Horace Heffner wrote: > > On Aug 16, 2009, at 1:13 PM, Stephen A. Lawrence wrote: > >> I looked this over, and two questions occurred to me. >> >> 1) What's the power factor when it's running? > > > It's DC. Power factor is 1.
Uhhhhhhhh.... Right. You did say the applied voltage was DC (or anyway that's generally what motorcycle batteries produce) ... but for some reason I didn't realize the motor drew constant current. For some reason I assumed its draw varied as it turned, and also assumed the back EMF was non-constant. I guess I had an image involving brushes somewhere in the back of my mind! Sigh... :-[ > > >> It's got a back EMF, due >> apparently to a hysteresis effect, > > > The hysteresis provides a field through which the current runs. This > provides the standard i L B force of the motor. The back emf is provided > by Faraday induction, as with most motors. The i L vector that > generates the torque(motive force) moves sideways through the B (or vice > versa) at velocity v, creating back emf epsilon, which by Faraday is: > > epsilon = d phi/dt = d (L * v * B)/dt > > If there is no motion there is no back emf. If there is motion v then > the back emf should be proportional to v. > > It makes the resistance *look* larger as a function of rpms because the > voltage drop across the motor gets bigger with rpms. Yes, I did understand this aspect (at least a little!). What was bothering me is that if the actual *resistance* varied with RPM, that would *look* *like* variation in the induced back EMF, which would confuse any attempt at understanding the relationship between the "useful" power draw and RPM. Assuming your model is correct, as you say somewhere, the back EMF should be proportional to i^2. On the other hand, "back EMF" due to resistance goes as i, so it should be easy to sort them out: Spin the shaft at constant RPM, and apply varying (relatively low) currents to it. (Seems a lot easier than the experiments you've been doing...) You said, later on in the email, > The resistance is not a function of rpm. It is a function of > temperature, but the back emf effects start out when the motor is cold > and continue when it is hot. Of course it will be a function of temp, but my thought was that it could also be a function of whether the shaft is spinning. In particular, if you're getting arcing in the bearings when the thing is in operation and spinning, it's a safe bet that the current's following a path when it's spinning that it's not taking when the motor is stationary, and maybe that's a higher resistance path. (But maybe that doesn't make sense, either, and it should actually be a lower resistance path, via the arcs?) But any difference would probably be an abrupt, on/off thing that showed up when it started to spin, and your scope traces for the "stopped runs" seem to show the back EMF dropping smoothly as the RPMS dropped (if I didn't totally misunderstand them). And I suspect I'm barking up the wrong tree anyway. > However, with this motor, B varies with v, and not linearly. B drops if > v is too high or too low. It likely is the case the motor is running > (especially when v is stabilized) in a range where B drops with > increased rpms, thus as v grows B drops nearly proportionately. This, > if solidly confirmed, is good confirmation of my hysteresis model of the > motor's performance. > > This is my mental model of the thing anyway. It makes sense, and your diagrams, on pp 15-16 of HullMotor.pdf, made its (assumed) operation pretty clear. > > >> and on the face of it that word >> "hysteresis" suggests the EMF and current may not be in phase. > > It's DC, voltage and current have to be in phase because there is only > one phase. Yeah, I finally got that bit :-[ > >> In >> particular, if the power factor varies as the RPM changes that could >> have odd effects. (You may have mentioned that somewhere; if so, >> apologies, I overlooked it.) > > The main thing that varies is the resistance of the nichrome resistors > as they heat up. > >> >> 2) Is there any way to measure the *resistance* of the running motor? > > On page 12 of: > > http://www.mtaonline.net/~hheffner/HullMotor.pdf > > You can see the voltage drop of the single nichrome resistor > configuration was 0.063 ohms. > > More recent experiments are documented at: > > http://www.mtaonline.net/%7Ehheffner/HullMotor2.pdf > > Note especially the experiment on page 6, and the wiring diagram for it > in Fig. 6. This experiment still used the 0.063 ohm cold resistor at > that time. > > Here are the traces for the motor being stopped: > > http://www.mtaonline.net/%7Ehheffner/HullVAmotorStop.jpg > > The current through the resistor starts out at 5 V. The current is thus > (5 V)/(0.063 ohms) = 79 A, The voltage drop through the motor starts > out at about 2.8 V, so the resistance Rmotor = (2.8 V)/(79 A) = 0.036 ohms. > > I added some shunts to the nichrome resistor to increase current, at the > risk of blowing the battery. Here are traces from a regular run, and > here are traces from the stopped run using the new resistance: > > http://www.mtaonline.net/~hheffner/HullShuntRun1.jpg > > http://www.mtaonline.net/~hheffner/HullShuntStop1.jpg I'm not sure I understand what I'm seeing in HullShutStop1.jpg. Voltage drop across the motor goes from about 3.5 volts down to about 1.6 volts before you pull the plug -- is that during *braking* of the shaft, so that it slows from full speed down to 0 RPM? That's what I think you're saying; but is that correct? > > I then jacked the starting speed way up and got this: > > http://www.mtaonline.net/~hheffner/HullShuntHighRPM2.jpg > > > >> I'm wondering if it's the same as when it's stopped, > > > I don't think the resistance goes up very much because the motor doesn't > heat up a lot like the nichrome wires. Eh -- good point. It's dropping around 3 volts, at 100 amps or more, it should be getting pretty hot if that's going into heat. But wait, let us think this through. Stopped voltage drop looks like about 1 to 1.5 volts on the scope, running drop looks like about 3 volts, difference is 1.5 to 2 volts. So that's the back EMF, right? Around 1.5 to 2 volts? If that's *NOT* going into heat, then it's going into mechanical energy (or RFI...). If you're putting, what, 140 amps through this, I think you said in the paper, then that's in the ballpark of 250 watts, or about a third of a horsepower. A third of a horsepower -- going into an unloaded shaft? Hmmm.... I think this had better be showing up as heat somewhere, if the energy books are to balance! > In any case, when the run starts > the motor is cold, and thus the resistance is known. The lack of > back-emf change with rpms is clear right from the cold start. Assuming > the battery is fully charged, calculating the back emf is easy, it is > just the running voltage drop minus the stopped voltage drop across the > motor. Since the voltage drop across the motor doesn't seem to change > much despite rpms or run time, it seems to be a bit of a mystery, at > least to me. > > > >> but off hand I >> don't see how to tease apart the effects of the back EMF from the >> effects of the (possibly varying) running resistance. > > >> Not sure why >> resistance would be a function of RPM, of course, but on the other hand >> I'm not sure it wouldn't be, either! > > The resistance is not a function of rpm. It is a function of > temperature, but the back emf effects start out when the motor is cold > and continue when it is hot. > > >
