What will be the maximum coefficient of performance (COP) for the vapour absorption cycle if Tg is generator temperature, Tc is environment temperature and Te is refrigerated space temperature?

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SSC JE Mechanical 4 Dec 2023 Official Paper - II
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  1. \(\frac{T_g\left(T_c-T_e\right)}{T_c\left(T_g-T_e\right)}\)
  2. \(\frac{T_c\left(T_g-T_e\right)}{T_g\left(T_c-T_e\right)}\)
  3. \(\frac{T_g\left(T_c-T_e\right)}{T_e\left(T_g-T_c\right)}\)
  4. \(\frac{T_e\left(T_g-T_c\right)}{T_g\left(T_c-T_e\right)}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{T_e\left(T_g-T_c\right)}{T_g\left(T_c-T_e\right)}\)
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Detailed Solution

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Explanation:

  • In the case of VARS, the COP can be remembered directly.
  • Just apply a heat engine between the upper two temperatures and a refrigerator between lower two

Live Test 2 (1-50) images Q14b

\(\begin{array}{l} {\left( {COP} \right)_{VARS}} = {\eta _{carnot,engine}} \times {\left( {COP} \right)_{carnot,ref}}\\ = \left( {1 - \frac{{{T_O}}}{{{T_G}}}} \right)\left( {\frac{{{T_E}}}{{{T_O} - {T_E}}}} \right)\\ = \left( {\frac{{{T_G} - {T_O}}}{{{T_G}}}} \right)\left( {\frac{{{T_E}}}{{{T_O} - {T_E}}}} \right) \end{array}\)

\(\therefore COP=\frac{{{T_E}\left( {{T_G} - {T_o}} \right)}}{{{T_G}\left( {{T_o} - {T_E}} \right)}}\)

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