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Railways NTPC (Technical Ability) Refrigeration and Air-Conditioning Question Bank Refrigeration and Air-Conditioning

  • question_answer
    A refrigerator storage is supplied with 3600 kg of fish at a temperature of\[27{}^\circ C\]. The fish has to be cooled to \[23{}^\circ C\] for preserving it for a long period without deterioration. The cooling takes place in 10 hours. The specific heat of fish is 2.0 kJ/kg K above freezing point of fish and 0.5 kJ/kg K below freezing point of fish, which is\[-\,3{}^\circ C\]. The latent heal of freezing is 230 kJ/kg. What is the power to drive the plant if the actual COP is half that of the ideal COP?

    A) 30 kW              

    B) 15 kW

    C) 12 kW              

    D) 6 kW.

    Correct Answer: C

    Solution :

    \[{{T}_{1}}=237+27=300\,K\]   \[{{T}_{2}}=237-23=250\,K\]     \[{{(COP)}_{\operatorname{Re}f}}=\frac{{{T}_{2}}}{{{T}_{1\,}}-{{T}_{2}}}=\frac{250}{300-250}=\frac{250}{50}=5\] \[Q=\frac{3600}{3600\times 10}[2\times \left( 27+3 \right)+0.5(-\,3+23)+230]\]\[\,\,=\,\,30kJ/s\] Actual \[COP\,\,=\,\,\frac{5}{2}\,\,=\,\,2.5\] \[W=\frac{30}{2.5}=12\,\,kW\]


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