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Question on Ohm's law and motors

by sharnrock
Tags: motors
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Dickfore
#19
Jun26-10, 11:07 PM
P: 3,014
Let's look at a very crude model of an electric motor: the railgun. (scheme attached)



there are two conducting rails and a cylindrical conductive bar can roll on them freely. There is also a magnetic field perpendicular to the plane of the rails. The load is drawn by the rolling bar.

There are two governing equations for the motor: the Kirchoff loop equation and Newton's Second Law.

Let's start with Newton's Second Law for the bar. There are two forces acting on the bar:

Ampere's force: [itex]F_{a} = B l I[/itex]
external force (opposite of what the force with which the bar pulls the load) - it is an external parameter [itex]F[/itex]

The bar has mass [itex]m[/itex]

The acceleration of the bar is [itex]a = dv/dt[/itex]

So, Second Newton's Law gives:

[tex]
m \, \frac{dv}{dt} = B L I - F
[/tex]

Let's consider Kirchoff's Law now.
There are two voltage sources acting in the loop:
external voltage [itex]V[/itex] which is an external force
induced EMF: [itex]E_{i} = -B L v[/itex]

The resistance of the whole circuit is [itex]R[/itex].

The current through the circuit is [itex]I[/itex].

Second Kirchoff's Rule gives:

[tex]
V - B L v = R I
[/tex]

The mechanical variable for the system is [itex]v[/itex] - the velocity of the bar (or, in rotary geometries the angular velocity of the rotor). The electric variable of the motor is the current passing through it [itex]I[/itex]. The external parameters are the generated force [itex]F[/itex] (or, generated torque in rotary geometries) and the applied voltage [itex]V[/itex]. The internal parameters are the mass of the bar [itex]m[/itex] (or, the moment of inertia of the rotor in rotary geometries), the length of the bar [itex] L [/itex], the magnetic field [itex]B[/itex] and the total Ohmic resistance of all the conducting elements (rails and bar) [itex]R[/itex].

If you wish, you can eliminate the mechanical variable [itex]v[/itex] completely and get an equation involving only the electric state variable [itex]I[/itex]. Namely, solve the second equation with respect to [itex]v(t)[/itex]:

[tex]
v(t) = \frac{V(t) - R I(t)}{B L}
[/tex]

differentiate it with respect to time (keeping in mind that the external voltage as well as the current might vary in time)

[tex]
\frac{dv}{dt} = \frac{V'(t) - R I'(t)}{B L}
[/tex]

and substitute this into the first equation:

[tex]
\frac{m}{B L} \left(V'(t) - R I'(t)\right) = B L I(t) - F(t)
[/tex]

That is, we get a first order linear differential equation:

[tex]
\frac{d I(t)}{d t} + \frac{(B L)^{2}}{m R} \, I(t) = \frac{1}{R} \left(V'(t) + \frac{B L}{m} F(t)\right)
[/tex]

How is this equation like Ohm's Law? If anything, it looks like there is a capacitative element:

[tex]
\begin{array}{l}
v_{\mathrm{ef}} - \frac{Q}{C_{\mathrm{ef}}} = R_{\mathrm{ef}} \, I \\

I = \frac{d Q}{d t}
\end{array}
[/tex]

[tex]
\frac{d I}{d t} + \frac{1}{R_{\mathrm{ef}} C_{\mathrm{ef}}} I = \frac{V'_{\mathrm{ef}}(t)}{R_{\mathrm{ef}}}
[/tex]

and, taking the effective resistance to be the same as the Ohmic resistance of the true motor (which can be done, by taking [itex]F = 0[/itex] - free running motor), then we see that the effective voltage applied on this circuit is:

[tex]
V_{\mathrm{ef}} = V(t) + \frac{B L}{m} \int_{t_{0}}^{t}{F(t') \, dt'}
[/tex]

i.e. it is time dependent even if the external parameters V(t) and F(t) are constant, and, the effective capacitance is:

[tex]
C_{\mathrm{ef}} = \frac{m}{(B L)^{2}}
[/tex]
Attached Thumbnails
motor.png  
sharnrock
#20
Jun26-10, 11:16 PM
P: 14
Quote Quote by Dickfore View Post
...

i.e. it is time dependent even if the external parameters V(t) and F(t) are constant, and, the effective capacitance is:

[tex]
C_{\mathrm{ef}} = -\frac{m}{(B L)^{2}}
[/tex]

which is negative!
ok, I get what you're saying, it's not a resistor. Very nice.
stevenb
#21
Jun26-10, 11:18 PM
P: 697
A rail gun is not a good model to use for an induction motor; however I understand the motivation to use a simple illustration to make a point.

Generally, we shouldn't expect to get an Ohm's law relation. However, it is a basic fact that real work must show up as a resistive term on the electrical side. Reactive power can be inductive or capacitive loading. But real work looks resistive.

