DC Series motor: current vs load....

[peter010]

Hiz

lets assum we have a load fixed on the roter of a the 'DC series motor' in the attached photo, where:

Vt: DC source voltage (constant)
Lf: field's inductive resistance (will be neglected)
Rf: field's resistant
Ra: Armature resistance
Ia= Armature current, If: field current
M: back emf (Ea)

and the related functionl equation is:

kQw (Ea)= (Vt-Ia(Ra+Rs)/kQ ; //Q:flux

Now, if the load is doubled, then by applying kerckhoff laws, the armature current must be reduced. Anyhow, this seems to me contradicts with this motor characteristic of which that current increases as torque or load increases (kindly find the second attached photo), in other words: Torque = KQI.

Its really confusing! I spent one day trying to sort it out uselessly.

Appreciate your help to understant this.

 

[cnh1995]

Now, if the load is doubled, then by applying kerckhoff laws, the armature current must be reduced

How?

 

[peter010]

How?

I am presuming or imagining the load as an extra resistance RL in the circuit. Once this resistance get doubled in the circuit, the current will be affected inversely.

If this approach does not make since, then kindly how it should be, or what will happen if I doubled that load?

regards

 

[cnh1995]

If this approach does not make since, then kindly how it should be, or what will happen if I doubled that load?

If you increase the load, the speed of the motor will drop. What will happen to the back emf? Supply voltage and rotor resistance are constant.

 

[peter010]

If you increase the load, the speed of the motor will drop. What will happen to the back emf? Supply voltage and rotor resistance are constant.

back emf will drop down and thus geting more current in the curcuit. AWWW

Thank you very much )))))

 

[Windadct]

I would also point out - if you double the resistance in a circuit you are not "doubling the load"

P = V² / R -- double R and you cut the Power in half.

To double the load - for a fixed voltage you need to cut the resistance in half - and / or think - adding another resistor in parallel - that is truly "doubling the load"

 

[peter010]

I would also point out - if you double the resistance in a circuit you are not "doubling the load"

P = V² / R -- double R and you cut the Power in half.

To double the load - for a fixed voltage you need to cut the resistance in half - and / or think - adding another resistor in parallel - that is truly "doubling the load"

oh ok.. May I ask you a favour please,, can u add this resistance R of load to the attached DC serries cercuit and then attach it. This will help me so much for imaging its place in the curcuit )

 

[Windadct]

My point being that you seemed to imply in your comment "if the load is doubled, then by applying kerckhoff laws, the armature current must be reduced" - that doubling the load meant increasing or doubling the resistance (impedance) in the circuit. I was not saying to add resistance to the circuit as you drew it.

The electric circuit does not care what the "load" is, the resistors convert to heat the motor converts to mechanical energy.

 

[peter010]

My point being that you seemed to imply in your comment "if the load is doubled, then by applying kerckhoff laws, the armature current must be reduced" - that doubling the load meant increasing or doubling the resistance (impedance) in the circuit. I was not saying to add resistance to the circuit as you drew it.

The electric circuit does not care what the "load" is, the resistors convert to heat the motor converts to mechanical energy.

oh thanks alot. I thought you was referring to double the mechanical load by expressing it metaphorically by a resistor. :)

 

[peter010]

Acually, now i am discussing it please from another perspective, what if I used a pulse width modulation device to reduce the main votage to a certain limit, let's say Vnew=70% Vold. What will happen to the back emf? Can I then predict or calculate precisly the value of back emf at same load ?

 

[NascentOxygen]

PWM uses a pulse rate sufficiently high that the motor speed doesn't measurably change during the power ON and power OFF intervals.

Back emf in a series DC motor is a function of both (i) rotor speed, and (ii) field strength. Although the applied voltage is being pulsed ON and OFF, inductance of the rotor-field path should see this current held approximately constant, with its level decreasing as the controller lowers the speed.

 

[peter010]

PWM uses a pulse rate sufficiently high that the motor speed doesn't measurably change during the power ON and power OFF intervals.

Back emf in a series DC motor is a function of both (i) rotor speed, and (ii) field strength. Although the applied voltage is being pulsed ON and OFF, inductance of the rotor-field path should see this current held approximately constant, with its level decreasing as the controller lowers the speed.

Actually, is it possible to calculate the new current value and the new back emf value after decreasing the source voltage?

I tried to figure this out, but i could not predict or calculate the new value of the current since i can't depend then on the original characteristic chart of this motor.

 

[NascentOxygen]

Remember, armature resistance is a constant, in calculations.

I'm sure a google search will find what you need. Perhaps start with Series DC Motors here: http://www.electricaleasy.com/2014/07/characteristics-of-dc-motors.html

 

[jim hardy]

I tried to figure this out, but i could not predict or calculate the new value of the current since i can't depend then on the original characteristic chart of this motor.

you are aware that because the field is in series with the armature ,
flux is a function of armature current
and i can't tell by inspection whether you accounted for that in your equation

kQw (Ea)= (Vt-Ia(Ra+Rs)/kQ ; //Q:flux

What's Q ?

the two basic equations for a DC motor are
counter EMF Ea = KΦRPM,
Torque = (same)K X 7.04 X Φ X I_armature , torque in foot pounds
so
Torque = K X 7.04 X I_armature² X K' , where K' relates I_armature to Φ (which inductance does nicely if you know #turns)

i think you'll have to find K and K' by a curve fit ?

:Power out the shaft is to first approximation Ea X Ia

old jim

 

[peter010]

you are aware that because the field is in series with the armature ,
flux is a function of armature current
and i can't tell by inspection whether you accounted for that in your equation

What's Q ?

the two basic equations for a DC motor are
counter EMF Ea = KΦRPM,
Torque = (same)K X 7.04 X Φ X I_armature , torque in foot pounds
so
Torque = K X 7.04 X I_armature² X K' , where K' relates I_armature to Φ (which inductance does nicely if you know #turns)

i think you'll have to find K and K' by a curve fit ?

:Power out the shaft is to first approximation Ea X Ia

old jim

Thanks for responding

Actually, I can best describe my case by saying: After reducing the source voltage, I have then two unknowns: the current and the EMF.

[Before reducing the voltage, I was able to identify the current value for my motor's circuit from this motor's characteristic graph, but after reducing the the source voltage, the characteristic graph is not applicable any more and thus I can't use it to identify the new current value]

So, that is, how to solve the circuit for the two unknowns?? )

 

[NascentOxygen]

I was able to identify the current value for my motor's circuit from this motor's characteristic graph,

Can you post a clear photo of that graph?

 

[jim hardy]

So, that is, how to solve the circuit for the two unknowns?? )

i see two equations and two unknowns...