2 Plug Flow Reactor Adiabatic

8/2/2019 2 Plug Flow Reactor Adiabatic

1/33


2/33

(1.44)

The energy balance in a tubular reactor is given in eq. (1.44)

by:

For an adiabatic reactor, . Usually, there is a negligible

amount of work done on or by the reacting mixture.

(1.46)

0Q

0XFdTCTHdTCFWQ0A

T

TpR

0

Rx

n

1i

T

Tpii0AS

R0i

n1i

T

Tpii

T

TpR

0

Rx

0iR

dTCXdTCTH


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The circumflex denotes that the heat capacities are evaluated

at some mean temperature value between TR and T.

R

T

Tp

pTT

dTC

C R (1.48)

For the case of constant or mean heat capacities:

Rp

T

Tp

TTCdTCR

(1.47)


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In a similar fashion we can write the integral involving i andCpi in eq. (1.46) as

0ipiiTT

piiTTCdTC

0i

The circumflex denotes that the heat capacities are evaluated

at some mean temperature value between TR and T.

0i

T

Tpi

piTT

dTC

C 0i

n

1i0ipiiRpR

0

RxTTCXTTCTH

(1.49)

For constant or mean heat capacity, eq. (1.46) becomes:

(1.50)

(1.51)


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Eq. (1.51) can be solved for T

n

1i 0ipiiRpR

0

Rx TTC

XTTC

TH

n

1i0ipii

n

1ipiiRppR

0

RxTCTCXTCTXCXTH

Rp

n

1i0ipiiR

0

Rxp

n

1ipii

TCXTCTHXCXCT

p

n

1ipii

Rp

n

1i0ipiiR

0

Rx

CXC

TCXTCTHX

T

(1.52)


6/33

This equation can be combined with the differential mole

balance derived from eq. (1.1):

(1.1) T,XrF

dX

dV

A

0Ato obtain the temperature, conversion, and concentration

profiles along the length of the reactor.

One way of accomplishing this combination is to use eq.

(1.52) to construct a table of T as a function of X.

Once we have T as a function of X, we can obtain k(T) as a

function of X and hence rA as a function of X alone.


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The elementary reversible gas-phase reaction is carried out in

a PFR in which pressure drop is neglected and pure A entersthe reactor:

A BMole balance:

Rate law:

with

and

dXr

FdV

A

0A

C

B

AAK

CCkr

T1T1REexpkk 11

T

1

T

1

R

HexpTKK

1

0

Rx

1CC

(a)

(b)

(c)

(d)


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Stoichiometry: Gas, = 0, P = P0

TT

X1CC 00AA

CB

AAKCCkr

T

TXCC 0

0AB

(1.46) n

1i

T

Tpii

T

TpR

0Rx

0iR

dTCXdTCTH

(e)

(f)

(g)

Eqs. (a) through (g) and (1.52) can easily be solved using

either Simpson's rule or an ODE solver.

Combine:

Energy balance (eq. 1.46):


9/33

Normal butane, C2H4, is to be isomerized to isobutane in a plug-

flow reactor. The reaction is to be carried out adiabatically in theliquid phase under high pressure using essentially trace amounts of

a liquid catalyst which gives a specific reaction rate of 3 1.1 h1 at

360 K. Calculate the PFR volume necessary to process 100,000

gal/day (163 kg mol/h) of a mixture 90 mol % n-butane and 10 mol

% i-pentane, which is considered an inert. The feed enters at 330 K.

