TITLE.PM5

610 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th13-1.pm5

p 3 = p 2 × V

V

2

3

F

HG

I

KJ

γ

= 12 × V

V

1

4

F

HG

I

KJ

γ

Q V

V

V

V

4

1

3

2

=

L

N

M

O

Q

P

= 12 ×^1

6

F^14

HG

I

KJ

.

= 0.97 bar. (Ans.)

Hence

pTT

p

pTT

p

112

2

334

4

===

=

===

=

U

V

||

W

||

18 bar

12 bar

0.97 bar

1.46 bar

(Ans. )

683 K

333.2 K

(ii)Change in entropy :

Change in entropy during isothermal expansion,

S 2 – S 1 = mR loge

V

V

2

1

F

HG

I

KJ

=

pV

T

11

1

loge

V

V

2

1

F

HG

I

KJ

Q pV mRT

mR pV

T

=

=

L

N

M

M

O

Q

P

or P

=

18 10 018

10 683

5

3

××

×

.

loge (1.5) = 0.192 kJ/K. (Ans.)

(iii)Mean thermal efficiency of the cycle :

Heat supplied, Qs = p 1 V 1 loge

V

V

2

1

F

HG

I

KJ

= T 1 (S 2 – S 1 )

= 683 × 0.192 = 131.1 kJ

Heat rejected, Qr = p 4 V 4 loge

V

V

3

4

F

HG

I

KJ

= T 4 (S 3 – S 4 ) because increase in entropy during heat addition

is equal to decrease in entropy during heat rejection.

∴ Qr = 333.2 × 0.192 = 63.97 kJ

∴ Efficiency, η =

QQ

Q

sr

s

−

= 1 –

Q

Q

r

s

= 1 –

63.97

131.1

= 0.512 or 51.2%. (Ans.)

(iv)Mean effective pressure of the cycle, pm :

The mean effective pressure of the cycle is given by

pm = Work done per cycle

Stroke volume

V

V

3

1

= 6 × 1.5 = 9

Stroke volume, Vs = V 3 – V 1 = 9V 1 – V 1 = 8V 1 = 8 × 0.18 = 1.44 m^3

∴ pm = ()QQ J

V

sr

s

−× = ()QQ

V

sr

s

−× (^1) ()Q J= 1
= ()131.1 63.97^10
44 10
3
5
−×

1. ×

= 0.466 bar. (Ans.)