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Question on Ohm's law and motorsby sharnrock
Tags: motors 
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#19
Jun2610, 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] 


#20
Jun2610, 11:16 PM

P: 14




#21
Jun2610, 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. 


#22
Jun2710, 12:27 AM

Sci Advisor
P: 4,016

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. 


#23
Jun2710, 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 noone 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. 


#24
Jun2710, 07:22 AM

P: 14




#25
Jun2710, 07:30 AM

P: 14




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