Efficiency of a combustion motor

[Karol]

Homework Statement

The efficiency of a heat machine is $E=\frac{Q_2-Q_1}{Q_2}$ where Q₂ is the heat inserted because of the explosion and Q₁ is the heat that leaves through the exhaust.
It says in the book that E can be derived from the volumes ratio V₁/V₂:
$$E=1-\frac{1}{\left(V_1/V_2 \right)^{\gamma-1}}$$ why?

Homework Equations

Adiabatic expansion: $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$

The Attempt at a Solution

$$E=1-\frac{1}{\left(V_1/V_2 \right)^{\gamma-1}}=\frac{T_2-T_1}{T_2}$$
But V₂ and T₂ have nothing to do with Q₂, the heat from the explosion. also i don't know what's the connection between V₁ or T₁ and the heat lost through the exhaust.
I have just started studying the second law, maybe the book is excluding some explanations, it's an old book for high school but the level is quite high

 

[CWatters]

Not my field but perhaps look up Carnot Efficiency. That relates efficiency and temperature.

 

[BvU]

Note that E is the maximum efficiency, achievable only if all process steps are carried out reversibly, i.e. entropy is conserved. Then Q₂ = T₂dS₂ and idem 1 (with a - sign). Over all $\Delta S = Q_2/T_2 - Q_1/T_1$ and only $Q_2 - Q_1$ can be drawn off as work.

 

[siddharth23]

Note that E is the maximum efficiency, achievable only if all process steps are carried out reversibly, i.e. entropy is conserved.

You're right. But mostly actual cycles aren't considered till you study the subject in depth. And I think there is an easier way for this proof.

But V2 and T2 have nothing to do with Q2, the heat from the explosion. also i don't know what's the connection between V1 or T1 and the heat lost through the exhaust.

If you know the the changes in temperatures (ΔT), you can use the equations-
Q = mC_PΔT (constant pressure) or Q = mC_VΔT (constant volume)
Knowing the relation between temperature and volume, you can introduce 'V' into the equation.

I must say though, it isn't really too easy for a starter.

 

[Karol]

Thanks i solved but it's long development so i won't write it here

 

[siddharth23]

No problemo!