l12: nonisothermal reaction engineering

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

L12-1

Review: Thermochemistry for Nonisothermal Reactor Design

kA B A A

A0

dX rdV F

A Ar kC

FAXA = 0.7

Mole balance:

Rate law:

Stoichiometry: A A 0

0

A A0 A

F C v

C C (1 X )

A0 AA

A0 0

C (1 X )dXdV C

k

ERTk Ae

Arrhenius Equation

A A

E 1 1R T T1

01

dX (1 X )dV

k exp

Need relationships: X T V

Consider an exothermic, liquid-phase reaction operated adiabatically in a PFR (adiabatic operation- temperature increases down length of PFR):

FA0

The energy balance provides this relationship

E 1 1R T T1

1k k exp


L12-2

Rate of accum of energy in

system

Rate of work done

by syst

energy added to syst by

mass flow in

energy leaving syst by mass flow out

Heat in= - + -

n ni i i i sin outi 1 i 1

W FPV FPV W

n nssyyssi iin outi 1 i 1

i i

ˆdE Q F E E FW

dt

Review: Terms in Energy Balance

WS: shaft workP : pressure

Flow workiV specific volume

i iE U Internal energy is major contributor to energy term

s i

n nssyyssi i in i out

i 1 ii i

1

ˆdEQ - F( ) - F( )

dtW PU UPV V

i i iH U PV n n

s i0 i0 i ii 1 i 1

0 Q W F H FH

Steady state:

Accum of energy in system

shaft work

Energy & work added by flow in

Energy & work removed by flow out

Heat in=0= - + -


L12-3

Review: Relate T to Conversion

i i0 i A0 A i A0 i i AF F F X F F X i0i

A0

Fwhere

F

If XA0=0, then:

n ns i0 i0 i

ii

1 i 10 Q W F H HF Steady state:

n nssyyss

s i0 i A0 ii 1 i 1

A0 i i AFˆdE

Q W H F Hdt

X

in nssyyss

s ii

A0 0 i i i A0 Ai 1 i 1

ˆdEQ W H H H FF X

dt

n nsys

s i0 i i A0 i Ai 1 i 1

i iA0 A0dE

Q W H H HFF XFdt

RXH T heat of reaction

Total energy balance (TEB)

0 at steady state

Multiply out:

n n

s A0 i0 i i RX A0 Ai 1 i 1

0 Q W F H H H T F X

Accum of energy in system

shaft work

Energy & work added by flow in

Energy & work removed by

flow out

Heat in= - + -


L12-4

Review: Q in a CSTR CSTR with a heat exchanger, perfectly mixed inside and outside of reactor

T, X

FA0

T, X

Ta

Ta

The heat flow to the reactor is in terms of:• Overall heat-transfer coefficient, U• Heat-exchange area, A•Difference between the ambient temperature in the heat jacket, Ta, and rxn temperature, T

aQ (UA T T)


L12-5

Integrate the heat flux equation along the length of the reactor to obtain the total heat added to the reactor :

A Va aQ U(T T)dA Ua(T T)dV

adQ Ua(T - T)dV

Heat transfer to a perfectly mixed PFR in a jacket

a: heat-exchange area per unit volume of reactor

For a tubular reactor of diameter D, a = 4 / D

For a jacketed PBR (perfectly mixed in jacket):

ab b

1 dQ dQ Ua (T T)dV dW

Heat transfer to a PBR

Review: Tubular Reactors (PFR/PBR):

AaV


L12-6

L12: Nonisothermal Reactor DesignSteady–state total energy balance (TEB):

i0n nssyys

is

s A0 i RX A0 Ai 1 i 1

ˆdE0 Q HW F H T

dH F X

t

At a particular temperature: Ti i R piTR

H H (T ) C dT no phase change

TT i

i i0

i R pi i R piT TR R0 H (T ) C dT H (T ) C dTH H

Tnssyyss

s A0 i p,i RX A0 Ai 1Ti0

ˆdEQ W F C dT H T F X

dt

For a SS nonisotherm flow reactor:

