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Nice thread. I have been pondering on some DC motor theory and perhaps you can help me throw some light on an issue. Speed of a DC motor is proportional to emf and inverse proportional to magnetic flux (field current). I have e.g. experienced the following: If the field circuit in a separately excited DC motor is opened, the motor will rev up until it probably may break. Why is this so? I mean, I am familiar with the equations, but why is it that a decrease in magnetic flux till cause the speed to increase; and why will the motor rev up till chaos with flux=0? I can't intuitively see why the flux would limit the speed. Isn't it the same flux that initially causes the armature to rotate?
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Dec 4, 2011 22:22
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Please tell me about artificial neutral points... Let's say we have a wye-connected three-phase transformer with the neutral point connected to earth and a phase-to-phase voltage of 400VAC and 230VAC phase-to-neutral (a common configuration in Europe). Now let's get rid of the earth connection. If the loads are balanced we will still have 230VAC phase-to-neutral, the neutral now being called an artificial neutral point since it hasn't got a fixed reference point as the earth connection has been disconnected. Now, if the three-phase loads are unbalanced due to excess output on one phase, what effect will this have on the neutral point and the voltage on the other phases with respect to the neutral? I've heard all the time that the voltage on a single phase with respect to the artificial neutral point will vary with the difference of loads, but I can't really understand why. According to KCL the current will be zero in the middle point of the wye regardless of the loads. What am I missing?
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Jan 6, 2012 10:34
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grover posted:The neutral would be left to float, and could be any voltage. There is a magnetic component of transformers that is generally ignored, but serves to keep balanced 3-phased loads of floating delta (and wye) somewhat close to centered on ground. It's not enough to overcome a severe imbalance, though. Why would it be left to float and why could it take any voltage? What's the theory behind it, the mathematics?
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Jan 6, 2012 18:46
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grover posted:Using 120/208 as an example, the mathematics are that you will always have 120V between the neutral and each of the 3 phases, and 208V between each of the phases, but there is nothing keeping the neutral at 0V with respect to ground. If the neutral floats up to +100V with respect to ground, all the other phases will be 100V higher, too, with respect to ground. The relative voltages between them will be unchanged, though. I see! So the phase to neutral potential will not vary with changed loads. That sounds pretty obvious in fact, since otherwise the light would keep changing in brightness in hospitals and Norway and stuff that use the IT system. I still don't understand the calculations behind why the neutral to ground potential changes though. Say you have a wye-configured 120/208 secondary with the phase current I1 = I2 = I3. The sum of the currents will be 0. But if you have asymmetrical loads, say I1 = I2 = x, I3 = 2x; the sum current till be greater than 0. But how does this affect the voltage with respect to ground? Why will the neutral point of the symmetrically loaded transformer "see" a potential equal to ground, when the only connection it has to ground is capacitive and (perhaps) negligable? I do understand the risks (and the benefits) of not having a fixed reference point, but I really can't manage to grasp HOW the ungrounded neutral correlates to ground when they are not electrically connected (well, except cap). Is it that you build up an electrical field not part of the electrical field of earth, so that the potential between these fields are more of a "random" thing than something you can predict? Having the electrical components grounded makes it easy to predict what may happen, but leaving the buildup of electric fields to nature makes it tough to deal with.
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Jan 7, 2012 08:15
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