# Does the thermal efficiency depend on the reversibility of the engine when isochoric or isobaric processes are involved?

I'm confused on the efficiency of a thermal engine (with an ideal gas) in the case in which it is reversible or not reversible, in particular where the ideal gas follows isochoric or isobaric processes.

Infact during any isochoric and isobaric processes (reversible or not) $$Q_{\mathrm{isochoric}}=n c_v \Delta T$$ $$Q_{\mathrm{isobaric}}=n c_p \Delta T$$

The heats depend only on the temperatures and not on the process. So it really seems to me that the reversibility or irreversibility does not change the efficiency of the heat engine at all.

Consider the cycle in the diagram, made of 2 isochoric and 2 isobaric processes. The cycle is travelled clockwise. The efficiency is $$\eta=1-\frac{|Q_{C\rightarrow D}+Q_{B\rightarrow C}|}{Q_{A\rightarrow B}+Q_{D\rightarrow A}}=1-\frac{|nc_p(T_{D}-T_{C})+nc_v(T_{C}-T_{B})|}{nc_p(T_{B}-T_{A})+nc_v(T_{A}-T_{D})}\tag{I}$$

Consider the two following cases

1. All the processes in the diagram are reversible

2. One of more of the processes is not reversible

Does $\eta$ changes between case 1 and 2?

On the one hand my answer would be no, as said before, because heats are all functions of temperatures only both in case 1. and 2..

On the other hand this does not makes sense, because Carnot theorem requires that in case 1 (in wich all processes are reversible) the efficiency is the same of a Carnot engine working betweeen highest and lowest temperature (in this case $T_{D}$ and $T_{B}$), that is, in case 1., efficiency should be $$\eta_{\mathrm{reversible}}=1-\frac{T_D}{T_B}\tag{II}$$ Which is not equal to $(\mathrm{I})$.

To sum up, it seems to me that, in this case, $\eta$ does not depend on the reversibility of the engine. I don't see this dipendence which, nevertheless, must be there, because of Carnot theorem.

So how does $\eta$ depend of the reversibility of processes in this one case?

The sign convention used is the one in picture

There are basically two possible answers, depending on the character of the irreversibility.

So let's think: What would an irreversible process look like? Here are some examples...

1. The hot reservoir is much hotter than the ideal gas, and/or the cold reservoir is much colder than the ideal gas.

• This does not change the engine efficiency. It merely increases the Carnot limit for the situation. In other words, the engine falls short of its potential efficiency, not because the efficiency went down, but because the potential efficiency went up.
2. During the isochoric step, while holding the volume constant, we heat the ideal gas so fast that it creates sound waves.

• Sound waves are a true efficiency loss. Energy in the form of sound waves is exiting the whole engine and getting absorbed in the walls and ceiling. This kind of energy leakage violates the formula $\eta = 1-Q_{Emitted}/Q_{Absorbed}$, because that formula assumes that all energy flux is either thermal energy or work. (The real definition of $\eta$ is $\eta \equiv W/Q_{Absorbed}$, so you can see how that works.)
3. Same as #2, but now the ideal gas container has sound-proof walls.

• Now the waves get dissipated into thermal energy within the ideal gas. So this is actually just a special case of #1. Remember, when the reservoir is much hotter than the gas, entropy is getting created. As part of this entropy-creation process, you might or might not have processes going on like the formation and dissipation of sound waves.

First, for an isochoric process, $Q=nc_v\Delta T$ is not true in general, because you can increase internal energy of a system by other means (paddle work, for example). Let us assume that there is no such process involved in your case, so indeed $Q=nc_v\Delta T$ for isochoric process. You are right in saying that this relation holds whether or not the isochoric heating process is reversible.

The trouble comes with the isobaric process. For a reversible process, first law may be written as $Q=\Delta H-\int_1^2 dp~ V$, for change from state 1 to 2, and $\Delta H=nc_p\Delta T$ is the enthalpy change. For a reversible process, pressure $p$ of the system is well defined at every point along the process path taken by the system, so the integral is meaningful and can be evaluated. If the path is such that $p=$constant, then we have $Q=nc_p\Delta T$.

But for an irreversible process from state 1 to 2, even though pressure at initial and final states are well defined because they are equilibrium states, during the intermediate stage, intensive thermodynamic quantities such as pressure, temperature etc. are not well defined, because here the system is not in an equilibrium state. So you will have to just remain with the more primitive form of first law i.e. $Q=\Delta U+W$, and you cannot conclude $Q=nc_p\Delta T$ just because pressures are equal at initial and final state, because work done is no more given by $\int_1^2 dV~ p$ (in fact there is no continuous path in $p-V$ diagram over which the integral may be defined).

