DC self excited shunt motors 1

[Rafae123]

Hi All, I have a 5kW DC shunt, self excited motor that is developing short circuits in the field winding. We know this from the resistance measurement (it has decreased a bit)and the fact that the filed current has increased slightly. I was wondering if I could insert some resistance in the circuit to bring the desired current back to spec e.g. 1.5amps, if this would buy me some time before replacing the motor. I understand that I cannot do this for ever, but can some one please explain why, relating to necessary field flux etc, maybe back to basics?

 

[waross]

A resistance in the field circuit will make the motor run faster. That may lead to overloading and armature damage as well. It probably won't buy any time.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Rafae123]

Thank you, with all due respect, I understand that, this will hold true if the motor's field had no developing shorts, my question was around that, due to the fact that the field HAS developed some shorts, I am trying to compensate for this which is in fact trying to bring bag the field to close to the original value. I want to know, in simple speak how the loss of ampere turns will eventually affect the performance of the motor. Cheers for your prompt response.

 

[melspuds]

What is the motor doing, that makes you believe that its a shorted shunt winding.
1)Change speed
2)Did the start (have enough torque to start)
3)Any other problems with the motor

How did you take measurements?
1)Were the brushes lifted
2)Is the motor a pure shunt or a compound
3)was the motor temperature hotter that normal
4)Was the motor disconnected from the controller

Motor nameplate
1)Armature voltage and current
2)Shunt voltage and current
3)If compound, what is the voltage and current

Any other info would help
Dave

 

[waross]

If the dimensions of a DC coil are kept the same, then the Amp turns depends on the applied voltage and on the resistance of one turn of wire.
Do a simple sample calculation. Assume that the average resistance of one turn is one Ohm and the applied voltage is one Volt.
1 turn = 1 Ohm. At one Volt, 1 Amp times 1 turn = 1 Amp turn.
2 turns = 2 Ohms. At 1 Volt, 1/2 Amp times 2 turns = 1 Amp turn.
10 turns = 10 Ohms. At 1 Volt 1/10 Amp times 10 Turns = 1 Amp turn.
If turns are shorted with a good connection, then as turns are shorted out of the circuit, the current will increase and the Amp turns will be substantially the same but the coil will run hotter.
In real life things are not that simple.
Generally as the insulation breaks down the short from turn to turn will be a higher resistance than a dead turn to turn short.
The current will tend to divide between the shorted turn and the short circuit. The current through the turn to turn short will create heat. The Amp turns will not change as much as you may think.
If the field Amp turns are reduced, the motor will run faster.
To echo melspuds;
"What is the motor doing, that makes you believe that its a shorted shunt winding. "

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Rafae123]

Awsome, thank you waross. I can accept this answer, cheers

 

[electricpete]

"What is the motor doing, that makes you believe that its a shorted shunt winding.

op said this was based on slight decrease in resistance. I wonder if temperature was considered.

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(2B)+(2B)' ?

 

[panter]

rafae,
suggestions of colleagues is that measuring winding resistance is not sufficient reason to conclude that you have motor with shortened winding.
You have to give a DC voltage and measure the voltage drop at each excitation coil.
If you suspect a shortened winding even better is to give AC voltage and measure the voltage drop on each pole.
If that test demonstrate significant deviations that is good reason and you're sure to take off your motor and send it to reparation .
Good luck