# For irreversible processes , why is $\frac{Q_1}{T_1} < \frac{Q_2}{T_2}$?

In the book of Kondepudi & Prigogine, Modern Theormodynamics, at page 97, it is stated that

All real heat engines that go through a cycle in finite time must involve irreversible processes such as flow heat due to a temperature gradient. They are efficient. Their efficiency $$\eta^{\prime}$$ is less than the efficiency of a reversible heat engine, i.e. $$\eta^{\prime}=1-\left(Q_{2} / Q_{1}\right)<1-\left(T_{2} / T_{1}\right)$$ This implies $$T_{2} / T_{1} whenever irreversible processes are involved. Therefore, while the equality (3.2 .4) is valid for a reversible cycle, for the operation of an irreversible cycle that we encounter in reality we have the inequality $$\frac{Q_1}{T_1} < \frac{Q_2}{T_2}.$$

However, my understanding lead to the same last equality but with a $$>$$ sign.

Note: For the notation: the index $$i$$ is the same as $$1$$, and $$out$$ is the same as $$2$$.

Let me explain:

The efficiency of a engine is defined as $$\eta = W^{out} / Q_i$$, so if the engine is ideal, i.e all the processes are reversible, so that the difference between the $$Q_i$$ and $$Q_{out}$$ is equal to the $$W^{out}$$, i.e all the heat left in the system converted to work, then we have

$$\eta = W^{out} / Q_i = \frac{Q_i - Q_{out}}{Q_i} = 1- Q_{out}/Q_i.$$ However, is the engine is not ideal, i.e there are processes that are irreversible, then you are not converting all the left heat to work, hence $$W^{out} < Q_i - Q_{out}$$, so $$\eta = W^{out} /Q_i < 1 - Q_{out} / Q_i.$$

The author derives (at page 97) that, $$\eta = 1 - T_{out}/T_i$$, so this implies

$$1 - T_{out} / T_i < 1 - Q_{out} / Q_i,$$ for irreversible processes, but this implies $$Q_{out} / Q_i < T_{out} / T_i,$$ and arranging them accordingly leads to

$$\frac{Q_{out}}{T_{out}} < \frac{Q_i}{T_i}$$ exact opposite of what the author derives.

Also, intuitively speaking, the author also quotes Carnot's as (for Carnot engines)

[...]there should not occur any change of temperature which may not be due to a change of volume.

This means that, in an irreversible process, there might be some temperature changes due to heat, and this lowers $$Q_2$$ while increasing $$T_2$$, so that the ratio $$Q_2 / T_2$$ becomes lower; this is totally compatible with the result that I obtained, and not compatible with the author derives.

Question: So, what is wrong in my argument ?

For the notation:

• The first inequality cited from the book is absolutely correct. Since the second inequality is mathematically identical to the first, it must also be correct. So there must be some error in what you did. I will look at your analysis step by step to see if I can find out where the error occurs. If I can find it, I'll post an answer. – Bob D Aug 30 at 18:17
• @BobD I would really appreciate it – onurcanbektas Aug 30 at 18:24
• I'm working on it right now. Found a conceptual error at the outset, but haven't yet finished the analysis. Stand by. – Bob D Aug 30 at 18:33
• I have found the error. See my answer. Hope it works for you. Let me know if you have any difficulty understanding the why of the error. – Bob D Aug 30 at 18:53

## 1 Answer

The first inequality cited from the book is absolutely correct. Since the second inequality is mathematically identical to the first, it must also be correct. So there must be some error in what you did. The following is a step by step review of your analysis in an attempt to find out where.

The efficiency of a engine is defined as $$\eta = W^{out} / Q_i$$

That is correct. The thermal efficiency of any heat engine, whether or not it is reversible, is the net work done divided by the gross heat added. So far, so good.

so if the engine is ideal, i.e all the processes are reversible, so that the difference between the $$Q_i$$ and $$Q_{out}$$ is equal to the $$W^{out}$$, i.e all the heat left in the system converted to work

Be careful, because this is not quite true.

It is correct that the work output is the difference between the heat added and the heat rejected. But that's true regardless of whether or not the engine is "ideal" or whether or not the processes are reversible. The difference is when the engine is not "ideal" or the process is irreversible, the heat rejected, $$Q_{out}$$ will be greater than the $$Q_{out}$$ for the ideal reversible process. That means $$W_{out}$$ will be less for the same $$Q_{in}$$ in a irreversible process than an ideal reversible process which, in turn, means the efficiency will be less.

then we have

$$\eta = W^{out} / Q_i = \frac{Q_i - Q_{out}}{Q_i} = 1- Q_{out}/Q_i.$$

Yep, that's what you'll have. But like I said, this equation is true whether or not the cycle is reversible.

However, is the engine is not ideal, i.e there are processes that are irreversible, then you are not converting all the left heat to work, hence $$W^{out} < Q_i - Q_{out}$$,

No, this is where you are going wrong, and it stems from the original misconception about the relationship between work out and net heat in. The work output will always equal the heat added minus the heat rejected. But for an irreversible process, the heat rejected, $$Q_{out}$$ will always be greater than $$Q_{out}$$ for a reversible process. That means less net heat available to do work. Bottom line, $$W_{out}$$ for an irreversible cycle will always be less than $$W_{out}$$ for a reversible process with the same $$Q_{in}$$, meaning less efficiency.

Since the rest of your math follows the wrong assumption, you have reached the wrong conclusion about the second inequality of the authors book.

In conclusion, you are probably wondering just why $$Q_{out}$$ is greater for an irreversible process than it is for a reversible process. Let me explain.

When a cycle involves irreversible processes entropy is generated in the cycle. In order to complete any cycle all of the thermodynamic properties of the working fluid have to return to their original values. That includes entropy since it is a property of the system. Since an irreversible process generates entropy, i.e., adds entropy to the system, the cycle needs to get rid of this additional entropy to return to its original state. The only way to get rid of the entropy is to transfer it to the surroundings. And the only way to transfer entropy out of the system is by heat transfer. So this heat transfer out due to the entropy generation has to be added to the heat transfer out that would occur in the reversible process. Result: $$Q_{out}$$ for the irreversible cycle is greater than $$Q_{out}$$ for the reversible cycle.

Hope this helps.

• For some reason my comment is deleted, but anyway; thanks a lot again. – onurcanbektas Aug 31 at 10:46
• I thought I saw a comment then it wasn’t there. Something about increasing $Q_{in}$ – Bob D Aug 31 at 11:15
• I deleted that; I understood after reading all of your answer. – onurcanbektas Aug 31 at 11:20
• . Got it. Thanks – Bob D Aug 31 at 11:22