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2006 Spring Thermodynamics Homework 14 Solution 1. Consider the Carnot cycle occurring in a piston-cylinder device containing refrigerant-134a with operating conditions given below: Process A: Isothermal heat addition at TH = 30 C to convert saturated liquid to saturated vapor Process B: Isentropic and adiabatic expansion to TL = -20 C Process C: Isothermal heat removal at -20 C Process D: Isentropic and adiabatic compression back to the initial state With R-134a as the working fluid for this cycle calculate the thermal efficiency using th = Wnet Q in ME 201 Compare this to the ideal Carnot cycle efficiency given by Carnot = 1 TL TH Solution: We shall make our calculations process by process Process A Working Fluid: R-134a(compressible) System: Closed System Process: Isothermal State 1 State 2 T1 = 30 C T2 = 30 C P1 = 770.64 kPa P2 = 770.64 kPa 3 v1 = 0.0008421 m /kg v2 = 0.026622 m3/kg u1 = 92.93 kJ/kg u2 = 246.14 kJ/kg s1 = 0.34789 kJ/(kg K) s2 = 0.91879 kJ/(kg K) phase: sat.liq. phase: sat.vap. italicized values from tables Initial State: Fixed 1 ME 201 Thermodynamics Spring 2006 Final State: Fixed Wsh = 0 Q = ???? Wbnd = ???? st 1 Law: u2 - u1 = q - wbnd Pdv: wbnd = P(v2-v1) Both states are fixed, so we can go to the R-134a tables and look up the remaining properties. Since our process is both isothermal and isobaric due to the phase change, the boundary work is given by w A = P1 (v 2 - v1 ) = (770.64)(0.026622 - 0.0008421) = 19.867 kJ/kg The heat transfer will be given by the 1st law, or q A = w A + u 2 - u1 = 19.867 + 246.14 - 92.93 = 173.077 kJ/kg Process B Working Fluid: R-134a(compressible) System: Closed System Process: Isentropic and Adiabatic State 2 State 3 T2 = 30 C T3 = -20 C P2 = 770.64 kPa P3 = 132.82 kPa 3 v2 = 0.026622 m /kg v3 = 0.142611 m3/kg u2 = 246.14 kJ/kg u3 = 212.6639 kJ/kg s2 = 0.91879 kJ/(kg K) s3 = 0.91879 kJ/(kg K) phase: sat.vap. phase: 2 phase, x = 0.968 italicized values from tables Initial State: Fixed Final State: UNKNOWN Wsh = 0 QB = 0 Wbnd = ???? 1st Law: u3 - u2 = -wbnd Since our process is isentropic we have s3 = s2 = 0.91879 kJ/(kgK) Going to the tables at -20 C, we find sf = 0.10463 kJ/(kgK) and sg = 0.94564 kJ/(kgK) so that we have a two phase mixture with 0.91879 - 0.10463 x3 = = 0.968 0.94564 - 0.10463 2 ME 201 Thermodynamics Spring 2006 The remaining properties can then be calculated. The boundary work is given by the first law w B = u 2 - u 3 = 246.14 - 212.6639 = 33.47 kJ/kg Process C Working Fluid: R-134a(compressible) System: Closed System Process: Isothermal State 4 State 3 T3 = -20 C T4 = -20 P3 C = 132.82 kPa P4 = v4 = v3 = 0.142611 m3/kg u3 = 212.6639 kJ/kg u4 = s3 = 0.91879 kJ/(kg K) s4 phase: 2 phase, x = 0.968 phase: italicized values from tables Initial State: Fixed Final State: Fixed Wsh = 0 qC = ????? Wbnd = ???? st 1 Law: u4 - u3 = q - wbnd Pdv: wbnd = P(v4-v3) Immediately, we see that we do not have enough information to fix State 4. We will continue on to Process D and return later to this process. Process D Working Fluid: R-134a(compressible) System: Closed System Process: Isentropic and Adiabatic State 4 State 1 T4 = -20 C T1 = 30 C P4 = 132.82 kPa P1 = 770.64 kPa 3 v4 = 0.043127 m /kg v1 = 0.0008421 m3/kg u4 = 81.34492 kJ/kg u1 = 92.93 kJ/kg s1 = 0.34789 kJ/(kg K) s4 = 0.34789 kJ/(kg K) phase: 2 phase, x = 0.289 phase: sat.liq. italicized values from tables Initial State: UNKNOWN Final State: Fixed 3 ME 201 Thermodynamics Spring 2006 Wsh = 0 qD = 0 Wbnd = ???? st 1 Law: u1 - u4 = -wbnd Since our process is isentropic we recognize that s4 = s1 = 0.34789 kJ/(kg K) Going to the tables at -20 C, we find sf = 0.10463 kJ/(kgK) and sg = 0.94564 kJ/(kgK) so that we have a two phase mixture with 0.34789 - 0.10463 = 0.289 x4 = 0.94564 - 0.10463 The remaining properties can then be calculated. We have a compressible substance the boundary work is given by the first law w D = u 4 - u1 = 81.34492 - 92.93 = - 11.585 kJ/kg Now returning to process C since our state 4 is now fixed. Process C Working Fluid: R-134a(compressible) System: Closed System Process: Isothermal State 4 State 3 T3 = -20 C T4 = -20 C P3 = 132.82 kPa P4 = 132.82 kPa v4 = 0.043127 m3/kg v3 = 0.142611 m3/kg u3 = 212.6639 kJ/kg u4 = 81.34492 kJ/kg s3 = 0.91879 kJ/(kg K) s4 = 0.34789 kJ/(kg K) phase: 2 phase, x = 0.968 phase: 2 phase, x = 0.289 italicized values from tables Initial State: Fixed Final State: Fixed Wsh = 0 qC = ????? Wbnd = ???? st 1 Law: u4 - u3 = q - wbnd Pdv: wbnd = P(v4-v3) Since our process is both isothermal and isobaric due to the phase change, the boundary work is given by w C = P3 (v 4 - v 3 ) = (132.82)(0.043127 - 0.142611) = - 13.214 kJ/kg 4 ME 201 Thermodynamics Spring 2006 The heat transfer will be given by the 1st law, or q C = w C + u 4 - u 3 = - 13.214 + 81.34 - 212.66 = - 144.53 kJ/kg Our thermal efficiency is given by w net th = q in The net work is given by w net = w A + w B + w C + w D = 19.867 + 33.47 + (-13.214) + (-11.585) = 28.545 kJ/kg Since heat is only added during process A q in = q A = 173.077 kJ/kg So that 28.545 = 0.1649 173.077 The Carnot cycle efficiency is T (-20 + 273) Carnot = 1 - L = 1 TH (30 + 273) = 0.165 th = 5

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