流体力学与传热习题参考解答英文.docx

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流体力学与传热习题参考解答英文

1.Waterispumpedataconstantvelocity1m/sfromlargereservoirrestingonthefloortotheopentopofanabsorptiontower.Thepointofdischargeis4meterabovethefloor,andthefrictionlossesfromthereservoirtothetoweramountto30J/kg.Atwhatheightinthereservoirmustthewaterlevelbekeptifthepumpcandeveloponly60J/kg?

22

U22P2u,R

w=Z2g+-hf(Z0+1丄)

22

ZZ2Z!

431m

2.Thefluid(density1200kg/m)ispumpedataconstantrate20m3/hfromthelargereservoirtotheevaporator.Thepressureabovethereservoirmaintainsatmospherepressureandthepressureoftheevaporatorkeeps200mmHg(vacuum).Thedistancebetweenthelevelofliquidinthereservoirandtheexitofevaporatoris15meterandfrictionallossinthepipeis120J/kgnotincludingtheexitofevaporator,whatisthepumpeffectiveworkandpowerifthediameterofpipeis60mm?

乙g+^

hf

2.67x104N/m2

hf120J/kg

2005

10F2x1.013x105

760

1200Kg/m360

U2

V

A

20

1.97m/s

3600*

n/4*0-062

Z1

0

Z2

15

2.67x10

1.972

W

15x9.81

120246.88J/kg

1200

2

NW

Q

246.88x1200x20/3600=1646W

hf=6.5U2

whereUisthevelocityinthepipe,find

a.watervelocityatsectionA-A'.

b.waterflowrate,inm3/h.

\L

8mAK

23

V=UA=2.9n/4x01x360082m/h

1Z1g+U/2

U22

Z2g+2

hf

U1

0

12Z1

6m

Z20

hf

6.5U2

6x9.81

u2

6.5U2

2

U2.9m/s

4.Waterpassesthroughthevariablepipe.Thevelocityinthesmallpipeis2.5m/s.TheverticalglasstubesareinsertedrespectivelyatthesectionAandBtomeasurethepressure(seefig.)Ifthefrictionlossbetweentwosectionis15J/kg,whatisthewatercolumndiffereneebetweentwoglasstubes?

Bytheway,drawtherelativeliquidcolumnheightoftwotubesintheFig.

a乙g+U;

bZbg+U;hf

2

乙二Zb

・・2

・・2

ab

Ub

Uahhf

1.232/2

2.52/21512.63

2

2f

12.63

3

Pa&

Rg

R=

3

1.29x10m

9.8x10

UaAaUbAb

Ub2.5*(33/47)21.23m/s

5.Acentrifugalpumptakesbrine(density1180kg/m3,viscosity1.2cp)fromthebottomofasupplytank

anddeliversitintoanothertank.Thelinebetweenthetanksis300mof25mmdiameterpipe(innerdiameter).Theflowrateis2m3/h.Inthisline,therearetwogatevalves,fourelbows(90o)andonereturnbend,whatisthefrictionlossiftheroughnessofpipeis0.025mm?

lU2

hfhfsthflocal4fdV

v2m3/h=1.2cp=1.2x10-3Pa.s

k=0.025mmk/d=0.025/25=0.001

kc=0.4kl=1kr=2x0.07=0.14

kel4x0.753kre15-2.2

uv/A2/(3600x/4x0.0252)1.13m/s

Re山2.78x104f0.063

2

hf4x0.0063x300/0.025x1.132/2+

(0.4+1+2x0.07+4x0.7+1.5)x1.132/2=197.86J/kg

 

1000

V0uos00.61x/4x0.00012x2gx0.6x(13600

5.8x10m3/s

7.Waterflowsthroughapipewithadiameterdi100mmasshowninfigure.

a.whenthevalveisclosed,Ris600mmandhequals1500mm.Whilethevalveopenspartially,R=400mmandh=1400mm,f=0.00625(Finningfactor)andkc=0.5(contractioncoefficient),whatistheflowrateofwater,inm3/h?

b.Ifthevalveopensfully,whatisthepressureofsection2-2',inN/m2?

