Chemical Process Equipment Design Solution Manual

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Solutions Manual for Chemical ProcessEquipment Design(2017)byRichard TurtonJoseph A. ShaeiwitzCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-1Chapter 11.021321sfWezguudPfirst term: enthalpy termsecond term:kinetic energy, often written as (v)2)/2third term: potential energyfourth term: frictional lossesfifth term: shaft work2.mass flowrate constant, so, if larger pipe is 1 and smaller pipe is 2222111vAvAm, whereAis cross sectional area for flow, if density constant,then2211vAvA, so if area goes down, velocity must go up in pipe 23.Pressure head is a method for expressing pressure in terms of the equivalent heightof a fluid, where the pressure head is the pressure at the bottom of a that heightof that fluid. From the mechanical energy balance, the value is obtained by dividingevery term byg, such that all terms have units of length.4.SinceAvm, if the density is constant, since the cross-sectional area is constant,the velocity must be constant.Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-25.vAvmmass in = mass out, so if single pipe, mass flowrate must remain constantif density remains constant, volumetric flowrate remains constantif density and area remain constant, velocity must remain constant6.forcesviscousforcesinertialReinertial forces keep fluid flowing, viscous forces resist fluid flowCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-37.all of the following have similar shapefRelaminar flowturbulent flowflow in pipeCDRelaminar flowturbulent flowflow pastsubmerged objectfRelaminar flowturbulent flowflow in packed bedCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-48.pipe length: linear relationship, as pipe gets longer, friction increases proportionallyvelocity in pipe (or flowrate): in turbulent flow, friction increases as square ofvelocity or flowrate, linear in laminar flowpipe diameter:strong inverse relationship; in turbulent flow, friction goes withd-5;in laminar flow it isd-49.series: mass flowrate constant, pressure drops additive10.parallel: mass flowrates additive, pressure drops equal11.forcompressibleflow,densitynotconstantaspressurechanges,somustdoindicated integral021321sfWezguudPfor incompressible flow, density constant, so first term is simplified0213sfWezguuP12.frictional drag due to “skin friction,” which is contact with solid object/surfaceform drag is from energy loss due to flow around object (fluid changing directiontakes energy)13.in a packed bed,void fraction = volume of bed not occupied by solid (void space)/total volume of bedCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-514.total volume is volume of bed if emptysolid volume is total volume of solids in bedvoid volume is volume of bed not occupied by solidstotal volume = solid volume + void volume15.sphericity = surface area of sphere/surface area of particle, both having samevolume16.in a manometer, where the pressure drop is expressed as a positive numberhgPflowingfluidfluidmanometer)(, whereh= is the difference in manometer fluidheightSo, for a large pressure drop, the height difference (and hence the height of themanometer required) decreases if the manometer fluid is dense, like mercury.However, for very small pressure drops, accuracy is lost due to the small heightdifference in a mercury manometer, so a less dense manometer fluid is better.17.That is the flowrate at which the available net positive suction head equals therequired net positive suction head. For higher flowrates, the fluid will vaporize uponentering the pump, causing cavitation, which damages the pump.However, it isphysically possible to operate at higher flowrates.18. For a centrifugal pump, this is where the control valve is wide open, so there isminimal pressure drop across the control valve. At this point, since the control valveis wide open, the flowrate is at its maximum possible value. Therefore, it is physicallyimpossible to operate at higher flowrates.Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-619.The mechanical energy balance across a pump or compressor (neglecting any heightdifference between suction and discharge) reduces todPWsSince vapor densities are 2-3 orders-of-magnitude lower than liquid densities, moreshaft work is required, so the cost of power increases proportionally.20.Centrifugalcompressorscannotachieveverylargecompressionratios(outletpressure/inlet pressure), so staging is needed to get large compression ratios.Positive displacement compressors can have larger compression ratios.Energyincompressionisminimizedwithisothermalcompression,whichisnotpossible, since compressing a gas causes the temperature to increase. Isothermaloperation could be approached with an infinite number of compression/intercoolingstages with infinitesimal pressure and temperature increases, which is a nice limitingcase,butimpossible.Stagingcompressorswithintercoolingisanattempttoapproach the limiting case in a practical way. The economics of a process determinesthe number of stages to use.There is also the problem that if the temperature in a compressor stage increasestoomuch,thesealswillgetdamaged,whichisagoodreasonthekeepthecompression ratio low.Companion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-721.For fully developed turbulent flow, it is assume that the friction factor has reachedits asymptotic value. The proportionalities are52DvLPa.Sincethepressuredropisproportionaltoflowratesquared,doublingtheflowrate increases the pressure drop by a factor of four.b.Since the pressure