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
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