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

Question

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 evaporator. Assuming a single cylinder, single-acting compressor operating at 1000 r.p.m. with stroke equal to 1.2 times the bore.
Determine : (i) The power required, and
(ii) The cylinder dimensions.
Properties of ammonia are :

Saturation temperature, °C	Pressure bar	Enthalpy (kJ/kg)		Entropy (kJ/kg K)		Specific volume (m³/kg)		Specific heat (kJ/kg K)
		Liquid	Vapour	Liquid	Vapour	Liquid	Vapour	Liquid	Vapour
- 10 35	0.9157 13.522	154.056 366.072	1450.22 1488.57	0.82965 1.56605	5.7550 5.2086	— 1.7023	0.417477 0.095629	— 4.556	2.492 2.903

(UPSC)

Accepted Answer

Given : From the table above:
[latex] h_{2} = 1450.22 kJ/kg ; h_{3} ^{ ′} = 1488.57 kJ/kg ; h_{f_4} [/latex] = 366.072 kJ/kg ;
[latex] h_{f_4} ^{ ′} = h_{1} = h_{f_4} [/latex] – 4.556 (308 – 303)
= 366.07 – 4.556 (308 – 303) = 343.29 kJ/kg

Also [latex] s_{3} = s_{2}[/latex]

or [latex] s_{3} ^{′} + c_{p} \ln (\frac{T_{3}}{T_{3} ^{′}}) [/latex] = 5.755

5.2086 + 2.903 [latex]\ln (\frac{T_{3}}{308}) [/latex] = 5.755

or [latex]\ln (\frac{T_{3}}{308}) = \frac{5.755 - 5.2086}{2.903} [/latex] = 0.1882

[latex]\frac{T_{3}}{308} = e^{0.1882} [/latex] = 371.8 K

Now, [latex] h_{3} = h_{3} ^{′} + c_{p} (T_{3} – T_{3} ^{′}) [/latex]
= 1488.57 + 2.903 (371.8 – 308) = 1673.8 kJ/kg

Mass of refrigerant, [latex] \dot{m} = \frac{50}{ h_{2} - h_{1}} = \frac{50}{1450.22 - 343.29} [/latex]
= 0.04517 kJ/s

(i) Power required :
Power required = [latex] \dot{m} (h_{3} – h_{2})[/latex]
= 0.04517 (1673.8 – 1450.22) = 10.1 kW.

(ii) Cylinder dimensions :

[latex] \dot{m} = \frac{π}{4} D² × L × \frac{N}{60}[/latex] × 0.417477 = 0.04517 (calculated above)

or [latex] \frac{π}{4} D² × 1.2 D × \frac{1000}{60}[/latex] × 0.417477 = 0.04517

or D³ = [latex] \frac{ 0.04517 × 4 × 60}{π × 1.2 × 1000 × 0.417477 } [/latex] = 0.006888

∴ Diameter of cylinder, D = [latex] (0.006888)^{1/3}[/latex] = 0.19 m.

and, Length of the cylinder, L = 1.2D = 1.2 × 0.19 = 0.228 m.

Question 12.20: A food storage locker requires a refrigeration capacity of 5...

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