Steady State Steady State NonisothermalNonisothermal ...libvolume2.xyz/biotechnology/semester7/bioreactor... · Steady State Steady State NonisothermalNonisothermal Reactor Design

Steady State Steady State NonisothermalNonisothermal

Reactor DesignReactor Design

DickyDicky DermawanDermawanwww.dickydermawan.net78.netwww.dickydermawan.net78.net

[email protected]@gmail.com

ITKITK--330 Chemical Reaction Engineering330 Chemical Reaction Engineering

RationaleRationale

� All reactions always accompanied by heat effect: exothermic reactions vs. endothermic reactions

� Unless heat transfer system is carefully designed, reaction mass temperature tend to change

� Design of heat transfer system itself requires the understanding of this heat effect

� Energy balance is also needed, together with performance equations derived from mass balance

ObjectivesObjectives

� Describe the algorithm for CSTRs, PFRs, and PBRs that are not operated isothermally.

� Size adiabatic and nonadiabatic CSTRs, PFRs, and PBRs.

� Use reactor staging to obtain high conversions for highly exothermic reversible reactions.

� Carry out an analysis to determine the Multiple Steady States (MSS) in a CSTR along with the ignition and extinction temperatures.

� Analyze multiple reactions carried out in CSTRs, PFRs, and PBRs which are not operated isothermally in order to determine the concentrations and temperature as a function of position (PFR/PBR) and operating variables

Why Energy Balance?Why Energy Balance?

Imagine that we are designing a nonisothermal PFR for a Imagine that we are designing a nonisothermal PFR for a first order liquid phase exothermic reaction:first order liquid phase exothermic reaction:

Performance Performance equation:equation:

0A

A

F

r

dV

dX −=

Kinetics:Kinetics: =− Ar k AC⋅

The temperature will increase with conversion down

the length of reactor

−⋅⋅=T

1

T

1

R

Eexpkk

1

a1

Stoichiometry:Stoichiometry: 0υ=υ A0A CF ⋅υ=

0A00A CF ⋅υ=)X1(CC 0AA −⋅=

Combine:Combine:

0

X1

υ

−

−⋅⋅=

1

a1

T

1

R

Eexpk

dV

dX

T

1)V,T(XX =

)V(TT =

)X(TT =)V(XX =

Energy BalanceEnergy Balance

∑=

⋅+⋅+⋅+⋅+⋅=⋅n

1i

0I0I0D0D0C0C0B0B0A0A0i0i HFHFHFHFHFHF:In

∑=

⋅+⋅+⋅+⋅+⋅=⋅n

1i

IIDDCCBBAAii HFHFHFHFHFHFOut

At steady state:

dt

EdHFHFWQ

sysn

1i

ii

n

1i

0i0is =⋅−⋅+− ∑∑==

&&

∑=

+−n

1i

sWQ &&0iF 0iH ∑

=

−n

1iiF iH 0=

Consider generalized reaction:

DCBAad

ac

ab +→+

I0AI

ad

D0AD

ac

C0AC

ab

B0AB

0AA

FF

)X(FF

)X(FF

)X(FF

)X1(FF

Θ⋅=

+Θ⋅=

+Θ⋅=

−Θ⋅=

−⋅=

Upon substitution:

( )ABab

Cac

Dad

A0 HHHHXF- −−+⋅⋅

∑∑==

⋅−⋅n

1i

ii

n

1i

0i0i HFHF( ) ( ) ( )

( ) ( )

−⋅Θ+−⋅Θ+

−⋅Θ+−⋅Θ+−⋅=

CI0IID0DD

C0CCB0BBA0A0A HHHH

HHHHHHF

∑∑==

⋅−⋅n

1i

ii

n

1i

0i0i HFHF ( )∑=

−⋅Θ⋅=n

1i

i0ii0A HHF )T(HXF Rx0A ∆⋅⋅−

Energy Balance (cont’)Energy Balance (cont’)

∑∑==

⋅−⋅n

1i

ii

n

1i

0i0i HFHF ( )∑=

−⋅Θ⋅−=n

1i

0iii0A HHF )T(HXF Rx0A ∆⋅⋅−

∫ ⋅+=T

T

piRoii

R

dTC)T(HH

From thermodynamics, we know that:

