The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05m3 and 32C. The clearance is 8% and 15kJ are added per cycle. Determine the mean effective pressure.

Respuesta :

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]

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