Question

In a vapour absorption refrigerator, the temperature of evaporator and ambient are 10°C and 30°C respectively. If the COP of the system is 2, estimate the generator temperature.
1. 90°C
2. 85°C
3. 80°C
4. 75°C

Answer 1

Correct Answer - Option 3 : 80°C

Concept:

\(C.O{.P_{VARS}} = {{\rm{\eta }}_{{\rm{carnot}}}} \times {\rm{C.O}}{{\rm{.P}}_{{\rm{carnot}}}} \)

\(C.O.P_{VARS}= \frac{{{{\rm{T}}_{\rm{g}}} - {{\rm{T}}_{\rm{o}}}}}{{{{\rm{T}}_{\rm{g}}}}} \times \frac{{{{\rm{T}}_{\rm{e}}}}}{{{{\rm{T}}_{\rm{o}}} - {{\rm{T}}_{\rm{e}}}}}\)

where, T_g = Generator Temperature; To = Ambient Temperature; Te = Evaporator Temperature

Calculation:​

Given:

T_o = 30° C + 273 = 303 K, Te = 10° C + 273 = 283 K,

COP_VARS = 2, and Tg = ??

\(C.O.P_{VARS}= \frac{{{{\rm{T}}_{\rm{g}}} - {{\rm{T}}_{\rm{o}}}}}{{{{\rm{T}}_{\rm{g}}}}} \times \frac{{{{\rm{T}}_{\rm{e}}}}}{{{{\rm{T}}_{\rm{o}}} - {{\rm{T}}_{\rm{e}}}}}\)

 

\(2 = \frac{{{{\rm{T}}_{\rm{g}}} - 303}}{{{{\rm{T}}_{\rm{g}}}}} \times \frac{{283}}{{303 - 283}}\)

∴ T_g = 352.87 = 352.87 - 273 = 79.87 = 80° C​.