∫ ⋅+=0i

R

T

T

piRoi0i dTC)T(HH Thus: )TT(C

~dTCHH 0ipi

T

T

pi0ii

0i

−⋅=⋅=− ∫

0i

T

T

pi

piTT

dTC

C~ 0i

−

⋅

=

∫

( )RpRoRxRx TTC)T(H)T(H −⋅∆+∆=∆

R

T

T

pi

piTT

dTC

C R

−

⋅∆

=∆

∫

)T(H)T(H)T(H)T(H)T(H RoDR

oDa

bR

oDa

cR

oDa

dR

oRx −⋅−⋅+⋅=∆

pApBab

pCac

pDad

p CCCCC −⋅−⋅+⋅=∆

∑=

−⋅⋅Θ⋅−=n

1i

0pii0A )TT(C~

F

Energy Balance (cont’)Energy Balance (cont’)

∑∑==

⋅−⋅n

1i

ii

n

1i

0i0i HFHF ( )∑=

−⋅Θ⋅−=n

1i

0iii0A HHF )T(HXF Rx0A ∆⋅⋅−Upon substitution:

∑∑==

⋅−⋅n

1i

ii

n

1i

0i0i HFHF ( )]TTC)T(H[XF RpRoRx0A −⋅∆+∆⋅⋅−

Finally;.

0HFHFWQn

1iii

n

1i0i0is =⋅−⋅+− ∑∑

==

&&

( ) 0TTC)T(HXF)TT(C~

FWQ RpRoRx0A

n

1i

0ipii0As =−⋅∆+∆⋅⋅−−⋅⋅Θ⋅−− ∑=

&&

So what?

Energy Balance (cont’)Energy Balance (cont’)For adiabatic reactions:

The energy balance at steady state becomes:

After rearrangement:

0Q =&

When work is negligible: 0Ws =&

( )[ ] 0TTC)T(HXF)TT(C~

F RpRoRx0A

n

1i

0ipii0A =−⋅∆+∆⋅⋅−−⋅⋅Θ⋅− ∑=

( )[ ]RpRoRx

n

1i

0ipii

TTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ

=∑=

This is the X=X(T) we’ve been looking for!

Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR Design

)X1(CC 0AA −⋅=

Case A: Sizing: X specified, calculate V (and T)

Performance equation:

Kinetics:

Stoichiometry:

Combine:

A

0A

r

XFV

−

⋅=

=− Ar k AC⋅

−⋅⋅=T

1

T

1

R

Eexpkk

1

a1

)X1(Ck

XFV

0A

0A

−⋅⋅

⋅=

Solve the energy balance for T

( )[ ]RpRoRx

n

1i

0ipii

TTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ

=∑=

Calculate k

Calculate V using combining equation


)X1(CC 0AA −⋅=

Case B (Rating): V specified, calculate X (and T)


Kinetics:

Stoichiometry:

Mole balance:

A

0A

r

XFV

−

⋅=

=− Ar k AC⋅

−⋅⋅=T

1

T

1

R

Eexpkk

1

a1

)X1(Ck

XFV

mb0A

mb0A

−⋅⋅

⋅=

Energy balance:( )[ ]RpR

oRx

n

1i

0ipii

ebTTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ

=∑=

Find X & T that satisfy BOTH the material balance and energy balance,

viz. plot Xmb vs T and Xeb vs T in the same graph: the intersection is the solution


Example: P8-5A

The elementary irreversible organic liquid-phase reaction:

A + B → C

is carried out adiabatically in a CSTR. An equal molar feed in A and B enters at 27oC, and the volumetric flow rate is 2 L/s.