Additional information:

HRx = -6900 J/mol* butaneCpn-B = 141 J/mol.K

Cpi-B= 141 J/mol.KCpi-P = 161 J/mol.K

Activation energy = 65.7 kJ/mol

KC = 3.03 at 60C

CA0

= 9.3 g mol/dm3 = 9.3 kg mol/m3

EXAMPLE 1.2


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SOLUTION

Reaction:

Mole balance:

Rate law:

with

dXr

FdV

A

0A

C

B

AA

K

CCkr

T

1

T

1

R

Eexpkk

1

1

(a)

(b)

(c)

T1

360

1

31.8

700,65exp1.31k

T

1

360

17906exp1.31k


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T

1

T

1

R

HexpTKK

1

0

Rx

1CC

(d)

T1

333

1

31.8

6900exp03.3K

C

T133313.830exp03.3KCStoichiometry (liquid phase, v = v0):

X1CC0AA

XCC 0AB

(e)

(f)


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C

0A0AA

K

XCX1Ckr

Combine:

XK

111kCr

C

0AA

(g)

(h)

(1.46)

n1i

T

Tpii0A0A

T

TpR

0

Rx

0iR

dTCFXFdTCTH

Energy balance (eq. 1.46):

1

F

F

0A

0A

A

with:

0

F

F

0A

0B

B


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0K.molJ141141CCC pApBp 1111.0

9.0

1.0

F

F

0A

0I

I

(i)

Eq. (46) becomes:

n1i

0ipiiR

0

RxTTCXTH

n1i

pii

R

0

Rx

0

C

XTHTT

with: 161111.0141CCCCpIIpBBpAA

n

1ipii

= 158.9 J/mol . K


14/33

Eq. (i) becomes:

X43.43330T (j)

molJ900,6THR

0

Rx

At equilibrium rA = 0, equation (b) becomes:

C

e

0AAK

X1kC0r

C

C

eK1

KX (k)We will now integrate Equation (a) with Simpson's rule, but

first we have to calculate (FA0/-rA) as a function of X.


15/33

For example for X = 0.1

K34.3341.043.43330T 1

h77.534.334

1

360

17906exp1.31k

00.334.334

1

333

1

3.830exp03.3KC

75.0

0.31

0.3

K1

KX

C

C

e

h.m

kmol83.511.0

0.3

111

m

kmol3.9

h

77.5r

33A

33A0A m830.2h.mkmol83.51

htotalkmol163totalmolbutanemol9.0

r

F


16/33

X T k KC Xe - rA FA0/(-rA)

0 330.00 4.22 3.10 0.76 39.28 3.735

0.1 334.34 5.77 3.00 0.75 51.83 2.830

0.2 338.69 7.81 2.91 0.74 67.62 2.1700.3 343.03 10.49 2.82 0.74 87.19 1.682

0.4 347.37 14.00 2.73 0.73 111.12 1.320

0.5 351.72 18.54 2.65 0.73 139.93 1.048

0.6 356.06 24.39 2.58 0.72 174.02 0.843

0.7 360.40 31.87 2.51 0.71 213.62 0.687


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0

5

10

15

20

25

30

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

FA0

/(-rA

)

X

1 2 3 4 5 76


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2

yy

2

yy

2

yy

2

yy1.0Area 43322110

2

yy

2

yy

2

yy

2

yy43766554

2y

yyyyyy2

y1.0Area 7

654321

0

3m13.3


19/33

When the problem is to find the temperature profile along

the reactor, then the algorithm is as follows.

Mole balance:

Rate law:

Energy balance:

0A

A

F

r

dV

dX AA

Ckr (1.1)

For no work done on the system, , and adiabatic

operation, , the energy balance is written as eq. (1.46)

0WS

0Q

0XdTCTHdTC TT

pR

0

Rx

T

Tpii

R0i

(1.46)


20/33

Differentiating eqs. (1.46) with respect to V yields

dVdX

THdV

dT

C R0

Rxpii

0dV

dXdTC

dV

dTCX

T

Tpp

R

dV

dXdTCTH

dV

dTCXC

T

TpR

0

Rxppii

R

dV

dX

CXC

dTCTH

dV

dT

ppii

T

TpR

0

Rx

R

(1.53)


21/33

Combining eqs. (1.1) and (1.53) yields

ppii0A

T

TpR

0

RxA

CXCF

dTCTHr

dV

dTR

(1.54)

Here we have two equations, (1.1) and (1.54), that must be

solved simultaneously.

One method that can be used to solve the problem is Runge-Kutta method.


