TITLE.PM5

GAS POWER CYCLES 643

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\M-therm\Th13-3.pm5

Also during the compression operation 1-2,

p 1 V 1 γ = p 2 V 2 γ

or

p

p

V

V

2

1

1

2

=

F

HG

I

KJ

γ

= (16.14)1.4 = 49.14

or p 2 = p 1 × 49.14 = 0.9 × 49.14 = 44.22 bar

Pressure or explosion ratio, β =

p

p

3

2

65

44 22

=

.

= 1.47

Putting the value of r, ρ and β in eqn. (i), we get

ηdual = 1 –^1

16.14

47 76 1

14 1 47 1 47 4 76 1

14

()

()

. () ()

.

−

×−

−+ × −

L

N

M

M

O

Q

P

P

= 7 – 0.328

3.

.47 1.

243 1

0 564

−

+

L

N

M

O

Q

P = 0.6383 or 63.83%. (Ans.)
Example 13.24. An oil engine working on the dual combustion cycle has a compression
ratio 14 and the explosion ratio obtained from an indicator card is 1.4. If the cut-off occurs at 6
per cent of stroke, find the ideal efficiency. Take γ for air = 1.4.
Solution. Refer Fig. 13.19.
Compression ratio, r = 14
Explosion ratio, β = 1.4

If ρ is the cut-off ratio, then

ρ−

−

(^1) =
1
6
r 100
or
ρ−
−
1
14 1= 0.06
∴ρ = 1.78
Ideal efficiency is given by
ηideal or dual = 1 –
11
()^111
()
rγ ()()
βργ
− ββγρ
−
−+ −
L
N
M
M
O
Q
P
P
= 1 –
1
14
478 1
14 1 41 4 4781
14
()
()

. () ( )

.

−

×−

−+ × −

L

N

M

M

O

Q

P

P

= 1 – 0.348

3.

138 1

4 528

−

+

L

N

M

O
Q
P = 0.614 or 61.4%. (Ans.)
Example 13.25. The compression ratio for a single-cylinder engine operating on dual cycle
is 9. The maximum pressure in the cylinder is limited to 60 bar. The pressure and temperature of
the air at the beginning of the cycle are 1 bar and 30°C. Heat is added during constant pressure
process upto 4 per cent of the stroke. Assuming the cylinder diameter and stroke length as
250 mm and 300 mm respectively, determine :
(i)The air standard efficiency of the cycle.
(ii)The power developed if the number of working cycles are 3 per second.
Take for air cv = 0.71 kJ/kg K and : cp = 1.0. kJ/kg K
Solution. Refer Fig. 13.21.
Cylinder diameter, D = 250 mm = 0.25 m
Compression ratio, r = 9
Stroke length, L = 300 mm = 0.3 m