please someone tell me about the relationship of the reversible and irreversible nature of an engine and how is it related to its change in entropy

  • $\begingroup$ Are you asking about the change in entropy of the working fluid in an engine which undergoes a cycle? $\endgroup$ Aug 23 at 16:28
  • $\begingroup$ @ChetMiller , if there is a numerical based on a cyclic heat engine that receives a certain amount of heat from a reservoir , does a certain amount of work and rejects some amount of heat to the sink , given the heat , work done and temperature of the reservoir , surroundings and the sink how can we determine if is reversible , irreversible or impossible $\endgroup$ Aug 23 at 16:54
  • $\begingroup$ I don't understand what you mean by a "numerical." $\endgroup$ Aug 23 at 16:56
  • $\begingroup$ What are your thoughts on the answer to this question? $\endgroup$ Aug 23 at 16:58
  • $\begingroup$ @ChetMiller , it has something to do with entropy i guess , but im not sure how to relate them... $\endgroup$ Aug 23 at 17:03

please someone tell me about the relationship of the reversible and irreversible nature of an engine

A reversible engine is one that operates so slowly (quasi-statically) and without friction, that its working fluid is always in thermal and mechanical equilibrium with its surroundings, as applicable to the process(es) involved.

An irreversible engine does not operate slowly (quasi-statically) and/or does involve friction. The working fluid in an irreversible engine is not in thermal and/or mechanical equilibrium with its surroundings.

and how is it related to its change in entropy

An irreversible engine generates entropy. A reversible engine does not generate entropy, but can transfer entropy by transferring heat reversibly (e.g. a reversible isothermal process).

The total entropy change of the system between two different equilibrium states is the sum of the entropy transferred and entropy generated (if any) or

$$\Delta S_{sys}=S_{2}-S_{1}=\int_1^2 \frac{dQ}{T_B}+\sigma$$

Where the first term on the right is the entropy tansferred, $T_B$ is the temperature at the boundary (or interface) between the system and surroundings where heat transfer $dQ$ is occurring, and $\sigma$ is the entropy generated by an irreversible engine where $\sigma>0$.

For a reversible engine, $\sigma=0$, $T_B$ is the equilibrium temperature of the system, $T_{sys}$, and $dQ$ is a reversible transfer of heat.

Since entropy is a state function that does not depend on the process, the total entropy change of a system between two states for an irreversible process can be calculated by assuming any convenient reversible process between the same two states. The assumed process need not bear any resemblance to the actual process as long as the end states are the same. For the assumed process one then applies the equation

$$\Delta S_{sys}=S_{2}-S_{1}=\int_1^2\frac{dQ_{rev}}{T_{sys}}$$

Where $dQ_{rev}$ is a reversible transfer of heat (heat transfer where the temperature difference between the working fluid and surroundings is infinitesimal).

Combining the two equations

$$\Delta S_{sys}=\int_1^2\frac{dQ_{rev}}{T_{sys}}=\int_1^2 \frac{dQ}{T_B}+\sigma=S_{tran}+S_{gen}$$

In the case of a cycle, the change in entropy of the engine working fluid is always zero, regardless of whether it is a reversible or irreversible engine. That's because entropy is a state function having a unique value for every equilibrium state. Since a cycle begins and ends at the same equilibrium state, the entropy change of the system is necessarily zero for any cycle regardless of the process(es), i.e., $\Delta S_{sys}=0$.

If the engine is reversible, then the entropy change of the surroundings in the cycle will also be zero. If the engine is irreversible, the entropy generated by the engine is tansferred to the surroundings in the form of heat for a closed system, and/or mass transfer for an open system for an increase in the entropy of the surroundings.

Hope this helps.

  • $\begingroup$ thank you , the explanation was very helpful $\endgroup$ Aug 24 at 3:28

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