Power Plant Engineering

GAS TURBINE POWER PLANT 299

(2) Heat supplied per hg of air circulation.

(3) The thermal efficiency of the cycle. Mass of the fuel may be neglected.

Solution. Pl = 1 bar

P 2 = 5 bar

P 3 = 5 – 0.1 = 4.9 bar

P 4 = 1 bar

T 1 = 20 + 273 = 293 K

T 3 = 680 + 273 = 953 K

ηcompressor = 85%

ηturbine = 80%

ηcombustion = 85%

For air and gases: cP′ = 1.024 kJ/kgK

y = 1.4

Power developed by the plant,

P = 1065 kW

(1) The quantity of air circulation, ma =?

For isentropic compression 1 – 2,

2

1

T

T

=

(1)/

2

1

p

p

γ− γ







=

5 (1.4 1) / 1.4

1

−



 = 1.584

T 2 = 293 × 1.584 = 464 K

Now,

ηcompressor =^21

21

(T T )

(T′ T )

−

−

= 0.85

0.85 =

2

(464 293)

(T′ 293)

−

−

T 2 ′ = 494 K

For isentropic expansion process 3 – 4,

4

3

T

T =

(1)/

4

3

P

P

γ− γ







=

1 (1.4 1) /1.4

4.9

−







= 0.635

T 4 = 953 × 0.635 = 605 K

Now, ηturbine =

34

34

(T T )

(T T )

− ′

− = 0.80

0.85

0.80

=^4

(953 T )

(953 605)

− ′

−

3

5 b

ar

4.^9

ba

r

1 b

ar

2 2 ′

(^2931)
953
T(K)
4
4 ′
s
Fig. 9.31