Why Might the Efficiency of a Stirling Engine Differ from a Carnot Engine?

[Psi-String]

Halliday says that the efficiency of an ideal Stirling engine is lower than that of a ideal Carnot engine?? But why??

It seems to me that there efficiency are both $$\epsilon =1-\frac{T_L}{T_H}$$

Though Halldiay also say that this equation do not apply to Stirning engine but only to Carnot

Though Stirling engine involves two isochoric process, so unlike carnot which the two isothermal process are connected by adiabatic process, the entropy do change between the two temperature in the Stirling Cycle.

But from $$\Delta S = nR ln \frac{V_f}{V_i} + nC_V ln \frac{T_f}{T_i}$$

we can know that the entropy change in the two isochoric process of Stirling cycle canceled out. and still

$$\frac{|Q_H|}{T_H} = \frac{|Q_L}{T_H}$$ just like Carnot Cycle

So I think efficiency $$\epsilon=1-\frac{T_L}{T_H}$$ can apply to Stirling


Am I wrong??

 

[Andrew Mason]

Halliday says that the efficiency of an ideal Stirling engine is lower than that of a ideal Carnot engine?? But why??

It seems to me that there efficiency are both $$\epsilon =1-\frac{T_L}{T_H}$$

Though Halldiay also say that this equation do not apply to Stirning engine but only to Carnot

Though Stirling engine involves two isochoric process, so unlike carnot which the two isothermal process are connected by adiabatic process, the entropy do change between the two temperature in the Stirling Cycle.

But from $$\Delta S = nR ln \frac{V_f}{V_i} + nC_V ln \frac{T_f}{T_i}$$

we can know that the entropy change in the two isochoric process of Stirling cycle canceled out. and still

$$\frac{|Q_H|}{T_H} = \frac{|Q_L}{T_H}$$ just like Carnot Cycle

So I think efficiency $$\epsilon=1-\frac{T_L}{T_H}$$ can apply to Stirling

Am I wrong??

Yes. The efficiency of a Carnot engine is $\epsilon=1-\frac{T_L}{T_H}$ because $\Delta S = 0$

Efficiency is work/heat flow: $\eta = W/Q_h$.
Because W = Qh - Qc $\eta = 1 - Q_c/Q_h$

Since the heat flow into the Carnot engine is isothermal: $\int dS = \int dQ/T = Q_h/T_h$ and for the heat flow out: $\int dS = Q_c/T_c$. If the change in entropy is 0:

$$Q_c/T_c = Q_h/T_h$$ or $$Q_c/Q_h = T_c/T_h$$

If $\Delta S \ne 0$ then you cannot equate Tc/Th to Qc/Qh.

AM

 

[Psi-String]

But isn't $$\Delta S=0$$ for Stirling cycle?
Stirling engine is a closed path, doesn't that also imply that the entropy change in one stirling cycle is also zero??

 

[Andrew Mason]

But isn't $$\Delta S=0$$ for Stirling cycle?
Stirling engine is a closed path, doesn't that also imply that the entropy change in one stirling cycle is also zero??

No. The Stirling engine is a real engine. If you store the work output you cannot run the cylcle in reverse to get back to the original conditions without adding more work than it produced on the forward cycle. Since it is not reversible, $\Delta S \ne 0$

The Carnot is reversible. It operates at constant thermodynamic equilibirum using infinitessimal temperature differerences and slow adiabatic compressions and expansions. If you store the work output, you can use that work to reverse the cycle by an infinitessimal change in conditions.

AM

 

[Psi-String]

I see. thanks