TITLE.PM5

GAS POWER CYCLES 649

dharm

\M-therm\Th13-3.pm5

∴ T 2 = 300 × 2.954 = 886.2 K

p

p

v

v

2

1

1

2

=F 1514

HG

I

KJ

=

γ

(). ⇒ p 2 = 44.3 p 1

Constant pressure process 2-3 :

p

T

p

T

2

2

3

3

=

or T 3 = T 2 ×

p

p

3

2

= 886.2 ×

70

44 3

1

1

p

. p
= 1400 K
Also, Heat added at constant volume = Heat added at constant pressure ...(Given)
or cv (T 3 – T 2 ) = cp (T 4 – T 3 )
or T 3 – T 2 = γ (T 4 – T 3 )

or T 4 = T 3 +

TT 32 −

γ

= 1400 +

1400 886 2

14

−.

.

= 1767 K.

Constant volume process 3-4 :

v

T

v

T

3

3

4

4

= ⇒ v

v

T

T

4

3

4

3

1767

1400

== = 1.26

Also,

v

v

v

v

4

3

4

115

=

(/ )

= 1.26 or v 4 = 0.084 v 1

Also, v 5 = v 1

Adiabatic expansion process 4-5 :

T

T

v

v

v

v

4

5

5

4

1

1

1

14 1

0 084

=

F

HG

I

KJ

=

F

HG

I

KJ

γ−−

.

.

= 2.69

∴ T 5 = T^4

269

1767

269

656 9

..

==.K

∴ηair-standard =

Work done

Heat supplied =

Heat supplied Heat rejected

Heat supplied

−

= 1 – Heat rejected

Heat supplied

= 1 –

cT T

cT T c T T

v

v p

()

()()

51

32 43

−

−+ −

= 1 –

()

()()

TT

TT TT

51

32 43

−

−+ −γ

= 1 –

()

()()

656.9 300

1400 886.2 4 1767 1400

−

−+ −1.

= 0.653 or 65.3%. (Ans.)

Reasons for actual thermal efficiency being different from the theoretical value :

1. In theoretical cycle working substance is taken air whereas in actual cycle air with fuel
   acts as working substance.


2. The fuel combustion phenomenon and associated problems like dissociation of gases,
   dilution of charge during suction stroke, etc. have not been taken into account.