Plug Flow Reactor Equilibrium Conversion

7/25/2019 Plug Flow Reactor Equilibrium Conversion

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The highest conversion that can be achieved inreversible reactions is the equilibrium conversion.

For endothermic reactions, the equilibrium conversion

increases with increasing temperature up to amaximum of 1.0.

For exothermic reactions the equilibrium conversion

decreases with increasing temperature.


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EXOTHERMIC REACTIONS

To determine the maximum conversion that can be achieved inan exothermic reaction carried out adiabatically, we find the

intersection of the equilibrium conversion as a function of

temperature:

T

TpR

0

Rx

n

1i

T

Tpii

R

0

dTCTH

dTC

X (1.57)

with temperature-conversion relationships from the energy

balance (eq. 1.46). For Ti0 = T0,

C

C

eK1

KX


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Graphical solution of equilibrium and energy balance equations

to obtain the adiabatic temperature and the adiabatic equilibrium

conversion Xe


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EXAMPLE 1.5

For the elementary solid-catalyzed liquid-phase reaction:

A Bmake a plot of equilibrium conversion as a function of

temperature. Determine the adiabatic equilibrium

temperature and conversion when pure A is fed to the

reactor at a temperature of 300 K.

Additional information:

molcal000,40K298H0

A

molcal000,60K298H0

B

K.molcal50CAp K.molcal50C

Bp

K298at000,100Ke


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SOLUTION

e

B

AA

K

CCkrRate law:

Equilibrium, rA = 0, so:

C

Be

AeK

CC

Stoichiometry (v = v0) yields

C

e0A

e0AKXCX1C

TK1

TKX

e

e

e

(a)

(b)


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Equilibrium constant:

T

1

T

1

R

HexpTKTK

1

0

Rx

1ee

molcal000,20HHH 0

A

0

B

0

Rx

T

1

298

1

987.1

000,20exp000,100TK

e

T

298T78.33exp000,100TK

e

T298T78.33exp000,1001

T298T78.33exp000,100X

e

Substituting eq. (c) into (b)

(c)

(d)


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T (K) Ke Xe

298 100000.00 1.0000

300 79835.65 1.0000

325 6042.58 0.9998

350 661.28 0.9985

375 97.20 0.9898

400 18.16 0.9478

425 4.13 0.8051

450 1.11 0.5257

475 0.34 0.2545

500 0.12 0.1058


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For a reaction carried out adiabatically, the energy balance

reduces to

THTTC

THTTC

X

Rx

0p

Rx

0pii A

300T105.2000,20

300T50X

3

T Xe

300 0.0000

350 0.1250

400 0.2500

450 0.3750

500 0.5000

(e)


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0.00

0.20

0.40

0.60

0.80

1.00

1.20

300 350 400 450 500 550 600

X

T (K)

TK1

TKX

e

e

e

TH

TTCX

Rx

0p

e

A

0.40

460


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For a feed temperature of 300 K, the adiabatic equilibrium

temperature is 460 K, and the corresponding adiabatic

conversion is 0.40.

Higher conversions than those shown in the previous figure

can be achieved for adiabatic operations by connecting

reactors in series with inter-stage cooling:


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Increasing conversion by inter-stage cooling


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What conversion could be achieved in Example 1.5 if two

interstage coolers were available that had the capacity to coolthe exit stream to 350 K? Also determine the heat duty of

each exchanger for a molar feed rate of A of 40 moles. Assume

that 95% of equilibrium conversion is achieved in each reactor.

The feed temperature to the first reactor is 300 K.

EXAMPLE 1.6

SOLUTION

In Example 1.5, for an entering temperature of 300 K the

adiabatic equilibrium conversion was 0.40.

For 95% of equilibrium conversion, the conversion exiting

the first reactor is 0.38.


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The exit temperature from the first reactor is found from a

rearrangement of Equation (e):

300T105.2X 3

45230038.0400300X400T

We now cool the gas stream exiting the reactor at 460 K

down to 350 K in a heat exchanger.

The gas stream is then sent to the second reactor.

The first reactor


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T02 = 350 and X1 = 0.38

The energy balance leads to

The second reactor

(f)

Intersecting eq. (f) with eq. (d) yields in the maximum/

equilibrium conversion in the second reactor:

61.0X2

e

95% of Xe2 = X2 = 0.58

The corresponding exit temperature is

20122 TXX400T

35038.0X400T22

43035038.058.0400T2


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T03 = 350 and X2 = 0.58

The energy balance leads to

The third reactor

(g)

Intersecting eq. (g) with eq. (d) yields in the maximum/

equilibrium conversion in the second reactor:

78.0X3e

95% of Xe3 = X3 = 0.75

The corresponding exit temperature is

30233 TXX400T

35058.0X400T33

41835058.075.0400T3


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0.00

0.20

0.40

0.60

0.80

1.00

1.20

300 350 400 450 500 550 600

X

T (K)


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ENDOTHERMIC REACTIONS

Another example of the need for inter-stage heat transfer in a

series of reactors can be found when upgrading the octanenumber of gasoline. The more compact the hydrocarbon

molecule for a given number of carbon atoms, the higher the

octane rating. Consequently, it is desirable to convert straight-

chain hydrocarbons to branched isomers, naphthenes, andaromatics.

The reaction sequence is


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The first reaction step (k1) is slow compared to the second

step, and each step is highly endothermic. The allowable

temperature range for which this reaction can be carried outis quite narrow: Above 530C undesirable side reactions occur

and below 430C the reaction virtually does not take place. A

typical feed stock might consist of 75% straight chains, 15%

naphthas, and 10% aromatics.

One arrangement currently used to carry out these reactions

is shown in the figure below. Typical sizes of the reactors are

on the order of 10 to 20 m high and 2 to 5 m in diameter. A

typical feed rate of gasoline is approximately 200 m3/h at 2

atm. Hydrogen is usually separated from the product stream

and recycled.


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V or W


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Plug Flow Reactor Equilibrium Conversion

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