Efficiency of a Carnot engine is given as 80%. If the cycle direction be reversed, what will be the value of C.O.P. of reversed Carnot Cycle?
A. 1.25
B. 0.8
C. 0.5
D. 0.25
Answer: Option D
Solution(By Examveda Team)
Efficiency of a Carnot engine is given as 80% or $$\eta $$ = 0.80If the cycle direction be reversed
The relation between the effecicy and C.O.P. is defined by formula
$${\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = \frac{{1 - \eta }}{\eta } = \frac{{1 - 0.80}}{{0.80}} = 0.25$$
This 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
Reversed Carnot cycle means refrigeration only so answer is correct