Modern Engineering Thermodynamics

490 CHAPTER 13: Vapor and Gas Power Cycles
EXAMPLE 13.9 (Continued )
Solution
Using the Stirling cycle diagram shown in Figure 13.38, we can carry out the following analysis.
a. Since T 2 = T 3 , the ideal gas equation of state gives p 3 = p 2 (V 2 /V 3 ), where
V 2 = V 1 = Piston displacement – V 3 = 0.011 – 0.001 = 0.010 m 3 . Then, p 3 = (0.100 MPa)(0.0100/0.00100) = 1.00 MPa.
b. Since T 1 = T 4 , the ideal gas equation of state gives p 4 = p 1 (V 1 /V 4 ), where V 4 = V 3 = 0.001 m 3 .
Then, p 4 = (0.300 MPa) × (0.0100/0.00100) = 3.00 MPa.
c. Again using the ideal gas equation of state, we find that m = p 2 V 2 /RT 2 and
d. The ideal gas equation of state gives
m = ð0:100 MPaÞð1000 kPa/MPaÞð0:0100 m3 Þ
ð0:286 kJ/kgKÞð30:0 + 273:15 KÞ
T 1 = p 1V 1
mR
= ð0:300 MPaÞð1000 kPa/MPaÞð0:0100 m3 Þ
ð0:0115 kgÞð0:286 kJ/kg . KÞ
e. Equation (13.19) gives the Stirling cold ASC thermal efficiency of this engine as
= 0:0115 kg
= 912 K
ðη T Þ Stirling
= 1 − T L
= 1 − T 2
= 1 −
30:0 + 273:15 K
= 0:668 = 66:8%
T H T 1 912 K
cold ASC
Exercises
25. Determine what the maximum displaced piston pressure would be in Example 13.9 if the piston displacement is
increased from 0.0110 m 3 to 0.0200 m 3 . Assume all the other variables remain unchanged. Answer: p 3 = 1.90 MPa.
26. If the maximum power piston design pressure is increased from 0.300 MPa to 0.800 MPa in Example 13.9,
determine the corresponding design heat addition temperature. Assume all the other variables remain unchanged.
Answer: T 1 = T 4 = 2430 K.
27. Determine the Stirling cold ASC thermal efficiency in Example 13.9 if the piston displacement is reduced from 0.0110 m 3
to 8.00 × 10 –3 m 3 . Assume all the other variables remain unchanged. Answer: (η T ) Stirling cold ASC = 66.7% (no change).
Though the Stirling cycle engine did not compete well with alternate gas power cycle engines after about 1880,
its potential for high thermal efficiency (and consequently low fuel consumption) plus the low noise and low
air pollution traits of an external combustion process caused renewed interest in the late 20th century in its
applicability for automotive use.
13.14 ERICSSON CYCLE
In 1833, the Swedish-born engineer John Ericsson (1803–1889) developed a different type of hot air, reciprocating,
external combustion engine, which could operate on either an open or closed loop cycle. Ericsson’s engine
also used a thermal regenerator, but it differed from Stirling’s in that the constant volume regeneration process
was replaced by a constant pressure regeneration process. Therefore, the Ericsson cycle consists of two isothermal
processes and two isobaric (constant pressure) processes, as shown in Figure 13.40.
In 1839, Ericsson moved to America and continued to develop his engine. His large engines (up to 300 hp, with
pistons 14 ft in diameter) were very inefficient and could not compete economically with existing steam engine
technology. However, his small engines were reasonably successful and several thousand were sold by 1860. By
1880, the popularity of his engine had dropped off, and it was considered to be obsolete technology until modern
gas turbine power plants came into being in the mid 20th century. The Ericsson cycle is approximated by an open
loop gas turbine that has multistage compressor intercooling (to approximate T L = constant) and multistage turbine
reheating (to approximate T H = constant) along with thermal regeneration, as shown in Figure 13.41.
The Ericsson cycle thermal efficiency is given by
ðη T Þ Ericsson
= ð _W out Þ net Q
= _ H − j _Q L j
_Q H
_Q H
where (see Figure 13.40a), for a reversible ASC engine, we can write
_Q H = _mT H ðs 1 − s 4 Þ
= 1 − j _Q L j
_Q H