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Physics
Classical Physics
Thermodynamics
The entropy of a Carnot cycle and the efficiency equation
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[QUOTE="Chestermiller, post: 6005678, member: 345636"] 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 [I]transfer of entropy[/I] 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 [URL='https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/']change in entropy[/URL] 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 [I]entropy generated[/I] 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. [/QUOTE]
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Physics
Classical Physics
Thermodynamics
The entropy of a Carnot cycle and the efficiency equation
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