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

08 1495808237 205000 btuh energy available from the

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Unformatted text preview: from condensation of acetone with ∆Hvap = 237 Btu/lb and a molecular weight of 58.08 = 14.9(58.08)(237) = 205,000 Btu/h Energy available from the cooling of evaporated water from 90oF to 80oF = 22(18)(0.44)(90-80) = 2,000 Btu/h Total available energy = 205,000 + 2,000 = 207,000 Btu/h Energy required - Energy available = 423,000 - 207,000 = 216,000 Btu/h This energy must come from cooling of the water absorbent from 90oF to T , and condensed acetone from 78oF to T. Therefore, using a CP of 0.53 for liquid acetone, 216,000 = 14.9(58.08)(78-T) + 1,722(18)(1.0)(90-T) Solving, T = temperature of exiting liquid = 83oF. The mol% acetone in the entering gas = 15/702 x 100% = 2.14 %. This is outside of the explosive limits range of 2.5 to 13 mol%. Exercise 2.17 Subject: Volumetric flow rates of entering and exiting streams of an adsorber for removing nitrogen from subquality natural gas. Given: Temperature and pressure of feed gas and two product gases. Composition of the feed gas. Specification of 90% removal of nitrogen and a 97% methane natural gas product. Assumptions: Applicability of the Redlich-Kwong equation of state. Find: Volumetric flow rates of the entering and exiting gas streams in actual ft3/h. Analysis: Removal of nitrogen = 0.9(176) = 158.4 lbmol/h Nitrogen left in natural gas = 176 - 158.4 = 17.6 lbmol/h Methane in natural gas product = 17.6(97/3) = 569.1 lbmol/h Material balance summary: Component Nitrogen Methane Totals Temperature, oF Pressure, psia lbmol/h: Feed gas 176 704 880 70 800 Waste gas 158.4 134.9 293.3 70 280 Natural gas 17.6 569.1 586.7 100 790 Using the ChemCAD simulation program, the following volumetric flow rates are computed using the Redlich-Kwong equation of state: Stream Feed gas Waste gas Natural gas Actual volumetric flow rate, ft3/h 5,844 5,876 4,162 Exercise 2.18 Subject: Estimation of partial fugacity coefficients of propane and benzene using the R-K equation of state. Given: From Example 2.5, a vapor mixture of 39.49 mol% propane and 60.51 mol% benzene at 400oF and 410.3 psia. Assumptions: Applicability of the Redlich-Kwong equation of state. Find: Partial fugacity coefficients Analysis: From Example 2.5, the following conditions and constants apply, where the Redlich-Kwong constants, A and B, for each component are computed from Eqs. (2-47) and (2-48) respectively. T = 477.6 K P = 2829 kPa ZV = 0.7314 R = 8.314 kPa-m3/kmol-K A = 0.2724 B = 0.05326 From Eqs. (2-47) and (2-48), aP ai (2819) a Ai = 2i 2 = =i 2 2 RT (8.314) (477.6) 5593 bP bi (2819) Bi = i = = 0.7099bi RT (8.314)(477.6) Component, i Propane Benzene a, kPa-m6/kmol2 836.7 2,072 b, m3/kmol 0.06268 0.08263 Ai 0.1496 0.3705 Bi 0.04450 0.05866 The Redlich-Kwong equation for partial vapor fugacity coefficients is given by Eq. (256). Applying it to propane (P) and benzene (B), using the above values for constants and conditions, φPV = exp (0.7314 − 1) 0.04450 0.2724 0.1496 0.04450 0.05326 − ln ( 0.7314 − 0.05326 ) − 2 − ln 1 + 0.05326 0.05326 0.2...
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