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Ch 8: Steady – State Non-isothermal Reactor DesignEnergy Balances, Rationale and Overview

Calculate the volume necessary to achieve a conversion, X, in a PFR for a first-order, exothermic reaction carried out adiabatically.

The combined mole balance, rate law, and stoichiometry yield:

For an adiabatic, exothermic reaction the temperature profile might look something like this:

To solve this equation we need to relate X and T. We will use the Energy Balance to relate X and T. For example, for an adiabatic reaction, , the energy balance can be written in the form

set X, then calculate T, -VA, and , increment X, then plot vs. X

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1. Adiabatic (Q = 0); CSTR, PFR, Batch and PBR:

2. CSTR with heat exchanger, UA(Ta-T) and large coolant flow rate.

3 . PFR/PBR with heat exchange

3A. In terms of conversion, X

3B. In terms of molar flow rates, Fi

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4. For Multiple Reactions

5. Coolant Balance

These are eq’ns we will use to solve rxn engineering problems with heat changes.

3. Energy Balance

Typical units for each term are J/s; i.e. Watts

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Energy Balance

1st Law of Thermodynamics:

dE = dQ – dW (for a closed system)

Energy of the system

Heat flow to the system

Work done by the system on the surroundings

system}open an {for E F - E F WQdtdE

outout

flowmassby system

the toaddedenergy of rate

inin 321+−=••

Evaluation of the Work Term:

s1

i

~

i1

i

~

i

sf

WV P FV P F

WWW

+⋅⋅+⋅⋅−=

+=

∑∑==

•

n

i out

n

i in

flow shaft

∑∑==

••

⋅+⋅−+=n

iout

n

iins

1ii

1ii

sys H FH FWQdt

dE

Combining with i

~

ii VP U H +=

5

(1)

1. Replace Ei by Ei=Hi-PVi

2. Express Hi in terms of enthalpies of formation and heat capacities 3. Express Fi in terms of either conversion or rates of reaction4. Define ∆HRX

5. Define ∆CP

6. Manipulate so that the overall energy balance is either in terms of the Equations above 1.A, 1.B, 2, 3A, 3B, or 4 depending on the application

Step 1:Substitute

and,

into equation (1) to obtain the General Energy Balance Equation.

General Energy Balance:

For steady state operation:

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We need to put the above equation into a form that we can easily use to relate X and T in order to size reactors. To achieve this goal, we write the molar flow rates in terms of conversion and the enthalpies as a function of temperature. We now will "dissect" both Fi and Hi.

Flow Rates, Fi

For the generalized reaction:

In general,

ad

ac

ab- 1-

FF where X) ( F F

DCBA

A0

i0iiiA0i

====

=Θ+Θ=

νννν

ν

(2)

(3)

[ ]

[ ]

∑∑∑

∑∑

∑∑

∆Θ=−

+=∆

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⋅⋅Θ=

+Θ+=⋅⋅

= ∆=

==

X F (T)H - )H - (H F H F H F

H - H ab - H

ac H

ad H

X F )H ( - )H - (H F

........... )H - (H )H - (H F F H - F H

Then

A0RXii0iA0iii0i0

ABCDRX

n

1iA0

H

ii

n

1iiii0A0

n

1iBBB0AA0A0ii

n

1ii0i0

RX

434 21ν (4)

(5)

(6)

.Hin included be it will place, takeschange phase a If

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

:(6) and (1) combining ;0dtdEst -st for

RX

n

1iA0RXii0iA0

∆

=∆Θ+

⎭⎬⎫

⎩⎨⎧ =

∑=

•• (7)

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Assuming no phase change:

QiRo

ii H )(TH H ∆+=

Enthalpy of formation at TR(ref. temp.)

Change in enthalpy when T is changed from TR to T.

If there is a phase change:

dT C )(TH H

change) phase no (if dT C H

dT C )(TH dT C )(TH H

T

TpiR

oii

T

TpiQi

T

Tipl,mmi

T

Tips,R

oii

R

2

1

R

M

R

∫

∫

∫∫

⋅+=

⋅=∆

+∆+⋅+=

(8)

(9)

(10)

(11)

2iiip T T C ⋅+⋅+= γβα

0 X F (T)H -]T -[T C F W - Q

:(7) into Substitute

]T -[T CdT C H H

n

1iA0RXi0piiA0

i0pi

T

Tpii0i

i0

=∆⋅⋅Θ+

⋅=⋅=−

∑

∫

=

••(13)

(12)

Combine eq’n (11) & (5)

pApBpCpDp

Ro

ARo

BRo

CRo

Do

RX

Rpo

RXRX

C - C ab - C

ac C

ad C

)(TH - )(TH ab - )(TH

ac )(TH

ad H

where

)T - (T C H H

+=∆

+=∆

∆+∆=∆ (14)

(15)

(16)

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[ ] 0 X F )T - (T C H -]T -[T C F W - Qn

1iA0Rp

oRXi0piiA0 =⋅∆+∆⋅⋅Θ+ ∑

=

••(17)

Combine (13) and (14)

[ ] )T - (T C H-)T - (T C

X 0Q

0W

Rpo

RX

iopiis

∆+∆

Θ=

⎪⎭

⎪⎬⎫

=

= ∑•

•

Adiabatic Operation:

T

Xfor an exothermic, adiabatic rxn for an exo. rxn why does X

increase?

