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

Property constants in example 23 assumptions ideal

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Unformatted text preview: d reflux and distillate. Property constants in Example 2.3. Assumptions: Ideal gas and ideal gas and liquid solutions. Find: Condenser duty in kJ/h. Analysis: For the thermodynamic path, cool the overhead from 331 K to 325 K. Then condense at 325 K. Because this temperature change is small, compute vapor specific heats at the two temperatures and take the arithmetic average. From Eq. (2-35) and the constants in Example 2.3, for vapor heat capacity in J/kmol-K and temperature in K, o CPEB = −43098.9 + 707.151T − 0.481063T 2 + 130084 × 10 −4 T 3 . o CPS = −28248.3 + 615.878T − 0.40231T 2 + 0.993528 × 10 −4 T 3 Solving, Comp. EB S kg/h 77,500 2,500 MW 106.17 104.15 kmol/h 729.975 24.003 Vapor heat capacity, J/kmol-K 331 K 325 K Avg. 142,980 140,380 141,680 132,130 132,830 133,980 The sensible vapor enthalpy change, Qsensible from 331 K to 325 K is, Qsensible = i o ni CPi ∆T = i o ni CPi (331 − 325) = 729.975(141,680)(331 − 325) + 24.003(133,980)(331 − 325) = 620,560,000 J / h = 620,560 kJ / h For the latent heat of condensation, use Eq. (2-41) to estimate the molar heats of vaporization of at 325 K for the two components, using the vapor pressure constants given in Example 2.3. Exercise 2.11 (continued) vap ∆HEB = 8314 325 2 − −7440.61 325 2 + 0.00623121 + −9.87052 5 + 6(4.13065 × 10−18 ) 325 325 = 40,740,000 J / kmol = 40,740 kJ / kmol ∆HSvap = 8314 325 2 − −914107 . 325 2 + 0.0143369 + −17.0918 5 + 6(18375 × 10−18 ) 325 . 325 = 42,440,000 J / kmol = 42,440 kJ / kmol The latent heat of condensation, Qlatent , in kJ/kmol is, Qlatent = ni ∆Hivap = 729.975(40,740) + 24.003(42,440) i = 30,760,000 kJ / h The total condenser duty = Qsensible + Q latent = 620,000 + 30,760,000 = 31,380,000 kJ/h Exercise 2.12 Subject: Thermodynamic properties of a benzene (B) -toluene (T) feed to a distillation column. Given: Temperature, pressure, and component flow rates for the column feed. Property constants, critical temperature Assumptions: Phase condition is liquid (needs to be verified). Ideal gas and liquid solution. Find: Molar volume, density, enthalpy, and entropy of the liquid feed. Analysis: The feed is at 100oF and 20 psia. From Fig. 2.4, since the vapor pressures of benzene and toluene are 3.3 and 0.95 psia, respectively, the feed is a subcooled liquid. T = 311 K. From Eqs. (4) , Table 2.4 and (2-38), υL = i Mi ρL i υL = = Mi − 1− 2/ 7 AB 78.11 311 − 1− 562 B υL = T Tc 2 /7 304.1(0.269) 92.14 T 290.6(0.265) Total flow rate = 415 + 131 = 546 kmol/h Benzene mole fraction = 415/546 = 0.760 311 − 1− 593.1 2/ 7 = 0.0905 m3 / kmol = 01083 m3 / kmol . Toluene mole fraction = 1-0.76 = 0.240 From Eq. (4), Table 2.4 for a mixture (additive volumes), υL = 0.76(0.0905) + 0.24(0.1083) = 0.0948 m3/kmol = 1.52 ft3/lbmol Mixture molecular weight = M = 0.76(78.11) + 0.24(92.14) = 81.48 From Eq. (4), Table 2.4 for conversion to density, ρL = M/υL = 81.48/1.52 = 53.6 lb/ft3 Exercise 2.13 Subject: Liquid density of the bottoms from a distillation column. Given: Temperature, pressure, component flow rates Assumptions: Ideal liquid solution so th...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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