Answer:
[tex]P_m=181.42 KPa[/tex]
Explanation:
[tex]P_1=101.325 KPa,V_1=0.05m^3,T_1=32C[/tex],K=1.35
Clearance is 8%.
Heat added=15 KJ
We know that compression ratio [tex]r=1+\dfrac{1}{C}[/tex]
[tex]r=1+\dfrac{1}{0.08}[/tex]
r=13.5
[tex]r=\dfrac{V_1}{V_2}[/tex]
[tex]13.5=\dfrac{0.05}{V_2}[/tex]
[tex]V_2=3.7\times 10^{-3}[/tex]
We know that efficiency of otto cycle
[tex]\eta =1-\dfrac{1}{r^{k-1}}[/tex]
[tex]\eta =1-\dfrac{1}{13.5^{1.35-1}}[/tex]
[tex]\eta =0.56[/tex]
[tex]\eta =\dfrac{W}{Q}[/tex]
W is the work out put and Q is the heat addition.
[tex]0.56 =\dfrac{W}{15}[/tex]
W=8.4 KJ
We know that Work =Mean effective pressure x swept volume.
Here swept volume [tex]V_s=V_1-V_2[/tex]
[tex]V_s=0.05-3.7\times 10^{-3}[/tex]
[tex]V_s=0.0463 m^3[/tex]
Noe by putting the values
Work =Mean effective pressure x swept volume.
[tex]8.4=P_m\times 0.0463[/tex]
[tex]P_m=181.42 KPa[/tex]