614 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-1.pm5

Let compression ratio, rc (= r) =

v

v

1

2

and expansion ratio, re (= r) = v

v

4

3

(These two ratios are same in this cycle)

As

T

T

2

1

= v

v

1

2

F^1

HG

I

KJ

−γ

Then, T 2 = T 1. ()rγ−^1

Similarly,

T

T

3

4

=

v

v

4

3

F^1

HG

I

KJ

−γ

or T 3 = T 4. ()rγ−^1

Inserting the values of T 2 and T 3 in equation (i), we get

ηotto = 1 – TT

Tr Tr

41

4

1

1

1

−

.( )γγ−−− .( )

= 1 – TT

rTT

41

1

41

−

−γ ()−

= 1 –
1
()rγ−^1 ...(13.3)
This expression is known as the air standard efficiency of the Otto cycle.
It is clear from the above expression that efficiency increases with the increase in the value
of r, which means we can have maximum efficiency by increasing r to a considerable extent, but
due to practical difficulties its value is limited to about 8.
The net work done per kg in the Otto cycle can also be expressed in terms of p, v. If p is
expressed in bar i.e., 105 N/m^2 , then work done

W =

pv 33 pv 44 pv 22 pv 11

11

−

−

− −

−

F

HG

I

γγKJ × 10

(^2) kJ ...(13.4)
Also
p
p
3
4
= rγ =
p
p
2
1
∴
p
p
3
2

p
p
4
1
= rp
where rp stands for pressure ratio.
and v 1 = rv 2 = v 4 = rv 3 Q
v
v
v
v
(^1) r
2
4
3
L ==
N
M
O
Q
P
∴ W =
1
1
44 33 11
44
11 22
γ− 11
−
F
HG
I
KJ
−−
F
HG
I
KJ
L
N
M
M
O
Q
P
P
pv pv
pv
pv pv
pv

1
1
44 3 11
4
11
2
γ− 1
−
F
HG
I
KJ
−−
F
HG
I
KJ
L
N
M
M
O
Q
P
P
pv p
pr
pv p
pr

v (^1) pr pr
γ− 1 4 ()()γγ−−^1 −−^1111 −