化工单元操作

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1.52?105?0.247?105?18.41mtotalheadsofpumpisH?0.4?1000?9.81efficiencyofpumpis??Ne/N

QH?g26?18.41?1000?9.81??1.3kW36003600N=2.45kW

Thenmechanicalefficiency

1.3?100%?53.1%??2.45Theperformanceofpumpis

Flowrate，m3/h26Totalheads，m18.41Shaftpower，kW2.45Efficiency，%53.1

2.2Wateristransportedbyapumpfromreactor,whichhas200mmHgvacuum,tothetank,in

2whichthegauge

pressureis0.5kgf/cm2,

10masshowninFig.Thetotalequivalentlength

1ofpipeis200m

includingalllocalfrictionalloss.Thepipelineis?57×3.5mm,theorificecoefficientof

Coandorificediameterdoare0.62and25mm,respectively.Frictionalcoefficient?is0.025.Calculate:DevelopedheadHofpump,inm(thereadingRofUpressuregaugeinorificemeteris168mmHg)

Solution:

Equation(1.6-9)

2Rg（?f??）C00.622?0.168?9.81(13600?1000)V??044?1000?d0??25?1???1????50??D?

0.62?6.44?4.12m/s?0.9375

Massflowrate

sinceNe?

3.14?0.0252?1000?2.02kg/s42)Fluidflowthroughthepipefromthereactortotank,theBernoulliequationisasfollowsforV1=V2m?VoSo??4.12?H?p2?p1??z?Hf?g200?1.013?105?75707Pa760?z=10m

?p?0.5?9.81?104??p/?g=7.7m

Therelationbetweentheholevelocityandvelocityofpipe

22?d0??1?V?V0???4.12????1m/s?2??D?Frictionloss

lu220232?0.025??5.1mHf?4fd2g0.052?9.81so

H=7.7+10+5.1=22.8m

2.3.Acentrifugalpumpistobeusedtoextractwaterfromacondenserinwhichthevacuumis640mmofmercury,asshowninfigure.Attherateddischarge,thenetpositivesuctionheadmustbeatleast3mabovethecavitationvaporpressureof710mmmercuryvacuum.IflossesinthesuctionpipeaccountedforaheadofHg1.5m.Whatmustbetheleastheightoftheliquidlevelinthecondenserabovethepumpinlet?Solution:

Fromanenergybalance,

p?pv?Hf?NPSHHg?o?g

Where

Po=760-640=120mmHgPv=760-710=50mmHg

UseoftheequationwillgivetheminimumheightHgas

po?pv?Hf?NPSHHg??g

（0.12?0.05)?13600?9.81??1.5?3??3.55m1000?9.81

2.4Sulphuricacidispumpedat3kg/sthrougha60mlengthofsmooth25mmpipe.Calculatethedropinpressure.Ifthepressuredropfallsbyonehalf,whatwillthenewflowratebe?

?Densityofacid1840kg/m3?Viscosityofacid25×10-3Pas

Solution:

Velocityofacidinthepipe:

mu?volumetricflowratem3?????3.32m/s22?cross?sectionalareaofpipe0.785?d0.785?1840?0.025d24d?u0.025?1840?3.32?6109

25?10?3Reynoldsnumber:

Re???fromFig.1.22forasmoothpipewhenRe=6109,f=0.0085pressuredropiscalculatedfromequation1.4-9

lu2603.322hf??4f?4?0.0085?450J/kg

?d20.0252?p?p?450?1840?827.5kPa

orfrictionfactoriscalculatedfromequation1.4-25

2lu2603.322?0.2lu?0.2hf??4f?4?0.046Re＝4?0.046?6109?426J/kg?d2d20.0252?p?p?426?1840?783.84kPa

ifthepressuredropfallsto783.84/2=391.92kPa

????plu2?0.2?p???391920?4?0.046Re?＝4?0.046??????2d2???1840??4?0.046?1840??3??25?10??0.21.8?0.2lu1.8?1.22d

60u1079.891.8?`u1.220.0120.025so

u?1.8391920?0.0121.8?4.36?2.27m/s

1079..89newmassflowrate=0.785d2uρ=0.785×0.0252×2.27×1840=2.05kg/s

2.4Sulphuricacidispumpedat3kg/sthrougha60mlengthofsmooth25mmpipe.Calculatethedropinpressure.Ifthepressuredropfallsbyonehalfonassumptionthatthechangeoffrictionfactorisnegligible,whatwillthenewflowratebe?

