Refrigerant 134a is being held in a rigid, spherical tank (r=0.5m). Initially, t

Question

Refrigerant 134a is being held in a rigid, spherical tank (r=0.5m). Initially, the R-134a has a temperature of 263K and a specific internal energy of 110.6kJ/kg. A thermal reservoir (T=308K) is brought in contact with the tank, transferring enough heat to the tank to raise the pressure inside to 4 bar. Calculate the entropy changes of the refrigerant and the thermal reservoir, and total change in entropy for this process.

Explanation / Answer

Assume carnot heat engine, receive energy as heat transfer from hot reservoir and release heat to cold reservoir, while giving out work to surrounding. Qh | | --> W | Qc Hot reservoir, dS_h = (dQ_h / Th)rev ?Sh = Qh/Th Cold reservoir, dS_c = (dQ_c/Tc) ?Sc = Qc/Tc Reversible process, nett entropy change is zero, ?s_total=0 ?S_total = 0 ?Sh + ?Sc = 0 Qh/Th + Qc/Tc = 0, Qc < 0 (heat released to surrounding) Qh/Th = Qc/Tc Qh/Qc = Th/Tc Tc/Th = Qc/Qh ...... (1) Carnot efficiency, we already assumed a heat engine process, Eff = W_cycle / Qin Eff = Qh - Qc / Qh Eff = 1 - Qc/Qh, From (1) Eff = 1 - Tc/Th Carnot Efficiency is an engine undergoing a reversible power cycle while operating between temperatures of "hot" reservoir and "cold" reservoir, thus Carnot cycle depends only on the two temperatures. While any engine depends on other parameters not just temperatures only, hence less efficient.

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