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The condenser of a refrigeration system steadily admits $\dot{m} = 50 g/s$ of saturated R134a with a pressure $P_1 = 650kPa$ and a quality $x_1 = 0.9$. The condenser rejects $\dot{Q} = 7kW$ of heat to the surroundings at T = $20^{o}$C so that the refrigerant comes out of the condenser at a lower quality $x_2$. Assume there is no pressure drop along the condenser $P_2 = P_1$

Identify all the sources of irreversibility.

Calculate the total rate of entropy generated by this process $\dot{S}_{gen}$.

Calculate the rate of lost work for this process, $\dot{W}_{lost}$.

Calculate the rate of lost work $\dot{W}_{lost,HT}$ due to the entropy generation associated with irreversible heat transfer only

$$\dot{m}\cdot h_{in} = \dot{m}\cdot h_{out} + \dot{Q}_{out}$$ $$h_{out} = h_{in} - \frac{\dot{Q}_{out}}{\dot{m}}$$

Here I used the calculated value of enthalpy and the given value of P to fix the state to calculate entropy at both the inlet and the outlet. $$\dot{S}_{m,in} + \dot{S}_{gen} = \dot{S}_{m,out} + \dot{S}_{Q,out}$$ $$\dot{S}_{gen} = \dot{m}\cdot s_{m,out} + \frac{\dot{Q}_{out}}{T} - \dot{m}\cdot s_{m,in}$$ $$\dot{S}_{gen} = (0.05\frac{kg}{s})(0.8608\frac{kJ}{kg-K} - 0.39\frac{kJ}{kg-K}) + \frac{7kW}{293.15K} = 0.04742\frac{kJ}{K-s}$$ $$\dot{W}_{lost} = T\cdot \dot{S}_{gen} = 13.9kW$$ This is where I get lost. What are the other irreversible processes? I thought that the only irreversible process here was heat transfer out of the system. I know that if I change my system boundaries to enclose the entire system then the $\dot{W}_{in}$ has to be equal to $\dot{Q}_{out}$, but that doesn't seem to be what the question is asking. What am I missing/doing wrong?

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  • $\begingroup$ In your condenser calculation, if the quality coming out is lower than the quality coming in, how come the entropy coming out is higher than the entropy coming in? $\endgroup$ – Chet Miller Oct 17 '18 at 15:31
  • $\begingroup$ To answer your question “what are the other irreversible processes?” I believe the main source of irreversibility is in the throttling (expansion valve) process. It can be regarded as a "free expansion". The work lost is the work required to return the system to its state prior to the expansion. $\endgroup$ – Bob D Oct 18 '18 at 15:53
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As I said in my comment, you made a mistake in the substitution in the equation for $\dot{S}_{gen}$. The actual amount of entropy generation and lost work is very little in the condenser because the temperature of the condensing vapor is very close to the surroundings temperature of 20 C. The condenser is operating very close to reversible operation.

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