drop is proportional to diameter to the negative fifth power,increasing the diameter by 25% changes the pressure drop by 1.25-5= 0.33, sothe pressure drop decreases by a factor of three. Note that the friction factormay change slightly, since the roughness/pipe diameter value will change slightly.This is ignored in all parts of this problem.c.25Dv, so, for constant pressure drop, the flowrate increase significantlyd.it is exactly a proportional increasee.5.0Lv, so the flowrate decreases, but it is a one-half-power decrease, so thedecrease in flowrate is less than the increase in lengthf.subscript 1 is for the original, long segment; subscript 2 is for the shorter parallelsegments; take the ratio of pressure drops in both cases, noting that thediameters are constant, that the length of pipe 2 is half of pipe 1, and that theflowrates in each pipe are one-half of the original, so the pressure drop goesdown by a factor of eight; note that minor losses due to the parallel piping areneglected125.0)5.0(5.02525121221212DDvvLLPPCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-822.For laminar flow, the proportionalities are4DvLPa.Sincethepressure drop isproportional toflowrate,doubling theflowratedoubles the pressure drop.b.Since the pressure drop is proportional to diameter to the negative fourth power,increasing the diameter by 25% changes the pressure drop by 1.25-4= 0.41.c.4Dv, so, for constant pressure drop, the flowrate increase significantly, morethan for turbulent flowd.it is exactly a proportional increasee.1Lv, so the flowrate decreases in proportion to the increase in lengthf.subscript 1 is for the original, long segment; subscript 2 is for the shorter parallelsegments; take the ratio of pressure drops in both cases, noting that thediameters are constant, that the length of pipe 2 is half of pipe 1, and that theflowrates in each pipe are one-half of the original, so the pressure drop goesdown by a factor of four; note that minor losses due to the parallel piping areneglected25.0)5.0(5.04241121212DDvvLLPPCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-923.AvvvAvmpipe 1–area from Table 1.1, density givenm/s26.3m1065.21/sm00706.0/sm00706.0kg/m850kg/s62431113311Avvmvpipe 2–area from Table 1.1m/s66.1m.1063.79/sm0106.0kg/s9.01/s)m0106.0)(kg/m850(24-32223322Avvvmpipe 3–area from Table 1.1kg/s4.50/s)m00529.0)(kg/m850(m/s00529.0)m10m/s)(13.13032.4(33332-4333vmAvvkg/s51.1044321mmmmmpipe 4–area from Table 1.1m/s59.2m1069.47/sm0124.0/sm0124.0kg/m850kg/s51.102434443344AvvmvCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-1024.Avmassume= 1000 kg/m3for all sectionskg/s3.94m/s)3)(m1013.13(kg/m1000kg/s2.79m/s)5)(m10574.5(kg/m1000243222243111vAmvAma.kg/s73.6434321mmmmmmb.m/s818.0)m1019.82(kg/m1000kg/s73.6243444Amvc.243333m1085.30m/s)18.2(kg/m1000kg/s73.6AmAarea is closest to 2.5-in, schedule-40 pipe, and that pipe has a slightly larger xs area,so it is a good choiceCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-1125.42iidA, soA4-in= 0.0873 ft2,A3-in= 0.0491 ft2,A2-in= 0.0218 ft2,Atank= 19.63 ft2a.in tank/secft0.3927ft/sec)(0.02ft63.1932vb.at constant density, volumetric flowrates balanceft/sec5.4ft0.0491-ft/sec)4(ft0.0218-ft/sec)8(ft0.0873/secft39.0332223ininoutintankvvvv26.0212mWezgvPspfassume inlet and outlet at same pressure since no information provideduniform pipe diameter, so kinetic energy term zeropipe length not needed, since frictional loss givengal/min487)gal/ft487sec/min)(60(lb/ft62.4lb/sec7.67lb/sec7.670sec))(hp/lbfthp)(5500.8(20-/lblbft80ft)50()seclb/(lbft32.2ft/sec2.3233ff2f2.vmmCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-1227.0212mWezgvPtsfinlet and discharge both at atmospheric pressure, so pressure term is zeroinlet velocity is zero, since reservoir is like tank, so level is assumed constantmultiply 20 m of head bygto get frictional loss in correct unitsM W84.7W1084.7smkg1084.70)kg/m/s)(1000m25(80.0)m/sm(9.8110m)50(m/s81.920m/s)1(632633222ssWW28.0212mWezgvPspfP= 0, since both tanks are open to the atmospherevelocities both zero since tank levelsm6.70)kg/ms)(1000h/3600/h(1m10J/s)0.75(100-J/kg3.5m)10)(m/s81.9(332zzCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-1329.0212mWezgvPtsfpressure termiszero since reservoiranddischarge are both at atmosphericpressurefrictional loss assumed zero, since nothing statedlocation 1 is reservoir, location 2 is dischargehp37230lb/sec)65000(55.0sec))/(hplbft550()/seclb/lbft32.2(2sec/ft013.26ft)60(/seclb/lbft32.2ft/sec32.2ft/sec26.134ft)(10lb/ft62.4lb/sec65000f2f2222f2232ssWWv30.0212sfWezgvPThis is actually a nozzle problem, so only pressure and kinetic energy terms remain.must look up vapor pressure of water at 25°C, which is 3.168 kPalocation 1 isbefore the orifice, soP1= 34,500 kPa,P2= 3.168 kPam1042.1m/s)67.262(4mL10ms/min60mL/min250m/s67.26202/sm10305.5kg/m1000N/m34500-3168m/s10305.54m)1.0(mL)10/s)(m/60mL/min(min25042222263222224223242631ddvAvvvvCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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1-1431.0212mWezgvPspfP= 0, since both tanks are open to the atmospherevelocities both zero since tank levelsdensity of water in lb/gal = (62.4 lb/ft3)(ft3/7.48 gal) =8.3 lb/gala.hp37.10lb/gal)sec)(8.3in/60gal/min)(m30(sec))(hp/lbfthp)(5500.8(-/lblbft200ft)928873()seclb/(lbft32.2ft/sec2.32ff2f2ssWWb.without pump, need sufficient head from reservoir to overcome frictionft728928-zft2000/lblbft200ft)()seclb/(lbft32.2ft/sec2.32f2f2zzzCompanion to Chemical Process Equipment Design, ISBN-13: 9780133804478.

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