This is from E balance, not mole balance.

)T - (T CH Rpo

RX ∆>∆

so X vs T is linear!!!

Adiabatic Tubular Reactor

Rearrange (18):

[ ]∑∑

∆⋅+⋅Θ

⋅∆⋅+⋅Θ+⋅=

ppii

Rpopiio

RX

C X CT C X T CH-X

T (19)

(20) T) (X,r- dVdX F AA0 =

Combine with differential mole balance:

To obtain T, X and conc’n profiles along the reactor!

Use (19) to construct a table of T vs X.

Obtain k(T) as a function of X -rA as a function of X.

Look at Table 8-2A.

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TFA0

To

FAe

Te

∑ ⋅ ii HF ∑ ⋅ ii HF

V V+∆V

0 Ws assume V T) - (Ta Ua T) - (TaA U Q =∆⋅⋅=∆⋅=∆••

Steady – State Tubular Reactor with Heat Exchange:

0H FH FQVViiVii =⋅−⋅+∆ ∑∑ ∆+

•

•

∆Q

(1)

The heat flow to the reactor

Overall heat transfer coefficient (U)

Heat exchange area (∆A)

Difference between ambient temperature (Ta) and reactor temperature (T)

T)-(Ta V Ua T) - (TaA UQ ⋅∆⋅=∆⋅=∆•

(2)

DVA 4= Diameter of reactor

If ∆V 0

dVdHF - H

dVdF

T)-(Ta Ua

(3)n Eq' diff.

)(-r r dVdF :balance mole

dV)H (F d

T)-(Ta Ua

iii

i

Aiii

ii

∑∑

∑

⋅=

==

⋅=

ν

(3)

(4)

(5)

dVdT C

dVdH

dT C H

pii

pii

⋅=

⋅=

(6)

Combine (5) & (4) & (6)

dVdTCF - )(-r H T)-(Ta Ua piiA

H

ii

RX

⋅⋅⋅= ∑∑∆

434 21ν (7)

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Rearrange:

∑ ⋅−∆⋅

=pii

removedHeat generatedHeat

RXA

C FTa) -(T UaHr

dVdT

4 84 76484 76

(8)

X) ( F F iA0i ν+Θ⋅=

( ) { }PFR afor XC X F

Hr T) -(Ta Ua dVdT

piA0

RXA

∑ ⋅∆++Θ⋅∆⋅+

=νi

Substitute into (8)

for a PBR {dW = ρb dV}( )

∑ ⋅

∆⋅+⋅=

pii

RXAb

C F

H'r T) -(Ta Ua

dWdT ρ

These eq’ns will be coupled with mole balance eq’nsA0

A

Fr-

dVdX

=

Tao

FA0, To

V V + ∆V

R1R2

Reactants

Heat transfer fluid

The fluid will keep the rxn temperature constant for endo/exo – thermic rxns

You might have

A. Co – current Flow

B. Counter – current Flow

Balance on the Coolant Heat Transfer Fluid:

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FA, T

V V + ∆V

FA, T

Ta: coolant temperature

mc, Hc mc, Hc

0 V Ta)-(T UaH m -H m

0 Q E -E

VVcc

Vcc

conductionVV

outV

in

=∆+

=+

∆+

••

•

∆+

••

Divide by ∆V and take limit as ∆V 0

A. Co – Current Flow

The Energy Balance

Cm

Ta) - (T Ua dVdT

dVdTC

dVdH

0 Ta) - (T Ua dVdHm

p

aap

c

c

⋅=⇒=

=+⋅

•

•

c

cexothermic

endothermic

Tao

Tao

V

TFAo, To

V V + ∆V

Ta2 Ta

V = 0 V = Vf

Cm

T) - (Ta Ua dVdT

pc

a

⋅= •

c

At the entrance X = 0; V = 0; Ta = Ta2

At the exit V = Vf; Ta = Tao

The sol’n to the counter-current flow problem to find {T X} is a trial & error procedure.

