Lecture 65 : Vapour Compression Refrigeration Cycle

[@ankursrivastava9074]

Salute to prof s k som and suman chakrabaroty sir

[@ShailendraKumar-pv8ok]

You make so easy.

[@samhaazratechnicalsupport]

Very niece sir ji ❤❤❤

[@diplomatutorial3256]

Perfect

[@deguzmanmichaelv.2210]

Thank you sir

[@preritsharma8489]

Here I present my solution for the given problem :
State 1: saturated vapour
T1=263.15 K= -10°C, P1=219.12 kPa, h1=347.13 kJ/kg, s1=1.559 kJ/kg K
State 2: superheated vapour
P2=1 MPa, T2= 60°C, h2=381.8 kJ/kg
State 3: saturated liquid
P3=P2=1 MPa, T3=42°C, h3=240 kJ/kg
State 4: liquid vapour mixture
P4=P1=219.12 kPa, h4=h3=219.13 kJ/kg
State 2s:hypothetical state(if process 1 to 2 would have been isoentropic)
P2s=P2=P3=1 MPa, s2s=s1=1.559 kJ/kgK
T2s=48°C, h2s= 373.5 kJ/kg
Now, ql=h1-h4=107.13 kJ/kg
qh=h2-h3=141.8 kJ/kg
win=h2-h1=34.67 kJ/kg
COP = ql/win = 3.08
n(ise)=(h2s-h1) /(h2-h1) =0.76