Question

A reversible heat pump is driven by the work output of a reversible heat engine, as shown in the figure below. If we assume ideal devices, then what will be the ratio of the total power ${\dot{Q}}_{L1}+{\dot{Q}}_{H2}$ heats the house to the power from the hot energy source ${\dot{Q}}_{H1}$ in terms of the temperature

• A. $\frac{T_{room}}{T_H}+\frac{T_H(T_H-T_{room})}{T_{room}(T_{room}-T_{amb})}$
• B. $\frac{T_{room}}{T_H}+\frac{T_{room}(T_H-T_{room})}{T_H(T_{room}-T_{amb})}$
• C. $\frac{T_{room}}{T_H}+\frac{T_{room}(T_{room}-T_{amb})}{T_H(T_H-T_{room})}$
• D. $\frac{T_{room}}{T_H}+\frac{T_H(T_{room}-T_{amb})}{T_{room}(T_H-T_{room})}$

Answer 1

The correct option is B $\frac{T_{room}}{T_H}+\frac{T_{room}(T_H-T_{room})}{T_H(T_{room}-T_{amb})}$

As the given information


Ratio of $\frac{\dot{Q}{L}_1+{\dot{Q}}_{H_2}}{{\dot{Q}}_{H_1}}=\frac{{\dot{Q}}_{L_2}}{{\dot{Q}}_{H_1}}+\frac{{\dot{Q}}_{H_2}}{{\dot{Q}}_{H_1}}....\left(1\right)$

For reversible beat engine

$\Rightarrow \frac{{\dot{Q}}_{L_1}}{{\dot{Q}}_{H_1}}=\frac{T_{room}}{T_H}....\left(2\right)$

$(COP{)}_{HP}=\frac{{\dot{Q}}_{H_2}}{W}=\frac{T_{room}}{(T_{room}-T_{amb})}$

${\dot{Q}}_{H_2}=\frac{W\left(T_{room}\right)}{(T_{room}-T_{amb})}$

$\frac{{\dot{Q}}_{H_2}}{W}=\frac{T_{room}}{(T_{room}-T_{amb})}......\left(3\right)$

${\eta }_{HE}=\frac{W}{{\dot{Q}}_{H_1}}=\left(\frac{T_H-T_{room}}{T_H}\right).....\left(4\right)$

From (3) and (4)

$\frac{{\dot{Q}}_{H_2}}{{\dot{Q}}_{H_1}}=\frac{T_{room}}{T_H}\left(\frac{T_H-T_{room}}{T_{room}-T_{amb}}\right)....\left(5\right)$

From (1), (2) and (5)

$\frac{Q_{L_1}+Q_{H_2}}{Q_{H_2}}=\frac{T_{room}}{T_H}+\frac{T_{room}}{T_H}\left(\frac{T_H-T_{room}}{T_{room}-T_{amb}}\right)$

Option (b) is correct.