Assume a coolant temp at the entrance (Ta2)

Solve ODEs to calculate X, T and Ta as a function of V:

Find Ta(V = Vo)

a2

a2oaa2oa

Tanother assume Else

T)V(VT T-)V(VT If ==→<= ε

B. Counter – Current Flow

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Equilibrium Conversion

As T X {For endothermic rxns}

As T X {For exothermic rxns}

Exothermic Rxns:

rxn}order {1st Kc 1

Kc Xe+

=

From Le Chaltlier’s Law

(Kc as T if ∆H<0) exothermic γ)(RTK

K pc =

To find the max X in an exothermic rxn carried adiabatically:

To1To

Xe

Xe1

Adabatic temperature

To1 >To

Energy balance

⎭⎬⎫

⎩⎨⎧

∆−

⋅⋅Θ=∑

X T as T tochanged is T if

)T(H)T-(T C

X

e

o1o

RXN

opiie

11R

)(THexp )(TK )(TK

0C if

T R)T-T(C )(TH

T R(T)H

dTdlnK

21

RRXNo

1p2p

~

2R

~

RRXNo

2RXNp

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆=

=∆

⋅∆+∆

=⋅

∆=

TT

p

p

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Xe

Xeb

To increase the conversion in an exothermic rxn, use multiple reactors with interstate cooling:

For Endothermic Rxn (you need heating)

Xe

3

2

1

350 500 600

X

T, K

As T Xe but the (-rA) decreases!

(so the conversion is achieved at the end of the reactor)

So there must be an optimum temperature to achieve max X.

Curve A: Rxn rate slow, rxn dictated by rate of rxn and reactor volume

As T , r , X

Curve B: Rxn rate very rapid. Virtual equilibrium reached in X dictated by equilibrium conversion

Optimum Feed Temperature

Adiabatic Reactor of fixed size

Reversible & Exothermic Rxn T ; X as 0H if

XCH-T T

rx

pA

rxo

<∆

⋅∆

=

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Adiabatic Rxn Algorithm

Suppose

1. Choose X

Calculate T

Calculate k

Calculate T/To

Calculate CA

Calculate CB

Calculate KC

Calculate -rA

2. Increment X and then repeat calculations.

3. When finished, plot vs. X or use some numerical technique to find V.

Levenspiel Plot for anexothermic, adiabatic reaction.

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Consider:

For an exit conversion of 70%For an exit conversion of 40%PFR Shaded area is the volume.

For an exit conversion of 70%For an exit conversion of 40%

CSTR Shaded area is the reactor volume.

We see for 40% conversion very little volume is required.

CSTR+PFR

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For an intermediate conversion of 40% and exit conversion of 70%

Looks like the best arrangement is a CSTR with a 40% conversion followed by a PFR up to 70% conversion.

Evaluating the Heat Exchanger Term

Energy transferred between the reactor and the coolant:

Assuming the temperature inside the CSTR, T, is spatially uniform:

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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:

Since the coolant flow rate is high, Ta1 Ta2 Ta:

Multiple Steady States (MSS)

where Heat generated term

Heat removed term

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R(T):

Varying Entering Temperature

Vary non-adiabatic parameter κ:

if you increase FA0 (molar flow rate) or decrease heat – exch area then κ will decrease

Increase To

Slope = Cpo (1+κ)

T

R(T)

κ decrease

T

R(T)κ=∞

κ= 0

Ta To

κκ

κ

+⋅+

=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=

1T T T

C FUa

a0c

p0A0

Low E

High EG(T)

T

increasing TG(T)

T

G(T)

For a first order rxn:

kkX⋅+

⋅=

ττ

1

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Extinction temp

Ignition temp

Lower st-st

Upper st-st

Ignition – Extinction Curve:

The point of intersection of G(T) and R(T) give Tst-st.

By plotting Tst-st vs To, we obtain ignition – extinction curve.

As To Tst-st

Runaway Rxns in a CSTR: R(T),

G(T)

Tc T* Tangency pointκκ+⋅+

=1

T T T a0c

E

*2

c*

rcTR T-TT ⋅

==∆

Reactor T

if this diff. is exceeded, transition to the upper st – st will occur.

At this high temp, it is undesirable or even dangerous.

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PFR

rxn) single a(for C F

(T))H)(-(-rT)-(T UadVdT

m

1ipii

RxnAa

∑=

∆+=

When q multiple rxns occur with m species:

[ ]species :j

rxn : i

C F

(T)H-)(-rT)-(T Ua

dVdT

m

1jpjj

1ijRxn,ija

∑

∑

=

=

∆+=

q

i

Ex:

[ ] [ ]

C FC FC F(T)H-)(-r(T)H-)(-rT)-(T Ua

dVdT

pCCpBBpAA

Rxn,2B2BRxn,1A1Aa

++∆+∆+

=

C B

B A

:1Rxn

2

1

k

k

⎯→⎯

⎯→⎯

CSTR

[ ] [ ] rxn) single a(for 0 Vr(T)H -]T -[T C F W - Qn

1iARXN0piiA0s =⋅∆⋅⋅Θ+ ∑

=

••

0 (T)HrV-]T -[T C F T)-UA(T

0 (T)HrV-]T -[T C F W - Q

n

1i 1ijRXN,ij0piiA0a

n

1i 1ijRXN,ij0piiA0s

=∆⋅⋅⋅Θ+

=∆⋅⋅⋅Θ+

∑ ∑

∑ ∑

= =

= =

••

q

i

q

i

When q multiple rxns take place0