written 2.5 years ago by
stanzaa37 ♦ 1.0k

•
modified 2.5 years ago

$T_E=T_{sat_{Pe}}=12℃,T_C=T_{sat_{Pc}}$=28℃
Capacity=10 TR, x1=0.95
$h1=h_{f_{12℃}}+ x1×h_{fg_{12℃}}$
$h_{fg_{12℃}}=h_gh_f=1447.74144.929$
=$1302.811 \frac{kJ}{kg} (from \ psychrometric \ chart \ booklet \ for \ N_3 )$
$h1=144.929+0.95×1302.811$
$=1382.6\frac{kJ}{kg}$
Now mark point 1 on ph chart of ammonia corresponding to 12°C and h1. Then draw constant entropy line from point 1upwards to intersect horizontal line of 28°C. This is point 2. Draw a horizontal from this point towards left and where this line intersects constant temperature line of 23°C in subcooled region is point 3.
Condition of refrigerant at outlet of compressor i.e. point 2
$T2=67°C, h2=1598 kJ/kg, v2=0.15 m3/kg, P2=10.5 \ bar$
Condition of refrigerant at inlet to evaporator i.e. point 4
$T4=12°C, h4=310 kJ/kg, P4=2.65$ bar
COP of given system = $\frac{R.E.}{Wc}=\frac{h1h4}{h2h1}$
COP=4.98
Also, $Q ̇_a=m ̇_R×(h1h4)$
For 10 TR capacity, $10 x 3.5= m ̇_R×(1382.6310)$
mass flow rate,$m ̇_R=0.0326 \frac{kg}{sec}$
Now, power required = $m ̇_R×(h2h1)$
∴power required=7.0287 kW