- #1
zsolt2
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Homework Statement
Hello I have a bit of a problem solving this
A 20 kW, 500 V d.c. shunt wound motor draws a current of 45 A when running at full load with a speed of 600 rev min–1. On no load, the current drawn from the supply is 5 A. If the armature resistance is 0.3 Ω and the shunt field resistance is 220 Ω, calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full load value, assuming that the flux per pole does not change.
Homework Equations
Armature current: Ia=20000/500=40A
Field current= 500/220=2.3A
The Attempt at a Solution
Now first i don't understand where is the missing 2.7A or is the field current=5A?
If the torque falls half of its value the armature current falls to half as well Ia=20A
Speed:
On full load E=V-Ia x Ra=500- 40 x 0.3=488V
When Ia falls to half= E=500-20 x 0.3=494V
if the back e.m.f. increases the speed increases: 494/488=1.012 new speed= 600 x 1.012=607 rev/min
Efficiency:
V x Ia=Ia^2 x Ra + E x Ia
500 x 20=20^2 x 0.3 + 494 x 20=10000W
Can anyone tell me if I am right so far
Thank you