An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kPa and 20∘C . If the maximum temperature in the cycle is not to exceed 2200 K. (Take y for air = 1.4 and cp for air = 1.005 kJ/kgK) then find the thermal efficiency

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Bollacasaep22 · Accepted Answer

c    p    =  1.005  k  J      /    k  g  K  ,         c    v    =  0.718  k  J      /    k  g  K  ,     R  =  0.287  k  J      /    k  g  K  ,         γ    =  1.4        T    2    =      T    1    (            V      1              V      2            )                  γ            −      1        =  (  293  K  )  (  20      )          0.4        =  971.1  K    Process 2-3: constant pressure heat addition,                      P        3                    V        3                    T      s        =                    P        2                    V        2                    T      2        ⇒            V      3              V      2        =            T      3              T      2        =            2200      K              971.1      K        =  2.265    Process 3-4: isentropic expansion,        T    4    =      T    3    (            V      3              V      4            )                  γ            −      1        =      T    3    (            2.265              V        2                    V      4            )                  γ            −      1        =      T    s    (      2.265    r        )                  γ            −      1          =  (  2200  )  (      2.265    20        )          0.4                 [  ∵      V    4    =      V    1    ]    =  920.6  K        q          i      n        =      h    3    −      h    1    =      c    p    (      T    3    −      T    2    )  =  (  1.005  )  (  2200  −  971.1  )  =  1235.04     k  J      /    k  g        q          o      u      t        =      u    4    −      u    1    =      c    v    (      T    4    −      T    1    )  =  (  0.718  )  (  920.6  −  293  )  =  450.6     k  J      /    k  g              ω              n      e      t      ,      o      u      t        =      q          i      n        −      q          o      u      t        =  1235  −  450.6  =  784.4     k  J      /    k  g              η              t      h        =                    ω                    n        e        t        ,        o        u        t                    q              i        n              =  63.5  %

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