Dynamically induced hell - Page 1
Forum section: electronics / beginners

author: Connoiseur
author_rank: Regular Contributor
timestamp: December 22, 2015, 10:39:06 am
content: Someone asked me this very basic but mind boggling questionAssume a current transformer with a pure resistive burden such that the core is just at the verge of saturation. Now it is moved along the current carrying AC transmission line with some velocity ‘v’. Will the core saturate and what will be the voltage waveform across the burden?I'm totally confused about two things:First I think, the core will not get saturated because the field lines are cylindrical at any instant of time. The core is moving such that there is no change in flux (due to no radial movement b/w wire and core), therefore d?/dt = 0 and hence emf=0. Thus waveform is sinusoidal.On the other hand I think that emf = Blvsin?,Where, B= flux density; l=length of conductor i.e. axial length of toroid; v=velocity of movement; ? angle b/w flux & conductorTherefore the core will get saturated due to another AC component (magnitude proportional to velocity) and the waveform will be chopped sinewave.Which assumption is correct (or neither) because they seem quite contradictory? Any help would be greatly appreciated.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2015, 11:19:07 am
content: The second formula is a very special case. It's completely lacking vectors (namely, B, v and l are properly all vector quantities!), and uses only one angle. So it seems to assume velocity perpendicular to B, to get the maximum value of the remaining vector product. (E is also a vector quantity, so E = v x l x B makes sense.)(EMF typically being a scalar, means taking the magnitude. Which is only appropriate when a direction has been solved for, which usually means integrating along a winding or something like that. Properly, the path has to be a closed loop, which a segment isn't even!)Most important is that the CT's winding cannot move through the flux, because the flux is largely contained within the core, which moves with it. That might help your mental picture.If we talk about the situation without a core, i.e., a Rogowski coil, then we should talk about each segment of the winding.1. We can ignore the solenoidal component, because this component of l is parallel to the magnetic field, so l x B = 0. (Consider a conventional, single pass winding on a toroid form. Consider what the winding looks like if the form thickness goes to zero: rather than a toroidal helix, it's a toroid itself. And therefore makes one turn through space, parallel to the B-field from the current-being-sensed wire. This solenoidal component can be opposed by making a two layer winding where the winding pitch reverses at the end, so it doubles back on itself, or by placing the lead-in wire beneath the winding-in-progress, so both leads end up at the same end of the winding when complete.)2. That leaves two directions. For a toroid of rectangular cross section, we have four segments: (1) the innermost segment, which is parallel to the central wire, closer to it, and going in the same direction (say); (2) the far end, where the turns are radial, rising from the inside to the outside; (3) the outermost segment, which is parallel to the central wire, farther away from it, and going in the opposite direction; and (4) the radial returns from outside back to inside.3. The radials (2) and (4) are equal and opposite, so can be ignored as well. They are displaced by the height of the toroid, but the field is uniform along the axis, so that makes no difference. (If the field were conical somehow, there would be a difference. But it's rather hard to get a decreasing current along a wire.*)(*Unless the wire is delivering a displacement current into free space, i.e., its electrical length is comparable to the frequency of AC going through it. But outside of antennas, we don't need to make this modification to Kirchoff's.)4. That leaves the inner and outer segments. The inner segments (1) are closer to the central wire, and therefore see more induction than (3) outside. (This is how a current transformer normally operates.)If we displace the assembly with axial velocity, then we have l x B = 0 for the solenoid component (still), v x l = 0 for the inner and outer segments (1) and (3), and v x l x B != 0 for the radials (2) and (4) but they still oppose exactly, for the same reasons.So the correct answer is again 0, as well it should be.Note that I've applied the second formula as such, taking some liberty with its definition (using the more general vector form, hopefully to be more illuminating than cryptic), and applied it to the same problem, straight line segments of length l. The segments are arranged into a winding, so that we can properly sum (integrate) a (scalar) EMF for the total winding, which is what we are looking for.Tim

author: Connoiseur
author_rank: Regular Contributor
timestamp: December 22, 2015, 11:52:08 am
content: Thanks for the reply but I’m afraid I didn’t quite get you.This is actually a hypothetical question. I hope the image will make it clearer.Quote from: Connoiseur on December 22, 2015, 10:39:06 amPS: I typed in word and pasted here in order to get Greek letters theta and phi but it automatically changed to qs mark somehow so, it’s(d phi)/dt andBlv sin theta not blvsin?.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2015, 12:10:52 pm
content: Yeah, this forum is terribly anglocentric on its character settings...Here's the segments labeled, and a cross section. Since it's radially symmetric, we only need to look at one turn (assumed in plane); they're all the same. And any segments of wire that join one turn to another are aligned parallel to the toroid core, so fit into 1.Tim

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 22, 2015, 03:27:10 pm
content: I feel compelled to note that this was a really interesting question!