Consider air in a cylinder undergoes a cycle with four processes. The process fr

Question

Consider air in a cylinder undergoes a cycle with four processes. The process from A to B is isometric and pressure doubles during this process. The process from B to C is isobaric and volume doubles during the process. The process from C to D is adiabatic. Then the air returns to initial state A through an isothermal process from D to A. Assuming air is an ideal gas and has a specific heat at constant volume cv = 5 / 2R, determine the following quantities: Determine the change of entropy per unit mass for each process.

Explanation / Answer

Change in entropy in process 1-2 is ds1-2 = Cp ln(T2/T1) - R ln (P2/P1)

For isometric process 1-2, T2/T1 = P2/P1.

Hence, ds1-2 = (Cp-R) ln(P2/P1) = Cv ln2 = 5/2*R *0.693 = 5/2*287*0.693 = 497.33 J/kg-K

Change in entropy in process 2-3 is ds2-3 = Cp ln(T3/T2) - R ln (P3/P2)

For isobaric process 2-3, T3/T2 = V3/V2 and P3/P2 = 1.

Hence, ds2-3 = Cp ln(V3/V2) - R ln 1 = Cp ln 2 = (5R/2 + R) ln2 = 7R/2*0.693 = 7*287*0.693/2 = 696.27 J/kg-K

Change in entropy in process 3-4 is ds3-4 = 0 since process is adiabatic.

Since in a cycle net entropy change is zero. Hence, ds1-2 + ds2-3 + ds3-4 + ds4-1 = 0

Or, 497.33 + 696.27 + 0 + ds4-1 = 0

Or, ds4-1 = 1193.6 J/kg-K

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