Densityofacid1840kg/m3Viscosityofacid25×10-3PaFrictionfactorf?0.0056?0.500forhydraulicallysmoothpipeRe0.32Solution:

Writeenergybalanceequation:

2p1u12p2u2?z1??H??z2??hf?g2g?g2g?plu2H?hf????gd2g?4d2u??3

u?3?412??3.32m/s22?d?3.14?0.025?18403.32?0.025?1840?6115?325?10Re?f?0.0056?0.5000.5?0.0056??0.00870.320.32Re6115?plu2603.322H?hf????4?0.0087?46.92

?gd2g0.0252?9.81Δp=46.92×1840×9.81=847.0kpa

2.6ThefluidispumpedthroughthehorizontalpipefromsectionAtoBwiththeφ38?2.5mmdiameterandlengthof30meters,shownasfigure.Theorificemeterof16.4mmdiameterisusedtomeasuretheflowrate.OrificecoefficientCo＝0.63.thepermanentlossinpressureis3.5×104N/m2,thefrictioncoefficientλ=0.024.find:

(1)WhatisthepressuredropalongthepipeAB?

(2)WhatistheratioofpowerobliteratedinpipeABtototalpowersuppliedtothefluidwhentheshaftworkis500W,60?ficiency?(Thedensityoffluidis870kg/m3)

solution：

22uApAuAzAg???w?zAg????hf

?2?2pApA?pB?lu2?p0??hf???d2?2Ao?16.4?????0.247A?33?u0?C01?0.24722gR??????0.632?9.81?0.6?13600?870???8.5m/s

??0.97870∴u=(16.4/33)2×8.5=2.1m/s

302.12?3.5?104?76855N/m2∴pA?pB???hf?0.024?8700.0332(2)

Ne?Wm??p?2du??76855?0.785?0.0332?2.1?138W?4so

theratioofpowerobliteratedinfrictionlossesinABtototalpowersuppliedtothefluid138?100％＝46％

500?0.6

3.2Asphericalquartzoseparticle（颗粒）withadensityof2650kg/m3settles（沉淀）freelyinthe20℃air,trytocalculatethemaximumdiameterobeyingStocks’lawandtheminimumdiameterobeyingNewton’slaw.

Solution:

ThegravitysettlingisfollowedStocks’law,somaximumdiameterofparticlesettledcanbecalculatedfromRethatissetto1

Ret?dcut???1,then

ut?

?dc?

equation3.2-16fortheterminalvelocity

2dc(?S??)g??dc?18?solvingforcriticaldiameter

dc?1.2243?2(?S??)g

Checkuptheappendix

Thedensityof20℃airρ=1.205kg/m3andviscosityμ=1.81×10-3N·s/m2

(1.81?10?3)2dc?1.2243(2650?1.205)?1.205?5.73?10?5m?57.3?m

whenReynoldsnumber≥1000,theflowpatternfollowsNewton’slawandterminalvelocitycanbecalculatedbyequation3.2-19

ut?1.75gdp??p????1

criticalReynoldsnumberisRet??ut?dc??1000，2

rearrangingtheequation2gives

ut?1000?3??dccombinationofequation1withequation3

?(?S??)gdc1000??1.74?dc??(?S??)?solvingforcriticaldiameter

??32.33dc

?2

(1.81?10?3)2??32.33dc(2650?1.205)?1.205?1.512?10?3m?1512um

3.3Itisdesiredtoremovedustparticles50micronsindiameterfrom226.5m3/minofair,usingasettlingchamberforthepurpose.Thetemperatureandpressureare21oCand1atm.Theparticledensityis2403kg/m3.Whatminimumdimensionsofthechamberareconsistentwiththeseconditions?(themaximumpermissiblevelocityoftheairis3m/s)solution:

