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Assuming the velocities to be low and the process to

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Assuming the velocities to be low and the process to be adiabatic, find the requiredmass flow rate of air through the turbine.Solution:C.V. Turbine, no heat transfer, no'KE, no'PEEnergy Eq.6.13:hin= hex+ wTIdeal gas so use constant specific heat from Table A.5wT= hin- hex#Cp(Tin- Tex)= 1.004[50 - (-30)] = 80.3 kJ/kgW.= m.wTm.= W./wT= 0.1/80.3 =0.00125 kg/s6.23A small turbine, shown in Fig. P 6.23, is operated at part load by throttling a 0.25kg/s steam supply at 1.4 MPa, 250qC down to 1.1 MPa before it enters the turbineand the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhausttemperature (and quality if saturated).Solution:C.V. Throttle, Steady,q = 0 and w = 0. No change in kinetic or potentialenergy. The energy equation then reduces toEnergy Eq.6.13:h1= h2= 2927.2from Table B.1.3C.V. Turbine, Steady, no heat transfer, specific work: w =1100.25= 440 kJ/kgEnergy Eq.:h1= h2= h3+ w = 2927.2h3= 2927.2 - 440 = 2487.2 kJ/kgState 3:(P, h)Table B.1.2h < hg2487.2 = 191.83 + x3× 2392.8Tv123T =45.8qC,x3=0.959
6-176.24A liquid water turbine receives 2 kg/s water at 2000 kPa, 20oC and velocity of 15m/s. The exit is at 100 kPa, 20oC and very low velocity. Find the specific workand the power produced.Solution:Energy Eq.6.13:h1+12V21+ gZ1= h2+12V22+ gZ2+ wTProcess:Z1= Z2andV2= 0State 1:Table B.1.4h1= 85.82State 2:Table B.1.1h2= 83.94(which is at 2.3 kPa so weshould add'Pv = 97.7×0.001 to this)wT= h1+12V21²h2= 85.82 + 152/2000 – 83.94 =1.99 kJ/kgW.T= m.×wT= 2×1.9925 =3.985 kWNotice how insignificant the specific kinetic energy is.6.25Hoover Dam across the Colorado River dams up Lake Mead 200 m higher thanthe river downstream. The electric generators driven by water-powered turbinesdeliver 1300 MW of power. If the water is 17.5qC, find the minimum amount ofwater running through the turbines.Solution:C.V.: H2O pipe + turbines,Continuity:m.in= m.ex;Energy Eq.6.13:(h+V2/2 + gz)in= (h+V2/2 + gz)ex+ wTWater states:hin#hex;vin#vexsoTHDAMLakeMeadNow the specific turbine work becomeswT= gzin- gzex= 9.807 × 200/1000 = 1.961 kJ/kgm.= W.T/wT=1300×103kW1.961 kJ/kg= 6.63×105kg/sV.= m.v = 6.63×105× 0.001001 =664 m3/s
6-18(1)Compressors fans6.26A compressor in a commercial refrigerator receives R-22 at -25oC, x = 1. The exitis at 800 kPa, 40oC.Neglect kinetic energies and find the specific work.Solution:C.V. Compressor, steady state, single inlet and exit flow.For this device we also assume no heat transfer andZ1= Z2From Table B.4.1 :h1= 239.92 kJ/kgFrom Table B.4.2 :h2= 274.24 kJ/kgEnergy Eq.6.13 reduces towc= h1- h2= (239.92 – 274.24) =-34.3 kJ/kg6.27The compressor of a large gas turbine receives air from the ambient at 95 kPa,20qC, with a low velocity. At the compressor discharge, air exits at 1.52 MPa,430qC, with velocity of 90 m/s. The power input to the compressor is 5000 kW.Determine the mass flow rate of air through the unit.Solution:C.V. Compressor, Steady state, single inlet and exit flow.Energy Eq.6.13:q + hi+Vi2/2 = he+Ve2/2 + wHere we assumeq#0andVi#0so using constant CPofrom A.5-w = CPo(Te- Te) +Ve2/2 = 1.004(430 - 20) +(90)22 × 1000= 415.5 kJ/kgm.

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