# Electrical Machines DC Shunt Motor

Tags:
1. Dec 3, 2016

### Edges_sharp

• New poster has been reminded to use the Homework Help Template in schoowork posts here
An 8.25 kW shunt DC motor is supplied with a terminal DC voltage of 300 V. The armature resistance of the motor Ra is 0.25 , and the field resistance Rf is 40 . The field winding consists of 1500 turns. An adjustable resistance Radj is connected in series to the field winding. Radj may be varied over the range from 0 to 80  and is currently set at 20 . Armature reaction may be ignored in this machine. The magnetization curve for this motor, taken at a speed of 1200 rpm, is shown in Figure Q1.
(i) What is the speed of the motor when operating at rated load with an armature current of 30 A?
(ii) If the motor is now unloaded with no changes in terminal voltage or Radj, what is the no-load speed of the motor?
(iii) Iron core losses are 250 W. What are the copper losses and mechanical losses in the motor at rated load? (ignore brush and stray losses)
(iv) What is the efficiency of the motor at rated load?
(v) If the terminal voltage is reduced to 260 V and the armature current stays at 30 A, calculate the required Radj in order to restore the speed of the motor to the value in (i).

I am having difficulty with Question 1(I) of this example. I have tried finding the internal voltage, then finding "If" to then work out however many Ampere Turns that will produce and read the Internal Generated voltage from the graph. If provided with the no load speed I would not find this an overly difficult question however due to the omission of this information I am struggling quite a bit. I am not asking for the full solution to all of the questions. However I would really appreciate a nudge in the right direction with 1(I).

Many Thanks

File size:
32.6 KB
Views:
40
2. Dec 3, 2016

### Staff: Mentor

Hi Edges_sharp.

Have you calculated the field current at the settings in use?

3. Dec 3, 2016

### Edges_sharp

Yes I am getting it to be 4.875A

4. Dec 3, 2016

### Staff: Mentor

How did you calculate that?

5. Dec 3, 2016

### Edges_sharp

Not sure if my method is right but using Ea=VT-IaRa
Ea=300-30*0.25=292.5

Then to calculate field current I did If=Ea/Rf=292.6/60=4.875

6. Dec 3, 2016

### Staff: Mentor

I'm puzzled by your symbol after the 40? Does it look like Ω on your screen? It doesn't on mine. Maybe it's a windows character you copied?

7. Dec 3, 2016

### Edges_sharp

Yes it is copied and should be ohms

8. Dec 3, 2016

### Staff: Mentor

Rf of 60 Ω is right. But in a shunt machine the field current is determined by the terminal voltage.

You should always have a model circuit at hand when attempting problems on motors.

9. Dec 3, 2016

### Edges_sharp

Thanks very much :)

10. Dec 3, 2016

### Staff: Mentor

It looks like a tiny subscript 2 to me.

If you click on the $\Sigma$ symbol at the top of the editor window in your browser then a line of Greek letters will appear lower down from where you can click on one to have it included in what you're typing. (If you are using the mobile app....then I have no idea what it does.)

11. Dec 4, 2016

### cnh1995

I believe mechanical output of the motor is 8.25 kW when it is drawing 30 A armature current. This is not possible with a supply voltage of 300V.

Last edited: Dec 4, 2016
12. Dec 4, 2016

### Edges_sharp

I am only giving the information present in the question

13. Dec 4, 2016

### Staff: Mentor

Power engineers are known for their healthy disdain of exactitude.

14. Dec 4, 2016

### cnh1995

Well, in my opinion, you can solve the question only by "assuming" the magnetization curve to be linear, while still using the values on the non-linear graph.