Another minor point. Carnot theorem is applicable to an engine working between only two heat reservoirs (at different temperatures). In your case there are several.

On the other hand this does not makes sense, because Carnot theorem requires that in case 1 (in wich all processes are reversible) the efficiency is the same of a Carnot engine working betweeen highest and lowest temperature (in this case $T_{D}$ and $T_{B}$), that is, in case 1., efficiency should be $$\eta_{\mathrm{reversible}}=1-\frac{T_D}{T_B}\tag{II}$$

You are forgetting a really important point. Carnot's theorem states that:

• All heat engines between two heat reservoirs are less efficient than a Carnot heat engine operating between the same reservoirs.
• Every Carnot heat engine between a pair of heat reservoirs is equally efficient, regardless of the working substance employed or the operation details.

The formula for the maximum efficiency of an engine between two heat reservoirs is

$$\eta_{max} = 1-\frac{T_C}{T_H}$$ where $T_C<T_H.$

Carnot's theorem is about engines working between two heat reservoirs: two reservoirs, no more, no less.

The only way to make an engine work between two reservoirs is to perform two isothermal transformations and two adiabatic transformations: any other transformation will introduce other heat sources in your engine. For example in your case you have a non-adiabatic transformations cutting isotherms in the $PV$ plane, thus exchanging heat with the external environment.

See picture below: every color represent a different isotherm, i.e. a different reservoir you are exchanging heat with.

Carnot's cycle only exchanges heat with two reservoirs (the curves crossing the orange isotherm are adiabatic, so that $\delta Q=0$ for those):

Remember than the definition of the efficiency is the ratio between the work "extracted" or transferred to the machine and the heat taken from the hot source.

$\eta=\frac{W}{Q_h}$

For a reversible process this expression is equal to

$\eta=1+\frac{Q_c}{Q_h}$

But for an irreversible process it is not. The reason is that the work done by the gas, is no longer equal to the work transferred to the machine. For instance, imagine your gas doing a quasistatic irreversible process, in which it pushes a piston with friction. The gas will still make a work $W$ as it expands, but the work received by the rest of the machine, $W'$, will be diminished by the amount of heat dissipated by the friction. Thus, instead of $W$ in the first equation you must use $W'$, and the rest of the equalities will no longer follow.

• Thanks for the answer! I report the formula for efficiency of thermal engine as it is on textbook with the sum of heats of each single process $_i$ in the cycle $$\eta=1-\frac{|\sum Q_{\mathrm{EMITTED},i}|}{\sum Q_{ABSORBED,i}}$$ ($Q_{\mathrm{EMITTED},i}$ is negative, so there is the absolute value and minus sign). And this holds true for any thermal engine (reversible or not). Commented Jul 11, 2016 at 12:58
• I understand what you mean in the example with friction but here I'm just considering the cycle in the picture, whithout looking at how it is done in reality and how much work is actually taken from engine. Suppose that all the work can be taken from it, so $W'=W=\sum Q_i$. What I would like to consider is the case in which the cycle is reversible and the case in wich it is not. What bothers me is that, in the case of isochoric or isobarics, efficiency, according to previous definition, seems not to change between the two cases. Is there a way to understand this? Commented Jul 11, 2016 at 13:00
• yes, that is the part where we disagree, that you cannot watch just the gas cycle, but I might be wrong, it is an interesting question and I hope there will be more posted answers
– user65081
Commented Jul 11, 2016 at 15:47

Infact during any isochoric and isobaric processes (reversible or not) $$Q_{\mathrm{isochoric}}=n c_v \Delta T$$ $$Q_{\mathrm{isobaric}}=n c_p \Delta T$$

I think this is wrong. Because $Q$ is a path function. True formula (I think) is as below: $$\Delta U=n c_v \Delta T$$ And it is same for reversible and irreversible processes (if $\Delta T$ is same).

For calculating $Q$ you should use the first law of thermodynamics $$\Delta U=Q-W\;\Longrightarrow\;Q=\Delta U+W$$ And $W$ is different for reversible and irreversible processes.

From "THERMODYNAMICS An Engineering Approach, Fifth Edition, by YUNUS A. CENGEL and MICHAEL A. BOLES"