Question #125503

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.

Expert's answer

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

The condenser pressure(Pcon)=11.82 bar=11.82x10^{5} Pa

The evaporator pressure(Peva)= 1.64 bar=1.64x10^{5} Pa

** **

**From Property Table of R-22:**

**At point 2:** Refrigerant is dry saturated at

P=1.64 bar, from Table h_{2}=238kJ/kg

T2=-30^{o}C, entropy(s_{2})=979 J/KgK

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

P=11.82 bar, from Table for the entropy value entropy(s_{3}) =979 J/KgK. Temperature T3=63.7^{o}C.

At pressure P=11.82 bar and Temp(T)= 63.7^{o}C

Enthalpy(h_{3}) =288KJ.

**At point 4:** Refrigerant is subcooled by10^{ o}C , P=11.82 bar, ant T=Tcond-10^{ o}C=29.7^{ o}C-10^{ o}C=19.7^{ o}C

From Table: Enthalpy(h_{4}) =68.8kJ At pressure P=11.82 bar.

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

Therefore, h_{4}=h_{1}=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

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