An air-standard cycle is executed in a closed system with 1 kg of air, and it co

Question

An air-standard cycle is executed in a closed system with 1 kg of air, and it consists of the following three processes: (a) Isentropic compression from 100 kPa, 27oC to 700 kPa. (b) p=constant heat addition to initial specific volume. (c) v=constant heat rejection to initial state. (i) Calculate the maximum temperature and efficiency. (ii) Show the cycle on T-s and p-v diagrams. Assume cold air-standard. What if: How would the conclusion change if (a) isentropic compression took place from 100 kPa, 27oC to 500 kPa.?

Explanation / Answer

States { State-1: Air; Given: { p1= 100.0 kPa; T1= 27.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; m1= 1.0 kg; } State-2: Air; Given: { p2= 700.0 kPa; s2= "s1" kJ/kg.K; Vel2= 0.0 m/s; z2= 0.0 m; m2= "m1" kg; } State-3: Air; Given: { p3= "p2" kPa; v3= "v1" m^3/kg; Vel3= 0.0 m/s; z3= 0.0 m; m3= "m1" kg; } } Analysis { Process-A: b-State = State-1; f-State = State-2; Given: { T_B= 25.0 deg-C; } Process-B: b-State = State-2; f-State = State-3; Given: { T_B= 25.0 deg-C; } Process-C: b-State = State-3; f-State = State-1; Given: { T_B= 25.0 deg-C; } } Tmax=1827.9oC, Thermal Efficiency=18.46% and MEP=452.4 kPa. Tmax=1227.6oC, Thermal Efficiency=16.36% and MEP=286.4 kPa

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