- #1

zsolt2

- 20

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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