Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Find a and b component flow rates in feed permeate and

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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 fractions. 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 = (1) 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) Analysis: (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: γ −1 γ P Tout = T1 out (2) P1 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 adsorption. 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 regenera...
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