(a) Calculate the CSTR volume necessary to achieve 85% conversion

(b) Calculate the conversion that can be achieved in one 500 L CSTR and in two 250 L CSTRs in series

mol/kcal 41)K273(H

mol/kcal 15)K273(H

mol/kcal 20)K273(H

oC

oB

oA

−=

−=

−=

cal/mol.K 30C

cal/mol.K 15C

cal/mol.K 15C

pC

pB

pA

=

=

=

cal/mol 10000E

K 300at 01.0k

a

smolL

=

=⋅

mol/L 1.0C 0A =

Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR DesignCase A: Sizing: X specified, calculate V (and T)


Kinetics:

Stoichiometry:

Combine:

A

0A

r

XFV

−

⋅=

=− Ar k BA CC ⋅⋅

−⋅⋅=T

1

T

1

R

Eexpkk

1

a1

20A

022

0A

0A

)X1(Ck

X

)X1(Ck

XFV

−⋅⋅

⋅υ=

−⋅⋅

⋅=

Energy balance:

( )[ ]RpRoRx

n

1i

0ipii

TTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ

=∑=

Calculate k

Calculate V using combining equation

)X1(CC 0AA −⋅=

)X1(C)X(CC 0ABB0AB −⋅=⋅ν−Θ⋅=

Kcal/mol 301515CCC~

pBBpA

n

1i

pii ⋅=+=⋅Θ+=⋅Θ∑=

cal/mol 6000- kcal/mol 6152041HHH)273(H oB

0A

oC

oRx =−=++−=−−=∆

0151530CCCC pBpApCp =−−=−−=∆

K47020085.0300T200

300T

)6000(

)300T(3085.0 =⋅+=⇒

−=

−−−⋅

=

smol

L 317.4

470

1

300

1

987.1

10000exp01.0k

⋅=

−⋅⋅=

L 175)85.01(1.0317.4

85.02V

2=

−⋅⋅

⋅=


)X1(CCC 0ABA −⋅==

( )[ ]RpRoRx

n

1i

0ipii

ebTTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ

=∑=

Case B (Rating): V specified, calculate X (and T)


Kinetics:

Stoichiometry:

Mole balance:

A

0A

r

XFV

−

⋅=

−⋅⋅=T

1

T

1

R

Eexpkk

1

a1

2mb0A

mb0

)X1(Ck

XV

−⋅⋅

⋅υ=

Energy balance:

=− Ar k BA CC ⋅⋅

2mb

mb

)X1(1.0T

1

300

1

987.1

10000exp01.0

X2500

−⋅⋅

−⋅⋅

⋅=

200

300T

)6000(

)300T(30Xeb

−=

−−

−⋅=

0

0.2

0.4

0.6

0.8

1

300 350 400 450 500

Xmb

Xeb

T 300 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 482 484 485 490 500

Xmb 0.172 0.245 0.325 0.406 0.482 0.552 0.613 0.666 0.711 0.750 0.783 0.810 0.834 0.854 0.871 0.885 0.898 0.908 0.918 0.919 0.921 0.922 0.926 0.933

Xeb 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.910 0.920 0.925 0.950 1.000

Application to Adiabatic PFR/PBR DesignApplication to Adiabatic PFR/PBR Design

T

T

P

P

X1

X1CC 0

0

0AA ⋅⋅⋅ε+

−⋅=

( )[ ]

TTC)T(H

)TT(C~

XRpR

o

Rx

n

1i

0ipii ⇒

−⋅∆+∆−

−⋅⋅Θ=

∑=

Example for First Order Reaction


Kinetics:

Stoichiometry:

Pressure drop:

−⋅⋅=T

1

T

1

R

Eexpkk

1

a1

Energy balance:

=− Ar k AC⋅

for PFR/small ∆P: P/P0 = 1

)X1(P/P

P

T

T

2dW

dP

0

0

0

⋅ε+⋅⋅⋅α

−=

)X1(CC0AA

−⋅=Gas liquid

0A

A

F

r

dW

dX −=

[ ]

[ ]

[ ]

p

n

1i

pii

n

1i

0piiRp

o

Rx

n

1i

0piiRp

o

Rxp

n

1i

pii

n

1i

n

1i

0piipiiRpp

o

Rx

CXC~

TC~

TCXHX

T

TC~

TCXHXTCXTC~

TC~

TC~

TCXTCXHX

∆⋅+⋅Θ

⋅⋅Θ+⋅∆⋅+∆−⋅=

⋅⋅Θ+⋅∆⋅+∆−⋅=⋅∆⋅+⋅⋅Θ

⋅⋅Θ−⋅⋅Θ=⋅∆⋅+⋅∆⋅−∆−⋅

∑

∑

∑∑

∑ ∑

=

=

==

= =

)X(TT =

Combine:

