Hydrocarbon Processing - October 2021 - 56

Hydrogen
TABLE 2. Data for outlet wet mol fraction
Inlet, kmol/hr Outlet, kmol/hr Outlet wet mol fraction
H2
N2
CH4
C2
C3
H6
H8
CO2
CO
H2
O
20
50
900
20
10
2,910
223.76
49.72
907.47
62.16
0.37
2,785.32
TABLE 3. Inlet/outlet moles for Case 2
Inlet moles, kmol/hr
H2
Naphtha carbon
O
H2
CO2
CO
CH4
20
80 x 5.586 = 446.88
3 x 446.88 = 1,340.64
TABLE 4. Data for outlet wet mol fraction
Inlet,
H2
Naphtha carbon
O
H2
CO2
CO
CH4
kmol/hr
20
446.88
1,340.64
0.056
0.012
0.225
0.015
0.000093
0.691
A similar balance is done in Case 2 for treated naphtha feed.
Case 2 feed is 80 kmol/hr of naphtha (estimated C number of
5.586), an S/C ration of 3 mol/mol, an outlet pressure of 24.1
bara (23.8 atma), an inlet T of 450°C and an outlet T of 477°C.
The outlet composition (dry basis) reported by the laboratory
was the following:
* H2
* CH4
= 21.5%
= 54%
* CO = 0.5%
* CO2
= 24%.
The unknowns include the outlet dry flowrate and the outlet
steam, and it is not known how much steam has been consumed
in the reactions to make products.
To reiterate, d is the outlet dry flowrate, and s are the moles
Outlet moles, kmol/hr
0.215d
s
0.24d
0.005d
0.54d
of steam at the outlet. The inlet and outlet moles for Case 2 are
shown in TABLE 3. To calculate d using carbon balance, Eq. 15
is used:
446.88 = 0.24d + 0.005d + 0.54d
d = 569.3 kmol/hr
To calculate s using oxygen balance, Eq. 16 is used:
1340.64 = s + 2 × 0.24 × 569.3 + 0.005 × 569.3
s = 1,064.53 kmol/hr
The total wet outlet flow (TABLE 4) is calculated using Eq. 17:
569.3 + 1064.53 = 1,633.83 kmol/hr
Outlet,
kmol/hr
122.39
1,064.53
136.63
2.85
307.41
Outlet wet
mol fraction
0.0749
0.6517
0.0836
0.00174
0.1882
The total wet outlet flow (TABLE 2) is calculated in Eq. 12:
1,243.11 + 2,785.3 = 4,028.4 kmol/hr
× ([CO] [H2
]3)/[CH4
][H2
O] = 23.92
To calculate the SMR equilibrium constant, Eq. 18 is used:
Kp (SMR) = P2
× ([CO] [H2
]3)/[CH4
23.82 × (0.00174 × 0.07493
3.38 × 10-3
:
Ln (1/Kp) = 0.2513Z4
27.1337Z - 3.2770
where:
Z = (1000/T)-1
(12)
To calculate the SMR equilibrium constant, Eq. 13 is used:
Kp (SMR)=P2
(0.000093 × 0.0563
:
)/(0.225 × 0.691) = 6 × 10-5
×
(13)
Eq. 13 can be substituted with Eq. 14 to obtain equilibrium
temperature in °K2
Ln (1/Kp) = 0.2513Z4
0.58101Z2
where:
Z = (1,000/T)-1
-T is in °K
Ln (1/Kp) = Ln (1/6 × 10-5
) = 9.721
The equation can be solved using Excel: Z = 0.485; Teq
= 400.2 - 400.2 = 0°C.
=
673.4°K (400.2°C). Therefore, the approach to SMR equilibrium
in the pre-reformer is measured outlet temperature - equilibrium
temperature = T − Teq
56 OCTOBER 2021 | HydrocarbonProcessing.com
K. R. RAMAKUMAR is a Senior Technical Service Engineer
for Johnson Matthey Catalyst Technologies, and is based in Dubai.
He is a chartered chemical engineer with 16 yr of experience in
the downstream oil and gas industry. Prior to joining Johnson
Matthey in 2014, Mr. Ramakumar worked in refineries in India and
the UAE and was involved in operations, technical services, and
the commissioning of hydrogen and hydroprocessing units.
- 0.3665Z3
+ 27.1337Z - 3.277
-
(14)
Part 2. Part 2 will be featured in the November issue of Hydrocarbon
Processing.
LITERATURE CITED
1
2
Riazi, M. R., Characterization and Properties of Petroleum Fractions, ASTM
International, West Conshohocken, Pennsylvania, 2005.
Twigg, M. V., Catalyst Handbook, Second Edition, CRC Press, Boca Raton, Florida,
1996
][H2
O] =
)/(0.1882 × 0.6517) =
(18)
Eq. 18 can be substituted with Eq. 19 to obtain the equilibrium
temperature in °K2
- 0.3665Z3
-T is in °K
Ln (1/Kp) = Ln (1/3.38 x 10-3
) = 5.691
The equation can be solved using Excel: Z = 0.333; Teq
= 477 − 476.9 = 0.1°C.
= 750°K
(476.9°C). Therefore, the approach to SMR equilibrium in the
pre-reformer is T − Teq
- 0.58101Z2
+
(19)
(17)
(15)
(16)
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