How does field rheostat resistance affect the speed of a DC shunt motor?

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Discussion Overview

The discussion revolves around the effect of field rheostat resistance on the speed of a DC shunt motor. Participants explore the relationship between field current, torque, and motor speed, considering both theoretical and practical implications.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that increasing the field rheostat resistance from 10Ω to 100Ω results in a speed increase from 1500 rpm to 1700 rpm, questioning if the motor would start at 1700 rpm again.
  • Another participant argues that with lower field current, the starting torque is reduced, leading to less acceleration and a lower steady-state speed.
  • Some participants propose that the motor will draw more armature current to produce the necessary torque, regardless of the field current.
  • Mathematical relationships are introduced, such as RPM being related to counter EMF and torque, with participants discussing how these equations reflect the motor's behavior under different conditions.
  • There is a suggestion that the acceleration rate does not affect the final speed but rather how quickly the motor reaches that speed.
  • One participant acknowledges errors in their reasoning, particularly regarding the assumption of constant back EMF and the relationship between power, torque, and speed.
  • Another participant expresses appreciation for the collaborative effort to clarify the concepts, indicating a shared learning experience.

Areas of Agreement / Disagreement

Participants express differing views on whether the motor can start at a higher speed with reduced field excitation. There is no consensus on the implications of field rheostat resistance on motor performance, with multiple competing perspectives remaining unresolved.

Contextual Notes

Participants note that their discussions involve assumptions about motor behavior, such as the constancy of back EMF and the relationship between torque and speed, which may not hold in all scenarios. The mathematical derivations presented are based on hypothetical values and may not reflect real-world conditions.

Who May Find This Useful

Individuals interested in electrical engineering, motor control, and the dynamics of DC motors may find this discussion relevant, particularly those exploring the effects of field control on motor performance.

cnh1995
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Suppose a motor is run at 1500 rpm and field rheostat is at minimum position (say 10Ω). Then the field rheostat resistance is increased to 100Ω and the speed obtained is 1700 rpm. Now the motor is shut down and started again. Will it now run at 1700 rpm?
My logic: No it won't, because:
the field current is less⇒starting torque is less⇒load torque is same⇒torque difference is less⇒acceleration is less⇒steady state speed is less.

Please correct me if I'm wrong.(I recently had an argument with my friend with whom I'm going to give a presentation on this method.He disagrees and my professor backed him.)
I don't understand how it will run at higher speed when started at lower field current. By that logic, at no field current, the motor should run at infinite speed. Am I missing something?
 
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cnh1995 said:
I don't understand how it will run at higher speed when started at lower field current.

What is nature of the source for armature current?
The motor will simply draw more armature current to make whatever torque is necessary.

Long thread on that here:
https://www.physicsforums.com/threa...c-motor-increase-when-flux-is-reduced.804006/
cnh1995 said:
By that logic, at no field current, the motor should run at infinite speed.
It'll try. Remember this picture?
pot.com%2F-_pGliNArb3Q%2FUPwvsdT_RHI%2FAAAAAAAAO1s%2FXhTXTlDj64U%2Fs640%2Fimage_1358694250175816.jpg
 
So it won't run at 1700 rpm and will draw more current to produce same torque right?
 
cnh1995 said:
So it won't run at 1700 rpm and will draw more current to produce same torque right?
Or will it just accelerate more gently? Looks to me like you've already established (by measurement) the rpm for that particular applied voltage, field current and torque.
cnh1995 said:
when the field rheostat resistance is increased to 100Ω and the speed obtained is 1700 rpm.

does your observation agree with the math ?

Go back to your basic motor equations

Counter EMF = KΦ X RPM
so RPM = Counter EMF / KΦ
RPM = (Vapplied - Iarmature X Rarmature ) / KΦ

Torque = 7.04 K X Φ X Iarmature (Torque in foot pounds)
Iarmature = Torque / 7.04KΦ

Substitute that Iarmature into RPM formula

RPM = (Vapplied - (Torque/7.04KΦ) X Rarmature) / KΦ

RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2first term is unloaded speed
second term is how much slower it runs because of load torque.

field rheostat sets KΦ,
load's speed-torque curve sets torque...

Play with the algebra ? What's RPM/Torque ?
 
jim hardy said:
Or will it just accelerate more gently? Looks to me like you've already established (by measurement) the rpm for that particular applied voltage, field current and torque.

does your observation agree with the math ?

Go back to your basic motor equations

Counter EMF = KΦ X RPM
so RPM = Counter EMF / KΦ
RPM = (Vapplied - Iarmature X Rarmature ) / KΦ

Torque = 7.04 K X Φ X Iarmature (Torque in foot pounds)
Iarmature = Torque / 7.04KΦ

Substitute that Iarmature into RPM formula

RPM = (Vapplied - (Torque/7.04KΦ) X Rarmature) / KΦ

RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2first term is unloaded speed
second term is how much slower it runs because of load torque.

field rheostat sets KΦ,
load's speed-torque curve sets torque...

Play with the algebra ? What's RPM/Torque ?
Whatever values I mentioned are not experimentally obtained. I just took some random values, just to reduce wording.

