TITLE.PM5

GAS POWER CYCLES 625

dharm

\M-therm\Th13-2.pm5

For the adiabatic (or isentropic) process 3-4

p 3 V 3 γ = p 4 V 4 γ

p 4 = p 3 × V

V

3

4

F

HG

I

KJ

γ

= p 3 ×^1

r

F

HG

I

KJ

γ

= 21.48 ×^1

55

14

.

F.

HG

I

KJ

= 1.97 bar. (Ans.)

Also

T

T

4

3

=

V

V

3

4

F^1

HG

I

KJ

−γ

=

1 1

r

F

HG

I

KJ

−γ

=

1

5

14 1

5.

F

HG

I

KJ

.−

= 0.505

∴ T 4 = 0.505 T 3 = 0.505 × 1172 = 591.8 K. (Ans.)

V 4 = V 1 = 0.45 m^3. (Ans.)

(ii)Percentage clearance :

Percentage clearance

= V

V

V

VV

c

s

=

−

2

12

× 100 =

0 081

0 45 0 081

.

..−

× 100

= 21.95%. (Ans.)

(iii)Efficiency :

The heat rejected per cycle is given by

Qr = mcv(T 4 – T 1 )

= 0.517 × 0.71 (591.8 – 303) = 106 kJ

The air-standard efficiency of the cycle is given by

ηotto =

QQ

Q

sr

s

−

= 210 106

210

− = 0.495 or 49.5%. (Ans.)

Alternatively :

ηotto = 1 –^11

()rγ−

= 1 –^1

(.) 55 1.4−^1

= 0.495 or 49.5%. (Ans.)

(iv)Mean effective pressure, pm :

The mean effective pressure is given by

pm =

W

V

QQ

s VV

(work done) sr

(sweptvolume) ( )

= −

12 −

= ()

(.. )

210 106 10

0 45 0 081 10

3

5

−×

−×

= 2.818 bar. (Ans.)

(v)Power developed, P :
Power developed, P = Work done per second
= Work done per cycle × number of cycles per second
= (210 – 106) × (210/60) = 364 kW. (Ans.)
Example 13.14. (a) Show that the compression ratio for the maximum work to be done
per kg of air in an Otto cycle between upper and lower limits of absolute temperatures T 3 and T 1
is given by

r =

T

T

3

1

F 1/ 2 ( 1)

HG

I

KJ

−γ

L

N

M

M

M

O

Q

P

P

P