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