Answer to Question #295550 in Molecular Physics | Thermodynamics for Harsh

Question

A heat pump working on the Carnot cycle takes in heat from a reservoir at 5°C and delivers

heat to a reservoir at 60°C. The heat pump is driven by a reversible heat engine which takes

in heat from a reservoir at 840°C and rejects heat to a reservoir at 60°C. The reversible heat

engine also forgives a machine that absorbs 30 kW. If the heat pump extract 17 kJ/s from

the 5°C reservoir, determine :

a) The rate of heat supply from the 840°C source

b) The rate of heat rejection to the 60°C sink.

Answer 1

Solution;

The combine heat engine and heat pump is shown below;

(a)

Efficiency of the heat engine;

$\eta=1-\frac{T_2}{T_1}=1-\frac{333}{1113}=0.7$

Let $Q_1$ be the heat supplied from the from the reservoir at 840°C

Also the efficiency of the heat engine;

$\eta_E=\frac{Work done}{Heat supplied}=\frac{Q_1-Q_2}{Q_2}$

$W_E=Q_1-Q_2=\eta_EQ_1$

The net work output of the combined heat engine and heat pump is ;

$W=W_E-W_p=30kW$

Therefore;

$W_p=W_E-30kW=0.7Q_1-30$

The C.O.P of heat pump ;

$C.O.P_p=\frac{T_2}{T_2-T_3}=\frac{333}{333-278}=6.05$

Also the C.O.P of the heat pump is;

$C.O.P_p=\frac{Heat supplied}{Work supplied}=\frac{Q_4}{W_p}$

$W_p=\frac{Q_4}{C.O.P_p}=\frac{17}{6.05}=2.81$

Hence;

$2.81=0.7Q_1-30$

$Q_1=46.87kW$

(b)

For the heat engine;

$Q_2=Q_1-W_E=Q_1-0.7Q_1$

$Q_2=46.87-0.7(46.87)=14.061kW$

For the heat pump;

$Q_3=Q_4+W_p$

$Q_4=17+2.81=19.81kW$