Answer to Question #125503 in Mechanical Engineering for Moeez

Question

A refrigerant plant of 100 TR capacity uses R-22 as refrigerant. The condensing and evaporating pressure are 11.82 bar and 1.64 bar, respectively. The refrigerant enters the compressor at dry saturated state, whereas it leaves the condenser, subcooled by 10°C. Actual COP is 70% of its theoretical value. Find the following:

a)               Theoretical and actual COP

b)              Mass flow in kg/s

c)               Compressor power.

Answer 1

The cooling capacity=100TR=100x3.516=351.6 kW

The condenser pressure(Pcon)=11.82 bar=11.82x10⁵ Pa

The evaporator pressure(Peva)= 1.64 bar=1.64x10⁵ Pa

 

 

From Property Table of R-22:

At point 2:  Refrigerant is dry saturated at

P=1.64 bar, from Table h₂=238kJ/kg

T2=-30^oC, entropy(s₂)=979 J/KgK

At point 3:  Refrigerant is superheated having same entropy as that of point 2, therefore entropy(s₃)=s₃=979 J/kgK

P=11.82 bar, from Table for the entropy value entropy(s₃) =979 J/KgK. Temperature T3=63.7^oC.

At pressure P=11.82 bar and Temp(T)= 63.7^oC

Enthalpy(h₃) =288KJ.

At point 4:  Refrigerant is subcooled by10 ^oC , P=11.82 bar, ant T=Tcond-10 ^oC=29.7 ^oC-10 ^oC=19.7 ^oC

From Table: Enthalpy(h₄) =68.8kJ At pressure P=11.82 bar.

At point 1:  Refrigerant undergoes constant enthalpy process in expansion valve.

Therefore, h₄=h₁=68.8kJ

a)      Theoretical COP=(h2-h1)/(h3-h2)=(238-68.8)/(288-238)=3.384

Actual COP=0.70x Theoretical COP=0.7x3.384=2.368

b)     Mass flow rate (m) is given as,

TR=mass flow rate(m)x(h2-h1)

              Mass flow rate(m)= TR /(h2-h1)=351.6/169.2=2.078 kg/s

c)      Compressor power=mass flow rate (m)x(h3-h2)=2.078x(288-238)=103.9kW