Understanding Back EMF in Electric Motors

[annamal]

Homework Statement

A 120-V, series-wound motor has a field
resistance of 80 ohms and an armature resistance
of 10 ohms. When it is operating at full speed, a
back emf of 75 V is generated. What is the the power output of the
motor?

Relevant Equations

P = IV
I = V/R

Current drawn by motor = $\frac{120-75}{90}$ = 0.5 Amps
The answer says the power output of the motor is 37.5 Watts. The only way I can get that answer is if P = IV = 0.5(75) = 37.5.
This implies that in an electric motor, the back emf is the voltage of the motor. This confuses me because shoudn't the voltage driving the motor be the source voltage minus back voltage which equals 45 volts not 75 volts?

 

[Delta2]

I am not sure at all here but I **THINK** motors have a power factor $\cos\phi$ which is not 1, and the power output is $P=VI\cos\phi$, V the source voltage of the motor (120 V in this problem) ,I the current, and $\phi$ the phase difference between the current and voltage.

 

[bob012345]

This confuses me because shoudn't the voltage driving the motor be the source voltage minus back voltage which equals 45 volts not 75 volts?

How would you define what the load on the motor is? And how would you define the efficiency of a simple motor then?

 

[haruspex]

Take a look at e.g. https://studyelectrical.com/2015/02/back-emf-and-significance-in-dc-motor.html. In particular, note "The electric work done in overcoming and causing the current to flow against Eb is converted into mechanical energy developed in the armature. "

 

[annamal]

Take a look at e.g. https://studyelectrical.com/2015/02/back-emf-and-significance-in-dc-motor.html. In particular, note "The electric work done in overcoming and causing the current to flow against Eb is converted into mechanical energy developed in the armature. "

So that makes the voltage's motor equal to the back emf (which is Eb)? No, it still seems like the motor's voltage is source voltage - back emf

 

[hutchphd]

The voltage supplied to the motor minus the resistive loss in the series wound stator and rotor equals the back emf. It all works out.
I don't know what the "voltage's motor" is...

 

[Delta2]

So that makes the voltage's motor equal to the back emf (which is Eb)? No, it still seems like the motor's voltage is source voltage - back emf

Source voltage- back emf is the voltage drop in the total ohmic resistance of the motor. I **think** we can view the motor as a coil that has ohmic resistance R, and inductive Reactance X( which of course cannot simply be $L\omega$ as in the case of a simple coil because a motor is a rotating coil).

Oh nvm I see now @hutchphd says almost the same thing in his earlier post...

 

[haruspex]

So that makes the voltage's motor equal to the back emf (which is Eb)? No, it still seems like the motor's voltage is source voltage - back emf

Try an analogy: pushing a block up a ramp against friction and gravity.
The friction represents the resistances, while the gravitational force component is the back emf. The useful work done is the work in overcoming gravity.
Or replace the motor with a battery to be recharged. The battery exerts a pdf opposing that of the power source. The useful work done is the charge pushed through multiplied by that opposing pdf.

 

[Delta2]

Try an analogy: pushing a block up a ramp against friction and gravity.
The friction represents the resistances, while the gravitational component is the back emf. The useful work done is the work in overcoming gravity.

Nice analogy but it would be interesting to see the full math behind the relationship that mechanical power of a motor is back EMF times current. Unfortunately for some reason I can't access the page you have linked, I know my ISP can be weird at times, I ll try again later.

My rough sketch of how the math could go : The mechanical power would be $$P=\vec{T}\cdot\vec{\omega}$$ where T the torque of the motor and $\omega$ its revolution rate. Its the rotational analogue of $P=\vec{F}\cdot\vec{v}$.

