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?
