42 ALTERNATING GENERATORS AND SYNCHRONOUS MOTORS. — 
terms of the output of the generator is a parabola; that 
is, if the load on the alternator is doubled, the armature 
copper loss is quadrupled. | 
The no load excitation loss can be termed a constant loss, 
but the excess of the excitation energy is approximately 
proportional to the generator load. 
The above rules hold good under condition of constant 
temperature. 
78. Efficiency.—The commercial efficiency can be expressed 
by the formula, 
Ww 
W+Wi+ Wn + Wit We ~ 
Efficiency = (8) 
in which 
= Output of the generator, 
W, = Iron losses, 
Wm = Field copper losses, 
W, — Armature copper losses and 
W, = Friction losses. 
79. Example %.—Supposing in a three phase generator of 
1600 KVA. normal rating, the following losses have 
been measured: Iron loss, 35 K.W.; excitation loss, 5 
K.W.; armature loss 11.5 K.W.; friction and ventilation | 
losses 13 K.W. 
Solution: The efficiency of the generator at full normal 
load according to Formula (8) will be: 
1.600 
~ 1,600 + 85 +5+11.54+13 0.96. 
That is, 96% of the input will be delivered as useful 
output. Ans. 
80. Example 8.—The energy losses of the same generator 
when working at one half normal load, are indicated in 
the test curves, Fig. 24. What will be the efficiency at 
one half load, that is, at 800 KVA? 
