# Calculating the maximum efficiency of an engine cycle

I'm a little confused about calculating the maximum efficiency of an engine cycle. From what I've learned the carnot cycle is connected by two adiabatic processes and two isothermal, and this is the most efficient engine cycle.

But on a recent homework assignment I was given an engine cycle that had two isothermal processes and two isobaric. I calculated the efficiency using the total amount of work, divided by the heat going into the engine:

$eff={|W_{total}|\over |Q_{in}|}$

And then I was instructed to find the maximum efficiency, I didn't know how to do this so a friend of mine told me to calculate it using the equation for the efficiency of a carnot cycle in terms of temperature:

$eff = 1 - {T_c\over T_h}$

This turned out to be the correct answer according to my professor, but I still don't understand why this gives the maximum efficiency. Why is this the case? Why can this be used to find the maximum when the original cycle didn't have any adiabatic processes?

All engines

1) That are reversible

2) and work between two temperatures

Have efficiency equal to $1-\frac{T_c}{T_h}$.

There is nothing special about the cycle you mentioned.

You are asked to find the maximum efficiency of your engine, but when will your engine have maximum efficiency?

When you make the engine reversible.

And what is the efficiency of a reversible engine that works between two temperatures?

Equal to $1-\frac{T_c}{T_h}$ (it does not matter, whether the cycle is the particular cycle you mentioned or not).

• With "All reversible engines working between two temperatures (source and sink) have same efficiency equal to the Carnot efficiency" do you mean the same maximum efficiency? Or is it just in general? Feb 19 '17 at 4:24
• I meant,there is nothing special about carnot cycle ,any reversible cycle has efficiency equal to $1-Tc/Th$
– Paul
Feb 19 '17 at 4:33
• @Grace :check my edit
– Paul
Feb 19 '17 at 4:40
• Unless you mean that heat exchange occurs only at two temperatures $T_c$ and $T_h$ in which case you are talking about a Carnot cycle, what you have written is wrong. Instead it is fairly easy to show that any reversible cycle which exchanges heat also at temperatures other than the maximum $T_h$ and minimum $T_c$ has lower efficiency than the the Carnot efficiency $1-T_c/T_h$. Feb 19 '17 at 18:34
• yes, that is it: only two temperatures at which heat is exchanged therefore in between those temperatures the cycle must be adiabatic (no heat exchange). The two fixed temperatures are the isothermals. Is this clear now? Feb 20 '17 at 3:11