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A gas undergoes a thermodynamic cycle consisting of three processes: process 1–2: constant volume, v = 0.028 m3, u2 - u1 = 26.4 kj process 2–3: expansion with pv = constant, u3 = u2 process 3–1: constant pressure, p = 1.1 bar, w31 = -10.5 kj there are no significant changes in kinetic or potential energy. (a) calculate the net work for the cycle, in kj. (b) calculate the heat transfer for process 2–3, in kj. (c) calculate the heat transfer for process 3–1, in kj. (d) is this a power cycle (depict 1) or a refrigeration cycle (depict 0)?

Respuesta :

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Process 1-2: V1 = V2 = 0.028 m^3; U2 - U1 = 26.4 kJ
Process 2-3: PV = Constant; U3 = U2
Process 3-1: P3 = P1 = 1.1 Bar; W31 = -10.5 kJ

Part (a): Net work for the cycle
W = W12 + W23 + W31
W12 = 0 kJ ( since volume is constant)
W23 = C*ln(v3/v2); but, W31 = P3(V1-V3) => V1-V3 = W31/P3 = -10.5/(1.1*10^5) = -9.55*10^-5 => V3 = V1+9.55*10^-5 = 0.0281 m^3
In addition, PV = C = P3V3 = 1.1*10^5*0.0281 = 3090.5

Then, W23 = 3090.5*ln (0.0281/0.028) = 11.02 kJ
Substituting;
W = 0+11.02-10.5 = 0.518 kJ

Part (b): Heat transfer in process 2-3
Q23-W23 = U3-U2, but U3 = U2 and thus U3-U2 = 0

Then,
Q23 = W23 =11.02 kJ

Part (c): Heat transfer for process 3-1
Q31-W31 = U1-U3 = U1-U2 = -(U2-U1) = -26.4
Q31 = -26.4+W31 = -26.4-10.5 = -36.9 kJ

Part (d): Power or refrigeration cycle?
Q_cycle = Q12 +Q23 +Q31 = 26.4+11.02-36.9 = 0.52 kJ
W_cycle = 0.52 kJ
Since, Q_cycle >0 and W_cycle > 0, this is a power cycle.

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