Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Both the Stirling and Ericsson cycles are totally reversible, as is the Carnot

cycle, and thus according to the Carnot principle, all three cycles must have

the same thermal efficiency when operating between the same temperature

limits:

(9–14)

This is proved for the Carnot cycle in Example 9–1 and can be proved in a

similar manner for both the Stirling and Ericsson cycles.

hth,Stirling
hth,Ericsson
hth,Carnot
 1 

TL

TH

506 | Thermodynamics

EXAMPLE 9–4 Thermal Efficiency of the Ericsson Cycle

Using an ideal gas as the working fluid, show that the thermal efficiency of

an Ericsson cycle is identical to the efficiency of a Carnot cycle operating

between the same temperature limits.

Solution It is to be shown that the thermal efficiencies of Carnot and

Ericsson cycles are identical.

Analysis Heat is transferred to the working fluid isothermally from an external

source at temperature THduring process 1-2, and it is rejected again isother-

mally to an external sink at temperature TL during process 3-4. For a

reversible isothermal process, heat transfer is related to the entropy change by

The entropy change of an ideal gas during an isothermal process is

The heat input and heat output can be expressed as

and

Then the thermal efficiency of the Ericsson cycle becomes

since P 1 
P 4 and P 3 
 P 2. Notice that this result is independent of

whether the cycle is executed in a closed or steady-flow system.

hth,Ericsson
 1 

qout

qin

 1 

RTL ln 1 P 4 >P 32

RTH ln 1 P 1 >P 22

 1 

TL

TH

qout
TL 1 s 4 s 32 
TLaR ln

P 4

P 3

b
RTL ln

P 4

P 3

qin
TH 1 s 2 s 12 
THaR ln

P 2

P 1

b
RTH ln

P 1

P 2

¢s
cp ln

Te

Ti

R ln

Pe

Pi

R ln

Pe

Pi

q
T ¢s

Stirling and Ericsson cycles are difficult to achieve in practice because

they involve heat transfer through a differential temperature difference in all

components including the regenerator. This would require providing infi-

nitely large surface areas for heat transfer or allowing an infinitely long time

for the process. Neither is practical. In reality, all heat transfer processes take

place through a finite temperature difference, the regenerator does not have

an efficiency of 100 percent, and the pressure losses in the regenerator are

considerable. Because of these limitations, both Stirling and Ericsson cycles

0

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