After all Covid-related hassle, a new micropost!
Do you know how to calculate the induced voltage – or coil fluxes, for that matter – from finite element analysis results?
Most of the time, nobody cares. After all, your software of choice is supposed to do that for you. But, it’s still worthwhile to educate oneself every now and then. Besides, it’s not that uncommon to get a bunch of raw data from a client, subcontractor, or a retired coworker, and having to figure out things yourself.
First, some basics.
By definition, coil flux is the surface integral of the flux density, over a surface enclosed by the coil:
Now, looks good on paper – at least to a high school physics student. In practice, there are a few complications
First, coils are sadly not infinitely thin. Instead, they usually consist of several series-turns of smaller or larger conductors. Thus, the outermost turns actually enclose a larger surface than the innermost ones. For very large conductors, such as hairpins, the inner part of a single conductor may enclose a significantly smaller area than the outermost part.
The second issue is related to computational accuracy. Most magnetics problems – by far the most – are solved (under the software hood) with the vector potential (A) formulation, defined by:
Since the flux density is now a derivative of our primary quantity A, it is considerably less accurate. Meaning, if we use B directly, our results won’t be as reliable as possible.
Solution in 2D
Getting rid of derivatives
The solution to both of these results is relatively obvious in 2D. The only special trick we have to know is the Stokes theorem: After plugging in in place of B, we can transform the above surface integral into a line integral without any derivatives:
In other words, we have replaced the surface integral by a line integral, across the surface boundary.
As this gets us rid of the derivative, we have solved the second issue.
Accounting for coil size
The first issue – large area of each coil side – is also easy to solve in 2D. Remember, we are speaking about a 2D solution, with the field entirely in the xy-plane. This means that the vector potential is entirely z-directional, and that the end-winding fluxes contribute nothing to the computed flux.
Now, remember that we are speaking about a coil. The coil consists of the active part – two z-directional slot-bound coil-sides inside the motor core – plus then the end-winding. As the end-winding does not contribute to the computed flux in 2D, we are left with just the two coil-sides.
Therefore, the above line integral is actually simplified into a simple evaluation of the vector potential at the coil sides:
The above expression still assumes the coil sides to be extremely thin. To account for the non-zero size, we can use a simple average:
In plain words, we get the flux linked by a single coil by calculating the average vector potential over each coil side, and then multiplying their difference by the problem length.
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