)X(k)X(TT

)T(kk

=

=)P,X(CC

)X(TT

)P,T,X(CCAA

AA =

=

=

)P,X(rr

])P,X[C],X[k(rr

AA

AAA

−=−

−=−

)P,X(g)P,T,X(gdW

dP

)P,X(f )r(fdW

dXA

==

=−= Thus

The combination results in 2 simultaneous differential equations

Sample Sample

Problem Problem

for for

Adiabatic Adiabatic

PFR PFR

DesignDesign

P8-6A

Sample Problem for Adiabatic PBR Sample Problem for Adiabatic PBR

DesignDesign

NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the ReactorHeat Transfer Rate to the Reactor

)TT(CFW1i0ipii0As−−⋅⋅Θ⋅−−∑=&

⇒

Rate of energy transferred between the reactor and the coolant:

The rate of heat transfer from the exchanger to the reactor:

−

−

−⋅⋅=

2a

1a

2a1a

TT

TTln

TTAUQ&

⇒Combining:

⇓⇐⇐

⇓

⇓⇒ ⇒

0HXF)TT(C~

FW Rx0A

n

1i

0ipii0As =∆⋅⋅−−⋅⋅Θ⋅−− ∑=

&Q&

NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the Reactor (cont’)Heat Transfer Rate to the Reactor (cont’)

)TT(CFW1i0ipii0As−−⋅⋅Θ⋅−−∑=&

⇒⇓

At high coolant flow rates the exponential term will be small,

so we can expand the exponential term as a Taylor Series, where the terms of second

order or greater are neglected:

Then:

0HXF)TT(C~

FW Rx0A

n

1i

0ipii0As =∆⋅⋅−−⋅⋅Θ⋅−− ∑=

&( )TTAU 1a −⋅⋅

The energy balance becomes:

SampleSample

Problem forProblem for

DiabaticDiabatic

CSTRCSTR

DesignDesign

P8P8--4B4B

Sample Problem for Diabatic CSTR DesignSample Problem for Diabatic CSTR Design

Application of Energy Balance to Diabatic Application of Energy Balance to Diabatic

Tubular Reactor DesignTubular Reactor Design

Heat transfer in CSTR: ( )TTAUQ 1a −⋅⋅=&

In PFR, T varies along the reactor:

( ) ( ) dVTTV

AUdATTUQ

V

a

A

a ⋅−⋅⋅=⋅−⋅= ∫∫&

( )TTaUdV

Qda −⋅⋅=

&

Thus:

D

4

L

LD a

areaktor tabung volume

reaktor tabung selimut luas

V

A

4D

2=

⋅

⋅⋅π=

==

⋅π

For PBR: dW1

dVW

VV

W

bbb ⋅

ρ=⇒

ρ=⇔=ρ

Thus: ( )TTaU

dW

Qda

b

−⋅ρ⋅

=&

Application of Energy Balance to Diabatic Tubular Application of Energy Balance to Diabatic Tubular

Reactor DesignReactor Design

The steady state energy balance, neglecting work term:

Differentiation with respect to the volume V:

( )TTaUdV

Qda −⋅⋅=

&

and recalling that

Or:

