Biomedical Engineering Reference
In-Depth Information
nearly all pressures. For their separation, then, it is necessary to use a system
such as a pressure swing adsorber (S3), as shown in
Figure 9.15
.
An important consideration is the additional water required to keep the
carbon dioxide dissolved while the hydrogen is being separated. The amount,
which may be considerable, can be expressed as the ratio of water to gaseous
product (R) on a weight basis. When pressure and R increase, the purification
of hydrogen increases but the amount of hydrogen in the gas phase
decreases. Therefore, we can recover more hydrogen with less purity or less
hydrogen with more purity. This depends on an adjustment of the pressure
and R. Example 9.1 illustrates the computation.
Example 9.1
Design a separator to produce 79% pure hydrogen from an SCWG operating at
250 bars of pressure. Assume the following overall gasification equation, which
produces hydrogen, methane, carbon dioxide, and carbon monoxide.
C
6
H
10
O
5
4
:
5H
2
O
4
:
5CO
2
7
:
5H
2
CH
4
0
:
5CO
1
5
1
1
1
Solution
We use the carbon dioxide solubility curve in
Figure 9.15
to design the separa-
tor. Here, at 250 bars of pressure and 25
C, we find the solubility of CO
2
to be
0.028 mole fraction. This implies that 1 mol of water is needed to dissolve
0.028 mol of carbon dioxide.
To separate gaseous hydrogen from liquid water, we reduce the ambient temper-
ature to 25
C. From
Figure 9.13
we find that the hydrogen solubility is only 0.0031
at 250 bars and 25
C, so (1
0.0031) or 0.9969 fraction of hydrogen produced will
be in the gas phase here. The gas may, however, contain other gases, so to ensure
that the hydrogen is 79% pure, we need to add water to the separator. If we know
the operating temperature, pressure, and weight ratio of the water to the gas mixture,
the amount of product in the liquid and vapor phases can be calculated according to
an equation of state. Here, we use
Figure 9.16
computed by the Peng
Robinson
equation. For 250 bars of pressure and a mole fraction of 79% in the gas phase, we
get R
80. Thus, the amount of water required is 80
(the mass of product gas).
5
3
From the
overall
gas
equation,
the mass
of
product
gas
is
4.5
44
7.5
2
1
16
0.5
28
243 g/mol of biomass. The mass of
3
1
3
1
3
1
3
5
water is 80
19.4 kg.
From the property table of water, we get the density of water at 25
C and
250 bars, which is 1008.5 kg/m
3
. The volume of water is 19.4 kg/1008.5 kg/m
3
243
19,440 g
3
5
5
5
0.0192 m
3
.
Volume of product gas
X
ð
nRT
=
P
Þ
5
10
2
3
kmol
314 kPa m
3
5
ð
4
:
5
1
7
:
5
1
1
1
0
:
5
Þ
3
3
ð
8
:
kmol
2
1
K
2
1
10
2
kPa
298 K
Þ=ð
250
Þ
3
3
00134 m
3
0
:
5
Therefore,
the
total
volume
of
biomass
that
is
gasified
is
0.0205 m
3
/mol.
0.0192
0.00134
1
5
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