Solution Manual for Transport Processes and Separation Process Principles, 5th Edition

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Solution Manual forTransport Processes andSeparation Process PrinciplesFifth EditionChristie John GeankoplisA. Allen HerselDaniel LepekAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.2-1)Eq. (1.3-4)k353.2°C273.15°C80.0580 C==+==°Eq. (1.3-2)°C80.01/1.8F32F176°F==°°=Eq. (1.3-3)°R°F460176460°R636°R=+=+=Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.2-2)Eq. (1.3-2)°C1 1.8°F321/1.81553268.33°C===Eq. (1.3-4)K°C273.1568.33273.15341.5K=+=+=Eq. (1.3-3)°R°F460155460615°R=+=+=Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.3-1)22MW O32.00,MW N28.02MW air0.21 32.000.79 28.0228.9 kg / kg molAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.3-2)22MW 28.0132.0044.0156.0 kg2COO2COmoles CO2.00 kg mol28.01kg/kg mol+==222moles O1.00,1.00 32.0032.0kg O2.00 44.088.0Kg COAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.3-3)GasgMWg molmol/secN22028.020.71380.1648 mol/secO28332.002.59380.5990CO24544.011.02250.2362Total148 g4.330/g mol1.000034.2g/g mol148gAverage mol wt34.2 kg/kg mol4.3301g mol==Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.3-4)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.3-5)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.4-1)P = 2.4 mm HgFrom Appendix A.11 atm = 760 mm Hg32.4 mm HgP3.1610atm760 mm Hg/atm1 atm = 33.90 × 12 in. H2O (at 4°C)22in.H O2.4Patm 33.90121.285in.H O760atm1 μm = 10−3mm32.4 mmP2, 400μm Hg10mm/μm1 atm = 1.01325 × 105Pa52.4PaPatm 1.0132510320 Pa760atmAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.4-2)V1= 65.0 ft3, T1= 460 + 90 = 550 °R, T2= 460 + 65 = 525 °RFrom Appendix A.1, 1 atm = 14.696 psiap1= 29.0 + 14.696 = 43.696 psia, p2= 75.0 + 14.696 = 89.696 psiap1= 43.696/14.696 = 2.975 atm, p2= 89.696/14.696 = 6.104 atmEq. (1.4-2)312122.975pVV65.030.24ftp6.104R = 0.7302 ft3· atm/lb mol · °R, MW of N2= 28.01Eq. (1.4-1) npV/RT2.97565.0 / 0.73025500.4815lb molDensity,r =0.4815()28.01()/ 30.24=0.446lbm/ft3Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.5-1)Basis: 10 000 kg/a feedTotal balance10 000 = S + WSolids balance0.38 (10 000) = 0.745 + (0)W2W1000051354865kg H O/dS5135kg/d of 74%solutionAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.5-2)Basis: 1000 kg/h “dry” fish cakeSolids balance0.20 WS 00.60 1000W3000 kg/h wet cake feedTotal balance2WS10003000S1000S2000 kg H O/hAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.5-3)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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(1.5-4)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.

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