Tn

s A0 i p,i RX A0 Ai 1Ti0

0 Q W F C dT H T F X

Goal: Use TEB to design nonisothermal steady-state reactors

Needs to be “simplified” before we can apply it to reactor design

n

s A0 i p,i i0 RX A0 Ai 1

0 Q W F C T T H (T)F X

Constant (average) heat capacities :

TpiTi0

C dT

Substitute(Hi – Hi0) = - (Hi – Hi0)


L12-7

Relating HRX(T) to H◦RX(TR) and

Overall Change in Heat Capacity

TRX R TR PRX H (T C dT)H T

nP i pi

i 1overall heat capaci Ct Cy:

Tn

s A0 i p,i A0 Ai 1

RT

PTi0

X TRR CH (0 Q W F C dT T FdT X)

nRX R i i R

i 1overall heat of reaction at reference t H T Hemp: T

nT

i pin

i i R TR ii 1 1RX H (T ) C dH TT

Only considering constant (average) heat capacities:

R

ns A0 i p X,i i0 A0 A

i 1R P RCH (T0 Q W F C T)T FTT X

Tn

s A0 i p,i A0 Ai 1Ti

RX0

0 Q W F C dT F XH T

T = reaction temp Ti0 = initial (feed) temp TR= reference temp


L12-8

Solving TEB for Conversion

n

s A0 i p,i i0 RX R P R A0 Ai 1

ˆ0 Q W F C T T H (T ) C T T F X

Rearrange to isolate terms with XA on one side of eq:

n

A0 i p,i i0 s RX R P R A0 Ai 1

ˆF C T T W Q H (T ) C T T F X

Solve for XA:

nA0 i p,i i0 s

i 1A

RX R P R A0

F C T T W QX

ˆH (T ) C T T F

Plug in Q for the specific type of reactor, and solve this eq simultaneously with design equation

Always start with this TEB:


L12-9

Solving TEB for XA for an Adiabatic Rxn

ns A0 i p,i i0 RX R P R A0 A

i 1ˆ0 Q W F C T T H (T ) C T T F X

Rearrange:

n


ˆF C T T Q W H (T ) C T T F X

Which term in this equation is zero because we’re solving for an adiabatic reaction?

a) dEsys/dtb) c) Ẇ

d) FA0

e) None of the aboveQ

When the reaction is adiabatic (Q=0):

n


ˆF C T T W H (T ) C T T F XQ

n


ˆF C T T W H (T ) C T T F X0


L12-10

Solving TEB for XA for an Adiabatic Rxn

ns A0 i p,i i0 RX R P R A0 A

i 1ˆ0 Q W F C T T H (T ) C T T F X

When shaft work can be neglected (Ẇ=0) and the reaction is adiabatic (Q=0):

Rearrange:

n

A0 i p,i i0 RX R P R Ai

s A01

ˆF C T T H (T ) C T T FQ XW

n

A0 i p,i i0 RX R P R A0 Ai 1

ˆF C T T H (T ) C T T F0 X0

Solve for XA:

nA0 i p,i i0

i 1A

RX R P R A0

F C T TX

ˆH (T ) C T T F

ni p,i i0

i 1A

RX R P R

C T TX

ˆH (T ) C T T

T = reaction temp Ti0 = initial (feed) temperature TR= reference temp

Solve this eq simultaneously with design equation Design eqs do not change, except k will be a function of T


L12-11

Nonisothermal Adiabatic Operation

ns A i pi i A RX R p Ri

ˆQ W F C ( T T ) F X H ( T ) C ( T T )

0 0 01

0

Constant or mean heat capacities

For a system with no shaft work ( ) & adiabatic operation ( ):0sW 0Q

ni pi i0

i 1

RX R p R

C (T T )X

ˆH (T ) C (T T )

Usually, p R RX RC ( T T ) H ( T )

