Editing Carnot heat engine (section)

== Efficiency of real heat engines ==
For a real heat engine, the total thermodynamic process is generally irreversible. The working fluid is brought back to its initial state after one cycle, and thus the change of entropy of the fluid system is 0, but the sum of the entropy changes in the hot and cold reservoir in this one cyclical process is greater than 0.

The internal energy of the fluid is also a state variable, so its total change in one cycle is 0. So the total work done by the system {{mvar|W}} is equal to the net heat put into the system, the sum of <math> Q_\text{H} </math> > 0 taken up and the waste heat <math> Q_\text{C} </math> < 0 given off:<ref name="PlanckBook">{{cite book |last=Planck |first=M. |title=Treatise on Thermodynamics |page=90 |quote=§90, eqs.(39) & (40) |publisher=Dover Publications |year=1945}}</ref>

{{NumBlk2|:|<math> W=Q=Q_\text{H}+Q_\text{C} </math> |LnSty=1px dashed |2}}

For real engines, stages 1 and 3 of the Carnot cycle, in which heat is absorbed by the "working fluid" from the hot reservoir, and released by it to the cold reservoir, respectively, no longer remain ideally reversible, and there is a temperature differential between the temperature of the reservoir and the temperature of the fluid while heat exchange takes place.

During heat transfer from the hot reservoir at <math>T_\text{H}</math> to the fluid, the fluid would have a slightly lower temperature than <math>T_\text{H}</math>, and the process for the fluid may not necessarily remain isothermal. 
Let <math>\Delta S_\text{H}</math> be the total entropy change of the fluid in the process of intake of heat.

{{NumBlk2|:|<math>\Delta S_\text{H}=\int_{Q_\text{in}} \frac{\text{d}Q_\text{H}}{T} </math> |LnSty=1px dashed |3}}

where the temperature of the fluid {{mvar|T}} is always slightly lesser than <math>T_\text{H}</math>, in this process.

So, one would get:

{{NumBlk2|:|<math> \frac{Q_\text{H}}{T_\text{H}}=\frac{\int \text{d}Q_\text{H}}{T_\text{H}} \leq \Delta S_\text{H} </math> |LnSty=1px dashed |4}}

Similarly, at the time of heat injection from the fluid to the cold reservoir one would have, for the magnitude of total entropy change <math> \Delta S_\text{C} </math>< 0 of the fluid in the process of expelling heat:

{{NumBlk2|:|<math> \Delta S_\text{C}\geqslant\frac{Q_\text{C}}{T_\text{C}}< 0 </math> |LnSty=1px dashed |5}}

where, during this process of transfer of heat to the cold reservoir, the temperature of the fluid {{mvar|T}} is always slightly greater than <math>T_\text{C}</math>.

We have only considered the magnitude of the entropy change here. Since the total change of entropy of the fluid system for the cyclic process is 0, we must have

{{NumBlk2|:|<math> \Delta S_\text{H}+\Delta S_\text{C} = \Delta S_\text{cycle} = 0</math> |LnSty=1px dashed |6}}

The previous three equations, namely ({{EquationNote|3}}), ({{EquationNote|4}}), ({{EquationNote|5}}), substituted into ({{EquationNote|6}}) to give:<ref name="FermiBook">{{cite book |last=Fermi |first=E. |title=Thermodynamics |page=47 |quote=below eq.(63) |publisher=Dover Publications (still in print) |year=1956}}</ref>

{{NumBlk2|:|<math>-\frac{Q_\text{C}}{T_\text{C}}\geqslant\frac{Q_\text{H}}{T_\text{H}}</math> |LnSty=1px dashed |7}}

{{incomprehensible|section|date=January 2024}}<!--In what follows ≤ and ≥ are treated as operators (binary functions on numbers), but they're relations (propositions about pairs of numbers). It doesn't make sense to add their "results" together, negate them, or compare them to 0. They're also not formatted properly with {{math}} or <math>-->
For  [ΔSh ≥ (Qh/Th)] +[ΔSc ≥ (Qc/Tc)] = 0

[ΔSh ≥ '''(Qh/Th)'''] = - [ΔSc ≥ (Qc/Tc)]

<nowiki>=</nowiki> [-ΔSc ''≤'' '''(-Qc/Tc)''']

it is at least '''(Qh/Th) ''≤'' (-Qc/Tc)'''

Equations ({{EquationNote|2}}) and ({{EquationNote|7}}) combine to give

{{NumBlk2|:|<math> \frac{W}{Q_\text{H}} \leq 1- \frac{T_\text{C}}{T_\text{H}} </math> |LnSty=1px dashed |8}}

To derive this step needs two adiabatic processes involved to show an isentropic process property for the ratio of the changing volumes of two isothermal processes are equal. 

Most importantly, since the two adiabatic processes are volume works without heat lost, and since the ratio of volume changes for this two processes are the same, so the works for these two adiabatic processes are the same with opposite direction to each other, namely, one direction is work done by the system and the other is work done on the system; therefore, heat efficiency only concerns the amount of work done by the heat absorbed comparing to the amount of heat absorbed by the system.

Therefore, (W/Qh) = (Qh - Qc) / Qh

<nowiki>=</nowiki> 1 - (Qc/Qh)

<nowiki>=</nowiki> 1 - (Tc/Th)

And, from ({{EquationNote|7}})

'''(Qh/Th) ''≤'' (-Qc/Tc)'''                      here Qc it is less than 0 (release heat) 

'''(Tc/Th) ''≤'' (-Qc/Qh)'''

'''-(Tc/Th) ≥ (Qc/Qh)'''

'''''1+''[-(Tc/Th)] ≥ ''1+''(Qc/Qh)'''    

1 - (Tc/Th) '''≥ (Qh + Qc)/Qh'''         here Qc<0, 

1 - (Tc/Th) '''≥ (Qh - Qc)/Qh'''

1 - (Tc/Th) '''≥ W/Qh'''

Hence,

{{NumBlk2|:|<math> \eta \leq \eta_\text{I} </math> |LnSty=1px dashed |9}}

where <math>\eta = \frac{W}{Q_\text{H}}</math> is the efficiency of the real engine, and <math>\eta_\text{I}</math> is the efficiency of the Carnot engine working between the same two reservoirs at the temperatures <math>T_\text{H}</math> and <math>T_\text{C}</math>. For the Carnot engine, the entire process is 'reversible', and Equation ({{EquationNote|7}}) is an equality. Hence, the efficiency of the real engine is always less than the ideal Carnot engine.

Equation ({{EquationNote|7}}) signifies that the total entropy of system and surroundings (the fluid and the two reservoirs) increases for the real engine, because (in a surroundings-based analysis) the entropy gain of the cold reservoir as <math>Q_\text{C}</math> flows into it at the fixed temperature <math>T_\text{C}</math>, is greater than the entropy loss of the hot reservoir as <math>Q_\text{H}</math> leaves it at its fixed temperature <math>T_\text{H}</math>. The inequality in Equation ({{EquationNote|7}}) is essentially the statement of the [[Clausius theorem]].

According to the second theorem, "The efficiency of the Carnot engine is independent of the nature of the working substance".