NEWEST PSFA FINAL EXAM COMPLETE ACTUAL EXAM 75 REAL QUESTIONS AND CORRECT DETAILED ANSWERS (CORRECT VERIFIED ANSWERS) LATEST UPDATED VERSION

Chapter 11 • Turbomachinery
p p
p




11.1 Describep thep geometryp andp operationp ofp ap humanp peristalticp PDPp whichp isp cher-
pishedpbypeverypromanticppersonponpearth.pHowpdoptheptwopventriclespdiffer?




Solution:p Clearlypweparepspeakingpofpthephumanpheart,pdrivenpperiodicallypbyptravellingpc
ompressionpofpthepheartpwalls.pOnepventriclepservespthepbrainpandptheprestpofpone’spextremities,pw
hilepthepotherpventriclepservesptheplungspandppromotespoxygenationpofpthepblood.p p Ans.



11.2 Whatpwouldpbeptheptechnicalpclassificationpofpthepfollowingpturbomachines:

(a) aphouseholdpfanp=panpaxialpflowpfan. Ans.p(a)
(b) apwindmillp=panpaxialpflowpturbine. Ans.p(b)
(c) anpaircraftppropellerp=panpaxialpflowpfan. Ans.p(c)
(d) apfuelppumppinpapcarp=pappositivepdisplacementppumpp(PDP). Ans.p(d)
(e) anpeductorp=papliquid-jet-pumpp(specialppurpose). Ans.p(e)
(f) apfluidpcouplingptransmissionp=papdouble-impellerpenergyptransmissionpdevice. Ans.p(f)
(g) appowerpplantpsteampturbinep=panpaxialpflowpturbine. Ans.p(g)



11.3 Ap PDPp canp deliverp almostp anyp fluid,p butp therep isp alwaysp ap limitingp very-high
viscositypforpwhichpperformancepwillpdeteriorate.pCanpyoupexplainpthepprobablepreason?

Solution: High-
viscosityp fluidsp takep ap longp timep top enterp andp fillp thep inletp cavityp ofp apPDP.pThuspapPDPppu
mpingphigh-viscositypliquidpshouldpbeprunpslowlyptopensurepfilling. Ans.



11.4 Anpinterestingpturbomachinepisptheptorquepconverterp[58],pwhichpcombinespbothpappu
mppandpapturbineptopchangeptorquepbetweenptwopshafts.pDopsomepresearchponpthispconceptpa
ndpdescribepit,pwithpapreport,psketchespandpperformancepdata,ptopthepclass.

Solution:p Aspdescribed,pforpexample,pinpRef.p58,ptheptorquepconverterptransfersptorquepTpfr
ompappumpprunnerptopapturbineprunnerpsuchpthatpppumppumppppturbineturbine.pMaximumpeffi
ciencypoccurspwhenpthepturbinepspeedppturbinepispapproximatelypone-
halfpofptheppumppspeedpppump.p Ans.

,810 SolutionspManualp •p FluidpMechanics,pFifthpEdition


11.5 Whatp typep ofp pumpp isp shownp inp Fi
g.pP11.5?pHowpdoespitpoperate?

Solution:p Thispispapflexible-
linerppump.pTheprotatingpeccentricpcylinderp
actspaspap“squeegee.”p Ans.

Fig.p P11.5



11.6 Fig.pP11.6pshowsptwoppointspaphalf-
pperiodpapartpinpthepoperationpofpappump.pWh

atptypepofppumppispthis?pHowpdoespitpwork?p
Sketchpyourpbestpguesspofpflowpratepversuspti
mepforpapfewpcycles.

Solution:p Thispispapdiaphragmppump.pAspt
hepcenterprodpmovesptopthepright,popeningpApa
ndpclosingpB,pthepcheckpvalvespallowpAptopfil
lpandpBptopdischarge.pThen,pwhenptheprodpmo
vesptopthepleft,pBpfillspandpApdischarges.pDep
endingpuponpthepexactposcillatorypmotionpofpth
epcenterprod,pthepflowpratepispfairlypsteady,pbein
Fig.p P11.6
gphigherpwhenptheprodpispfaster.p p Ans.


