CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A Cannot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is

    A)  900 K                   

    B)         600 K                   

    C)  700 K                   

    D)         800 K

    E)  550 K

    Correct Answer: B

    Solution :

    Cannot efficiency relation \[\eta =\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\] Case I      \[\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}=0.4\]                 \[{{T}_{1}}-{{T}_{2}}=0.4{{T}_{1}}\]                 \[{{T}_{2}}=0.6{{T}_{1}}\] Case II      \[\frac{T_{1}^{}-{{T}_{2}}}{T_{1}^{}}=0.5\]                 \[T_{1}^{}=\frac{0.6}{0.5}{{T}_{1}}=600\,K\]


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