Xenergy balance

Temperature

0sW0Q

CSTR, PFR, PBR, Batch

Adiabatic exothermic reactions


L12-12

Nonisothermal CSTRDesign equation (From mass balance) :

A0

A

F XV

r

nA0 i pi i0 A0 RX R p R

i 1s ˆF C (T T ) F X H (T ) C (T T )WQ 0

Energy balance: Coupled

With the exception of processes involving highly viscous materials,the work done by the stirrer can be neglected (i.e. ) 0sW

aQ UA(T T)With heat exchanger:

nA0 i pi i0 A0 RX R p R

i 1a ˆF C (T T ) F X HUA(T T) 0 (T ) C (T T ) 0

nai pi i0

i 1A0RX R p RˆX H (T ) C (T

UA(T T)C (T T T )

F)

na i pi i0 RX R p R

iA0 A0

1ˆUA(T T) C (T T ) X H (T ) C (T T )F F

RX

nai pi i0

i 1p R

A0R ˆX H (T ) C (

UA(T T)C (T T )

FT T )


L12-13

Application to CSTR

a) Solve TEB for T at the exit (Texit = Tinside reactor)b) Calculate k = Ae-E/RT where T was calculated in step ac) Plug the k calculated in step b into the design equation to calculate VCSTR

Case 1: Given FA0, CA0, A, E, Cpi, H°I, and XA, calculate T & V

a) Solve TEB for T as a function of XA

b) Solve CSTR design equation for XA as a function of T (plug in k = Ae-E/RT )c) Plot XA,EB vs T & XA,MB vs T on the same graph. The intersection of these 2

lines is the conditions (T and XA) that satisfies the energy & mass balance

Case 2: Given FA0, CA0, A, E, Cpi, H°I, and V, calculate T & XA

XA,EB = conversion determined from the TEB equationXA,MB = conversion determined using the design equation

XA

T

XA,EB

XA,MB

XA,exit

Texit

Intersection is T and XA that satisfies both equations


L12-14

Application to a Steady-State PFR

PFRFA0 FA

distance

TXA

Negligible shaft work (ẆS=0) and adiabatic (Q=0)

a) Use TEB to construct a table of T as a function of XA

b) Use k = Ae-E/RT to obtain k as a function of XA

c) Use stoichiometry to obtain –rA as a function of XA

d) Calculate:

XA AA0

A AXA0

dXV Fr X ,T


L12-15A first order reaction A(l) → B(l) is to be carried out adiabatically in a CSTR. Given A, E, T0, 0, CA0, and FA0, find the reactor volume that produces a conversion XA. The heat capacities of A & B are approximately equal, & ẆS=0.a) Solve TEB for T:

n

A0 i p,i i0 RX R P R A0 Ai 1

F C T T H (T ) C T T F X

n

s A0 i p,i i0 RX R P R A0 Ai 1

0 Q W F C T T H (T ) C T T F X

RX R P Rn

i p,i i0i

A1

HC T T (T ) C T T X

ni p,i i0

iRX R P

1i p,i PA RA AH (T ) C C TX XC C T XT T

Multiply out

ni p,i i

ni p,i P A 0RX R A P R A

i 1 i 1C C X H (T )X C T X CT TT

Factor out Tn

i p,i P A RX R A P R A p,A A0i 1

C C X H (T )X C T X C TT

A An

RX R P R i p,i i0i 1

ni p,i P

1A

i

XH (T ) C T C T

CT

C

X

X

Plug in values (∆Cp, ∆H°RX(TR), Cp,i) given in problem statement (look them up if necessary) & solve

Temp when specified XA is reached

0 0

Isolate T


L12-16A first order reaction A(l) → B(l) is to be carried out adiabatically in a CSTR. Given A, E, T0, 0, CA0, and FA0, find the reactor volume that produces a conversion XA. The heat capacities of A & B are approximately equal, & ẆS=0.a) Solve TEB for T of reaction when the specified XA is reached:

nRX R A P R A i p,i i0

i 1n

i p,i P Ai 1

H (T )X C T X C TT

C C X

b) Calculate k = Ae-E/RT where T was calculated in step (a) Look up E in a thermo book

c) Plug the k calculated for the reaction’s temperature when the specified XA is reached (in step b) into the design equation to calculate VCSTR

