Power Plant Engineering

GAS TURBINE POWER PLANT 303

ηc 1 =^21

21

(T T )

(T T )

′−

−

T 2 =^121

1

T(T T)+−′

ηc

=

298 (408 298)

0.83

+−

= 430.5 K

T 4 = T 2 = 430.5 K

Applying isentropic law to the process 6 – 7′

6

7

T

T′

=

(1)/

3

2

P

P

γ− γ







= (3)0.286 = 1.37 K

T 7 ′ =

1523

1.37

= 1111 K

ηt 1 =^67

67

(T T )

(T T )′

−

−

T 7 = T 6 – ηt 1 (T 6 – T 7 ′)

= 1523 – 0.83(1523 – 1111) = 1181 K

T 9 = T 7 = 1181 K (as equal work is developed by each turbine)

Wc = 2CPa (T 2 – Tl) = 2 × 1(430.5 – 298) = 266 kJ/kg

Wt = 2CPa(T 6 – T 7 ) = 2 × 1(1523 – 1181) = 687.5 kJ/kg

Wn = Wt – Wc = 687.5 – 266 = 421.5 kJ/kg

When the ideal regeneration is given, then

ε = 1 therefore T 5 = T 9 = 1181 K = T 7

QS (heat supplied) = 2Cpa(T 6 – T 5 )

= 2 × 1(1523 – 1181) = 684 kJ/kg

(1) Thermal η =

W

Q

n

s

=

421.5

684

= 0.615 = 61.5%

(2) Power developed by the plant = Wn × m = 421.5 × 16.5 = 6954.75 kW.
Example 5. A gas-turbine power plant generates 25 MW of electric power. Air enters the com-
pressor at 10°C and 0.981 bar and leaves at 4.2 bar and gas enters the turbine at 850°C. If the turbine
and compressor efficiencies are each 80%, determine

(1) The temperatures at each point in the cycle

(2) The specific work of the cycle

(3) The specific work of the turbine and the compressor

T 9

T 5

T 10

T 4

Fig. 9.34