Electrical Machines DC Shunt Motor

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Discussion Overview

The discussion revolves around the analysis of a shunt DC motor, specifically focusing on calculating various parameters such as speed, current, losses, and efficiency under different operating conditions. The scope includes theoretical calculations and problem-solving related to electrical machines.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in calculating the speed of the motor at rated load with a specified armature current, seeking guidance on the approach.
  • Another participant suggests calculating the field current based on the settings provided, leading to a value of 4.875 A.
  • There is a discussion about the calculation method for the internal voltage and field current, with some participants questioning the accuracy of the resistance values used.
  • Concerns are raised about the feasibility of the mechanical output of the motor given the supply voltage and armature current, indicating a potential inconsistency in the problem statement.
  • A participant proposes that solving the problem may require assuming a linear magnetization curve despite the non-linear nature of the graph provided.
  • Another participant shares their own calculations, including the magnetomotive force (mmf) and internal voltage, but expresses uncertainty about the next steps in the problem-solving process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or assumptions regarding the motor's performance. Multiple competing views and uncertainties remain regarding the interpretation of the problem and the values used in calculations.

Contextual Notes

Some calculations depend on assumptions about the linearity of the magnetization curve, and there are unresolved questions about the accuracy of resistance values and the feasibility of the given power output under specified conditions.

Edges_sharp
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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
 

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Hi Edges_sharp. :welcome:

Have you calculated the field current at the settings in use?
 
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Yes I am getting it to be 4.875A
 
Edges_sharp said:
Yes I am getting it to be 4.875A
How did you calculate that?
 
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
 
Edges_sharp said:
the field resistance Rf is 40 .
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?
 
Yes it is copied and should be ohms
 
Edges_sharp said:
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
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.
 
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Thanks very much :)
 
  • #10
Edges_sharp said:
Yes it is copied and should be ohms
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
Edges_sharp said:
An 8.25 kW shunt DC motor
Edges_sharp said:
What is the speed of the motor when operating at rated load with an armature current of 30 A?
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:
  • #12
I am only giving the information present in the question
 
  • #13
cnh1995 said:
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.
Power engineers are known for their healthy disdain of exactitude. :oldwink:
 
  • #14
Edges_sharp said:
I am only giving the information present in the question
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.
 
  • #15
Can someone please come back to this example.
I'm working on a similar problem and would appreciate the help. Thanks
 
Last edited:
  • #16
SK97 said:
Can someone please come back to this example.
I'm working on a similar problem and would appreciate the help. Thanks
Can you post your problem (along with your attempt)?
Edit: I see you already have.
 
  • #17
cnh1995 said:
Can you post your problem (along with your attempt)?
My problem is actually the same.

I have found I(f) to be 5A
therefore the mmf is 7500A.turns

and reading from the graph I saw the internal voltage at that point is around 275 V.

Now I'm not sure where to go from there, any help would be appreciated.
 

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