The C.O.P. of an absorption type refrigerator is given by (where T1 = Temperature at which the working substance receives heat, T2 = Temperature of cooling water and T3 = Evaporator temperature)
A. $$\frac{{{{\text{T}}_1}\left( {{{\text{T}}_2} - {{\text{T}}_3}} \right)}}{{{{\text{T}}_3}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}$$
B. $$\frac{{{{\text{T}}_3}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{{{\text{T}}_1}\left( {{{\text{T}}_2} - {{\text{T}}_3}} \right)}}$$
C. $$\frac{{{{\text{T}}_1}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{{{\text{T}}_3}\left( {{{\text{T}}_2} - {{\text{T}}_3}} \right)}}$$
D. $$\frac{{{{\text{T}}_3}\left( {{{\text{T}}_2} - {{\text{T}}_3}} \right)}}{{{{\text{T}}_1}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}$$
Answer: Option B
Solution(By Examveda Team)
The C.O.P. of an absorption type refrigerator is given by $$\frac{{{{\text{T}}_3}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{{{\text{T}}_1}\left( {{{\text{T}}_2} - {{\text{T}}_3}} \right)}}$$(Where T1 = Temperature at which the working substance receives heat, T2 = Temperature of cooling water and T3 = Evaporator temperature)
This Question Belongs to Mechanical Engineering >> Refrigeration And Air Conditioning
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
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