Question

A heat pump with refrigerant R22 is used for space heating between temperature limits of $-{20}^{\circ }Cand{25}^{\circ }C$. the heat required is 200 MJ/h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kgK. Other relevant properties are given below. The enthalpy (in kJ/kg) of the refrigerant at isentropic compressor discharge is

Answer 1

$S_2-S_2^{\prime }=C_{p}lo{g}_e\frac{T_2}{T_2^{\prime }}$

$S_1=S_2$ (for isentropic process)

$S_2=1.7841kJ/kgK$

$S_2=1.7183kJ/kgK$

$C_p=0.98kJ/kgK$

$\therefore lo{g}_e\frac{T_2}{T_2^{\prime }}=\frac{s_2-S_2^{\prime }}{C_p}$

$\frac{1.7841-1.7183}{0.98}=0.0671$
or
$T_2=1.0694\times T_2^{\prime }=1.0694\times 298$

= 318.695 K

$h_2-h_2^{\prime }=C_p(T_2-T_2^{\prime })$

$h_2=413.02+0.98(318.695-298)$

$h_2=433.3$ kJ\kg