Theequivalentlengthofthevalveis1.5mandtheFanningfactorfkeepsthesame?

(H2o=1000kg/m3,Hg=13600kg/m3)

2

2

g乙

u11

2

gZ2

u22

2

hf1-2

1

02

g(Hg

RH2oh)

39630N/m

2

2

lu

u

u1

0Z2=0

hf1-2

4f+k

c2.13u

d22

WecangetZ1fromthevalveclosed

h=1.5mR=0.6mZ1HggR/H2Oh6.66m

1

L玄

9.81x6.66u2/22.13u239630/1000

hf1-3

'and3-3',wehave

u=3.13m/sVh3600x/4x0.12x3.1388.5m/h

(2)whenthevalveopensfully,forsection1-1

2

2

d二

lin^B

jlJORinii

Forparallelpipeline

2.72

hfAhfBVtotaiVA+VB

hfA4fA

(l+^e)UA4x0.0075x10/0.053/2(2722)2

dA23600x/4x0.053"

0.333J/kg

(l+le)Ub22

hfB4fB

dB

—4x0.0045x2/0.3/2xuB0.333

2

uB2.36m/sVB=uBAB2.36x/4x0.232600m3/h

10.Aflatfurnacewallisconstructedof120mmlayerofsil-o-celbrick,withathermalconductivity0.08w/(moC),backedbya150mmofcommonbrick,ofconductivity0.8w/(moC),thetemperatureofinnerfaceofthewallis1400o,andthatoftheouterfaceis200C.

a.Whatistheheatlossthroughthewallinwpersquaremeter.

b.

Toreducetheheatlossto600w/m2byaddingalayerofcorkwithk0.2w/(moC)ontheoutsideofcommonbrick,howmanymetersofcorkarerequied?

0.120.15

b.600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m

13.Airatthenormalpressurepassesthroughthepipe(d20mm)andisheatedfrom20oCto100oC.Whatisthefilmheattransfercoefficientbetweentheairandpipewalliftheaveragevelocityofairis10m/s?

Thepropertiesofairat60oCareasfollows:

density1.06kg/m3,viscosity0.02cp,conductivity0.0289w/(moC),andheatcapacity1kJ/kg-K

ud10x0.02x1.0644

0.02x10

Re=l1.06x1010

2

39.66x0.0289心02=57.22w/m.k

14.Ahotfluidwithamassflowrate2250kg/hpassesthrougha25x2.5mmtube.Thephysicalpropertiesoffluidareasfollows:

k=0.5w/(moC),Cp=4kJ/kg-K,viscosity10-3N-s/m2,density1000kg/m3Find:

a.Heattransferfilmcoefficienthi,inw/(m2-K).

b.

Iftheflowratedecreasesto1125kg/handotherconditionsarethesame,whatistheih

c.Ifthediameteroftube(insidediameter)decreasesto10mm,andthevelocityukeepsthesameasthatofcasea,calculateh.

d.Whentheaveragetemperatureoffluidandquantityofheatflowpermeteroftubeare40oCand400w/m,respectively,whatistheaveragetemperatureofpipewallforcasea?

e.Fromthisproblem,inordertoincreasetheheattransferfilmcoefficientandenhanceheattransfer,whatkindsofmethodscanyouuseandwhichisbetter,explain?

Hint:

forlaminarflow,Nu=1.86[RePr]1/3

1/3

4

220.1

forturbulentflowNu=0.023Re0.8Pr1/3

hi25500X0.5183159w/m2k

Nu0.023Re0.8Pr1/3hi=6347w/m2k

⑷Q=hiAi(t-tw)=400=500x2x0.02(t-tw)

t=40Ctw39.41C

(5)theremethods:

increaseuorhiordecreasedThefirstisbetter

thepipeandthehotfluidgoesthroughtheoutside.Theinletandoutlettemperaturesofcoldfluidare20and80o,andtheinletandoutlettemperaturesofhotfluidare150and90,respectively.Thehi(filmcoefficientinsidepipe)is700w/(m2°C)andoverallheattransfercoefficientUo(basedontheoutsidesurfaceofpipe)is300w/(m2oC),respectively.Iftheheatlossisignoredandtheconductivityofpipewall(steel)istakenas45w/(moC),find:

(1)heattransferfilmcoefficientoutsidethepipeho?