( ) 0TTC)T(HXF)TT(C~

FQ RpRoRx0A

n

1i0ipii0A =−⋅+⋅⋅−−⋅⋅⋅− ∑

=∆∆Θ&

0dTC)T(HXFdTCFQT

TpR

oRx0A

T

Tpii0A

Ro

=

⋅+⋅⋅−⋅⋅⋅− ∫∫ ∑ ∆∆Θ&

Inserting

0dV

dXdTC)T(HF

dV

dTCXF

dV

dTCF

dV

Qd T

T

pRoRx0Ap0Apii0A

R

=⋅

⋅+⋅−⋅⋅⋅−⋅⋅⋅− ∫∑ ∆∆∆Θ

&

dV

dXFr 0AA ⋅=−

( )TTaU a −⋅⋅ ( )dV

dTCXCF ppii0A ⋅⋅+⋅⋅− ∑ ∆Θ ( ) )]T(H[r

RxA ∆−⋅−+ 0=

( ) ( )( )ppii0A

RxAa

CXCF

)]T(H[rTTaU

dV

dT

∆Θ

∆

⋅+⋅⋅

−⋅−+−⋅⋅=

∑

Coupled with 0A

A

F

r

dV

dX −=

)T,X(g=

)T,X(f=

Form 2 differential with 2

dependent variables X & T

Sample Problem for Diabatic Tubular Reactor Sample Problem for Diabatic Tubular Reactor

DesignDesign

Design for Reversible ReactionsDesign for Reversible Reactions

Endotermik: Endotermik: K naik dengan kenaikan T XXeqeq naik naik reaksikan

pada Tmax yang diperkenankan

K lnTRG ⋅⋅−=∆

2

Rx

TR

H

dT

K) (lnd

⋅=

∆

Eksotermik: Eksotermik: K turun dengan kenaikan T XXeqeq turun turun reaksikan

pada T rendah

→ →

→ →

Laju reaksi lambat pada T rendah!

Ada trade off antara aspek termodinamika dan kinetika

Xeq = Xeq (K)

= Xeq (T)

Design for Reversible HighlyDesign for Reversible Highly--Exothermic Exothermic

ReactionsReactions

--rrAA = = --rrAA (X,T)(X,T)

Generally:Generally: Higher X Higher X �� slower reaction rateslower reaction rate

Higher T Higher T �� faster ratefaster rate

At X = At X = XXeqeq : : --rrAA = 0= 0

Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic Exothermic

ReactionsReactions

#1#1 Starting with RStarting with R--free solution, between 0 free solution, between 0 dandan 100100ooC C

determine the equilibrium conversion of A for the elementary determine the equilibrium conversion of A for the elementary

aqueous reaction:aqueous reaction:

A A �� RR cal/mol 18000H

cal/mol 3375G0298

0298

−=

−=

∆

∆

The reported data is based on the following standard states of The reported data is based on the following standard states of

reactants and products:reactants and products: 1mol/LCC 0A

0R

==

Assume ideal solution, in which case:Assume ideal solution, in which case: CA

R

0AA

0RR

KC

C

C/C

C/CK ===

In addition, assume specific heats of all solutions are equal In addition, assume specific heats of all solutions are equal

to that of waterto that of waterCcal/g. 1C

0p =


Reactions:Reactions:

Reaction Rate in X Reaction Rate in X –– T DiagramT Diagram

k T( ) 0.0918exp 5859−1

T

1

298−

⋅

⋅:=

rA X T,( ) k T( )− CA0⋅ 1 X−X

K T( )−

⋅:=

Reaction Rate in The X Reaction Rate in The X –– T DiagramT Diagram

at Cat CA0A0 = 1 mol/L= 1 mol/L

0 10 20 30 40 50 60 70 80 90 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Suhu, C

Konversi

rA

− 0 01,

rA

− 0 025,

rA

− 0 05,

rA

− 0 1,

rA

− 0 25,

rA

− 0 5,

rA

− 1

rA

− 2

rA

− 4

Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic ReactionsExothermic Reactions::

Optimum Temperature ProgressionOptimum Temperature Progression

in Tubular Reactorin Tubular Reactor

#3#3

a.a. Calculate the space time needed for 80% conversion of a feed Calculate the space time needed for 80% conversion of a feed

starting with initial concentration of A of 1 mol/Lstarting with initial concentration of A of 1 mol/L

b.b. Plot the temperature and conversion profile along the length of Plot the temperature and conversion profile along the length of

the reactorthe reactor

Let the maximum operating allowable temperature be 95Let the maximum operating allowable temperature be 95ooCC

Design for Reversible Reactions: Heat EffectDesign for Reversible Reactions: Heat Effect

