Modern Engineering Thermodynamics

13.13 Stirling Cycle 489
Q H
4
1
4
T H = c
T
p
1
3 2
Q L
s
(a) T–s diagram
3 T L = c
2
V
(b) p–V diagram
Q regen
3 2
4 1
Constant
volume
regenerator
Q Q L H
Displacement piston
(c) Equipment schematic
Power piston
FIGURE 13.38
The Stirling cycle.
These entropy changes can be evaluated for the cold ASC using Eq. (7.36) of Chapter 7 as
and
s 1 − s 4 = c v ln ðT 1 /T 4 Þ+ R ln ðv 1 /v 4 Þ = R ln ðv 1 /v 4 Þ
s 2 − s 3 = c v ln ðT 2 /T 3 Þ+ R ln ðv 2 /v 3 Þ = R ln ðv 2 /v 3 Þ
Figures 13.38a and 13.38b show that T 2 = T 3 = T L , T 1 = T 4 = T H , and v 2 /v 3 = v 1 /v 4 , so that we have s 1 – s 4 = s 2 – s 3
and the preceding equations give
ðη T Þ Stirling
= 1 − T L /T H = 1 − T 2 /T 1 = 1 − T 3 /T 4 (13.19)
cold ASC
which is the same as the thermal efficiency of the Carnot cold ASC operating between the same two temperature
limits.
EXAMPLE 13.9
A new Stirling cycle engine using air as the working fluid is being designed
to power the small portable electric generator shown Figure 13.39. The
piston displacement is to be 0.0110 m 3 and the minimum volume in the
cylinder is to be V 3 = V 4 = 1.00 × 10 –3 m 3 . The design calls for a maximum
power piston pressure of p 1 = 0.300 MPa, a minimum displaced
piston pressure of p 2 = 0.100 MPa, and a displacer inlet temperature of
T 2 = 30.0°C. Use the Stirling cold ASC analysis shown in Figure 13.38 to
complete the design, determining
a. The displacer piston maximum pressure (p 3 ).
b. The power piston maximum pressure (p 4 ).
c. The mass of air in the engine (m).
d. The heat addition temperature (T 1 = T 4 ).
e. The Stirling cold ASC thermal efficiency of the engine.
FIGURE 13.39
Example 13.9.
(Continued )