The compression ratio of an air-standard Otto cycle is 9.5 Prior to the isentrop

Question

The compression ratio of an air-standard Otto cycle is 9.5 Prior to the isentropic compression process, the air is at 100 KPa, 35o C and 600 cm3 .The temperature at the end of the isentropic expansion process is    800 K. Using specific heat values at room temperature, determine

(A) the highest temperature and pressure in the cycle.

(B) The amount of heat transfered in, in KJ

(C) The thermal efficiency

(D) the mean effective pressure.

The answer in the back of the book states that for (A) 1969 K and 6072 KPa   (B) 0.59 KJ   (C) 59.4 percent and (D) 652 KPa

Explanation / Answer

In an Otto cycle, the first stage is the isentropic compression(where s1=s2), then the second stage is the heat addition(gain in temperature) and at constant volume(V2=V3), then the third stage is the isentropic expansion(where s3=s4) and then the fourth and last stage is the heat rejection(temperature drop) and at constant volume(V4=V1).

Given,

Compression ratio of an Otto cycle, r= 9.5

P1= 100kPa (starting of the Otto cycle)

T1= 35oC=308K

V1=600 cm3= 6*10-4 m3

T4=800K (ending of the Otto cycle)

From the Compression ratio,

r= Vmax/Vmin (or) V1/V2

Therefore by solving this we get V2= 6.32*10-5 m3

a) Now first we look at stage 1------->stage 2, which is an isentropic compression.

Therefore,

(T2/T1)=(V1/V2)k-1 where k=cp/cv and in this case k=1.4

Now using the values we have we can find T2, which by solving gives you T2=757.75K

And also (P2/P1)=(V1/V2)k where k=1.4

Therefore, solving this equation gives you P2=2338 kPa

Now we look at stage 3------>stage 4, which is isentropic expansion,

(T3/T4)=(V4/V3)k-1,where k=1.4

But V4=V1 and V3=V2

Using the values we have, T3=1969K

Now we look at stage 2------>stage 3, which is heat addition at constant volume.

(P2V2/T2)=(P3V3/T3)

Solving this from the given values give P3=6072kPa

Therefore the highest temperature is 1969K and the highest pressure is 6072kPa.

b) m=(P1V1/RT1)=6.788*10-4 kg

Therefore Heat transfered in the cycle, Qin=m(u3-u2)=mcv(T3-T2)

and after solving this we get Qin=0.59kJ

c) Now we look at stage 4----->stage 1, which is heat rejection at constant volume.

Heat transfered out of the cycle, Qout=m(u4-u1)=mcv(T4-T1)

and after solving this we get Qout=0.24kJ

Therefore, thermal efficiency, th=1-(Qout/Qin)=0.594=59.4%

d) Wnet,out=Qin-Qout=0.35kJ

MEP=Wnet,out/(Vmax-Vmin), where Vmax=V1 and Vmin=V2

Therefore, MEP=652kPa

I hope this helped :)

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