In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle temperature is 2031K. Determine the thermal efficiency.

Respuesta :

Answer:

[tex]\eta[/tex]=0.60

Explanation:

Given :Take [tex]\gamma[/tex]=1.4 for air

      [tex]P_1=100 KPa  ,T_1=300K[/tex]

  [tex]\frac{V_1}{V_2}[/tex]=r ⇒ r=16

As we know that  

   [tex]T_2=T_1(r^{\gamma-1})[/tex]

So [tex]T_2=300\times 16^{\gamma-1}[/tex]

  [tex]T_2[/tex]=909.42K

Now find the cut off ration [tex]\rho[/tex]

      [tex]\rho=\frac{V_3}{V_2}[/tex]  

         [tex]\frac{V_3}{V_2}=\frac{T_3}{T_2}[/tex]

[tex]\rho=\frac{2031}{909.42}[/tex]

 [tex]\rho=2.23[/tex]

So efficiency of diesel engine

[tex]\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}[/tex]

Now by putting the all values

[tex]\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}[/tex]

So [tex]\eta[/tex]=0.60

So the efficiency of diesel engine=0.60

     

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