Modern Engineering Thermodynamics

506 CHAPTER 13: Vapor and Gas Power Cycles
where m a and _m a are the mass of air in the cylinder and the cylinder’s air mass flow rate, respectively. The ASC
(i.e., reversible or indicated, see Table 13.2) thermal efficiency of any internal or external combustion engine can
now be written as
ðη T Þ ASC
= ð _W out Þ reversible
_Q in
= ð _W 1 Þ out
_Q fuel
= ð _W 1 Þ out
/ _m a
_Q fuel / _m a
where _Q in = _Q fuel is the heating value of the fuel. Combining these equations gives
mep = ðη
TÞ ASC
_Q fuel / _m a
= η
ð TÞ ASC
_Q fuel / _m fuel
v 2 − v 1 ðA/FÞðv 2 − v 1 Þ
where A/F = _m a / _m fuel is the air–fuel ratio of the engine. Now,
so Eq. (13.32) becomes
v 2 − v 1 = v 1 ðv 2 /v 1 − 1Þ = RT 1 ðCR − 1Þ/p 1
ð
ð _W 1 Þ out =
η TÞ ASC
_Q / _m ðDNp fuel 1 /CÞ
(13.34)
ðA/FÞðRT 1 ÞðCR − 1Þ
where D = nðV 2 − V 1 Þ = π 4 ðBoreÞ 2 × ðStrokeÞ × ðNumber of cylindersÞ is the total piston displacement of the
engine. Equation (13.34) allows us to determine the horsepower output of an ideal frictionless internal combustion
engine, and when actual dynamometer test data are available, Eq. (13.33) allows us to determine the
engine’s mechanical efficiency.
EXAMPLE 13.15
A six-cylinder, four-stroke Otto cycle internal combustion engine has a total displacement of 260. in 3 and a compression ratio of
9.00 to 1. It is fueled with gasoline having a specific heating value of 20.0 × 10 3 Btu/lbm and is fuel injected with a mass-based
air-fuel ratio of 16.0 to 1. During a dynamometer test, the intake pressure and temperature were found to be 8.00 psia and
60:0°F while the engine was producing 85.0 brake hp at 4000. rpm. For the Otto cold ASC with k = 1:40, determine the
a. Cold ASC thermal efficiency of the engine.
b. Maximum pressure and temperature of the cycle.
c. Indicated power output of the engine.
d. Mechanical efficiency of the engine.
e. Actual thermal efficiency of the engine.
Solution
a. From Eq. (13.30), using k = 1:40 for the cold ASC,
b. From Figure 13.48a,
ðη T Þ Otto
= 1 − CR 1−k = 1 − 9:00 −0:40 = 0:585 = 58:5%
cold ASC
_Q H = _Q fuel = ð _mc v Þ a ðT 1 − T 4s Þ = _m fuel ðA/FÞðc v Þ a ðT 1 − T 4s Þ
and
Since process 3 to 4s is isentropic, Eq. (7.38) gives
_Q / _m fuel
T 1 = T max = T 4s +
ðA/FÞ mass
ðc v
T 4s = T 3 CR k−1 = ð60:0 + 459:67Þð9:00Þ 0:40 = 1250 R
Þ a
Then,
T max =
20:0 × 10 3 Btu/lbm fuel
ð16:0 lbm air/lbm fuelÞ½0:172 Btu/ðlbm air .
+ 1250 R = 8520 R
RÞŠ
Since process 4s to 1 is isochoric, the ideal gas equation of state gives
p max = p 1 = p 4s ðT 1 /T 4s Þ