TITLE.PM5

616 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th13-1.pm5

Solution. Bore of the engine, D = 250 mm = 0.25 m

Stroke of the engine, L = 375 mm = 0.375 m

Clearance volume, Vc = 0.00263 m^3

Initial pressure, p 1 = 1 bar

Initial temperature, T 1 = 50 + 273 = 323 K

p (bar)

25 3

4

1

Adiabatics

2

1

V V(m )^3

VC S

Fig. 13.6

Maximum pressure, p 3 = 25 bar

Swept volume, Vs = π/4 D^2 L = π/4 × 0.25^2 × 0.375 = 0.0184 m^3

Compression ratio, r =

VV

V

sc

c

+ =0 0184 0 00263+

0 00263

..

. = 8.
(i)Air standard efficiency :
The air standard efficiency of Otto cycle is given by
ηOtto = 1 –
1
()rγ−^1 = 1 –

1

() 8 14 1. −

= 1 –

1

() 804.

= 1 – 0.435 = 0.565 or 56.5%. (Ans.)

(ii)Mean effective pressure, pm :

For adiabatic (or isentropic) process 1-2

p 1 V 1 γ = p 2 V 2 γ

or p 2 = p 1
V
V

1

2

F

HG

I

KJ

γ

= 1 × (r)1.4 = 1 × (8)1.4 = 18.38 bar

∴ Pressure ratio, rp =

p

p

3

2

25

18.38

= = 1.36

The mean effective pressure is given by

pm =

pr r r

r

1 p

(^111) 14 1
11
18 8 1 361
4181
[( )( )]
()()
[{( ) } ( )]
()()
γ.
γ
− −− −
−−
= ×−−
−−

1. 
   


2. 
   

   ... [Eqn. (13.6)]