Modern Engineering Thermodynamics

492 CHAPTER 13: Vapor and Gas Power Cycles
Figure 13.40b shows that T 2 = T 3 = T L , T 1 = T 4 = T H ,andp 1 /p 4 = p 2 /p 3 , so that the previous equations yields
s 1 – s 4 = s 2 – s 3 and the thermal efficiency becomes
ðη T Þ Ericsson
= 1 − T L /T H = 1 − T 2 /T 1 = 1 − T 3 /T 4 (13.20)
cold ASC
Thus, like the Stirling cold ASC, the Ericsson cold ASC has the same thermal efficiency as the Carnot cold ASC
operating between the same two temperature limits. The Stirling and Ericsson cycle engines of the 19th century
were large and costly, and their actual operating thermal efficiencies were quite poor. They were ultimately
replaced by a newer, more efficient engine technology introduced in the second half of the 19th century, the
internal combustion Otto and Diesel cycles, discussed later in this chapter. However, the high thermal efficiency
potential of the Stirling and Ericsson cycles produces periodically renewed interest in utilizing these cycles within
the framework of modern technology.
EXAMPLE 13.10
A new Ericsson cycle engine is being designed with a pressure ratio of PR = p 4 /p 1 = 2.85, a power piston outlet pressure of
p 1 = 0.500 MPa, a maximum volume of V 1 = 0.0110 m 3 , and a minimum volume of V 3 = 3.00 × 10 –3 m 3 . The engine will
contain 0.0500 kg of air. Use the Ericsson cold ASC analysis shown in Figure 13.40 to complete this design, determining
a. The compressor inlet pressure and volume (p 2 and V 2 ).
b. The power piston outlet pressure and inlet volume (p 4 and V 4 ).
c. The compressor outlet pressure (p 3 ).
d. The temperatures at the inlet and outlet of the power and displacer pistons (T 1 , T 2 , T 3 , and T 4 ).
e. The Ericsson cold ASC thermal efficiency of this engine.
Solution
Using the Ericsson cycle diagram shown in Figure 13.40 we can carry out the following analysis.
a. For this cycle, the compressor inlet pressure is the same as the power piston outlet pressure (see Figure 13.40), or
p 2 = p 1 = 0.500 MPa. The compressor inlet volume is V 2 = V 3 × (CR), where the isentropic compression ratio
CR = V 1 /V 4 and V 4 = mRT 4 /p 4 . Now, from Figure 13.40,
T 4 = T 1 = p 1V 1
mR
= ð0:500 MPaÞð1000 kPa/MPaÞð0:0110 m3 Þ
ð0:0500 kgÞð0:286 kJ/kg . KÞ
= 385 K
and
p 4 = p 3 = p 2 ðPRÞ = p 1 ðPRÞ = ð0:500 MPaÞð2:85Þ = 1:43 MPa
so
V 4 = mRT 4
p 4
= ð0:0500 kgÞð0:286 kJ/kg . KÞð385 KÞ
ð1:43 MPaÞð1000 kPa/MPaÞ
= 0:00385 m 3
and the isentropic compression ratio is
CR = V 1 0:0110 m3
=
V 4 0:00385 m = 2:86
3
Then, the compressor inlet volume is V 2 = V 3 × (CR) = (3.00 × 10 –3 )(2.86) = 0.00858 m 3 .
b. For this cycle, the power piston outlet pressure is the same as the compressor inlet pressure (see Figure 13.40):
p 4 = p 3 = 1.43 MPa, and from part a, V 4 = 0.00385 m 3 .
c. Since p 4 /p 1 = p 3 /p 2 = PR, the compressor outlet pressure is p 3 = p 2 (PR), where the isentropic pressure ratio is given as
PR = 2.86. Then, p 3 = 0.500(2.86) = 1.43 MPa.
d. Using the ideal gas equation of state, we have
T 3 = T 2 = p 2V 2
mR
= ð0:500 MPaÞð1000 kPa/MPaÞð0:00858 m3 Þ
ð0:0500 kgÞð0:286 kJ/kg . KÞ
e. The Ericsson cold ASC thermal efficiency for this engine is given by Eq. (13.20) as
= 300: K
ðη T Þ Ericsson
= 1 − T L
= 1 − T 2
= 1 − 300: K = 0:221 = 22:1%
T H T 1 385 K
cold ASC