2
$\begingroup$

In a Carnot engine the net entropy changein a cycle is zero. But in an irreversible engine operating between two temperatures the net entropy change in a cycle is positive. As I have understood, this means the irreversible engine tends to lose more heat at lower temperature than the Carnot engine. Why is it so?

$\endgroup$
  • $\begingroup$ Can you please elaborate the explanation $\endgroup$ – nayana v Feb 5 '16 at 10:43
1
$\begingroup$

Net entropy change means entropy change of the world (world means system plus environment). Carnot cycle is a reversible cycle. For a reversible cycle, world entropy change is zero. Because both of system and environment return to their initial states when cycle is completed. But, for an irreversible cycle, world entropy change (net entropy change) isn't equal to zero. Although the system returns to its initial state, but the environment doesn't. So, net entropy change won't be zero and according to Increase in Entropy Principle, it will be positive.

$\endgroup$
0
$\begingroup$

If the engine is operating in a cycle, the change in entropy must be zero, since entropy is a function of state. However, the heat taken in from the hot reservoir divided by the temperature of the hot reservoir must be less than the heat released to the cold reservoir divided by the temperature of the cold reservoir. This means that the net entropy entering the system during each cycle is negative, and equal in magnitude to the entropy generated within the system during each cycle. So the net change in entropy over a cycle is zero.

$\endgroup$
  • $\begingroup$ This answer is a little confusing, as you seem to be saying that net change in entropy is zero during an irreversible cycle, which is not true. Alternatively, it maybe seems like you are just repeating the original question. Can you expand your answer a little? $\endgroup$ – march Feb 4 '16 at 4:37
  • $\begingroup$ Yes, that is exactly what I am saying, and it is true for the engine and its contents experiencing the cycle. If the engine and its contents return to its initial state after any cycle is complete (either reversible or irreversible), then its entropy must be the same as at the start of the cycle. However, the change in entropy of the engine plus it's surroundings is not the same, and increases during each irreversible cycle. So the entropy of the surroundings must be increasing. The entropy is being generated within the system but is being transferred to the surroundings. $\endgroup$ – Chet Miller Feb 4 '16 at 12:37
  • $\begingroup$ Right. It's a little unclear that you are saying the engine entropy change is zero while the net entropy of the surroundings increases. I think the OP is asking why irreversibility requires this. Specifically, the OP is wondering why that entails more heat expelled than in the Carnot cycle. $\endgroup$ – march Feb 4 '16 at 15:10
  • $\begingroup$ It doesn't necessarily mean that more heat is expelled than in a Carnot cycle. The change can occur partly at the heat intake part of the cycle as well. Why don't you do a simple model calculation (analysis) in which the isothermal expansion and compression legs occur irreversibly at constant imposed pressure, and the remainder of the cycle is done adiabatically and reversibly to see how it all plays out? Anyhow, we haven't heard back from the OP yet, so we don't know whether his/her question has actually been answered yet. $\endgroup$ – Chet Miller Feb 4 '16 at 15:23
  • $\begingroup$ Sure, I'm aware of all of that. I guess I'm saying that it seems to me that the OP is asking about that very thing (heat expelled being larger, which is not necessary), whereas you seem to have answered a different question. But you're right: the OP hasn't commented, so perhaps I shouldn't be so pushy. $\endgroup$ – march Feb 4 '16 at 17:16
-1
$\begingroup$

In any process total entropy can not decrease. It's s second law of thermodynamics. But still why is it so? I can not give a good explanation. Any system consists of atoms, photons, etc., these particles move around, collide. Motion of each of them is described by laws of mechanics. It turns out that laws of mechanics have a very interesting property: thermodynamics laws can be deduced from them. It's very interesting because at the first glance it looks like they do not have anything in common!

So, the entropy can not decrease. It can only increase or remain the same. That immediately means that at reversible process entropy remains the same. (if it increases we could reverse the process and decrease it).

If during some process entropy increases this process must be irreversible. Otherwise we would reverse it and decrease the entropy.

"Reversible" and "entropy remains the same" means almost the same. Almost. There remains a possibility that there could be some process such that the entropy during this process remains constant, but it's still irreversible. I can not think of an example of such a process.

In you question you mention some irreversible engine and ask, why the net entropy increases. Hmm. That depends on type of engine!

In ideal Carnot engine gas contacts heater and gets energy from it when both gas and heater have same temperature. But the less is temperature difference the slower is the process of transferring the energy. In not ideal Carnot engine the gas is slightly cooler than the heater. Heat is comes from heater to a slightly cooler gas - and here the entropy increases. Same situation is when gas is cooled by cooler: temperature of gas must be higher that the temperature of cooler, heat transfers from a gas to cooler ans entropy increases.

$\endgroup$
  • $\begingroup$ How can entropy increase (I also want to know why does efficiency of engine decrease) if there is a finite difference in temperature of gas and the heater(and cooler) $\endgroup$ – nayana v Feb 5 '16 at 10:51
  • $\begingroup$ When some small amount of heat $Q$ is transferred away from a body at temperature $T$ it's entropy decreases by $Q/T$. But the same amount of heat come to a cooler body and it's entropy increases by $Q/T1$. Total difference is positive as $T1<T$. $\endgroup$ – lesnik Feb 5 '16 at 19:15
  • $\begingroup$ Now to your question about efficiency. The amount of work the engine produces in one cycle is the area of the cycle on PV diagram. But for not ideal Carnot engine the area would be smaller! Because the temperature (and hence the pressure) of the gas when it contacts the heater is less than in ideal case, and when gas contacts the cooler it's higher then in ideal case. $\endgroup$ – lesnik Feb 5 '16 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.