A vapour compression heat pump is driven by a power cycle having a thermal efficiency of 25%. For the heat pump, refrigerant-12 is compressed from saturated vapour at 2.0 bar to the condenser pressure of 12 bar. The isentropic efficiency of the compressor is 80%. Saturated liquid

Question

A vapour compression heat pump is driven by a power cycle having a thermal efficiency of 25%. For the heat pump, refrigerant-12 is compressed from saturated vapour at 2.0 bar to the condenser pressure of 12 bar. The isentropic efficiency of the compressor is 80%. Saturated liquid enters the expansion valve at 12 bar. For the power cycle 80% of the heat rejected by it is transferred to the heated space which has a total heating requirement of 500 kJ/min. Determine the power input to the heat pump compressor. The following data for refrigerant-12 may be used :

Pressure, bar	Temperature, °C	Enthalpy, kJ/kg		Entropy, kJ/kg K
Pressure, bar	Temperature, °C	Liquid	Vapour	Liquid	Vapour
2.0 12.0	– 12.53 49.31	24.57 84.21	182.07 206.24	0.0992 0.3015	0.7035 0.6799

Vapour specific heat at constant pressure = 0.7 kJ/kg K. (Anna University)

Pressure, bar	Temperature, °C	Enthalpy, kJ/kg		Entropy, kJ/kg K
Pressure, bar	Temperature, °C	Liquid	Vapour	Liquid	Vapour
2.0 12.0	- 12.53 49.31	24.57 84.21	182.07 206.24	0.0992 0.3015	0.7035 0.6799

Vapour specific heat at constant pressure = 0.7 kJ/kg K. (Anna University)

Accepted Answer

Heat rejected by the cycle = [latex] \frac{500}{0.8} [/latex] = 625 kJ/min.

Assuming isentropic compression of refrigerant, we have
Entropy of dry saturated vapour at 2 bar
= Entropy of superheated vapour at 12 bar

0.7035 = 0.6799 + [latex] c_{p} \ln \frac{ T}{(49.31 + 273)} = 0.6799 + 0.7 × \ln (\frac{ T}{322.31})[/latex]

or [latex] \ln (\frac{ T}{322.31}) = \frac{0.7035 - 0.6799 }{0.7}[/latex] = 0.03371

or T = 322.31 [latex](e)^{0.03371} [/latex] = 333.4 K

∴ Enthalpy of superheated vapour at 12 bar
= 206.24 + 0.7(333.4 – 322.31) = 214 kJ/kg
Heat rejected per cycle = 214 – 84.21 = 129.88 kJ/kg

Mass flow rate of refrigerant = [latex] \frac{625}{129.88} [/latex] = 4.812 kg/min

Work done on compressor = 4.812 (214 – 182.07)
= 153.65 kJ/min = 2.56 kW

Actual work of compresson = [latex] \frac{2.56}{η_{compressor}} = \frac{2.56}{0.8} [/latex] = 3.2 kW

Hence power input to the heat pump compressor = 3.2 kW.

Question 12.24: A vapour compression heat pump is driven by a power cycle ha...

Related Answered Questions

(i) What are the advantages of using an expansion valve instead of an expander in a vapour compression refrigeration cycle ? (ii) Give a comparison between centrifugal and reciprocating compressors. (iii) An ice-making machine operates on ideal vapour compression refrigeration ...

Verified Answer:

The evaporator and condenser temperatures of 20 tonnes capacity freezer are – 28°C and 23°C respectively. The refrigerant – 22 is sub-cooled by 3°C before it enters the expansion valve and is superheated to 8°C before leaving the evaporator. The compression is isentropic. A sixcylinder ...

Verified Answer:

A refrigeration system of 10.5 tonnes capacity at an evaporator temperature of – 12°C and a condenser temperature of 27°C is needed in a food storage locker. The refrigerant ammonia is sub-cooled by 6°C before entering the expansion valve. The vapour is 0.95 dry as it leaves the evaporator ...

Verified Answer:

A food storage locker requires a refrigeration system of 2400 kJ/min. capacity at an evaporator temperature of 263 K and a condenser temperature of 303 K. The refrigerant used is freon-12 and is subcooled by 6°C before entering the expansion valve and vapour is superheated by 7°C ...

Verified Answer:

A refrigerator operating on standard vapour compression cycle has a co-efficiency performance of 6.5 and is driven by a 50 kW compressor. The enthalpies of saturated liquid and saturated vapour refrigerant at the operating condensing temperature of 35°C are 62.55 kJ/kg and 201.45 ...

Verified Answer:

In a simple vapour compression cycle, following are the properties of the refrigerant R-12 at various points : Compressor inlet : h2 = 183.2 kJ/kg v2 = 0.0767 m³/kg Compressor discharge : h3 = 222.6 kJ/kg v3 = 0.0164 m³/kg Compressor exit : h4 = 84.9 kJ/kg v4 = 0.00083 m³/kg The piston ...

Verified Answer:

A refrigeration cycle uses Freon-12 as the working fluid. The temperature of the refrigerant in the evaporator is – 10°C. The condensing temperature is 40°C. The cooling load is 150 W and the volumetric efficiency of the compressor is 80%. The speed of the compressor is 720 r.p.m. Calculate ...

Verified Answer:

A food storage locker requires a refrigeration capacity of 50 kW. It works between a condenser temperature of 35°C and an evaporator temperature of – 10°C. The refrigerant is ammonia. It is sub-cooled by 5°C before entering the expansion valve by the dry saturated vapour leaving the ...

Verified Answer:

A Freon-12 refrigerator producing a cooling effect of 20 kJ/s operates on a simple cycle with pressure limits of 1.509 bar and 9.607 bar. The vapour leaves the evaporator dry saturated and there is no undercooling. Determine the power required by the machine. If the compressor operaters ...

Verified Answer:

In a standard vapour compression refrigeration cycle, operating between an evaporator temperature of – 10°C and a condenser temperature of 40°C, the enthalpy of the refrigerant, Freon-12, at the end of compression is 220 kJ/kg. Show the cycle diagram on T-s plane. Calculate : (i) The ...

Verified Answer: