Question 6.240E: A heat pump with COP = 4 uses 1 kW of power input to heat a ...

C.V.TOT.
Energy Eq.:      \dot{ Q }_{ L }+\dot{ W }=\dot{ Q }_{ H }

Entropy Eq.:    0=\frac{\dot{ Q }_{ L }}{ T _{ L }}-\frac{\dot{ Q }_{ H }}{ T _{ H }}+\dot{ S }_{\text {gen } CV \text { tot }}

From definition of COP:                 \dot{ Q }_{ H }=\operatorname{COP} \dot{ W }=4 \times 1   kW =4   kW

From energy equation:                    \dot{ Q }_{ L }=\dot{ Q }_{ H }-\dot{ W }=(4-1)  kW =3   kW

Flux into heat pump at 32 F:          \frac{\dot{ Q }_{ L }}{ T _{ L – HP }}=\frac{3}{491.7} \frac{ kW }{ R }= 0 . 0 0 6 1   kW / R

Flux out of heat pump at 120 F:    \frac{\dot{ Q }_{ H }}{ T _{ H – HP }}=\frac{4}{579.7} \frac{ kW }{ R }= 0 . 0 0 6 9   kW / R

Flux out into room at T _{ H } = 78 F:        \frac{\dot{ Q }_{ H }}{ T _{ H }}=\frac{4}{537.7} \frac{ kW }{ R }= 0 . 0 0 7 4 4   kW / R

Flux from outside at 60 F:              \frac{\dot{ Q }_{ L }}{ T _{ L }}=\frac{3}{519.7} \frac{ kW }{ R }=0.00577   kW / R

Comment: Following the flow of energy notice how the flux from the outside at 60 F grows a little when it arrives at 32 F this is due to entropy generation in the low T heat exchanger. The flux out of the heat pump at 120 F is larger than the flux in which is due to entropy generation in the heat pump cycle (COP is smaller than Carnot COP) and finally this flux increases due to entropy generated in the high T heat exchanger as the energy arrives at room T.