A0 A A0 A A0 A A0 0 A

A A A0 A A0 A

F X F X F X C XV V V V

r kC kC 1 X kC 1 X

0 A

A

XV

k 1 X


L12-17

Now, the first order reaction A(l) → B(l) is carried out adiabatically with and inlet temp of 300 K, CPA = 50 cal/mol∙K, and the heat of reaction = -20,000 cal/mol. Assume ẆS=0. The energy balance is:

ni pi 0

i 1EB

RX

C T TX

H T

A

ni pi p

i 1C 1 C

AP 0

EBRX

C T TX

H T

EB

50 T 300X

20000

From thermodynamicsXEB

T

From energy balance

RX RX R P RˆH T H (T ) C T T

nA0 i p,i i0 s

RX R P

iA

A0R

1ˆH (T ) C T

F C

T

T T W QX

F

0 0


L12-18The irreversible, elementary liquid-phase reaction 2A → B is carried out adiabatically in a flow reactor with ẆS=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be the temperature inside of a steady-state CSTR that achieved XA= 0.8? Extra info:E = 10,000 cal/mol CpA= 15 cal/mol•K CpB= 30 cal/mol•K CpI = 15 cal/mol•K ∆HA°(TR) = -20 kcal/mol ∆ HB°(TR) = -50 kcal/mol ∆HI°(TR) = -15 kcal/molk = 0.02 dm3/mol•s at 350 KStart with SS EB & solve for T:

ns A0 i p,i i0 RX A0 A

i 10 Q W F C T T H (T)F X

n

A0 i p,i i0 A0 Ai 1

RX0 0 0 F C T T H (T)F X

n

A0 i p,i i0 R APR 0Ri

X A1

HF C T T T TCT F X

n

i p,i i0 RX R P R Ai 1

C T T H (T ) C T T X

n ni p,i P A RX R A P R A i p,i i0

i 1 i 1C T C TX H (T )X C T X C T

Multiply out brackets & bring terms containing T to 1 side

nRX R A P R A i p,i i0

i 1n

i p,i P Ai 1

H (T )X C T X C TT

C C X


L12-19The irreversible, elementary liquid-phase reaction 2A → B is carried out adiabatically in a flow reactor with ẆS=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be the temperature inside of a steady-state CSTR that achieved XA= 0.8? Extra info:E = 10,000 cal/mol CpA= 15 cal/mol•K CpB= 30 cal/mol•K CpI = 15 cal/mol•K ∆HA°(TR) = -20 kcal/mol ∆ HB°(TR) = -50 kcal/mol ∆HI°(TR) = -15 kcal/molk = 0.02 dm3/mol•s at 350 K

Start with SS EB & solve for T:

ni p,i

i 1n

i p,i

A R

i 1

A i0

A

R PR

P

X X T X TC

CT

X

C

C

H (T )

B A

b CCp Cp pa C 0p 1 cal cal30 15

2 mol K mol KCp

n

i p,ii

B I1

Acal cal1 0 1 = 1 15 + 1 15

mocalC 30

ml K mo l Kl K o

d c bH T H T H T H TD R C RH TRX R B R A Ra a a

1 cal cal50,000 20,0002 mol m

H R lT

oX R

calH T 5000RX R mol


L12-20The irreversible, elementary liquid-phase reaction 2A → B is carried out adiabatically in a flow reactor with ẆS=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be the temperature inside of a steady-state CSTR that achieved XA= 0.8? Extra info:E = 10,000 cal/mol CpA= 15 cal/mol•K CpB= 30 cal/mol•K CpI = 15 cal/mol•K ∆HA°(TR) = -20 kcal/mol ∆ HB°(TR) = -50 kcal/mol ∆HI°(TR) = -15 kcal/molk = 0.02 dm3/mol•s at 350 K

