Unformatted text preview: ind: (a) and (b) Component flow rates in feed, permeate, and retentate on a diagram.
(c) Method to further separate the permeate.
Analysis: (a) and (b) Let: P = permeate flow rate
R = retentate flow rate
Total material balance: 8,000 = P + R
Ethanol material balance: 8,000(0.23) = (0.60) P + (0.10) R
Solving (1) and (2) simultaneously, P = 2,080 kg/h and R = 5,920 kg/h (1)
(2) The resulting material balance and flow diagram is: (c) Gas adsorption, gas permeation, or distillation to obtain ethanol and the azeotrope,
which can be recycled. Exercise 1.17
Subject: Separation of hydrogen from light gases by gas permeation with hollow fibers.
Given: Feed gas of 42.4 kmol/h of H2, 7.0 kmol/h of CH4, and 0.5 kmol/h of N2 at 40oC
and 16.7 MPa. Retentate exits at 16.2 kPa and permeate exits at 4.56 kPa. Gas heat
capacity ratio = γ = 1.4.
Assumptions: Membrane is not permeable to nitrogen. Reversible gas expansion with
no heat transfer between the retentate and permeate. Separation index is based on mole
Find: (a) Component flows in the retentate and permeate if the separation index, SP, for
hydrogen relative to methane is 34.13, and the split fraction (recovery), SF, for hydrogen
from the feed to the permeate is 0.6038.
(b) Percent purity of hydrogen in the permeate.
(c) Exit temperatures of the retentate and permeate.
(d) Process flow diagram with complete material balance
Analysis: (a) and (d)
Hydrogen in permeate = (0.6038)(42.4) = 25.6 kmol/h
Hydrogen in retentate = 42.4 - 25.6 = 16.8 kmol/h
Let: x = kmol/h of methane in permeate
Then, 7.0 - x = kmol/h of methane in retentate
From Eq. (1-4),
25.6 / 16.8
SP = 34.13 =
x / (7 − x )
Solving (1), x = 0.3 kmol/h of methane in the permeate
Methane in the retentate = 7.0 - 0.3 = 6.7 kmol/h
The resulting material balance and flow diagram is: Exercise 1.17 (continued)
(b) Percent purity of hydrogen in permeate = 100% x 25.6/25.9 = 98.8%
(c) For reversible adiabatic (isentropic) expansion, assuming an ideal gas,
the final temperature is given from thermodynamics by:
Tout = T1 out
where subscript 1 refers to upstream side, subscript out refers to downstream side, both
temperature and pressure are absolute, and γ is the gas heat capacity ratio.
For both the retentate and the permeate, T1 = 40oC = 313 K and P1 =16.7 MPa. For the retentate, Pout = P3 = 16.2 MPa. From (2), Tout 16.2
= T3 = 313
16.7 1.4 −1
1.4 = 310 K = 37 o C For the permeate, P2 = 4.56 MPa. From (2), 4.56
T2 = 313
16.7 1.4 −1
1.4 = 216 K= − 57 o C Exercise 1.18
Subject: Natural gas is produced when injecting nitrogen into oil wells. The nitrogen is
then recovered from the gas for recycle.
Given: 170,000 SCFH (60oF and 14.7 psia) of gas containing, in mol%, 18% N2, 75%
CH4, and 7% C2H6 at 100oF and 800 psia. Recover the N2 by gas permeation followed by
The membrane is selective for nitrogen. The adsorbent is selective for methane. The
adsorber operates at 100oF, and 275 psia during adsorption and 15 psia during
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