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

  • question_answer
    A refrigerating system operating on reverse Brayton refrigeration cycle is used for maintaining 250 K. If the temperature at the exit of constant pressure cooling is 300 K and rise in the temperature of air in the refrigerator is so K, then the network of compression will be (assume air as the working substance with \[{{c}_{p}}=\,\,1\,\,KJ\,per\,kg\,per{}^\circ C)\]

    A) 250 kJ/kg                      

    B) 200 kJ/kg

    C) 50 kJ/kg            

    D) 25 kJ/kg

    Correct Answer: D

    Solution :

    Network of compression for the reversed Brayton refrigeration cycle\[=({{h}_{2}}-{{h}_{1}})-({{h}_{3}}-{{h}_{4}})\] Now, \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{3}}}{{{T}_{4}}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,{{T}_{2}}=250\times \frac{300}{200}=375\,K\]             Net work \[={{c}_{p}}[({{T}_{2}}-{{T}_{1}})-({{T}_{3}}-{{T}_{4}})]\]                         \[\,\,=0.1\,[(375-250)-(300-200)]\] \[\,\,=1[125-100=1\times 25=25\,kJ/kg]\]


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