Power Plant Engineering

290 POWER PLANT ENGINEERING

Heat supplied to the system = KP(Tl – T 4 )

Heat rejected from the system = Kp(T 2 – T 3 )

where Kp = Specific heat at constant pressure,

Work done = Heat supplied – Heat rejected

= KP(Tl – T 4 ) – Kp(T 2 – T 3 )

Thermal efficiency (η) of Brayton Cycle

η =

Work done

Heat Supplied

= 11 4 2 3

14

[K {(T T ) (T T )}]

[K (Tp T )]

−−−

−

η = 1 –^23

14

(T T )

(T T )

−

−

...(1)

For expansion 1-2

1

2

T

T

=

(1)/

1

2

P

P

γ− γ







T 1 = T 2

(1)/

1

2

P

P

γ− γ







For compression 3-4

4

3

T

T =

(1)/

4

3

P

P

γ− γ





=

(1)/

1

2

P

P

γ− γ







T 4 = T 3

(1)/

1

2

P

P

γ− γ







Substituting the values of Tl and T 4 in equation (1), we get

η = 1 –^23

(1)/ (1)/

11

23

22

(T T )

TTPP

PP

γ− γ γ− γ

−

 

 −

 



η = 1 – (1)/^23

(^123)
2
(T T )
P
(T T )
P
γ− γ
−

 −
