Page 656 - Basic Electrical Engineering

Iron, friction, and windage losses = 1100 − 5 − 440 = 655 W

These losses are constant losses and are same at any load. This means, on full
load these losses will remain at 655 W.

At full-load, I L = 40 A

I a = I − I = 40 − 2 = 38 A
f
L
2
2
I R a = (38) × 0.2 = 289 W
a
2
2
I R f = 2 × 110 = 440 W
f
Iron, friction, and windages losses = 655 W

Total losses = 289 + 440 + 655 = 1384 W

Efficiency

Example 7.9 A four-pole dc generator has 1000 conductors. The flux per

pole is 25 mWb. Calculate the EMF induced when the armature is lap
connected and run at 1500 rpm. At what speed the generator must be driven

to produce the same EMF with the armature winding wave connected?

Solution:

Case I

−3
P = 4, Z = 1000, ϕ = 25 × 10 Wb
A = 4, N = 1500 rpm