Now the Torque we expect it to be proportional to current so ${T}=kI$ (because the magnetic force is $dF=I\vec{dl}\times \vec{B}$) and the back emf we expect to be proportional to $\omega$ (we know for example that a loop rotating in a magnetic field $\vec{B}$ has EMF $\mathcal{E}=B\omega A\sin\omega t$) so if back EMF$E_b=k\omega$, then $$P=kI\omega=E_bI$$

 

[annamal]

The voltage supplied to the motor minus the resistive loss in the series wound stator and rotor equals the back emf. It all works out.
I don't know what the "voltage's motor" is...

I meant the motor's voltage.

 

[annamal]

Nice analogy but it would be interesting to see the full math behind the relationship that mechanical power of a motor is back EMF times current. Unfortunately for some reason I can't access the page you have linked, I know my ISP can be weird at times, I ll try again later.

My rough sketch of how the math could go : The mechanical power would be $$P=\vec{T}\cdot\vec{\omega}$$ where T the torque of the motor and $\omega$ its revolution rate. Its the rotational analogue of $P=\vec{F}\cdot\vec{v}$.

Now the Torque we expect it to be proportional to current so $T=kI$ (because the magnetic force is $dF=I\vec{dl}\times \vec{B}$) and the back emf we expect to be proportional to $\omega$ (we know for example that a loop rotating in a magnetic field $\vec{B}$ has EMF $\mathcal{E}=B\omega A\sin\omega t$) so if back EMF$E_b=k\omega$, then $$P=kI\omega=E_bI$$

But how do you know the voltage calculated is the back voltage and not the source voltage?

 

[Delta2]

But how do you know the voltage calculated is the back voltage and not the source voltage?

Not sure I perfectly understand your question, but I don't see the reason why the source voltage would depend on the rotation rate $\omega$ of the motor.

 

[haruspex]

You have a source emf $E_s$, a back emf $E_b$ and a resistance R.
You correctly calculated that the current is $(E_s-E_b)/R$. So the power delivered by the source is $E_s(E_s-E_b)/R$ and the power consumed by the resistance is $(E_s-E_b)^2/R$. That leaves $E_s(E_s-E_b)/R-(E_s-E_b)^2/R=(E_s-E_b)E_b/R=IE_b$ as the power used by the motor.

 

[Delta2]

That's simply brilliant @haruspex, a clever shortcut using conservation of energy! Now why I didn't think of this ...

 

[bob012345]

You have a source emf Es, a back emf Eb and a resistance R.
You correctly calculated that the current is (Es−Eb)/R. So the power delivered by the source is Es(Es−Eb)/R and the power consumed by the resistance is (Es−Eb)2/R. That leaves Es(Es−Eb)/R−(Es−Eb)2/R=(Es−Eb)Eb/R=IEb as the power used by the motor.

The efficiency of the motor can be defined as $\eta=\frac{P_{out}}{P_{in}}=\frac{E_b I}{E_s I}=\frac{E_b}{E_s}$

then $E_b$ approaches $E_s$ in well designed motors. If this were a water wheel the wheel provides a torque to the load axle but the axle provides a counter torque to the wheel. At constant speed, except for losses there is no net work done on the wheel itself thus the power is transmitted through the wheel to the load. The back emf does the same function in the motor. It allows the transmission of the power through the motor to the load.

 

[bob012345]

I've been having a real difficult time with Latex editing just going away if I make an edit to a post since yesterday.

 

[Delta2]

I tried to do a small editing and i didnt have any problem. What browser and operating system do you have? I know these things haven't changed since yesterday but anyway just asking.

 

[bob012345]

I tried to do a small editing and i didnt have any problem. What browser and operating system do you have? I know these things haven't changed since yesterday but anyway just asking.

I'm on a Mac and my browser is Chrome.

 

[Delta2]

I am on a PC with Microsoft Edge and Windows 10. Occasionally I run into some problems too but they are fixed , either by themselves!, or by a browser cache reset. Try to reset the browser cache.

 

[bob012345]

I am on a PC with Microsoft Edge and Windows 10. Occasionally I run into some problems too but they are fixed , either by themselves!, or by a browser cache reset. Try to reset the browser cache.

Thanks, I think that was it!