# Homework Help: DC Motor

1. Sep 12, 2016

### zsolt2

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Armature current: Ia=20000/500=40A
Field current= 500/220=2.3A

3. The attempt at a solution
Now first i dont 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

2. Sep 12, 2016

### cnh1995

That's not correct.
5kW is the mechanical power output of the motor and not electrical power input.
That's 2.2728 A(≈2.3 A).
So the armarure current would be simply supply current minus field current.

3. Sep 12, 2016

### zsolt2

So the armature current Ia=45-2.3=42.7
Then the power=V x Ia=500 x 42.7=21350W
And that 1350W is a fixed lost?
Ia on half torque = 42.7/2=21.35A
Speed:
E=V-ia x Ra=500-42.7 x 0.3=487V
Half torque
E=500-21.35 x 0.3=494V
The speed increases if the back e.m.f. grows (Ia decreases) 494/487=1.014
n2=n1 x 1.014=600 x 1.014=608 rev/min

4. Sep 13, 2016

### cnh1995

Looks good!

5. Sep 13, 2016

### zsolt2

Thank you!
And the efficiency :
Total power used: V x Ia = 500 x 21.35= 10675W
Losses:
Armature loss: Ia^2 x Ra =21.35^2 x 0.3=136.7W
Field loss: If^2 x Rf= 2.3^2 x 220=1163.8W
Brush losses: 2 x Ia= 2 x 21.35=42.7W
Fixed loss: 1350W
All losses:136.7+1163.8+42.7+1350=2693.2W
Is that looks right?

6. Sep 13, 2016

### cnh1995

I didn't qoute this part in my previous post. You should use the no load data to find the fixed loss.
You need not take brush drop into account if it is not mentioned in the problem.

7. Sep 13, 2016

### zsolt2

Ok
So the current drawn on no load is 5A the power consumed is 500 x 5=2500W
Field loss is 2.3^2 x 220=1163.8~ 1164W
Then the fixed loss: 2500-1164=1336W
Efficiency at half torque :
10675/10675+1336+1164+137=0.8
Efficiency =80%

8. Sep 13, 2016

### cnh1995

To be accurate, you should also subtract armature copper loss from this. Usually, because of low armature resistance, armature copper loss is neglected on no load but here, since nothing is mentioned as such, you should calculate it.

9. Sep 13, 2016

### zsolt2

Ok thank you very much for the help!!!!