TITLE.PM5

400 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th8-2.pm5

(b) Follows polytropic process with n = 1.4, and

(c) A constant temperature process (which completes the cycle).

Evaluate the following :

(i)The heat received in the cycle ;

(ii)The heat rejected in the cycle ;

(iii)Efficiency of the cycle.

Solution. Fig. 8.12 shows the cycle on p-V and T-s planes.

p(Pressure)

V(Volume)

12 p = Constant

3

pV = Const.n

pV = Const.

T(Temp.)

s (Entropy)

T = Const.

1

2

3

pV = Const.n

p = Const.

Fig. 8.12

Pressure, p 1 = 8 bar

Volume, V 1 = 0.035 m^3

Temperature, T 1 = 280 + 273 = 553 K

Pressure, p 2 = 8 bar (= p 1 )

Volume, V 2 = 0.1 m^3

Index, n = 1.4

To find mass of air, use the relation

p 1 V 1 = mRT 1

∴ m =

pV

RT

11

1

=

8 10 0 035

287 553

××^5

×

.

= 0.1764 kg

From p 2 V 2 = mRT 2

T 2 =

pV

mR

(^22) = 810 01
01764 287
××^5
×
.

. = 1580 K
Also, p 2 V 2 1.4 = p 3 V 3 1.4

and

T

T

2

3

=

p

p

2

3

F 14 1 14

HG

I

KJ

(.−)/.

But T 3 = T 1 as 1 and 3 are on an isothermal line.

∴^1580

553

=^8

3

0414

p

F

HG

I

KJ

./.