Page 561 - Handbook of Electrical Engineering

WORKED EXAMPLE FOR CALCULATING THE PERFORMANCE OF A GAS TURBINE 553

Therefore, from (2.17),

(273.0 + 950.0) × 0.50403 − (273.0 + 20.0)
η i = 1.0 −
(273.0 + 950.0) − ((273.0 + 20.0) × 1.984)
323.43
= 1.0 − = 0.496 per unit
641.69
Step 4. From (2.18),
581.31 1.0
T 2e = + 1.0 − × 293.0
0.85 0.85
◦
◦
= 632.18 K or 359.18 C.
Step 5. Also from (2.18),

T 4e = 616.43 × 0.87 + (1.0 − 0.87) × 1223.0
◦
◦
= 695.28 K or 422.28 C.
Step 6. From (2.20),

1223.0(1.0 − 0.50403) × 0.85 × 0.87 − 293.0(1.984 − 1.0)
η p =
1223.0 × 0.85 − 293.0(1.984 − 1.0 + 0.85)
160.25
= = 0.319 per unit
502.188
Step 7. From (2.27),
Let
1.4
d = =−1.75
2(1.0 − 1.4)
r pmax = (293.0/(1223.0 × 0.85 × 0.87)) d

= 7.187 per unit

F.5 DETAILED SOLUTIONS
Step 8. Initially convert the pressure drops into the SI system of measurement units of ‘bar’.

P 1 = 125.0/10200.0 = 0.01226 bar

And
P 4 = 50.0/10200.0 = 0.0049 bar

The combustion pressure drop in ‘bar’ is,

P 4 = r pt × P 4 × 0.04 = 11.0 × 1.0 × 0.04 = 0.44 bar