Modern Engineering Thermodynamics

504 CHAPTER 13: Vapor and Gas Power Cycles
where CR = v 3 /v 4s is the isentropic compression ratio. If theintaketemperatureandpressure(T 3 and p 3 )are
known, we can find u 3 and v r3 from the table. Then, if we know the compression ratio (CR), we can find
v r4 = v r3
CR and v r2 = v r1 × CR
We can now find u 4s and T 4s from the tables. However, to find u 1 , T 1 , u 2s , and T 2s , we need to know more information
about the system. Consequently, the heat of combustion (Q H /m = _Q H / _m ), maximum pressure (p 1 ), or
maximum temperature (T 1 ) in the cycle is usually given to complete the analysis.
For the Otto cold ASC,
Then,
j _Q L j = _mðu 2s − u 3 Þ = _mc v ðT 2s − T 3 Þ and _Q H = _mðu 1 − u 4s Þ = _mc v ðT 1 − T 4s Þ:
ðη T Þ Otto
cold ASC
The process 1 to 2s and process 3 to 4s are isentropic, so
= 1 − T
2s − T 3
= 1 − T
3 T2s /T 3 − 1
T 1 − T 4s T 4s T 1 /T 4s − 1
Since T 1 /T 4s = T 2s /T 3 ,
T 1 /T 2s = T 4s /T 3 = ðv 1 /v 2s Þ 1−k = ðv 4s /v 3
= ðp 1 /p 2s Þ ðk − 1Þ/k = ðp 4s /p 3 Þ
ðk − 1Þ/k
Þ 1−k
ðη T Þ Otto
= 1 − T 3 /T 4s = 1 − PR ð1−kÞ/k = 1 − CR 1−k (13.30)
cold ASC
where CR = v 3 /v 4s is the isentropic compression ratio and PR = p 4s /p 3 is the isentropic pressure ratio.
Since T 3 = T L but T 4s < T 1 = T H , the Otto cold ASC thermal efficiency is less than that of a Carnot cold ASC
operating between the same temperature limits (T 1 and T 3 ). Because the Otto cycle requires a constant volume
combustion process, it can be carried out effectively only within the confines of a piston-cylinder or other fixed
volume apparatus by a nearly instantaneous rapid combustion process.
EXAMPLE 13.14
The isentropic compression ratio of a new lawn mower Otto cycle gasoline engine is 8.00 to 1, and the inlet air temperature
is T 3 = 70.0°F at a pressure of p 3 = 14.7 psia. Determine
a. The air temperature at the end of the isentropic compression stroke T 4s .
b. The pressure at the end of the isentropic compression stroke before ignition occurs p 4s .
c. The Otto cold ASC thermal efficiency of this engine.
Solution
a. The isentropic compression ratio for an Otto cycle engine is defined as
from which we have
CR = v
3
= T 1
3
1−k
v 4s T 4s
T 4s = T 3
CR 1−k = T 3 × CR k − 1 = ð70:0 + 459:67 RÞð8:00Þ 0:40 = 1220 R
b. For the Otto cycle, the isentropic pressure and compression ratios are related by PR = CR k , where PR = p 4s =p 3 and
CR = v 3 /v 4s . Then,
p 4s = p 3 CR k = ð14:7 psiaÞð8:00Þ 1:40 = 270: psia
c. Equation (13.30) gives the Otto cold ASC thermal efficiency as
ðη T Þ Otto
= 1 − T 1−k
3
k
= 1 − PR = 1 − CR 1−k = 1 − ð8:00Þ 1−1:40 = 0:565 = 56:5%
T 4s
cold ASC