能量守恒定律基础

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1、EnergyConservation(Bernoulli’sEquation)IntegrationofEuler’sequationBernoulli’sequationFlowwork+kineticenergy+potentialenergy=constantpADxUndertheactionofthepressure,thefluidelementmovesadistanceDxwithintimeDtTheworkdoneperunittimeDW/Dt(flowpower)isEnergyConservation(cont.)Itisvali

2、dforincompressiblefluids,steadyflowalongastreamline,noenergylossduetofriction,noheattransfer.Examples:Determinethevelocityandmassflowrateofeffluxfromthecircularhole(0.1mdia.)atthebottomofthewatertank(atthisinstant).ThetankisopentotheatmosphereandH=4mH12p1=p2,V1=0EnergyEquation(con

3、t.)Example:Ifthetankhasacross-sectionalareaof1m2,estimatethetimerequiredtodrainthetanktolevel2.h(t)12First,choosethecontrolvolumeasenclosedbythedottedline.Specifyh=h(t)asthewaterlevelasafunctionoftime.02040608010001234time(sec.)waterheight(m)42.5e-007h()t1000tEnergyconservation(co

4、nt.)Generalizedenergyconcept:Energyadded,hA(ex.pump,compressor)Energyextracted,hE(ex.turbine,windmill)Energyloss,hL(ex.friction,valve,expansion)pumpturbineheatexchangercondenserhEhAhL,frictionlossthroughpipeshLlossthroughelbowshLlossthroughvalvesEnergyconservation(cont.)Examples:D

5、eterminetheefficiencyofthepumpifthepowerinputofthemotorismeasuredtobe1.5hp.Itisknownthatthepumpdelivers300gal/minofwater.pumpZ=15in12Mercury(m=844.9lb/ft3)water(w=62.4lb/ft3)1hp=550lb-ft/shE=hL=0,z1=z26-india.pipe4-india.pipeQ=300gal/min=0.667ft3/s=AVV1=Q/A1=3.33ft/sV2=Q/A2=7.54

6、ft/szoEnergyconservation(cont.)Example(cont.)FrictionallossesinpipingsystemP1P2Consideralaminar,fullydevelopedcircularpipeflowpP+dpwDarcy’sEquation:R:radius,D:diameterL:pipelengthw:wallshearstressEnergyConservation(cont.)Energy:E=U(internalthermalenergy)+Emech(mechanicalenergy)=

7、U+KE(kineticenergy)+PE(potentialenergy)Work:W=Wext(externalwork)+Wflow(flowwork)Heat:Qheattransferviaconduction,convection&radiationdE=dQ-dW,dQ>0netheattransferindE>0energyincreaseandviceversadW>0,doespositiveworkattheexpenseofdecreasingenergy,dE<0U=mu,u(internalenergyperunitmass)

8、,KE=(1/2)mV2,PE=mgzWflow=m(p/)Th