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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Document preview page 1

Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 1

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

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

Need to ace your assignments? Use Solution Manual for Transport Processes and Separation Process Principles, 5th Edition for detailed, easy-to-understand solutions to textbook questions.

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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 1 preview imageSolution 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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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 2 preview image
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 3 preview image(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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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 4 preview image(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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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 5 preview image(1.3-1)22MW O32.00,MW N28.02MW air0.21 32.000.79 28.0228.9 kg / kg molAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 6 preview image(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 COAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 7 preview image(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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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 8 preview image(1.3-4)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
Page 9 of 16
Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 9 preview image(1.3-5)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 10 preview image(1.4-1)P = 2.4 mm HgFrom Appendix A.11 atm = 760 mm Hg32.4 mm HgP3.1610atm760 mm Hg/atm1 atm = 33.90 × 12 in. H2O (at 4°C)22in.H O2.4Patm 33.90121.285in.H O760atm1 μm = 10−3mm32.4 mmP2, 400μm Hg10mm/μm1 atm = 1.01325 × 105Pa52.4PaPatm 1.0132510320 Pa760atmAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 11 preview image(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)312122.975pVV65.030.24ftp6.104R = 0.7302 ft3· atm/lb mol · °R, MW of N2= 28.01Eq. (1.4-1) npV/RT2.97565.0 / 0.73025500.4815lb molDensity,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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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 12 preview image(1.5-1)Basis: 10 000 kg/a feedTotal balance10 000 = S + WSolids balance0.38 (10 000) = 0.745 + (0)W2W1000051354865kg H O/dS5135kg/d of 74%solutionAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 13 preview image(1.5-2)Basis: 1000 kg/h “dry” fish cakeSolids balance0.20 WS 00.60 1000W3000 kg/h wet cake feedTotal balance2WS10003000S1000S2000 kg H O/hAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 14 preview imageAncillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 15 preview image(1.5-3)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Solution Manual for Transport Processes and Separation Process Principles, 5th Edition - Page 16 preview image(1.5-4)Ancillary material to Geankoplis, et al., Transport Processes and Separation Process Principles, 5/E, ISBN: 9780134181028.
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Chemical Engineering
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Transport Processes And Separation ...

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