TITLE.PM5

GAS POWER CYCLES 653

dharm
\M-therm\Th13-3.pm5

Compression ratio, rc = 9

Expansion ratio, re = 5

Number of cycles/sec. = 8

Cylinder diameter, D = 250 mm = 0.25 m

Stroke length, L = 400 mm = 0.4 m

Heat liberated at constant pressure

= 2 × heat liberated at constant volume

(i)Pressure and temperatures at all salient points :

For compression process 1-2,

p 1 V 1 n = p 2 V 2 n

∴ p 2 = p 1 × V

V

n

1

2

F

HG

I

KJ

= 1 × (9)1.25 = 15.59 bar. (Ans.)

A1so, T

T

V

V

n

2

1

1

2

1

=F 9 125 1

HG

I

KJ

=

−

(). − = 1.732

∴ T 2 = T 1 × 1.732 = 303 × 1.732

= 524.8 K or 251.8°C. (Ans.)

Also, cp(T 4 – T 3 ) = 2 × cv(T 3 – T 2 ) ...... (given) ...(i)

For constant pressure process 3-4,

T

T

V

V

r

r

TT

c

e

4

3

4

3

43

9

5

8

8

===

==

=

ρ Compression

Expansion ratio

1.

1.

ratio ( )

()

V

V

ie r V

V

V V V V V V

r

r

V

V

r

r

e

c

cc

e

5

4

5

3

3 4 1 3 1 2 5 4

1

1

(. ., )=×

=×

=×=

∴= =

L

N

M M M M M M M M M M M

O

Q

P P P P P P P P P P P

ρ

ρρ

ρ

Substituting the values of T 2 and T 4 in the eqn. (i), we get

1.0(1.8T 3 – T 3 ) = 2 × 0.71(T 3 – 524.8)

0.8T 3 = 1.42(T 3 – 524.8)

0.8T 3 = 1.42T 3 – 745.2

∴ 0.62T 3 = 745.2

T 3 = 1201.9 K or 928.9°C. (Ans.)

Also,

p

T

p

T

3

3

2

2

= ...... for process 2-3

∴ p 3 = p 2 ×

T

T

3

2

= 15.59 ×

1201 9

524 8

.

. = 35.7 bar. (Ans.)
p 4 = p 3 = 35.7 bar. (Ans.)
T 4 = 1.8T 3 = 1.8 × 1201.9 = 2163.4 K or 1890.4°C. (Ans.)
For expansion process 4-5,
p 4 V 4 n = p 5 V 5 n