Exercise. This is solution to exercise 1.16 in the book.
Solution. Based on the split fraction of H2 in the permeate, we can compute
the molar flowrate of hydrogen as:
FH2 ,permeate = 0.6038 · 42.4
=
25.6 kmol/h
FH2 ,non−permeate
=
16.8 kmol/h
Also, since the membrane is impermeable to nitrogen,
FN2 ,non−permeate = 0.5 kmol/h
Then, based on the separation index for hydrogen and methane,
SPH2 /CH4 =
FH2 ,permeate
FCH4 ,non−permeate
·
= 34.13
FH2 ,non−permeate
FCH4 ,permeate
Also,
FCH4 ,non−permeate + FCH4 ,permeate = 7
Hence,
FCH4 ,non−permeate = 6.7 kmol/h
FCH4 ,permeate = 0.3 kmol/h
Thus,
Fnon−permeate = 24 kmol/h
Fpermeate = 25.9 kmol/h
Hydrogen purity in the permeate gas is:
purity =
25.6
= 98.84%
25.9
If the expansions are reversible then adiabatic process, hence,
p1−γ
V1γ = p1−γ
V2γ
2
1
The partial pressure of the permeate gases in the feed flow is:
pF,P =
Fpermeate
25.9
P =
16.7 = 8.67 MPa
Ffeed
49.9
while for non-permeate gases:
pF,N P =
Fnon−permeate
24
P =
16.7 = 8.03 MPa
Ffeed
49.9
Hence,
TP = TF
TN P = TF
pF,P
pP
pF,N P
pN P
1−γ
γ
= 313
1−γ
γ
= 313
8.67
4.56
−0.2857
8.03
16.2
−0.2857
= 260.5 K
= 382.5 K