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Unformatted text preview: 729 A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle. AssumptionsThe heat pump operates steadily. AnalysisSince the heat pump is completely reversible, the combination of the coefficient of performance expression, first Law, and thermodynamic temperature scale gives 73.26)K294/()K283(11/11COPrevHP,=−=−=HLTTThe power required to drive this heat pump, according to the coefficient of performance, is then kW741.326.73kW100COPrevHP,innet,===HQW&&According to the first law, the rate at which heat is removed from the lowtemperature energy reservoir is kW26.96kW741.3kW100innet,=−=−=WQQHL&&&The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is kW/K0.340===ΔK294kW100HHHTQS&&and that of the lowtemperature reservoir is kW/K0.340−=−==ΔK283kW26.96LLLTQS&&The net rate of entropy change of everything in this system is kW/K=−=Δ+Δ=Δ340.340.totalLHSSS&&&as it must be since the heat pump is completely reversible. 10°C21°C HP 100 kW LQ&netW&752 R134a undergoes an isothermal process in a closed system. The work and heat transfer are to be determined.Assumptions1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium.AnalysisThe energy balance for this system can be expressed as )(12outinenergiesetc.potential,kinetic,internal,in Changesystemmassandwork,heat,by nsferenergy traNet outinuumUQWEEE−=Δ=−Δ=−4342143421The initial state properties are 13)A(TableKkJ/kg0134.1kJ/kg74.246C20kPa2401111⋅==⎭⎬⎫°==suTPFor this isothermal process, the final state properties are (Table A11) KkJ/kg42497.)62172.)(20.(30063.kJ/kg29.111)16.162)(20.(86.7820.C202222212⋅=+=+==+=+=⎭⎬⎫=°==fgffgfsxssuxuuxTTThe heat transfer is determined from kJ/kg172.4KkJ/kg)0134.142497.)(K293()(12in−=⋅−=−=ssTqThe negative sign shows that the heat is actually transferred from the system. That is, kJ/kg172.4=outqThe work required is determined from the energy balance to be...
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This homework help was uploaded on 04/16/2008 for the course ME 2322 taught by Professor Oler during the Spring '06 term at Texas Tech.
 Spring '06
 OLER

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