11.7 AppistonpPDPphaspap5-inpdiameterpandpap2-
inpstrokepandpoperatespatp750prpmpwithp92%pvolumetricpefficiency.p(a)pWhatpispthepdelivery,
pinpgal/min? p(b)pIfptheppumppdeliverspSAEp10Wpoilpatp20Cpagainstpapheadpofp50pft,pwhatphor

sepowerpisprequiredpwhenpthepoverallpefficiencypisp84%?

Solution: ForpSAEp10Wpoil,ptakeppp870pkg/m3pp1.69pslug/ft3.pThepvolumepdisplacedpis
p
p=p (5)2p(2)p=p39.3pin3,
4
⎛ in3 ⎞p⎛p 1pgalp ⎞p⎛ strokes⎞
 Qp=p|p39.3 |p | |p| 750 |⎠p (0.92p efficiency)
stroke ⎝ 231pin 3p⎠ ⎝p min
⎝ ⎠
or: Qpp117pgal/min Ans.p(a)

, Chapterp11p •p Turbomachinery 811
⎛p117 3p p ⎞
1.69(32.2)p
|⎝ ftp /s|p(50)
gQHp 449 ⎠ p ftlbfp
Powerp=p =p =p846p p550pp1.54php Ans.p(b)
 0.84 s



11.8 Apcentrifugalppumppdeliversp550pgal/minpofpwaterpatp20Cpwhenpthepbrakephorsepow
erpisp22pandpthepefficiencypisp71%.p(a)pEstimatepthepheadprisepinpftpandptheppressureprisepinppsi
.p(b)pAlsopestimatepthepheadprisepandphorsepowerpifpinsteadpthepdeliverypisp550pgal/minpofpga
solinepatp20C.

Solution: (a)pForpwaterpatp20C,ptakeppp998pkg/m3pp1.94pslug/ft3.pTheppowerprelationpis
⎛p550p ft3p⎞
(62.4)p|p H
ftlbfp gQHp ⎝449 sp |⎠p
Powerp=p22(550)p=p12100p =p = ,
s  0.71
or Hpp112pft Ans.p(a)
Pressureprisep pp=pgHp=p(62.4)(112)p=p7011ppsfpp144pp49ppsi Ans.p(a)

(b)pForpgasolinepatp20C,ptakeppp680pkg/m3pp1.32pslug/ft3.pIfpviscosityp(Reynoldspnumber
)pispnotpimportant,pthepoperatingpconditionsp(flowprate,pimpellerpsizepandpspeed)parepexactlypt
hepsamepandphencepthepheadpispthepsamepandptheppowerpscalespwithpthepdensity:
gasoline ⎛p680⎞p
Hpp112p ftp (ofpgasoline); Powerp=pPwater =p22p p15php Ans.p(b)
⎝| |⎠
water 998



11.9 FigurepP11.9pshowspthepmeasuredpperformancepofpthepVickerspInc.pModelpPVQ40ppi
stonppumppwhenpdeliveringpSAEp10Wpoilpatp180Fp(pp910pkg/m3).pMakepsomepgeneralpobs
ervationspaboutpthesepdatapvis-à-vispFig.p11.2pandpyourpintuitionpaboutpPDPpbehavior.

Solution: Thepfollowingparepobserved:
(a) ThepdischargepQpispalmostp linearlypproportionalptopspeedppandpslightlyplesspforpthep hig
herpheadsp(Hporpp).
(b) Thep efficiencyp (volumetricp orp overall)p isp nearlyp independentp ofp speedp p andp againps
lightlyplesspforphighpp.
(c) Thep powerp requiredp isp linearlyp proportionalp top thep speedp p andp alsop top thep headp Hp(
orpp). Ans.

, 812 SolutionspManualp •p FluidpMechanics,pFifthpEdition




Fig.p P11.9



11.10 SupposepthatptheppumppofpFig.pP11.9pisprunpatp1100pr/minpagainstpappressureprisep ofp
210pbar.p(a)pUsingpthepmeasuredpdisplacement,pestimateptheptheoreticalpdeliverypinpgal/min.p
Frompthepchart,pestimatep(b)pthepactualpdelivery;pandp(c)pthepoverallpefficiency.

Solution:
(a)p Fromp Fig.p P11.9,p thep pumpp displacementp isp 41p cm3.p Thep theoreticalp de
liverypis
⎛|1100 r ⎞p⎛ cm3p ⎞=p45100 cm3 =p45 L =p11.9 gal
Qp= |p|p41 |p Ans.p(a)
⎝ min ⎠p⎝ rp p ⎠ min min min