An air-standard power cycle with constant specific heats is executed in a closed system and is composed of the following five processes. Assuming constant specific heats at room temperature, calculate the mean effective pressure of the cycle, in kPa.

Elements Of Electromagnetics
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Thermodynamics
An air-standard power cycle with constant specific heats is executed in a closed system and
is composed of the following five processes. Assuming constant specific heats at room
temperature, calculate the mean effective pressure of the cycle, in kPa.
Properties of air: Cy=0.718 kJ/(kgK), C,=1.005 kJ/(kgK), R=0.287 kJ/(kgK), k=1.4
Process 1-2: Isentropic compression from P1=99 kPa to P2. The minimum temperature of
the cycle is T1=285 K. The compression ratio of the cycle is 8.0.
Process 2-3: Heat addition of q23=918 kJ/kg to the cycle at constant volume.
Process 3-4: Isentropic expansion from P3 to P4=233 kPa.
Process 4-5: Heat rejection of q45 from the cycle at constant volume.
Process 5-1: Heat rejection of q51 from the cycle at constant pressure.
Transcribed Image Text:An air-standard power cycle with constant specific heats is executed in a closed system and is composed of the following five processes. Assuming constant specific heats at room temperature, calculate the mean effective pressure of the cycle, in kPa. Properties of air: Cy=0.718 kJ/(kgK), C,=1.005 kJ/(kgK), R=0.287 kJ/(kgK), k=1.4 Process 1-2: Isentropic compression from P1=99 kPa to P2. The minimum temperature of the cycle is T1=285 K. The compression ratio of the cycle is 8.0. Process 2-3: Heat addition of q23=918 kJ/kg to the cycle at constant volume. Process 3-4: Isentropic expansion from P3 to P4=233 kPa. Process 4-5: Heat rejection of q45 from the cycle at constant volume. Process 5-1: Heat rejection of q51 from the cycle at constant pressure.
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