TITLE.PM5

GAS POWER CYCLES 633

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\M-therm\Th13-3.pm5

= 0.06 (V 1 – V 2 ) = 0.06 (15 V 2 – V 2 )

= 0.84 V 2 or V 3 = 1.84 V 2

∴ρ = V

V

V

V

3

2

2

2

=^184. = 1.84

Putting the value in eqn. (i), we get

ηdiesel = 1 –

1

415

84 1

14 1 84 1

14

1.

1.

() 1.

()

.

.

−

−

−

L

N

M

M

O

Q

P

P

= 1 – 0.2417 × 1.605 = 0.612 or 61.2%. (Ans.)

Example 13.18. The stroke and cylinder diameter of a compression ignition engine are

250 mm and 150 mm respectively. If the clearance volume is 0.0004 m^3 and fuel injection takes

place at constant pressure for 5 per cent of the stroke determine the efficiency of the engine.

Assume the engine working on the diesel cycle.

Solution. Refer Fig. 13.16.

Length of stroke, L = 250 mm = 0.25 m

Diameter of cylinder, D = 150 mm = 0.15 m

Clearance volume, V 2 = 0.0004 m^3

Swept volume, Vs = π/4 D^2 L = π/4 × 0.15^2 × 0.25 = 0.004418 m^3

Total cylinder volume = Swept volume + clearance volume

= 0.004418 + 0.0004 = 0.004818 m^3

Volume at point of cut-off, V 3 = V 2 +^5

100

Vs

= 0.0004 +

5

100

× 0.004418 = 0.000621 m^3

∴ Cut-off ratio, ρ =

V

V

3

2

000621

0004

=0.

0.

= 1.55

Compression ratio, r =

V

V

VV

V

1 s

2

2

2 0004

= + =0.004418 + 0.0004

0. = 12.04

Hence, ηdiesel = 1 –

11

1

1 1

4 12.04

55 1

(^1141) 55 1
14
γ
ρ
γ ρ
γ
() ()
()
.
.
r −−
−
−
L
N
M
M
O
Q
P
P
=−
×
−
−
L
N
M
M
O
Q
P

1. 
   

   P

   
   


2. 
   


3. 
   

   = 1 – 0.264 × 1.54 = 0.593 or 59.3%. (Ans.)
   Example 13.19. Calculate the percentage loss in the ideal efficiency of a diesel engine with
   compression ratio 14 if the fuel cut-off is delayed from 5% to 8%.
   Solution. Let the clearance volume (V 2 ) be unity.
   Then, compression ratio, r = 14
   Now, when the fuel is cut off at 5%, we have
   ρ−
   −

   
   

1

r 1

=^5

100

or ρ−

−

1

14 1

= 0.05 or ρ – 1 = 13 × 0.05 = 0.65

∴ρ = 1.65

ηdiesel = 1 –^11

1

1 1

4 14)

65 1

(^111) 65 1
1
γ
ρ
γ ρ
γ
() (
()
.4
.4
r −−
−
−
L
N
M
M
O
Q
P
P
=−
×
−
−
L
N
M
M
O
Q
P

1. P

1.

1.