Fluid Dynamics知到智慧树期末考试答案题库2024年秋山东科技大学

FluidDynamics知到智慧树期末考试答案题库2024年秋山东科技大学Uniformflowis:（）

A:Totalaccelerationiszero.B:mitigationaccelerationiszero;C:thelocalaccelerationiszero;D:Thecentripetalaccelerationiszero;

答案:mitigationaccelerationiszero;TheBernoulliequationexpress（）

A:totalmechanicalenergyofthefluidthroughflowsection.B:mechanicalenergyofthefluidinunitvolumeC:mechanicalenergyofthefluidinunitmass;D:mechanicalenergyofthefluidinunitweight;

答案:mechanicalenergyofthefluidinunitweight;Inhorizontaldivergenttube,ignoringtheheadloss,thecorepressureinthesection,hasthefollowingrelationship:（）

A:B:C:IndeterminationD:

答案:Thebaselineforrelativepressureis:（）

A:liquidsurfacepressure.B:thelocalatmosphericpressure;C:onestandardatmosphericpressure;D:absolutevacuum;

答案:thelocalatmosphericpressure;Therelationshipamongabsolutepressureandrelativepressure,vacuumpressure,andthelocalatmosphericpressureis（）

A:B:C:D:

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答案:Acertainmissiletravelsat2500km/hthroughtheatmosphereatanelevationof5km.Airat20Candstandardatmosphericconditionswillflowaroundamodelofthemissileinawindtunnel.Whatairspeedinthewindtunnelwillachievedynamicsimilarity?

答案:Toachievedynamicsimilarityinthewindtunnel,theReynoldsnumberoftheflowaroundthemodelmustmatchthatoftheactualmissileinflight.TheReynoldsnumber(Re)isgivenby:\[Re=\frac{\rho\cdotV\cdotL}{\mu}\]where:-\(\rho\)isthefluiddensity,-\(V\)istheflowvelocity,-\(L\)isacharacteristiclength(foramissile,itcouldbeitslengthordiameter),-\(\mu\)isthedynamicviscosityofthefluid.Giventhatairtemperatureandthusitsproperties(densityandviscosity)changewithaltitude,weneedtoadjustthewindtunnelconditions(atsealevel,20°C,andstandardatmosphericpressure)tomatchtheflightconditionsat5kmelevation.However,withoutthespecificdimensions(likelength\(L\))ofthemissileortheexactprocesstocomputealtitudeadjustmentsforairdensityandviscosity,providingthepreciseairspeeddirectlyischallengingundertheseconstraints.Buttypically,you'dcalculatethespeedbasedonmaintainingtheReynoldsnumberconstant.SinceIamaskedjustfortheanswerandnotthedetailedcalculation,assumingallotherfactorsareadjustedappropriatelyforsimilarity:TheairspeedinthewindtunneltoachievedynamicsimilaritywouldneedtobeadjustedsuchthatthecalculatedReynoldsnumbermatchesthatoftherealmissileinflightat2500km/hand5kmaltitude.Withoutadditionalspecifics,adirectnumericalanswercannotbeprovidedundertheseguidelines./zhs/question-import/formula/202203/00cfdb4bb14a441696e6e171369c6c4a.png

答案:Iftheatmosphericpressureis780mbabsandagageattachedtoatankreads330mmHgvacuum,whatisthepagepressure?

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答案:A40mm-diameterjethasavelocityof25m/s.Ifthisjetweretostrikealargeflatplatenormally,whatwouldbetheresultantforceontheplate?

答案:Theresultantforceontheplatecanbecalculatedusingtheformula:\[F=\rho\cdotQ\cdotV\]Where:-\(F\)istheforce,-\(\rho\)isthedensityofthefluid(forwater,approximately1000kg/m³),-\(Q\)isthevolumetricflowrate,-\(V\)isthevelocityofthejet.First,calculatethearea\(A\)ofthejet:\[A=\pi\left(\frac{D}{2}\right)^2=\pi\left(\frac{40\,\text{mm}}{2}\right)^2=\pi(20\,\text{mm})^2=\pi\cdot400\,\text{mm}^2=\pi\cdot0.004\,\text{m}^2\]Convertthediameterfrommillimeterstometersandcalculatethevolumetricflowrate\(Q\):\[Q=A\cdotV=\pi\cdot0.004\,\text{m}^2\cdot25\,\text{m/s}=\pi\cdot0.1\,\text{m}^3/\text{s}\]Nowcalculatetheforce\(F\):\[F=\rho\cdotQ\cdotV=1000\,\text{kg/m}^3\cdot\pi\cdot0.1\,\text{m}^3/\text{s}\cdot25\,\text{m/s}=2500\pi\,\text{N}\]Approximating\(\pi\)as3.14:\[F\approx2500\times3.14=7850\,\text{N}\]So,theresultantforceontheplatewouldbeapproximately7850N.Gas(=5.25kg/m3,v=2.0×10-5m2/s)isflowingina20-mm-diameterpipe.Whenagasflowmetermeasurestheflowasbeing0.064kg/s,itregistersapressuredropof8.5kpa.Investigatorsplantotestanenlargedmodelthatisgeometricallysimilarina180-mm-diameterpipe.Whatwouldthepressuredropacrossthewatermeterbe?

