The operating temperature of a cold storage is -2°C. The heat leakage from the surrounding is 30 kW for the ambient temperature of 40°C. The actual C.O.P. of refrigeration plant used is one fourth that of ideal plant working between the same temperatures. The power required to drive the plant is
A. 1.86 kW
B. 3.72 kW
C. 7.44 kW
D. 18.6 kW
Answer: Option D
Solution(By Examveda Team)
The operating temperature of a cold storage is -2°C. The heat leakage from the surrounding is 30 kW for the ambient temperature of 40°C. The actual C.O.P. of refrigeration plant used is one fourth that of ideal plant working between the same temperatures. The power required to drive the plant is 18.6 kWThis Question Belongs to Mechanical Engineering >> Refrigeration And Air Conditioning
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Nusselt number (NN) is given by
A. $${{\text{N}}_{\text{N}}} = \frac{{{\text{h}}l}}{{\text{k}}}$$
B. $${{\text{N}}_{\text{N}}} = \frac{{\mu {{\text{c}}_{\text{p}}}}}{{\text{k}}}$$
C. $${{\text{N}}_{\text{N}}} = \frac{{\rho {\text{V}}l}}{\mu }$$
D. $${{\text{N}}_{\text{N}}} = \frac{{{{\text{V}}^2}}}{{{\text{t}}{{\text{c}}_{\text{p}}}}}$$
In case of sensible heating of air, the coil efficiency is given by (where B.P.F. = Bypass factor)
A. B.P.F. - 1
B. 1 - B.P.F.
C. $$\frac{1}{{{\text{B}}{\text{.P}}{\text{.F}}{\text{.}}}}$$
D. 1 + B.P.F.
The undesirable property of a refrigerant is
A. Non-toxic
B. Non-flammable
C. Non-explosive
D. High boiling point
The desirable property of a refrigerant is
A. Low boiling point
B. High critical temperature
C. High latent heat of vaporisation
D. All of these
We know COP = TEMP OF COLD BODY/TEMP OF HOT BODY - TEMP OF COLD BODY.
So here COP = 271/(313-271) = 6.45.
Actual COP = 1/4th of Ideal COP = 1/4(6.45) = 1.6125.
We also know actual COP = Heat extracted/Work input.
1.6125 = 30/WI.
Therefore Power = 30/1.6125 = 18.6 K