
#1
Jan707, 10:42 AM

P: 79

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 [tex] \epsilon =1\frac{T_L}{T_H} [/tex] 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 [tex] \Delta S = nR ln \frac{V_f}{V_i} + nC_V ln \frac{T_f}{T_i} [/tex] we can know that the entropy change in the two isochoric process of Stirling cycle canceled out. and still [tex] \frac{Q_H}{T_H} = \frac{Q_L}{T_H} [/tex] just like Carnot Cycle So I think efficiency [tex] \epsilon=1\frac{T_L}{T_H} [/tex] can apply to Stirling Am I wrong?? 



#2
Jan707, 01:14 PM

Sci Advisor
HW Helper
P: 6,595

Efficiency is work/heat flow: [itex]\eta = W/Q_h[/itex]. Because W = Qh  Qc [itex]\eta = 1  Q_c/Q_h[/itex] Since the heat flow into the Carnot engine is isothermal: [itex]\int dS = \int dQ/T = Q_h/T_h[/itex] and for the heat flow out: [itex]\int dS = Q_c/T_c[/itex]. If the change in entropy is 0: [tex]Q_c/T_c = Q_h/T_h[/tex] or [tex]Q_c/Q_h = T_c/T_h[/tex] If [itex]\Delta S \ne 0[/itex] then you cannot equate Tc/Th to Qc/Qh. AM 



#3
Jan707, 05:05 PM

P: 79

But isn't [tex]\Delta S=0 [/tex] 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?? 



#4
Jan707, 08:26 PM

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P: 6,595

Carnot and Stirling engineThe 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 



#5
Jan807, 07:48 AM

P: 79

I see. thanks



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