Stirling Cycle

1
Stirling Cycle The following diagram describes
the Stirling cycle
( P1 V41 TH )
( P2 V23 TH )
1
q41 ? E41 gt 0

q12 - w12 gt 0

4
q23 ? E23 lt 0

2
q34 - w34 lt 0

3
( P4 V41 TL )
( P3 V23 TL )
2
The Stirling Cycle can be described as follows,
if we assume the working fluid to be an ideal gas
and each step to be carried out reversibly Step
12 In this step heat is absorbed by the gas
through the bottom plate and the gas expands
reversibly and isothermally at TH from V41 to
V23. Note the expansion occurs in the smaller
power cylinder. The expansion in the power
cylinder causes a flywheel to turn driving the
larger displacing piston down and forcing the
warmer gas above the displacing piston where it
comes into thermal contact with the cooler top
plate. Note the internal movement of the larger
displacing piston does not change the volume of
the gas. Because ? E 0 for a reversible
isothermal expansion of an ideal gas, the heat
absorbed by the gas is just balanced by the
expansion work done by the gas ? E12 0
q12 w12 q12 - w12 - - n R TH ln (
V23 / V41 ) gt 0 Step 23 The gas, now in
thermal contact with the cooler top plate, cools.
In this step heat is lost by the gas through the
top plate and the gas cools at constant volume
from TH to TL. Because the work in a constant
volume process is zero w23 - ? P
dV 0 the heat lost by the gas equates to a
decrease in the internal energy for the gas and
consequently a decrease in the gas
temperature q23 ? E23 n CV, m ( TL -
TH ) lt 0
3
Step 34 In this step the lower resulting
pressure of the cooler gas results in piston in
the power cylinder moving down compressing the
gas reversibly and isothermally at TL from V23 to
V41. The compression in the power cylinder causes
the flywheel to turn driving the larger
displacing piston up and forcing the cooler gas
to move below the displacing piston where it
comes into thermal contact with the warmer bottom
plate. Because ? E 0 for a reversible
isothermal compression of an ideal gas, the
compression work done on the gas is just balanced
by additional heat lost through the cooler top
plate ? E34 0 q34 w34 q34 -
w34 - - n R TL ln ( V41 / V23 ) lt
0 Step 41 The gas, now in thermal contact with
the warmer bottom plate, warms. In this step
heat is absorbed by the gas through the bottom
plate and the gas warms at constant volume from
TL to TH, completing the cycle. Again because
the work in a constant volume process is
zero w41 - ? P dV 0 the heat absorbed
by the gas equates to an increase in the internal
energy for the gas and consequently an increase
in the gas temperature q41 ? E41 n CV, m
( TH - TL ) gt 0 Do you expect the net work
done in one cycle of this reversible Stirling
engine to be positive or negative?
4
The net work done during one cycle of the
reversible Stirling engine just described is the
sum of the work in each of the four steps wnet
w12 w23 w34 w41 -
n R TH ln ( V23 / V41 ) 0 - n R TL ln ( V41
/ V23 ) 0 - n R TH ln ( V23 /
V41 ) 0 n R TL ln ( V23 / V41 )
- n R ln ( V23 / V41 ) (TH - TL ) For a
reversible Stirling engine operating between 20.0
C and 40.0 C and 2.00 cm3 and 5.00 cm3 and
utilizing 6.23 10-5 mols of an ideal gas, the net
work done will be equal to wnet - n
R ln ( V23 / V41 ) (TH - TL ) - (1.00 10-4
mols) (8.314 J/mol K) x ln ( 5.00 cm3 /
2.00 cm3 ) ( 313.2 K - 293.2 K ) - 0.00946
J This is 1/100,000 of the energy in one
match! How is the efficiency of the Stirling
cycle related to TH and TL?