Example Calculation

7/28/2019 Example Calculation

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Example 8 - 7 Production of Acetic Anhydride

Jeffreys1, in a treatment of the design of an acetic anhydride manufacturing facility, states that one

of the key steps is the vapor-phase cracking of acetone to ketene and methane:

He states further that this reaction is first order with respect to acetone and that the specificreaction rate can be expressed by

where k is in reciprocal seconds and Tis in Kelvin. In this design, it is desired to feed 8000 kg ofacetone per hour to a tubular reactor. The reactor consists of a bank of 1000 1-inch Schedule 40tubes. We will consider two cases:

1. The reactor is operated adiabatically.2. The reactor is surrounded by a heat exchanger where the heat-transfer coefficient is

110 J/m2AsAK and the ambient temperature is 1150 K.

The inlet temperature and pressure are the same for both cases at 1035 K, and 162 kPa(1.6 atm), respectively. Plot the conversion and temperature along the length of the reactor.

Solution

Let A = CH3COCH3, B = CH2CO, and C = CH4. Rewriting the reaction in symbols gives us

1. Mole Balance:

2. Rate law:

1G. V. Jeffreys, "A Problem in Chemical Engineering Design: The Manufacture of AceticAnhydride", 2

nded. (London: Institution of Chemical Engineers, 1964).

4233 CHCOCHCOCHCH +

Tk

222,3434.34)(ln =

CBA +

A

R

A rdV

Fd=

AA kCr =


2/7

3. Stoichiometry:

4. Combining yields:

RT

P

FFFF

Fk

dV

Fd

ACoBoAo

A

R

A

++=

2

5. Energy Balance:

CASE I. Adiabatic Operation

For no work done on the system, 0=

sW , and adiabatic operation, 0=Q (i.e., U = 0),

the energy balance becomes

( )

=

i

p

r

R iCF

rTatH

dV

Td

6. Calculation of mole balance parameters:

smolhkmolmolg

hkgFAo /3.38/9.137

/58

/8000===

( )

( )

T

AA

ATAA

ACoBoAoCBAT

AAoCoC

AAoBoB

F

Fy

RT

PyCyC

FFFFFFFF

FFFF

FFFF

=

==

++=++=+=

+=

2


3/7

7. Calculation of energy balance parameters:

( ) ( )

( )( )( )

( )( )

( )( )

( )( )

( ) ( ) ( )336223

33666

22

298 26

26

26

298

29810267.12981075.52988.6/80770

2981071.181095.301086.453

1

298077.00945.0183.05.0

29839.1304.2063.26/77.80

1071.18077.039.131

1095.300945.004.201

1086.45183.063.261

/77.80

298

+=

++

++

++++=

++

++

+

+=

+=

TTTmolJ

T

T

TmolkJ

dT

TT

TT

TT

molkJ

dTCKatHTatH

T

K

i

T

K

pirr i

( )( )( )( )( )( )( )

( )( )26

26

26

26

26

108.30115.08.6

1066.491715.043.33

1071.18077.039.13

1095.300945.004.20

1086.45183.063.26

TTF

TTF

TTFFF

TTFFF

TTF

CFCFCFCF

A

Ao

AAoCo

AAoBo

A

pCpBpA

i

piCBAi

+=

++

+++

++=

++=

Substituting the values calculated above into the two balance equations we may solve thematerial and energy balance equations simultaneously using the following MathCADprogram.

Example 8-7 Adiabatic Operation

First define all of the variables either as constants or functions

P 162 103:=

k T( ) exp 34.3434222

T

:=

CT T( )P

8.31 T:=


4/7

FAo 38.3:=

DHr T( ) 80770 6.8 T 298( )+ 5.75 10 3 T2 2982( ) 1.267 10 6 T3 2983( ):=

FICPI FA T,( ) FAo 33.43 0.1715 T+ 49.66 106 T2( ) FA 6.8 0.0115 T 3.8 10

6 T2( ):=

yA FA( )FA

2 FAo FA:=

Now give the initial conditions for the integrator

yFAo

1035

:=

and define the matrix that contains the two balance equations

D V y,( )

k y1( ) yA y0( ) CT y1( )

DHr y1

( ) k y1

( ) yA

y0

( ) CT

y1

( )

FICPI y0 y1,( )

:=

We now can start the integrator. We will integrate (as does Fogler) to a reactor volume of 5 m3.

Z Rkadapt y 0, 5, 100, D,( ):=

Change the molar flow rates for A (in the second column of Z) into conversion, then plot the results.

i 0 100..:=

Xi

FAo Z i 1,( )

FAo:=

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3Conversion Profile

Reactor volume (m**3)

C

onversionofA


5/7

0 1 2 3 4900

950

1000

1050

Temperature Profile

Reactor Volume (m**3)

T

emperature(K)

As was observed by Fogler, the reaction essentially stops due to the lowering of the temperaturein the adiabatic reactor. Lets go on to the case with heat transfer.

CASE II. Operation of a PFR with Heat Exchange

We can use the material balance portion from the prior case without change. The changes in thispart of the problem come only in the energy balance portion. Thus,

5. Energy Balance

( ) ( )

+=

i

p

hxr

R iCF

TTUArTatH

dV

Td

6. Parameter Evaluation

The one thing we will have to change is the fact that the reactor is composed of 1000individual tubes. Since the heat transfer area depends on the tube size we will need todo our analysis on one tube, then multiply by 1000 to get the total. Thus the initialmolar flow rate of A into a single tube is (38.3 mol/s)/1000 or


6/7

smolFAo /0383.0=

The surface area per volume in a cylindrical tube is 4/D so

11500266.0

44 === mmD

A

All of the other parameters remain as for the adiabatic case. We now can solve thematerial and energy balance equations simultaneously using the MathCAD programbelow

Example 8-7 Operation with heat exchange

First define all of the variables either as constants or functions

P 162 103:=

U 110:=

A 150:=

Thx 1150:=

k T( ) exp 34.3434222

T

:=

CT T( )P

8.31 T:=

FAo 38.3:=

DHr T( ) 80770 6.8 T 298( )+ 5.75 10 3 T2 2982( ) 1.267 10 6 T3 2983( ):=

FICPI FA T,( ) FAo 33.43 0.1715 T+ 49.66 106 T2( ) FA 6.8 0.0115 T 3.8 10

6 T2( ):=

yA FA( )FA

2 FAo FA:=

Now give the initial conditions for the integrator

yFAo

1035

:=

and define the matrix that contains the two balance equations

D V y,( )

k y1( ) yA y0( ) CT y1( )

DHr y1( ) k y1( ) yA y0( ) CT y1( )( ) U A Thx y1( )+

FICPI y0 y1,( )

:=


7/7

We now can start the integrator. We will integrate (as does Fogler) to a reactor volume of 1 m3.

Z Rkadapt y 0, 1, 100, D,( ):=

Change the molar flow rates for A (in the second column of Z, into conversion, then plot the result

i 0 100..:=

XiFAo Z i 1,( )

FAo:=

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8Conversion Profile


Conversion

ofA

0 0.2 0.4 0.6 0.8 11010

1015

1020

1025

1030

1035

1040

1045

1050Temperature Profile


Temperature(K)

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