Modern Engineering Thermodynamics

13.18 Otto Cycle 507
and, since the process 3 to 4s is isentropic,
ðk − 1Þ/k
T 4s /T 3 ðp 4s /p 3 Þ
or
p 4s = p 3 ðT 4s /T 3 Þ k/ðk−1Þ = ð8:00 psiaÞ 1250 R 1:40/0:40
= 172 psia
520 R
then,
c. Equation (13.34) gives the indicated power as
p max = ð172 psiaÞ½ð8520 RÞ/1250 RŠ = 1170 psia
j _W I j out = ð0:585Þð20:0 × 103 Btu/lbmÞð260: in 3 /revÞð4000: rev/minÞð1170 lbf/in 2 Þ/2
ð16:0Þ½0:0685 Btu/ðlbm . RÞŠð8520 RÞð9:00 − 1Þð12 in/ftÞð60 s/minÞ
1hp
= ð132,00 ft⋅lbf/sÞ
550 ft .
= 241 hp
lbf/s
d. Equation (13.33) gives the mechanical efficiency of the engine as
η m = ð _W B Þ out
ð _W I Þ out
= 85:0hp
241 hp
= 0:353 = 35:3%
e. Finally, the actual thermal efficiency of the engine can be determined from Eqs. (7.5) and (13.33) as
ðη T Þ Otto
= ð _W B Þ out
= ðη mÞð _W I Þ out
= ðη m Þðη T Þ
_Q
actual fuel
_Q Otto
fuel
cold ASC
= ð0:353Þð0:585Þ = 0:207 = 20:7%
Exercises
43. If the Otto cycle engine discussed in Example 13.15 has its compression ratio increased to 10.0 to 1, what
would be its new Otto cold ASC thermal efficiency? Assume all other variables remain unchanged. Answer:
(η T ) Otto cold ASC = 60.2%.
44. Find p max and T max for the Otto cycle engine discussed in Example 13.15 when the compression ratio is decreased from
9.00 to 8.00 to 1. Assume all other variables remain unchanged. Answer: p max = 1040 psia and T max = 8460 R.
45. Determine the indicated horsepower in Example 13.15 if the engine’s displacement is increased from 260. in 3 to
300. in 3 . Assume all other variables remain unchanged. Answer: (_W I ) out = 280. hp.
46. Determine the mechanical efficiency of the Otto cycle engine in Example 13.15 if the actual brake horsepower is 88.0 hp
instead of 85.0 hp. Assume all other variables remain unchanged. Answer: η m = 36.3%.
The previous example illustrates that the Otto cold ASC analysis generally predicts thermal efficiencies that are
far in excess of the actual thermal efficiencies. Typical Otto cycle IC engines have actual operating thermal efficiencies
in the range of 15−25%. The large difference between the cold ASC (which contains at least one isentropic
process) thermal efficiency and the actual thermal efficiency is due to the influence of the second law of
thermodynamics through the large number of thermal and mechanical irreversibilities inherent in this type of
reciprocating piston-cylinder engine. To improve its actual thermal efficiency, the combustion heat losses and
the number of moving parts in the engine must be reduced.
WHAT IS THE SMALLEST INTERNAL COMBUSTION ENGINE?
The Cox Tee Dee .010 model airplane engine (Figure 13.50) has the smallest internal combustion engine ever put into
production. This amazing little engine weighs just under an ounce and runs at 30,000 rpm. The fuel is 10–20% castor
oil plus 20–30% nitromethane mixed with methanol. With a bore of 0.237 in (6.02 mm) and a stroke of 0.226 in
(5.74 mm), it has a power output of about 5 W.
(Continued)