(2)thepipelengthrequiredforcounterflow,inm?

(3)whatisthepipelengthrequirediftheheatingmediumchangestosaturatedvapor(140oC)anditcondensestosaturatedliquidandotherconditionskeepunchanged?

(4)Whentheexchangerisusedforayear,itisfoundthatitcannotmeettheneedofproduction(theoutlettemperatureofcoldfluidcannotreach80°C),explainwhy?

11d0lmdo1230.002x23

h。

Vohidikdm300700x1945x21

1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21

2

h0642.9w/mk

LMTD=70°C=t1+12Q=UoAoTm=mcCp(Tcb-Tca)

2

300*2*0.023*70L=500/3600*1000*(80-20)

L=5.4m

8020

14020ln

14080

(d)scaleisformedontheoutside,V0isdecreased

16.

Waterflowsturbulentlyinthepipeof①25x2.5mmshelltubeexchanger.Whenthevelocityofwateruis1m/s,overallheattransfercoefficientUo(basedontheoutersurfaceareaofpipe)is2115w/(moC).Iftheubecomes1.5m/sandotherconditionskeepunchanged,Uois2660w/(moC).Whatisthefilmcoefficienthooutsidethepipe?

(Heatresistancesofpipewallandscaleareignored)

17.Waterandoilpassparallellythroughanexchangerwhichis1mlong.Theinletandoutlettemperaturesofwaterare15and40°C,andthoseofoilare150and100°C,respectively.Iftheoutlettemperatureofoil

decreasesto80PC,andtheflowratesandphysicalpropertiesandinlettemperaturesofwaterandoilmaintainthesame,whatisthepipelengthofnewexchanger?

(Heatlossandpipewallresistanceareneglected)

QWhChT1-T2WcCct2t1VAtm

WhChT1-T2'WcCct2't1

150

100

40

15

t2

50C

150

80

t2

15

L2

T1

T2'

tm1

150

8092.5

4OC

L1

T1

T2

tm2

150

1.85

10069.8

L2

1.85mL1=1m

tm1

92.5tm269.8

18.Airwhichpassesthroughthepipeinturbulentflowisheatedfrom20to80°C.Thesaturatedvaporat

mc1Cpc(Tcb

116.3oCcondensestosaturatedwateroutsidethepipe.Ifairflowrateincreasesto120%oftheoriginandinletandoutlettemperaturesofairstayconstant,whatkindofmethodcanyouemployinordertodothat?

(Heatresistaneeofpipewallandscalecanbeignored)

h12Ai

Tm2

mc2Cpc(Tcb

Tca)

hi2Tm2

mc2

1.2

1.20.8

Tm2

hi1Tm1

md

Tm1

Tm1

8020

ln(116.3

20)/(116.380)

Tca)

Tm2

ln(Th20)/(Th80)

8020

Th=118.5oC

19.

Waterflowsthroughthepipeofa①25x2.5mmshell-tubeexchangerfrom20to50oC.Thehotfluid(Cp1.9kJ/kgoC,flowrate1.25kg/s)goesalongtheshellandthetemperatureschangefrom80to30oC.Filmcoefficientsofwaterandhotfluidare0.85kw/(moC)and1.7kw/(moC).WhatistheoverallheattransfercoefficientUoandheattransferareaifthescaleresistancecanbeignored?

(theconductivityofsteelis45w/(moC).

W=1.25Kg/sCp=1.9Kj/kgC

21.Afilterpress(A=0.1m2)isusedforfilteringslurry.Thevacuuminsidethefilteris500mmHg.Oneliterfiltratecanbegotafterfilteringof5minand0.6moreliterfiltrateisobtainedafter5moremin.Howmuchfiltratewillbegotafterfilteringof5moremin?

22

forfilterpressV2VVe=KA

22

5min12Ve0.1x5K⑴

22

10min1.62x1.6Ve0.1x10K

(2)

From

(1)

(2),wecanseeVe=0.7K=48

.2

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