Start with SS EB & solve for T:

ni p,i

i 1n

i p,i

A R

i 1

A i0

A

R PR

P

X X T X TC

CT

X

C

C

H (T )

C 0p n

i p,ii 1

calC 30mol K

calH T 5000RX R mol

Acal30

molcal5000

Kca

moll30

mol K

X 294T

0

0

K

Acal cal5000 X 8820mol molT

cal30mol K

X 0.8AT 166.67K 0.8 294K

AX 0.8T 427.3K


L12-21The irreversible, elementary liquid-phase reaction 2A → B is carried out adiabatically in a flow reactor with ẆS=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be volume of the steady-state CSTR that achieves XA= 0.8? Extra info:E = 10,000 cal/mol CpA= 15 cal/mol•K CpB= 30 cal/mol•K CpI = 15 cal/mol•K ∆HA°(TR) = -20 kcal/mol ∆ HB°(TR) = -50 kcal/mol ∆HI°(TR) = -15 kcal/molk = 0.02 dm3/mol•s at 350 K

Solve the CSTR design eq for V at XA = 0.8 & T = 427.3K:

3dm 10,000cal mol 1 1 0.02 expmol s 1.987cal mol K 350K

Need at 427.3K:427.3

k k

3dmk 0.02 exp 2.60124

mol s

3dmk 0.2696mol s

A0 ACSTR

A

F XV

-r

A2

Ar kC A A0 AStoichiometry : C C 1 X

A0 0 AC

A0 A22STRCombine :

C 1

C XV

k X

3

CSTR 32

3

dm5 0.8sV

dm mol0.2696 1 1 0.8mol s dm

3CSTRV 370.9dm


L12-22The irreversible, elementary liquid-phase reaction 2A → B is carried out adiabatically in a flow reactor with ẆS=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. Use the 2-point rule to numerically calculate the PFR volume required to achieve XA=0.8? Extra info:E = 10,000 cal/mol CpA= 15 cal/mol•K CpB= 30 cal/mol•K CpI = 15 cal/mol•K ∆HA°(TR) = -20 kcal/mol ∆ HB°(TR) = -50 kcal/mol ∆HI°(TR) = -15 kcal/molk = 0.02 dm3/mol•s at 350 K• Use the energy balance to construct table of T as a function of XA

• For each XA , calculate k, -rA and FA0/-rA • Use numeric evaluation to calculate VPFR

XA T(K) k(dm3/mol•s) -rA(mol/dm3•s) FA0/-rA(dm3)0 294*

0.8 427.3* 0.2696**Calculated in CSTR portion of this problem

3dm 1 1k 0.02 exp 5032.7126K

mol s 350K 294

3dmk 0.00129

mol s

0.00129

22A A0 Ar k C 1 X

X 0 X 0A A

22

A A6 3

3dm mol molr 0.00129 1 1 0 r 0.00129

mol s dm dm s

0.00129

X 0.8 X 0.8A A

22

A A6 3

3dm mol molr 0.2696 1 1 0.8 r 0.010784mol s dm dm s

0.010784




XA T(K) k(dm3/mol•s) -rA(mol/dm3•s) FA0/-rA(dm3)0 294 0.00129 0.00129

0.8 427.3 0.2696 0.010784

A0 A0 0F C

A

3A0

A X 03

mol5F s 3876 dmmolr 0.00129

dm s

3

A0 3

mol dm molF 1 5 5dm s s

A

3A0

A X 0.83

mol5F s 463.6 dmmolr 0.010784

dm s

3876463.6




XA T(K) k(dm3/mol•s) -rA(mol/dm3•s) FA0/-rA(dm3)0 294 0.00129 0.00129 3876

0.8 427.3 0.2696 0.010784 463.6

10 8 0 0 80 1 1 02

0

X h2-point rule: f x dx f X f X where h X X h . h .X

3 30 8 3876 463 62

PFR.V dm . dm

31736 PFRV dm

l12: nonisothermal reaction engineering

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