The entropy of a Carnot cycle and the efficiency equation

  • #1
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I have my first question. It's about entropy in the Carnot cycle and I'll try to be direct.

The equal sign in the Carnot cycle efficiency equation is related to the fact that the total entropy doesn't change at the end of the whole cycle (being related to the fact that the heat exchanges occur in stages in which the auxiliary system is at same temperature that the respective thermal reservoir)? In other words, if the auxiliary system were at a different temperature from the hot reservoir, for example, would there be an additional growing of the entropy in the whole system, collaborating for the decreasing of the efficiency?

Thank you
 

Answers and Replies

  • #2
kith
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If you look at the formula for the change of entropy, [itex]\Delta S = \frac{Q}{T}[/itex], you can observe that the decrease in entropy in one body can only be exactly equal to the increase in entropy of the other body if both bodies have the same temperature [itex]T[/itex] as the heat [itex]Q[/itex] is flowing from one to the other. So yes, for maximum efficiency, the temperature of the working substance needs to be the same as the temperature of the reservoirs when heat is transferred.

(But since the spontaneous heat flow between two bodies is driven by a difference in temperature, reversible isothermal processes are approximations anyway.)
 
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  • #3
Chandra Prayaga
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If you are familiar with calculus, I refer you to "Concepts in Thermal Physics" by Blundell and Blundell, 2nd Ed, Ch 14, Ex 14.1, where the entropy changes in the reservoir and the system are calculated for the case where the starting temperatures are different.
 
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  • #4
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There are two types of irreversibility present in an irreversible Carnot cycle which contribute to entropy generation within the working fluid: finite temperature gradients associated with finite differences between the reservoir temperatures and the average working fluid temperature (the working fluid temperature is not spatially uniform) and finite velocity gradients associated with rapid deformation of the working fluid and related viscous dissipation of mechanical energy. There are also two mechanisms whereby the entropy of a working fluid can vary: entropy generation within the working fluid, and entropy transfer between the working fluid and the ideal reservoirs, via heat transfer at the boundary between the working fluid and the reservoirs.

In the case of an irreversible Carnot cycle, the transfer of entropy between the working fluid and the ideal reservoirs is given by ##\frac{Q_H}{T_H}-\frac{Q_C}{T_C}##, where ##Q_H## is the heat transferred from the hot reservoir to the working fluid at the hot reservoir boundary and ##Q_C## is the heat transferred from the working fluid to the cold reservoir at the cold reservoir boundary. In a cycle, the change in entropy of the working fluid over each cycle must be zero. Therefore, we must have that:
$$\Delta S=\frac{Q_H}{T_H}-\frac{Q_C}{T_C}+\sigma = 0$$where ##\sigma## is the entropy generated within the working fluid per cycle (as a result of finite temperature gradients and velocity gradients). In line with the 2nd law of thermodynamics, ##\sigma## must always be positive. The work done by the working fluid in the cycle is given by: $$W=Q_H-Q_C$$
Combining these equations yields the following for the efficiency of the irreversible Carnot cycle:
$$\eta=\frac{W}{Q_H}=\left(1-\frac{T_C}{T_H}\right)-\frac{\sigma T_C}{Q_H}$$The term in parenthesis on the right hand side is equal to the efficiency of the reversible Carnot cycle. The second term involving the irreversible entropy generation within the working fluid results in a reduction of the efficiency below that of the reversible Carnot cycle.
 
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  • #5
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If you are familiar with calculus, I refer you to "Concepts in Thermal Physics" by Blundell and Blundell, 2nd Ed, Ch 14, Ex 14.1, where the entropy changes in the reservoir and the system are calculated for the case where the starting temperatures are different.
Thank you, I checked the example, it's very interesting.
 
  • #6
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Thank you all, was very helpful to me and nice to understand.
 

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