Question 2.53: A heat pump working on a reserved carnot cycle takes in ener...
A heat pump working on a reserved carnot cycle takes in energy from a reservoir at 5°C and delivers it to another reservoir where temperature is 77°C. The heat pump derives power for its operation from a reversible engine operating within the higher and lower temperature 1077°C and 77°C. For 100 kJ/kg of energy supplied to reservoir at 77°C, estimate the energy taken from the reservoir at 1077°C.
The system is shown in Fig 2.58.
\begin{aligned}&T_1=273+1077=1350 \mathrm{~K} \\&T_2=T_4=273+77=350 \mathrm{~K} \\&T_3=273+5=278 \mathrm{~K}\end{aligned}
H.E.
\begin{aligned}\eta &=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1} \\\frac{Q_2}{Q_1} &=\frac{T_2}{T_1}\end{aligned}
H.P.
\begin{aligned}C O P &=\frac{Q_4}{Q_4-Q_3} \\&=\frac{T_4}{T_4-T_3} \\W_E &=W_P \\Q_1-Q_2 &=Q_4-Q_3 \\Q_1\left\lgroup1-\frac{T_2}{T_1}\right\rgroup &=Q_4\left\lgroup 1-\frac{T_3}{T_4}\right\rgroup\\Q_1\left\lgroup 1-\frac{350}{1350}\right\rgroup &=Q_4 \left\lgroup1-\frac{278}{350}\right\rgroup \\0.74074 Q_1 &=0.20571 Q_4 \\Q_4 &=3.6 Q_1\end{aligned}
\begin{aligned}Q_2 &=\frac{350 Q_1}{1350}=0.259 Q_1 \\Q_2+Q_4 &=100 \\(0.259+3.6) Q_1 &=100 \\Q_1 &=25.9 \mathrm{~kJ}\end{aligned}