A simple model for an induction motor, at least when it is stopped, is a transformer. A locked motor has a small resistance of the windings, but since no work is being done, the main load is inductive. When an induction motor is spinning rapidly in steady state and doing real mechanical work, then the electrical load looks mostly resistive. This resistance is frequency dependent and doesn't behave like a typical resistor. So in a sense the resistance has frequency dependence similar to reactance; however, there is no phase shift like a reactance and the load is mostly resistive.
vk6kro
#22
Jun27-10, 12:27 AM
Sci Advisor
P: 4,029
I did some tests on a 250 volt 50 Hz desk fan using a variac to provide variable input voltage.

Volts 115 V current 0.20 A Power factor 0.91 (stalled.)
Volts 156 V current 0.26 A Power factor 0.91
Volts 189 V current 0.26 A Power factor 0.93
Volts 207 V current 0.27 A Power factor 0.93
Volts 222 V current 0.28 A Power factor 0.92
Volts 240 V current 0.28 A Power factor 0.91

At no stage did the current increase with decreasing voltage, and the range of speed available was from 1500 RPM to zero. The power factor was basically unaffected.

This is probably a shaded pole motor. Industrial fans using squirrel cage motors would behave differently.

Notice that, by calculation, you can see that the impedance of the motor increased from 575 ohms at 115 volts to 857 ohms at 240 volts, but this does not translate into increased current at lower voltages because of the reduced supply voltage.
Snoogans
#23
Jun27-10, 02:57 AM
P: 36
Out of curiosity, what was the fan doing incorrectly for you to be troubleshooting. If it wasn't moving, a motor that doesn't get enough voltage to overcome starting friction will stall and act as a transformer with the secondary shorted, and appear to draw more current than it should.

You also said no-one could tell you why wires are rated by current carrying capacity. Thinner wires have a higher resistance per length, Power lost in the conductor = I^2*R which produces heat. The current rating is the point where the heat generated is within acceptable limits, more will result in insulator degradation and fires.
sharnrock
#24
Jun27-10, 07:22 AM
P: 14
Quote Quote by Snoogans View Post
Out of curiosity, what was the fan doing incorrectly for you to be troubleshooting. If it wasn't moving, a motor that doesn't get enough voltage to overcome starting friction will stall and act as a transformer with the secondary shorted, and appear to draw more current than it should.

You also said no-one could tell you why wires are rated by current carrying capacity. Thinner wires have a higher resistance per length, Power lost in the conductor = I^2*R which produces heat. The current rating is the point where the heat generated is within acceptable limits, more will result in insulator degradation and fires.
omg, it was a nightmare... We were replacing some fans at a paint facility for a nuclear power plant. The original fans were 240/480 3phase, and we showed up with two 240v split phase fans. Well, rather than just reordering the right fans we decided to just put these fans in and take power off the 120/208v 3phase panel that was right next to the 480 panel. So, we get the fans installed and run some new wire to them. We turn the first fan on, and it's running great. Move to the next fan flip the switch and... pow! The breaker trips. The fan starts up and won't come below 80 amps (14amps full load). We check the fan to make sure it's moving freely everything seems good. We leave and come back with more material, and now we can't get the first fan to turn on. It's doing the same thing. It won't hold all the time. It holds some of the time and runs normal, but not all the time. So, we change the breakers, the motor controls, the wire gets re-pulled with heavier gauge to prevent voltage drop. We do everything new, everything oversized. We can't find anything wrong other than the motor is a 240v motor on a 208v line. We even stuck a buck/boost transformer on the one fan to try and get it to hold. My boss is getting ready to order two new 200v motors, but before that we ask the utility company to have one of their electricians check their transformer to see if there are any loose wires that might be fluctuating the voltage. Well, it turns out that the transformer supplying that panel was supposed to be changed 2 years ago and has been sitting in the paint shop's corner since then. They've been having problems like this the whole time. It took us 3 weeks with 2 guys to install the 2 fans. Needless to say, we lost money on that job.
sharnrock
#25
Jun27-10, 07:30 AM
P: 14
Quote Quote by Snoogans View Post
You also said no-one could tell you why wires are rated by current carrying capacity. Thinner wires have a higher resistance per length, Power lost in the conductor = I^2*R which produces heat. The current rating is the point where the heat generated is within acceptable limits, more will result in insulator degradation and fires.
What I meant was.. I was trying to use logic in a round-a-bout way to get them to see my point. They know that every gauge with a specific insulation is rated for a specific amperage. My point was to explain that voltage doesn't create heat, amperage creates heat. If higher voltage drops amperage than putting 1000v into a small motor should make it run with less heat. It doesn't though, it will melt it because the amperage will shoot through the roof.


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