written 2.5 years ago by
stanzaa37 ♦ 1.0k

•
modified 2.5 years ago

$T_E=T_{sat_{Pe}}=15℃,T_C=T_{sat_{Pc}}=35℃$
From PH chart of R12,
We know that at point 1,
$h_1=181 \frac{kj}{kg};h_2=210 \frac{kJ}{kg};h_3=h_4=68 \frac{kJ}{kg}$
$COP=\frac{R.E}{Wc}=\frac{18168}{(210181}=3.896$
Also,
$Q_a=m ̇_r×(h_1h_4 )$
For 1 TR capacity,
$1×3.5=m ̇_r (18168)$
$m ̇_r=0.0309 \frac{kg}{sec}=1.8584 kg/min$
Also
$m ̇_r×v_1=piston \ displacement \ per \ TR ×η_vol$
$v_1=0.076 m^3⁄kg \ from \ ph \ chart$
Therefore,
piston displacement per $TR=0.1765 m^3⁄min$
Rate at which heat is rejected in condenser per TR
$Q_r=m_R×(h_2h_3 )$
$=0.0309×(210181)$
$=0.899\frac{ kJ}{sec}$
We know, ideal COP
$=\frac{T_E}{T_CT_E}$
$=\frac{258}{308258}$
=5.16