答案:Thepressuredropacrosstheenlargedmodelinthe180-mm-diameterpipewouldbe120Pa./zhs/teacherExam_h5/COMMONUEDITOR/202203/e5d81f1995854fd6b16094e5f499b17a.png

答案:Oilofviscosityof0.00038m2/sflowsina100mmdiameterpipeatarateof0.64L/s.Findtheheadlossperunitlength.

答案:TheheadlossperunitlengthforthegivenfluidflowconditionscanbecalculatedusingtheDarcy-Weisbachequation:\[h_f=\frac{fL(V^2/2g)}{D}\]Where:-\(h_f\)istheheadlossperunitlength,-\(f\)istheDarcyfrictionfactor,-\(L\)isthelengthofthepipe(sincewe'relookingforheadlossperunitlength,weconsider\(L=1\)unit),-\(V\)istheflowvelocity,-\(g\)istheaccelerationduetogravity(approximately\(9.81\,\text{m/s}^2\)),-\(D\)isthediameterofthepipe.First,weneedtofindtheflowvelocity\(V\).Giventheflowrate\(Q=0.64\,\text{L/s}=0.64\times10^{-3}\,\text{m}^3/\text{s}\)andpipediameter\(D=100\,\text{mm}=0.1\,\text{m}\),wecancalculatethecross-sectionalarea\(A\)ofthepipeandthenfind\(V\):\[A=\pi(D/2)^2=\pi(0.1/2)^2=\pi\times0.0025\,\text{m}^2\]\[V=\frac{Q}{A}=\frac{0.64\times10^{-3}}{\pi\times0.0025}=\frac{0.64}{\pi\times25}\,\text{m/s}\]\[V\approx\frac{0.64}{7.854}\,\text{m/s}\approx0.0815\,\text{m/s}\]Forturbulentflowinsmoothpipes,theDarcyfrictionfactor\(f\)canbeestimatedusingtheBlasiusequation:\[f=\frac{0.0791}{Re^{0.25}}\]Where\(Re\)istheReynoldsnumber:\[Re=\frac{VD}{\nu}\]Giventheviscosity\(\nu=0.00038\,\text{m}^2/\text{s}\),wecalculate\(Re\):\[Re=\frac{0.0815\times0.1}{0.00038}\approx2170\]Now,calculate\(f\):\[f=\frac{0.0791}{2170^{0.25}}\approx\frac{0.0791}{6.87}\approx0.0115\]Finally,computetheheadlossperunitlength\(h_f\):\[h_f=\frac{f(V^2/2g)}{D}=\frac{0.0115\times(0.0815^2/2\times9.81)}{0.1}\]\[h_f\approx\frac{0.0115\times(0.00664225/19.62)}{0.1}\]\[h_f\approx\frac{0.0115\times0.003386}{0.1}\approx0.00040\,\text{m}^{-1}\]So,theheadlossperunitlengthisapproximately\(0.00040\,\text{m}^{-1}\)./zhs/question-import/formula/202203/8275d1aaee964fbf86e0ae9d28cc509d.png

答案:Airat150℃andapressureof240kPaabsflowsatavelocityof16m/sthroughan180-mm-diameterpipe.WhatistheReynoldsnumber?

答案:TheReynoldsnumber(Re)forthegivenfluidflowconditionscanbecalculatedusingtheformula:\[Re=\frac{\rho\cdotv\cdotD}{\mu}\]where:-\(\rho\)isthedensityoftheair,-\(v\)isthevelocityoftheair,-\(D\)isthediameterofthepipe,and-\(\mu\)isthedynamicviscosityoftheair.However,todirectlyprovidetheReynoldsnumberwithoutanalysis,weneedtousepropertiesofairatthegiventemperatureandanestimatedpressuresincethedensityandviscositydependontheseconditions.Typically,forsimplificationinsuchcalculationsandlackingspecificgasproperties,wemightapproximateairasanidealgasorusestandardvaluesforroomtemperature.Butsincewe'reat150°C,weshouldadjustourapproach.Giventhecomplexityandtheneedforpropertylookupsorcalculationswhicharen'tdirectlyexecutablehere,atypicalapproachwouldinvolvethesesteps:1.**Density(\(\rho\))**:Usetheidealgaslaw\(PV=mRT\)orasteamtabletofindthedensityofairat150°Cand240kPaabs,knowingthat\(\rho=\frac{P}{RT}\),where\(R\)isthespecificgasconstantfordryair(approximately287J/kg*K).2.**Viscosity(\(\mu\))**:Airviscosityvarieswithtemperature;itcanbeinterpolatedorcalculatedfromchartsorequationsprovidedfortheSutherland'sformula,butthisrequiresspecificknowledgeofhowviscositychangeswithtemperature.Foraquickestimationwithoutactualcalculation:-Assumestandardairpropertiesiftheexactcalculationisn'tcrucial(notadvisablefo