H2Tech - Q1 2022 - 28

SPECIAL FOCUS ADVANCES IN HYDROGEN TECHNOLOGY
1,243.11 + 2,785.3 = 4,028.4 kmol/hr
(12)
To calculate the SMR equilibrium constant, Eq. 13 is used:
Kp (SMR)=P2
× ([CO] [H2
(0.000093 × 0.0563
:
]3)/[CH4
][H2
O] = 23.92
)/(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
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%.
TABLE 1. The inlet/outlet moles for Case 1
Inlet moles, kmol/hr
H2
N2
CH4
C2
C3
H6
H8
CO2
CO
H2
O
20
50
900
20
10
3 x [900 + 20 x 2 + 10 x 3] = 2,910
Outlet moles, kmol/hr
0.18d
0.04d
0.73d
0.05d
0.0003d
s
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
28 Q1 2022 | H2-Tech.com
223.76
49.72
907.47
62.16
0.37
2,785.32
0.056
0.012
0.225
0.015
0.000093
0.691
- 0.3665Z3
+ 27.1337Z - 3.277
-
(14)
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
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
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
][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.
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
= 750°K
(476.9°C). Therefore, the approach to SMR equilibrium in the
pre-reformer is T − Teq
Outlet moles, kmol/hr
0.215d
s
0.24d
0.005d
0.54d
- 0.58101Z2
+
(19)
(17)
(15)
(16)
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
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