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

Question

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 before leaving the evaporator coil. The compression of refrigerant is reversible adiabatic. The refrigeration compressor is two-cylinder single-acting with stroke equal to 1.25 times the bore and operates at 1000 r.p.m.

Properties of freon-12

Saturation temp, K	Absolute pressure, bar	Specific volume of vapour, m³/kg	Enthalpy, kJ/kg		Entropy, kJ/kg K
			Liquid	Vapour	Liquid	Vapour
263 303	2.19 7.45	0.0767 0.0235	26.9 64.6	183.2 199.6	0.1080 0.2399	0.7020 0.6854

Take : Liquid specific heat = 1.235 kJ/kg K ; Vapour specific heat = 0.733 kJ/kg K.
Determine :
(i) Refrigerating effect per kg.
(ii) Mass of refrigerant to be circulated per minute.
(iii) Theoretical piston displacement per minute.
(iv) Theoretical power required to run the compressor, in kW.
(v) Heat removed through condenser per min.
(vi) Theoretical bore and stroke of compressor.

Accepted Answer

The cycle of refrigeration is represented on T-s diagram on Fig. 35.
Enthalpy at ‘2’, [latex] h_{2} = h_{2} ^{′} + c_{p} (T_{2} – T_{2} ^{′})[/latex]
From the given table :
[latex] h_{2} ^{′} [/latex] = 183.2 kJ/kg
([latex] T_{2} – T_{2} ^{′}) [/latex] = Degree of superheat as the vapour enters the compressor = 7°C
∴ [latex] h_{2} [/latex] = 183.2 + 0.733 × 7 = 188.33 kJ/kg

Also, entropy at ‘2’, [latex]s_{2} = s_{2} ^{′} + c_{p} log_{e} \frac{ T_{2} }{ T_{2} ^{′}}[/latex]

= 0.7020 + 0.733 [latex]log_{e} (\frac{270}{263}) [/latex] = 0.7212 kJ/kg K

For isentropic process 2-3
Entropy at ‘2’ = Entropy at ‘3’

0.7212 = [latex]s_{3} ^{′} + c_{p} log_{e} (\frac{ T_{3} }{ T_{3} ^{′}})[/latex]

= 0.6854 + 0.733 [latex] log_{e} ( \frac{ T_{3} }{ 303})[/latex]

∴ [latex] log_{e} (\frac{ T_{3} }{ 303})[/latex] = 0.0488

i.e., [latex]T_{3} [/latex] = 318 K
Now, enthalpy at ‘3’, [latex] h_{3} = h_{3} ^{′} + c_{p} (T_{3} – T_{3} ^{′})[/latex]
= 199.6 + 0.733 (318 – 303) = 210.6 kJ/kg.

Also, enthalpy at [latex]4 ^{′}, h_{f_4} ^{′} = h_{f_4} – (c_{p})_{liquid} (T_{4} – T_{4} ^{′} ) [/latex] = 64.6 – 1.235 × 6 = 57.19 kJ/kg
For the process [latex]4 ^{′}[/latex] -1,
Enthalpy at [latex]4 ^{′} [/latex]= enthalpy at 1 = 57.19 kJ/kg
For specific volume at 2,

[latex] \frac{v_{2} ^{′} }{T_{2} ^{′} } = \frac{v_{2}}{T_{2}} [/latex]

∴ [latex] v_{2} = \frac{v_{2} ^{′} }{T_{2} ^{′} } × T_{2} = 0.0767 × \frac{270}{263} [/latex] = 0.07874 m³/kg

(i) Refrigerating effect per kg
= [latex]h_{2} – h_{1} [/latex] = 188.33 – 57.19 = 131.14 kJ/kg.

(ii) Mass of refrigerant to be circulated per minute for producing effect of 2400 kJ/min.

= [latex] \frac{2400}{131.14} [/latex] = 18.3 kg/min.

(iii) Theoretical piston displacement per minute
= Mass flow/min. × specific volume at suction
= 18.3 × 0.07874 = 1.441 m³/min.

(iv) Theoretical power required to run the compressor
= Mass flow of refrigerant per sec. × compressor work/kg

= [latex] \frac{18.3}{60} × (h_{3} – h_{2}) = \frac{18.3}{60} [/latex] (210.6 - 188.33 ) k J/s

= 6.79 kJ/s or 6.79 kW.

(v) Heat removed through the condenser per min.
= Mass flow of refrigerant × heat removed per kg of refrigerant
= 18.3 ( [latex] h_{3} – h_{f_4} ^{′} [/latex]) = 18.3 (210.6 – 57.19) = 2807.4 kJ/min.

(vi) Theoretical bore (d) and stroke (l) :
Theoretical piston displacement per cylinder

= [latex] \frac{Total displacement per minute }{Number of cylinder} = \frac{1.441}{2} [/latex] = 0.7205 m³/min.

Also, length of stroke = 1.25 × diameter of piston
Hence, 0.7205 = π/4 d² × (1.25 d) × 1000

i.e., d = 0.09 m or 90 mm.

and l = 1.25 d = 1.25 × 90 = 112.5 mm.

Question 12.25: A food storage locker requires a refrigeration system of 240...

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 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 ...

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: