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Derive equation 4.10for the efficiency of the Otto cycle.

Short Answer

Expert verified

The efficiency of the Otto cycle is η=1−V2V1γ−1

Step by step solution

01

Introduction

Efficiency (η)=Net work doneWnetHeat suppliedQ1

Heat supplied:

Q1=mCvT3-T2---(1)

Hear rejected:

Q2=mCvT4−T1−−−(2)η=Q1−Q2Q1=1−Q2Q1−−(3)From(1)and(2)η=1−mCvT4−T1mC4T3−T2η=1−T4−T1T3−T2−−(4)

02

Calculation

As it is an adiabatic process:

T2V2γ−1=T1V1γ−1T2T1=V1V2γ−1

Similarly,

T2V2γ−1=T1V1γ−1T3T4=V4V3γ−1=V1V2γ−1∴T2T1=T3T4

T3T2=T4T1−−−−(5)

Subtract the above equation by 1 both sides:

T3T2−1=T4T1−1T3−T2T2=T4−T1T1T1T2=T4−T1T3−T2=V2V1γ−1

Substitute the above equation in (4):

η=1-V2V1γ-1

03

Conclusion

Efficiency of the Otto cycle is η=1-V2V1γ-1.

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