ChE 201 Washington State University Chemical Process Principles and Calculations Voiland School of Chemical Engineering and Bioengineering Fall, 2009 Richard L. Zollars Problem 9.37, Felder and Rousseau Hydrogen is produced in the steam reforming of propane:

gHgCOvOHgHC 2283 733

The water-gas shift reaction also takes place in the reactor, leading to the formation of additional hydrogen:

gHgCOvOHgCO 222 The reaction is carried out over a nickel catalyst in the tubes of a shell-and-tube reactor. The feed to the reactor contains steam and propane in a 6:1 molar ratio at 125°C, and the products emerge at 800°C. The excess steam in the feed assures essentially complete consumption of the propane. Heat is added to the reaction mixture by passing a hot gas over the outside of the tubes that contain the catalyst. The gas is fed at 4.94 m3/mol C3H8, entering the unit at 1400°C and 1 atm and leaving at 900°C. The unit may be considered adiabatic.

Calculate the molar composition of the product gas, assuming that the heat capacity of the heating gas is 0.040 kJ/(mol·°C). SOLUTION Since we are bring asked to find a composition we can select a basis. Let’s pick one mole of propane entering the reactor. Thus there are 6 moles of H2O entering the reactor. Since the heat capacity of the heating gas is on a molar basis we also need the molar flow rate of the heating gas. Using the ideal gas law you get

mol

KKmol

atmLm

Lmatm

RT

PVn 98.35

15.167308206.0

100094.41

33

The inlet/outlet table would look like

nin inH nout

outH .ˆ

C3H8 1 1H 0 -

H2O (v) 6 2H n1

4H CO (g) - n2

5H

CO2 (g) - n3 6.H

H2 (g) - n4 7H

Heating Gas 35.98 3H 35.98

8H

We need to select reference states. Let’s use the elements (C, H2, and O2) at 25°C and the heating oil at 900°C as the reference states. Then

mol

kJH

H

dTTTTH

dTCHH HCpof

39.95415.88.103ˆ

251254

1071.31

251253

1011.1325125

2

1059.2225125068032.08.103ˆ

1071.311011.131059.22068032.08.103ˆ

ˆˆ

1

4412

338

225

1

125

25

3122851

125

25

83,1

For the remaining compounds you can use the enthalpy data in Table B.8. Thus

mol

kJdTCHH

mol

kJdTCHH

mol

kJdTCHH

vCOpof

vOHpof

vOHpof

39.8613.2452.110ˆˆ

78.21205.2983.241ˆˆ

43.23840.383.241ˆˆ

125

25

)(,5

800

25

)(2,4

125

25

)(2,2

mol

kJdTCHH

mol

kJdTCHH

gHpof

gCOpof

85.2285.220ˆˆ

14.35636.375.393ˆˆ

800

25

)(2,7

125

25

)(2,6

To get 3H use the heat capacity data given in the problem statement.

mol

kJdTCH gasHeatingp 209001400040.0ˆ

1400

900

,3

Thus the input/output table becomes

nin inH nout

outH .ˆ

C3H8 1 -95.39 0 - H2O (v) 6 -238.43 n1 -212.78 CO (g) - n2 -86.39 CO2 (g) - n3 -356.14 H2 (g) - n4 22.85 Heating Gas 35.98 20 35.98 0

For the overall system the energy balance would give

37.80685.2214.35639.8678.212

85.2214.35639.8678.2122098.3543.238639.951

ˆˆ

0

4321

4321

nnnn

nnnn

HnHn

H

outii

inii

This gives one equation in our four unknowns. The other three equations can be derived from balance on C, H and O. These are, respectively,

321

41

32

26:

2220826:

3:

nnnO

nnH

nnC

This gives the equations, in matrix form, as

37.806

6

20

3

85.2214.35639.8678.212

0211

2002

0110

4

3

2

1

n

n

n

n

Solving this gives

02.9

02.2

983.0

983.0

4

3

2

1

n

n

n

n

Thus there is a total of 0.983 + 0.983 + 2.02 + 9.02 = 13.01 moles in the stream leaving the reactor. The composition would be yH2O = 0.0756, yCO = 0.0756, yCO2 = 0.155, yH2 = 0.694.