GAS POWER CYCLES 615

dharm

\M-therm\Th13-1.pm5

=

v (^1) rpp 1
γ− 1 ()()γ− −−^141

pv rr
(^11) p
1
1
11
γ
γ
−
()()− −− ...[13.4 (a)]
Mean effective pressure (pm) is given by :
pm = pv^33 pv^44 pv^22 pv^11 vv 12
11
−
−
− −
−
F
HG
I
KJ
÷−
L
N
M
M
O
Q
P
γγ P
() bar ...(13.5)
Also pm =
pv rr
vv
p
11 1
12
1
11
γ
γ
−
−−
L
N
M
O
Q
P
−
()()−
()

pv rr
v v
r
(^11) p
1
1 1
1
11
γ
γ
−
−−
−
[( − ) ( )

pv rr
v r
r
(^11) p
1
1
1
11
1
γ
γ
−
−−
F −
HG
I
KJ
[( − ) ( )]
i.e., pm =
pr r r
r
1 p
(^111)
11
[( ) ( )]
()()
γ
γ
− −−
−−
...(13.6)
Example 13.7. The efficiency of an Otto cycle is 60% and γ = 1.5. What is the compression
ratio?
Solution. Efficiency of Otto cycle, η = 60%
Ratio of specific heats, γ = 1.5
Compression ratio, r =?
Efficiency of Otto cycle is given by,
ηOtto = 1 –^11
()rγ−
0.6 = 1 –
1
()r15 1.−
or
1
()r^05. = 0.4 or (r)0.5 =
1

4. = 2.5 or r = 6.25
   Hence, compression ratio = 6.25. (Ans.)
   Example 13.8. An engine of 250 mm bore and 375 mm stroke works on Otto cycle. The
   clearance volume is 0.00263 m^3. The initial pressure and temperature are 1 bar and 50°C. If the
   maximum pressure is limited to 25 bar, find the following :
   (i)The air standard efficiency of the cycle.
   (ii)The mean effective pressure for the cycle.
   Assume the ideal conditions.