( )[ ]RpRoRx

n

1i0ipiia

0A

TTC)T(H

)TT(C~

)TT(F

AU

X−⋅+−

−⋅⋅+−⋅⋅

=∑=

∆∆

Θ


ReactionsReactions:: CSTR PerformanceCSTR Performance


ReactionsReactions:: CSTR PerformanceCSTR Performance

#4#4 A concentrated aqueous AA concentrated aqueous A--solution of the previous solution of the previous examples, Cexamples, CA0A0 = 4 mol/L, F= 4 mol/L, FA0A0 = 1000 mol/min, is to be 80% = 1000 mol/min, is to be 80% converted in a mixed reactor. converted in a mixed reactor.

a.a. If feed enters at 25If feed enters at 25ooC, what size of reactor is needed?C, what size of reactor is needed?

b.b. What is the optimum operating temperature for this What is the optimum operating temperature for this purpose?purpose?

c.c. What size of reactor is needed if feed enters at optimum What size of reactor is needed if feed enters at optimum temperature?temperature?

d.d. What is the heat duty if feed enters at 25What is the heat duty if feed enters at 25ooC to keep the C to keep the reactor operation at its the optimum temperature?reactor operation at its the optimum temperature?

Interstage CoolingInterstage Cooling

Review on Energy Balance in CSTR OperationReview on Energy Balance in CSTR Operation

Bila term kerja diabaikan dan ∆HRx konstan:

∑=

−⋅⋅Θ⋅−−n

1i

0pii0As )TT(C~

FW&( )TTAU a −⋅⋅ 0HXF 0Rx0A =∆⋅⋅−

XF 0A ⋅ ( ) ( )

−⋅

⋅+−⋅⋅Θ⋅=∆−⋅ ∑

=a

0A

n

1i

0pii0A0Rx TT

F

AU)TT(C

~FH

Untuk CSTR:A

0A

r

XFV

−⋅

=

( )VrA ⋅− ( ) ( )

−⋅

⋅+−⋅⋅=∆−⋅ a

0A00p0A

0Rx TT

F

AU)TT(CFH

Pembagian kedua ruas dengan FA0:

( ) ⋅+−⋅=∆−⋅

⋅−0p00p

0Rx

0A

A C)TT(CHF

Vr

0p0A CF

AU

⋅⋅

( )aTT −⋅

0p0A CF

AU

⋅⋅

=κ1

TT

1

TT

CFAU

TAUTCFT a0

CFAU

aCFAU

0

0p0A

a00p0Ac

0p0A

0p0A

+κ⋅κ+

=+

⋅+=

⋅+⋅

⋅⋅+⋅⋅=

⋅⋅

⋅⋅

Multiple Steady Multiple Steady

State & Stability of State & Stability of

CSTR OperationCSTR Operation

1

TTT a0c +κ

⋅κ+=

κ

)TT( a0 ⋅κ+

( ) +−⋅=∆−⋅

⋅−)TT[(CH

F

Vr00p

0Rx

0A

A ( )]TT a−⋅

)1(Tc +κ⋅−+κ⋅⋅= )1(T[C 0p

)TT()1(C c0p −⋅+κ⋅=

−⋅κ+⋅= TT[C 0p ]

]

( )0RxHX ∆−⋅ )TT()1(C c0p −⋅+κ⋅=

)T(G )T(R=

A

0A

r

XFV

−

⋅=

( )[ ] TTC)T(H

)TT(C~

X

dengan Bandingkan

RpRoRx

n

1i0pii

−⋅∆+∆−

−⋅⋅Θ

=∑=

Review on Energy Balance in CSTR OperationReview on Energy Balance in CSTR Operation

Multiple Steady Multiple Steady State: Stability of CSTR State: Stability of CSTR

OperationOperation

Temperature Ignition – Extinction CurveFinding Multiple Steady State: Varying To

Upper steady state

Lower steady state

Ignition temperature

Extinction temperature

Runaway Reaction

Sample Problem on Multiple Steady State in Sample Problem on Multiple Steady State in

CSTR OperationCSTR OperationP8P8--17B17B

Documents

Steady State Steady State NonisothermalNonisothermal ...libvolume2.xyz/biotechnology/semester7/bioreactor... · Steady State Steady State NonisothermalNonisothermal Reactor Design