22/33

xz,xy,xfdx

dy

xz,xy,xgdx

dz

00yxy

00zxz

nnn1 z,y,xf.hk nnn1 z,y,xg.hl 2

lz,

2

ky,

2

hxf.hk 1

n

1

nn2 2l

z,2

ky,

2

hxg.hl 1

n

1

nn2


23/33

3n3nn4lz,ky,hxf.hk

3nnn4lz,ky,hxg.hl

4321 kk2k2k61k 4321 ll2l2l61l hxx

n1nkyy

n1nlzz

n1n

2l

z,2

ky,

2

hxf.hk 2

n

2

nn3 2l

z,2

ky,

2

hxg.hl 2

n

2

nn3


24/33

One of the key steps in producing acetic anhydride is the

vapor-phase cracking of acetone to ketene and methane:

CH3COCH3 CH2CO + CH4This reaction is first-order with respect to acetone and the

specific reaction rate can be expressed by:

where k is in reciprocal seconds and T is in kelvin.

In this design it is desired to feed 8000 kg of acetone per hour

to a tubular reactor. The reactor consists of a bank of 1000 1-inch schedule 40 tubes. The reactor is operated adiabatically.

The inlet temperature and pressure are at 1035 K and 162 kPa

(1.6 atm), respectively. Plot the conversion and temperature

along the length of the reactor.

EXAMPLE 1.3

T3422234.34kln


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SOLUTION

Let A = CH3COCH3, B = CH2C0, and C = CH4.

Rewriting the reaction symbolically gives us:

A B + CMole balance:

Rate law:

0A

A

F

r

dV

dX AA

Ckr (a)

(b)

A B C Total

Initial NA0 0 0 NA0

Reaction NA0 X

Final NA0 (1 X) NA0 X NA0 X NA0 (1 + X)


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0

0A

0AV

NC

V

X1N

V

N

C

0AA

A

If the gas mixture is assumed to behave as ideal gas:

RTN

PV

RTN

PV

t00t

0

0

0

0

00A

0A

0

00t

t VT

TX1V

T

T

N

X1NV

T

T

N

NV

T

T

X1

X1C

T

T

X1V

X1NC 00A0

0

0A

A

(c)

(d)

(e)

Combining (d) and (e) gives us:

(f)


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Combining eqs. (b) and (f) yields

T

T

X1

X1kC

r00A

A (g)

while combining eqs. (a) and (g) yields

TTX1 X1vkTTX1F X1kCdVdX 00

0

0A

0A (h)For no work done on the system, , and adiabatic

operation, , the energy balance leads to eq. (1.54)

0WS

0Q

ppii0A

T

TpR

0

RxA

CXCF

dTCTHr

dV

dTR

(1.54)


28/33

Calculation of mole balance parameters:

smol3.38hkmol9.137kmolkg58

hkg8000F0A

3330A

0Ammol8.18

mkmol0188.0

K1035K.kmol

m.kPa31.8

kPa162RTPC

sm037.2C

Fv

3

0A

0A

0


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Calculation of energy balance parameters:

a. The standard heat of reaction :R0Rx TH molkJ67.216TH

acetoneR

0

Rx

molkJ09.61THketeneR

0

Rx

molkJ81.74THmethaneR

0

Rx

67.21681.7409.61TH R0Rx molJ80770molkJ77.80


30/33

b. Cp :K.molJT1086.45T183.063.26C

26

pA

K.molJT1095.30T0945.004.20C

26

pB

K.molJT1071.18T077.039.13C

26

pC

8.663.2604.2039.13

ABC

0115.0183.00945.0077.0ABC

ABC

6108.3 2

pTTC

26

T108.3T0115.08.6


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c. i Cpi :

A

= 1

B = C = 0i Cpi = A CpA + B CpB + A CpB = CpA

The two equations are solved simultaneously using 2-

dimensional Runge-Kutta method.


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0.00

0.05

0.10

0.15

0.20

0.25

0 1 2 3 4

X

V (m3)


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900

925

950

975

1000

1025

1050

0 1 2 3 4

T

(K)

V (m

3

)

2 Plug Flow Reactor Adiabatic

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