TITLE.PM5

GAS POWER CYCLES 635

dharm
\M-therm\Th13-3.pm5

Solution. Refer Fig. 13.17.

p (bar)

2 3

Vs

Vs

8

100

Adiabatics

4

1 1 (27°C)

Vc

V(m )^3

Fig. 13.17

Cylinder diameter, D = 200 mm or 0.2 m

Stroke length, L = 300 mm or 0.3 m

Initial pressure, p 1 = 1.0 bar

Initial temperature, T 1 = 27 + 273 = 300 K

Cut-off =

8

100

Vs = 0.08 Vs

(i)Pressures and temperatures at salient points :

Now, stroke volume, Vs = π/4 D^2 L = π/4 × 0.2^2 × 0.3 = 0.00942 m^3

V 1 = Vs + Vc = Vs +

V

r

s

− 1

Q V V

c r

= s

−

L

N

M

O

Q

1 P

= Vs 1

1

11

+

−

F

HG

I

KJ

=

r −

r

r

× Vs

i.e., V 1 =
15
15 1

15

− 14

×=Vs × 0.00942 = 0.0101 m^3. (Ans.)

Mass of the air in the cylinder can be calculated by using the gas equation,

p 1 V 1 = mRT 1

m =

pV

RT

11

1

1 10^5 0 0101

287 300

= ××

×

.

= 0.0117 kg/cycle