Page 564 - Handbook of Electrical Engineering

556 HANDBOOK OF ELECTRICAL ENGINEERING

◦
The average value of T 3 and T 4e is 988.734 K. From the cubic expression in Table 2.1 for a fuel-to-air
ratio of 0.01, the revised value of C pt is,

C ptn = 1.0011 − 1.4117 × 10 −4 × 988.734
2
+ 5.4973 × 10 −7 × 988.734 − 2.4691 × 10 −10 × 988.734 3
= 1.160278

◦
Thenew valueof γ t is 1.895425 − 0.49296 × 1.160278 = 1.32345, and T 4 = 690.436 K. Now recal-
culate T 4e ,
◦
T 4e = 690.436 × 0.87 + (1.0 − 0.87) × 1223.0 = 759.67 K
◦
The new average value of T 3 and T 4e is 991.334 K.
Step 15. Recycle.
Repeat this iterative process from [step 14.1] until the variables settle at their stable values.
These eventually become,
C pt = C ptn = 1.16088

γ t = γ tn = 1.323156
◦
◦
T 4e = 991.455 K or 718.455 C
Step 16. The work done on the gearbox input shaft, from (2.32) is found as follows,

1.0 − 1.323156
1 − γ t
δ t = = =−0.24423
γ t 1.323156
U outea = 1.16088 × 1223.0 × (1.0 − 10.3743 −0.24423 ) × 0.87
= 537.592 − 340.062 = 197.53 kJ/kg

Step 17. Include the gearbox and generator losses.
The losses between the gearbox input shaft and the electrical terminals of the generator
U losses are,
U losses = (0.015 + (1.0 − 0.985)) × 12.0 = 0.36 MW

Hence the input to the gearbox is 12.0 + 0.36 = 12.36 MW. From sub-section 2.3 the mass flow of
the air–fuel mixture ‘m’is,
12.36 × 1000.0
W out
m = =
U outea 197.53
= 62.573 kg/sec = 225263 kg/hour

Step 18. Find the theoretical efficiency η pa .
From (2.20) the theoretical efficiency η pa can be found by using the appropriate pressure ratios
and ratios of the specific heats.