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

# Permeate exits the membrane unit at 20of and a low

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Unformatted text preview: tion. Permeate exits the membrane unit at 20oF and a low pressure. Two stages of compression with cooling are needed to deliver the permeate gas to the adsorber. The regenerated gas from the adsorber is compressed in three stages with cooling, and is combined with the retentate to give the natural gas product. Assumptions: The membrane is not permeable to ethane. The separation index, SP, defined by Eq. (1-4), is applied to the exiting retentate and permeate. Find: (a) Draw a labeled process flow diagram. (b) Compute the component material balance, based on the following data: Separation index, SP, for the membrane = 16 for nitrogen relative to methane. The adsorption step gives 97 mol% methane in the adsorbate with an 85% recovery based on the feed to the adsorber. The pressure drop across the membrane is 760 psi. The retentate exits at 800 psia. The combined natural gas product contains 3 mol% nitrogen. Place the results of the material balance in a table. Analysis: (b) Refer to the process flow diagram on next page for stream numbers. Let: ai = molar flow rate of N2 in lbmol/h in stream i. bi = molar flow rate of CH4 in lbmol/h in stream i ci = molar flow rate of ethane in lbmol/h in stream i Feed flow rate = 170,000 SCFM / 379 SCF/lbmole at SC = 448.5 lbmol/h a1 = 0.18(448.5) = 80.7, b1 = 0.75(448.5) = 336.4, c1 = 0.07(448.5) = 31.4 Because ethane does not permeate through the membrane, c3 = c6 = 31.4 and c2 = c4 = c5 = 0 Solve for a2, a3, a4, a5, a6, and b2, b3, b4, b5, b6 from 10 equations in 10 unknowns. Membrane selectivity: a2 / a3 SP = 16 = (1) b2 / b3 Component balances around the membrane unit: a2 + a3 = 80.7 b2 + b3 = 336.4 Component balances around the adsorber: a2 = a4 + a5 b2 = b4 + b5 Exercise 1.18 (continued) Component balances around the line mixer that mixes retentate with adsorbate gas: a6 = a3 + a5 b6 = b3 + b5 Methane purity in the adsorbate: b5 = 0.97(b5 + a5) Find: (b) (continued) Methane recovery: Mol% nitrogen in the final natural gas: b5 = 0.85 b2 a6 = 0.03(a6 + b6 + 31.4) All equations are linear except (1). Solving these 10 equations with a nonlinear equation solver, such as in the Polymath program, results in the following material balance table: Component Nitrogen Methane Ethane Total Stream 1 80.7 336.4 31.4 448.5 2 73.3 128.7 0.0 202.0 Flow rate, 3 7.4 207.7 31.4 246.5 lbmol/h 4 69.9 19.3 0.0 89.2 (a) Labeled process flow diagram 5 3.4 109.4 0.0 112.8 6 10.8 317.1 31.4 359.3 Exercise 1.19 Subject: Separation of a mixture of ethylbenzene, o-xylene, m-xylene, and p-xylene Find: (a) Reason why distillation is not favorable for the separation of m-xylene from p-xylene. (b) Properties of m-xylene and p-xylene for determining a means of separation. (c) Why melt crystallization and adsorption can be used to separate m-xylene from p-xylene. Analysis: (a) In the order of increasing normal boiling point: Component nbp, oR Ethylbenzene 737.3 Paraxylene 741.2 Metaxylene 742.7 Orthoxylene 751.1 Relative volatility 1.08 1.02 1.16 From t...
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