Will this train of thought work? For both the cases below, starting armature current will be same.
For case 1(rheostat at minimum):
Starting field current is higher ⇒Starting torque is higher⇒acceleration will be higher⇒steady state speed is higher⇒Steady state armature current is smaller.
For case 2(rheostat value increased):
Starting field current is smaller⇒Starting torque is smaller⇒acceleration will be smaller⇒steady state speed is smaller⇒steady state armature current is higher.
To produce the same load torque in both the cases,field flux*armature current product,i.e. Φ*Ia product should be same.
In case 1, Φ is more, Ia is less and in case 2, Ia is more,Φ is less, keeping the product constant.
So no way the motor should run at higher speed when started at smaller field excitation.

Is this logic okay? This can be proven by algebra.
 
cnh1995 said:
Will this train of thought work?
go back to those two equations, they won't let you down.
CounterEmf = KΦRPM
Torque = 7.04 KΦIarmature

cnh1995 said:
For case 1(rheostat at minimum):
Starting field current is higher ⇒Starting torque is higher⇒acceleration will be higher⇒steady state speed is higher reached more quickly ⇒Steady state armature current is smaller.

cnh1995 said:
For case 2(rheostat value increased):
Starting field current is smaller⇒Starting torque is smaller⇒acceleration will be smaller⇒steady state speed is smaller reached less quickly ⇒steady state armature current is higher.

Acceleration has nothing to do with final speed, only how quickly you reach it.

cnh1995 said:
Is this logic okay? This can be proven by algebra.
show that algebra ?

cnh1995 said:
So no way the motor should run at higher speed when started at smaller field excitation.

my algebra says
RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2

Halving KΦ doubles the first term and quadruples the second.
RPMcase1 = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2
RPMcase2 = 2Vapplied / KΦ - 4(Torque X Rarmature) / 7.04K2Φ2

RPMcase2 - RPMcase1 = Vapplied / KΦ - 3(Torque X Rarmature) / 7.04K2Φ2

That's the speed increase from halving excitation.
So long as Torque X Rarmature/7.04KΦ is less than 1/3 Vapplied, it'll be positive.
Speed increase will be zero when Vapplied / KΦ - 3(Torque X Rarmature) / 7.04K2Φ2 = 0
Vapplied / KΦ = 3(Torque X Rarmature) / 7.04K2Φ2
Rarmature = Vapplied X 7.04KΦ/(3 X Torque)
.
Assume some per-unit values
Torque = 1
Vapplied = 1
KΦ= 1

Rarmature = 1 X 7.04 X 1 / (3 X 1) = 234 % ?
Real motors have small Rarmature on purpose, a few percentbetter check my algebra, I'm challenged.
Show me yours?
 
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Actually i should have treated KΦ as a single term
and written

[Q RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04(KΦ)2
 
jim hardy said:
go back to those two equations, they won't let you down.
CounterEmf = KΦRPM
Torque = 7.04 KΦIarmatureAcceleration has nothing to do with final speed, only how quickly you reach it.show that algebra ?
my algebra says
RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2

Halving KΦ doubles the first term and quadruples the second.
RPMcase1 = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2
RPMcase2 = 2Vapplied / KΦ - 4(Torque X Rarmature) / 7.04K2Φ2

RPMcase2 - RPMcase1 = Vapplied / KΦ - 3(Torque X Rarmature) / 7.04K2Φ2

That's the speed increase from halving excitation.
So long as Torque X Rarmature/7.04KΦ is less than 1/3 Vapplied, it'll be positive.
Speed increase will be zero when Vapplied / KΦ - 3(Torque X Rarmature) / 7.04K2Φ2 = 0
Vapplied / KΦ = 3(Torque X Rarmature) / 7.04K2Φ2
Rarmature = Vapplied X 7.04KΦ/(3 X Torque)
.
Assume some per-unit values
Torque = 1
Vapplied = 1
KΦ= 1

Rarmature = 1 X 7.04 X 1 / (3 X 1) = 234 % ?
Real motors have small Rarmature on purpose, a few percentbetter check my algebra, I'm challenged.
Show me yours?
Well you are right as always! I made three major mistakes while thinking, which now make me want to kill myself...
1. I assumed back emf to be constant in both the cases, which actually isn't since the field is reduced. This was the reason I was saying speed won't increase..
2. Wrong understanding of starting torque.
3. I totally overlooked
P=2*pi*NT
If the field is reduced and load torque is same, that must draw more power from the source. So with T constant, N must increase with reduction in field.

Thats how my "train of thoughts" got derailed..

Thanks a lot for putting up with my nonsense..My apologies..!
 
Don't apologize at all - you made me straighten out my thinking !

Thanks for working through my ineloquent presentation.

It is heartwarming to see "The Light Come On"

I distinctly remember the morning Professor Grimm derived those two formulas for us boys , my immediate thought was "How Delightfully Intuitive ! " . Both fall out naturally from QVcrossB and right hand rule.
Complications from armature reaction you add in later.

Thanks for the feedback ! helps an old guy feel useful.

old jim
 
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