This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 534 Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be determined. Assumptions1This is a steadyflow process since there is no change with time. 2Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5The diffuser is adiabatic.PropertiesThe specific heat of air at the average temperature of (20+90)/2=55°C =328 K is cp= 1.007 kJ/kg⋅K (Table A2b). AnalysisThere is only one inlet and one exit, and thus mmm&&&==21. We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as outinenergiesetc.potential,kinetic,internal,in changeofRate(steady)systemmassandwork,heat,by nsferenergy tranet ofRateoutinEEEEE&&444 344 21&43421&&==Δ=b/2+2//2)+()2/(222211222211VhVhVhmVhm=+=+&&Solving for exit velocity, [ ] [ ]m/s330.2=⋅+=+=+=5.2225.21215.21212kJ/kg1/sm1000)K90K)(20kJ/kg007.1(2m/s)500()(2)(2TTcVhhVVp552Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined. Assumptions1This is a steadyflow process since there is no change with time.2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.PropertiesFrom the steam tables (Tables A4 through 6) kJ/kg3242.4/kgm0.029782C450MPa1013111==°==hTPvand kJ/kg2392.52392.10.92191.8192.kPa102222=×+=+===fgfhxhhxPAnalysis(a) The change in kinetic energy is determined from ( )kJ/kg1.95===Δ22222122/sm1000kJ/kg12m/s)(80m/s502VVke(b) There is only one inlet and one exit, and thus &&&mmm12==. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as outinenergiesetc.potential,kinetic,internal,in changeofRate(steady)systemmassandwork,heat,by nsferenergy tranet ofRateoutinEEEEE&&44 344 21&43421&&==Δ=bAIR 100 kPa 20°C 500 m/s 200 kPa 90°C STEAM m= 12 kg/s · P1= 10 MPa T1= 450°C V1= 80 m/s P2= 10 kPa x2= 0.92 V2= 50 m/s W · STEAM 20 kg/s 12.5 MPa 550°C 20 kg/s...
View
Full Document
 Spring '10
 CHEN
 Thermodynamics, Energy, kJ/kg, net energy transfer, R134a

Click to edit the document details