tocalculateterminalvelocityfromtheequation3.2-16

ut?2??p???gdp18?2??p???gdp

Thedensityof21℃airρ=1.205kg/m3andviscosityμ=1.81×10-5N·s/m2

ut?18?(50?10?6)2(2403?1.205)?9.81＝?0.181m/s?518?1.81?10Q?BLutso

BL?Q226.5??20.86m21ut60?0.181fromequation3.3-4

LH?uutthemaximumpermissiblevelocityoftheairis3m/sLH?30.181L?16.58H2setBtobe3m,thenfromequation1L=7mAndH=0.42m

3.4Astandardcycloneistobeusedtoseparatethedustofdensityof2300kg/m3fromthegas.Theflowrateofgasis1000m3/h,theviscosityofthegasis3.6?10-5N·s/m2,andthedensityis0.674kg/m3.Ifthediameterofcycloneis400mm,attempttoestimatethecriticaldiameter.

Solution:D=0.4m

B=D/4=0.1mh=D/2=0.2m

Q1000ui???13.9m/s

hB3600?0.2?0.1

Accordingtotheequation3.3-12forN=5：

9?B9(3.6?10?5)(0.1)dc???8?10?6m?8?m

?N(?p??)ui5?(2300?0)?13.9

3.6Afilterpressof0.1m2filteringareaisusedforfilteringasampleoftheslurry.Thefiltrationiscarriedoutatconstantpressurewithavacuum500mmHg.Thevolumeoffiltratecollectedinthefirst5minwasoneliterand,afterafurther5min,anadditional0.6literwascollected.Howmuchfiltratewillbeobtainedwhenthefiltrationhasbeencarriedoutfor15minonassumingthecaketobeincompressible?Solution:

Theequationfortheconstant-pressurefiltrationV2?2VVm?KA2t

5min1l.12?2Vm?K0.12?510min1.6l.1.62?2?1.6Vm?K?0.12?10solvingtheequationsaboveforVmandK

Vm?0.7andK=48

Fort?15minV2?2?0.7?V?48?0.12?15SolvingforV=2.073l

3.7Thefollowingdataareobtainedforafilterpressof0.0093m2filteringareainthetestPressuredifference/kgf/cm2filteringtime/sfiltrate/m31.05502.27×10-36609.10×10-33.5017.12.27×10-32339.10×10-3Calculate:

(1)filtrationconstantK,Vmatthepressuredifferenceof1.05

(2)iftheframeofthefilterisfilledwiththecakeat660s,whatisthefinalrateof

?dV?filtration??

?dt?E(3)andwhatisthecompressibleconstantofcaken?solution:

①fromequation3.4-19a

q2?2m?KtForpressuredifferencep?1.05㎏/㎝2

?2.27?10?3?2.27?10?3??0.0093???2?0.0093qm?K?501???9.1?10?3?9.1?10?3??0.0093???2?0.0093qm?K?6602??22solvingtheequations1and2gives

m3qm?0.03792mK?1.56?10?3m2/s

②

KA2?dV?=7.14?10?6m3/s????dt?E2(V?Vm)

Forpressuredifferencep?3.5㎏/㎝2

?2.27?10?3?2.27?10?3??K??1.713??0.0093???2?0.0093qm???9.1?10?3?9.1?10?3??K??2334??0.0093???2?0.0093qm??22solvingtheequations3and4gives

m3??0.03092qmm????p?????????p??1?nK??4.37?10?3m2/s

then

4.37?10?3?3.5?????31.051.56?10??

1?n

ln2.8ln3.33solvingforn=0.142

3.8Aslurryiffilteredbyafilterpressof0.1m2filteringareaatconstantpressure,theequationforaconstantpressurefiltrationisasfollows1?n?(q?10)2?250(t?0.4)

whereq=filtratevolumeperunitfilteringarea,inl/m2,t=filteringtime,inmincalculate:

(1)howmuchfiltratewillbegottenafter249.6mi