Question 6.242E: A farmer runs a heat pump using 2.5 hp of power input. It ke...

C.V. Hatchery, steady state.
Power:            \dot{ W }=2.5   hp =2.52544 .4 / 3600=1.767   Btu / s

To have steady state at 86 F for the hatchery

Energy Eq.:    0=\dot{ Q }_{ H }-\dot{ Q }_{\text {Loss }} \quad \Rightarrow \quad \dot{ Q }_{ H }=\dot{ Q }_{\text {Loss }}=20   Btu / s

C.V. Heat pump, steady state

Energy eq.:          0=\dot{ Q }_{ L }+\dot{ W }-\dot{ Q }_{ H } \quad \Rightarrow \quad \dot{ Q }_{ L }=\dot{ Q }_{ H }-\dot{ W }=18.233   Btu / s

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

C.V. From hatchery at 86 F to the ambient 50 F. This is typically the walls and the outer thin boundary layer of air. Through this goes \dot{ Q }_{\text {Loss }}.

Entropy Eq.:          0=\frac{\dot{ Q }_{ Loss }}{ T _{ H }}-\frac{\dot{ Q }_{ Loss }}{ T _{ amb }}+\dot{ S }_{ gen \text { walls }}