118 DIRECT-CURRENT ‘DYNAMOS AND MOTORS. 
(6) A certain 75 H. P. compound-wound motor takes 
140 amperes on a 440-volt circuit; its armature re- 
sistance ts .O6 ohm; series field resistance, .0O7 
ohm; shunt resistance, 293 ohms; compute its elec- 
trical efficiency. 
Solution.—The current employed for shunt excitation in 
this case is aa = 1.5 amperes; hence the current flowing 
in the armature, 140 — 1.5 = 138.5 amperes, and there- 
fore by formula (63): 
(440 x 140) — [(138.5" x .067) + (1.57 x 0) 
El. Ef. = 440 X 140 
_ 61,610 — 1945 _ 59,665 
* 61,610 61,610 7 Poo Ft 96.8%, 
84. Commercial Efficiency.—By the commercial effi- 
ctency, or net efficiency, of a dynamo-electric machine is 
meant the ratio of its output to its intake. The intake 
of a generator is the mechanical power required to drive ~ 
it, and is the sum of the total electrical power generated 
in the armature and of the power losses due to hysteresis, — 
eddy currents, and friction; the antake of a motor is the 
electrical power delivered to its terminals. The output 
of a generator is the electrical power available at its 
terminals; the output of a motor is the mechanical 
power available at its shaft, and consists in the useful 
power of the. armature diminished by the losses due to 
hysteresis, eddy currents, and friction. 
The commercial efficiency of a generator, therefore, is: 
PCE LR ves WwW 
Com. Eff. = wr W'+w, - W'+urtwetuwr 
Ww 
~ W+ w+ Wn + wy, t+ we tue?" 
and that of motor: 
Com, Eff. aes Ww" —s W'—u, Aa at wert we) 
