TITLE.PM5

GAS POWER CYCLES 647

dharm
\M-therm\Th13-3.pm5

For the constant volume process 2-3,

p

T

p

T

2

2

3

3

=

∴ T 3 = T 2 ×

p

p

3

2

= 874.1 ×
68
21 67.
= 2742.9 K. (Ans.)
Heat added at constant volume
= cv (T 3 – T 2 ) = 0.71 (2742.9 – 874.1) = 1326.8 kJ/kg
∴ Heat added at constant pressure
= Total heat added – heat added at constant volume
= 1750 – 1326.8 = 423.2 kJ/kg
∴ cp(T 4 – T 3 ) = 423.2
or 1.0(T 4 – 2742.9) = 423.2
∴ T 4 = 3166 K. (Ans.)
For constant pressure process 3-4,

ρ =

V

V

T

T

4

3

4

3

= =^3166

2742.9

= 1.15

For adiabatic (or isentropic) process 4-5,

V

V

V

V

V

V

V

V

V

V

5 r

4

5

2

2

4

1

2

3

4

=×=×=

ρ Q ρ=

F

HG

I

KJ

V

V

4

3

Also p 4 V 4 γ = p 5 V 5 γ

∴ p 5 = p 4 × V

V

4

5

F

HG

I

KJ

γ

= 68 ×

ρ γ

r

F

HG

I

KJ = 68 ×

115

9

.^14.
F
HG

I

KJ = 3.81 bar. (Ans.)

Again, T

T

5

4

=

V

Vr

4

5

(^11141)
115
9
F
HG
I
KJ
=GFH IKJ =FHG IKJ
γ− γ−−
ρ..
= 0.439
∴ T 5 = T 4 × 0.439 = 3166 × 0.439 = 1389.8 K. (Ans.)
(ii)Air standard efficiency :
Heat rejected during constant volume process 5-1,
Qr = cv(T 5 – T 1 ) = 0.71(1389.8 – 363) = 729 kJ/kg
∴ ηair standard
done

• supplied

Work

Heat

==QQ−

Q

sr

s

=

1750 729

1750

−

= 0.5834 or 58.34%. (Ans.)

(iii)Mean effective pressure, pm :

Mean effective pressure is given by

pm = Work done per cycle

Stroke volume

or pm =^1
34 3 11
44 55 22 11
V
pV V pV pV pV pV
s

()−+ −

−

− −

−

L

N

M

O

Q

γγP

VVrVVVVV V

Vr V

c cc

sc

15 2 3 4

1

== == =

=−

,,,

()

ρ Q r

VV

V

V

V

Vr V

sc

c

s

c

sc

= + =+

∴=−

L

N

M

M

M

O

Q

P

P

P

1

() 1