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

23 for pure component densities analysis from eq 4

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Unformatted text preview: at volume of mixing = 0 Find: Liquid density in various units using Fig. 2.3 for pure component densities. Analysis: From Eq. (4), Table 2.4, υL = xiυ iL = i υ iL = M ρL M ρ iL The calculations are summarized as follows using Fig. 2.3 for pure component densities, Comp. MW Flow rate, lbmol/h C3 iC4 nC4 iC5 nC5 44.09 58.12 58.12 72.15 72.15 2.2 171.1 226.6 28.1 17.5 Mole fraction, xi 0.0049 0.3841 0.5086 0.0631 0.0393 xiυ iL = 139.4 cm3/mol υL = Density, Fig. 2.3, g/cm3 0.2 0.40 0.43 0.515 0.525 MW = 59.5 i ρL = Molar volume, υι cm3/mol 220 145 135 140 130 MW 59.5 = = 0.427 g/cm 3 υL 139.4 = 427 kg/m3 = 26.6 lb/ft3 = 3.56 lb/gal = 149 lb/bbl where 1 bbl = 42 gal xi υ ι 1.1 55.7 68.7 8.8 5.1 Exercise 2.14 Subject: Condenser duty for a distillation column, where the overhead vapor condenses into two liquid phases. Given: Temperature, pressure, and component flow rates of overhead vapor and the two liquid phases. Assumptions: Ideal gas and ideal liquid solution for each liquid phase. Find: Condenser duty in Btu/h and kJ/h Analysis: Take a thermodynamic path of vapor from 76oC to 40oC and condensation at 40oC. For water, use the steam tables. Change in enthalpy from vapor at 76oC to 40oC and 1.4 bar = 1133.8 - 71.96 = 1,062 Btu/lb = 2,467,000 J/kg = 2,467 kJ/kg. For benzene, using data on p. 2-221 from Perry's 7th edition, change in enthalpy from vapor at 76oC to 40oC and 1.4 bar = 874 - 411 = 463 kJ/kg For isopropanol, using data on p. 2-179 from Perry's 7th edition, Average CP over the temperature range = 1.569 kJ/kg-K From data of p. 2-157 of Perry's 7th edition, Heat of vaporization at 40OC = 313 K = 770.4 kJ/kg Therefore, the enthalpy change of isopropanol = 1.569(76-40) +770.4 = 827 kJ/kg Condenser duty, QC = 2,350(2,467) + 24,600(463) + 6,800(827) = 22,810,000 kJ/h = 21,640,000 Btu/h Exercise 2.15 Subject: K-values and vapor tendency of light gases and hydrocarbons Given: Temperature of 250oF and pressure of 500 psia. Find: K-values in Fig. 2.8 and vapor tendency. Analysis: If the K-value is < 1.0, tendency is for liquid phase. If the K-value > 1.0, tendency is for vapor phase. Using Fig. 2.8, Component N2 H2S CO2 C1 C2 C3 iC4 nC4 iC5 nC5 K-value 17 3.1 5.5 8 3 1.5 0.71 0.35 0.38 0.10 Tendency vapor vapor vapor vapor vapor vapor liquid liquid liquid liquid Exercise 2.16 Subject: Recovery of acetone from air by absorption in water. Given: Temperature, pressure, phase condition, and component flow rates of feeds to and products from the absorber, except for exiting liquid temperature. Assumptions: Ideal gas and zero heat of mixing. Find: Temperature of exiting liquid phase. Potential for explosion hazard. Analysis: From the given component flow rates, water evaporates at the rate of 22 lbmol/h, and 14.9 lbmol/h of acetone is condensed. Take a thermodynamic path that evaporates water at 90oF and condenses acetone at 78oF. Energy to heat air from 78oF to 80oF = nCP∆T = 687(7)(80-78)=9,620 Btu/h Energy to heat unabsorbed acetone from 78oF to 80oF is negligible. Energy to vaporize water at 90oF = n∆Hvap = 22(1,043)(18) = 413,000 Btu/h Total required energy = 9,620 + 413,000 = 423,000 Btu/h Energy available...
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