

i

i

1

0 X1

pi

C H T

C

W T T r V

N

FQdTdt

RXRX P RRH T CH TT T pcalC 15 15 0

mol K

P PB PAbC C Ca

RX RX RH 0T H T

i0 A

n

pi p,A p,1

B00 Bi

C C CFF F

B0F 0n

pi p,Ai 1

i0 A0CF CF

SW 0 aQ - TTUA

Combine with EB:

ni A0

ipi

1psN NC C

RXcalH T 10,000mol

p,A iA

A0

0

s

0

p

H TRX R- T TdTdt

Ta CUA

C

T

N

F rAV


L16-25The elementary, liquid phase, exothermic reaction A →B is carried out in a 2 m3

CSTR that is equipped with a heat jacket. Pure A enters the reactor at 60 mol/min and a temperature of 310K. The coolant in the heat jacket is kept at 280 K. Provide all equations, all constants, the initial time, and the final time that must be entered into Polymath in order to plot temperature vs time for the first 20 min of reactor start-up. ΔHRX(TR) = -10,000 cal/mol CPA = CPS =15 cal/mol·K CpB=15 cal/mol·K ẆS=0 E = 20,000 cal/mol k = 1 min-1 at 400 K UA= 3200 cal/min•K u0= 300 L/minNeed equations for how T changes with time, CA changes with time, & k changes with T.

RXcalH T 10,000mol

p,A iA

A0

0

s

0

p

H TRX R- T TdTdt

Ta CUA

C

T

N

F rAV

UA 3200 aT 280

PAC 15 i0T 310

PSC 15

AAr kC

3 31000L 1V 2m V 00m 2 0 A

A0 00F

NV

u A

0A0

0F VN

u0 300u

A0F 60 20,000 1 1k 1 exp1.987 400 T

Use the mass balance to get eq for CA(t)

A0 AA

A

dF

NrF

dtV 0

AA0 A A0

dNC

dVC C

tk uu A0A A

A

Ct

ddC

CkC

t t

t 0 0 t f 20

0Vt u AA0 00FC u

Review: SS Nonisothermal Reactors

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Transcript of Review: SS Nonisothermal Reactors