# Efficiency in a thermodynamic cycle

Question

The area on a PV diagram within one thermodynamic cycle of a heat engine is 610J. THe total thermal energy transferred out of the gas in one cycle is 1300J. Determine the efficiency of the cycle.

Attempt (and subsequent lack of success) at solution

The area within the cycle is equivalent to the work done by the gas, which is 610J.

My first guess was just the ratio of the useful work done to the total, i.e $\frac{610}{1300+610}$ to give about 32%. However, it turns out that the answer is just $\frac{610}{1300}\approx 47\%$.

Thoughts

I thought that efficiency, intuitively, is the ratio of useful work to the total energy input

$$\text{efficiency}=\frac{\text{work}}{\text{thermal energy taken from hot reservoir}}=\frac{\Delta W}{Q_{hot}}$$

Alternatively, we could say that $$e=\frac{Q_{in}-Q_{out}}{Q_{in}}=1-\frac{Q_{out}}{Q_{in}}$$ These formulae seem intuitive, and I would be happy to apply them. But the question only gives us the value for $Q_{out}$, so we don't know $Q_{in}$ or $Q_{hot}$ (on another note, are these equal?)

I just don't see why the solution wants to find the ratio of work done to the thermal energy transferred out.

What you say is correct, you just misunderstand what data you are given.

There seems to be some confusion about the terms also, so let's clear it out while looking at the schematic of a heat engine (from this source): • $Q_{C}$ or $Q_{cold}$ is heat exchanged with the cold reservoir. For a heat engine, this is heat leaving the machine (can be considered "lost" or "wasted" or "unused").
• $Q_{H}$ or $Q_{hot}$ is heat exchanged with the hot reservoir. For a heat engine, this is heat entering the machine.
• $Q_{out}$ is heat leaving the machine - just another name for $Q_C$
• $Q_{in}$ is heat entering the machine - just another name for $Q_H$
• $W$ is work that the machine does - the useful energy so to speak.

Personally I always use the terms $Q_{in}$ and $Q_{out}$ because it intuitively makes more sense to me to think of energy in and out. But all are used in different litterature.

From the drawing, which illustrates energy conservation, it of course makes sense that $$Q_{in}=W+Q_{out}\quad\Leftrightarrow\quad W=Q_{in}-Q_{out}$$

Efficiency $e$ or $\mu$ is defined as the fraction of useful energy to the total inputted energy:

$$e=\frac{\text{useful energy (work)}}{\text{total energy inputted}}=\frac{W}{Q_{in}}=\frac{Q_{in}-Q_{out}}{Q_{in}}=1-\frac{Q_{out}}{Q_{in}}$$

All you said is true, there are just different "names" for the terms.

Now we just need to put in the data given.

The area on a PV diagram within one thermodynamic cycle of a heat engine is $610\;\mathrm{J}$.

Yes, this means that $W=610\;\mathrm{J}$

THe total thermal energy transferred out of the gas in one cycle is $1300\;\mathrm{J}$.

It is written in a tricky way but simply says that the energy taken from the gas and added to the engine is $1300\;\mathrm{J}$. So the inputted energy to the heat engine is $Q_{in}=1300\;\mathrm{J}$. In this case the hot reservoir (the heat source) is hot gas apparently. So we have $$e=W/Q_{in}=610/1300\approx 47\;\%$$.

• Aren't your bullet points contradictory? The first and third bullet points don't seem to be compatible. – surelyourejoking Oct 21 '15 at 11:10
• @surelyourejoking Oh yes, a typo. Corrected now. – Steeven Oct 21 '15 at 11:11
• "It is written in a tricky way but simply says that the energy taken from the gas and added to the engine is 1300 J". I follow everything up till here. How does "total thermal energy transferred out of the gas" become "energy taken from the gas and added to the engine"? – surelyourejoking Oct 21 '15 at 11:20
• @surelyourejoking Because the engine is not a gas. Simply consider the hot gas as the hot reservoir. Then surely, the heat inputted to the engine has to be taken from the hot reservoir e.i. it is heat taken out of the hot gas and added to the engine. – Steeven Oct 21 '15 at 11:22