Discussion Forum

Hi

for the axisymmetric case its 2*pi*r the multiplier to go from a cut view to the full solid, but in V4 the 2*pi is already introduced see the doc and the release notes (importing older files might get wrong values in V4). To be sure you get it right, check the full equations that COMSOL is using te the "equation sub-node" (turn on the equations in the "options preference"

for your case 2-3 I do not really udnerstand, and domain integration should work OK, so I'm obviously missing a point here

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Good luck
Ivar

Thank you Ivar for your answer.

First, I need to tell you that I'm using the version 3.5 a. I have a cilindric solid permanent magnet with radius equal to 5mm and height equal to 3mm, sorrounded by a single loop (coil) and I want to obtain the voltage induced at the coil due to the motion of the permanent magnet. I am using the Axissymmetry mode.

For this, I need to do the integration of the z component of Magnetic Flux Density, B (Bz_emqa) over the cross section of the magnet in order to get the magnetic flux through that section. Eventhough the cross section is the same through all the magnet, the magnetic flux density changes along z axis and also along r axis. I understand the voltage induced in the coil is given by

V_loop = -d(magnetic flux) /dt (the dervative respect to time of the magnetic flux) equation 1

I would like to do the integration of Bz_emqa over the cross section area, for ALL values of z, and later try to get the partial derivative of that integration respect to z and multiply it by the derivative of z respect to time like this

V_loop= -d(magnetic flux)/dt = - {partial derivative respect to z of[ int Bz_emqa*dA]} * [dz/dt] equation 2

If I understood well your answer, if I do, for example, the boundary integration in one end of the permanent magnet and multiply it by 2*pi* r I could get the integration over the area defined when I revolve that radius over 2*pi radians (the whole circle)

CASE A : If I just select go to POSTPROCESSING> BOUNDARY INTEGRATION> AND I SELECT ONE BOUNDARY OF THE MAGNET I GET THIS RESULT (SEE PICTURE 1 IN THE ATTACHMENT)

"Value of line integral: 4.300116e-4 [Wb/m], Expression: Bz_emqa, Boundary: 4

and if I multiply this by 2*pi *0.005m (the radius of the permanent magnet edge) I get

1.350921284e-5 [Wb]



CASE B: If I select the "compute surface integrals (for axisymmetric modes)" then I get.

7.460459e-6 [Wb] (SEE PICTURE 2 IN THE ATTACHMENT)


Because this results which I understand should yield the same answer are different, I used OPTIONS>EXTRUSION COUPLING VARIABLES>SUBDOMAIN VARIABLES and get the 3D representation of my 2D axisymmetric model (where By_3D is equivalent to Bz_emqa in 2D axisymmetric mode).

In this integration I obtained

7.305723e-6 [Wb] (SEE PICTURE 3 IN THE ATTACHMENT)

which is also different. Which of these 3 values for the magnetic flux through one end of the magnet (integration of the perpendicular component of B over the cross-section area) is correct?


2- I CAN OLNLY DO BOUNDARY INTEGRATION AT BOUNDARIES. HOW CAN I GET THE VALUE OF THIS INTEGRAL FOR ALL VALUES OF Z ALONG THE Z AXIS? CAN I JUST DRAW THE SOLID CYLINDRIC MAGNET AS A BUNCH OF PILED UP CYLINDRIC MAGNETS, ALL PILED UP, TOGETHER, AND DO THIS FOR ALL BOUNDARIES TO OBTAIN AT LEAST AN APPROXIMATION OF THE VALUES OF THE INTEGRAL FOR ALL Z VALUES?

3- IS THERE A WAY TO OBTAIN A RELATION (AT LEAST A FITTED CURVE) BETWEEN THIS INTEGRAL AND Z,R COORDINATES? oR IS THERE A WAY TO OBTAIN DIRECTLY THE partial derivative respect to z of [ int Bz_emqa*dA] ALONG THE Z AXIS?


THANK YOU VERY MUCH FOR YOUR HELP!!!

Hi

a partial reply for the 2*pi*r (as you hae many questions, and some are far more time demanding than others ;)

Do not forget that COMSOL is always working in 3D, but if you select 2D it assumes a default depth of 1[m] (slightly physics/application mode dependent though pls check carefully each time) and in 2D-axi it's the "loop length" = 2*pi*r (for each value of r) that is gicing the "third dimension".

Now if you integrate an edge in 2D axi you get a result as [something/m] because it is per "[1/m] loop length". If you check the tick box "evaluate surface integral", while having an integration over an edge, you are in fact simply multiplying internally by "2*pi*r*Your_expression" inside the integral (not 2*pi*0.03[m]).

In comsol the integrations are not clearly shown up. all equatins are related to a "finite element" hence a small dxdydz volume and it's implicit that everything (almost) is integrated over a volume or a surface or an edge length (or summed over points or node, the latter for reaction force reactf() calculations)

Try it out, with simple surfaces, (mesh the result and solve "get inital value" as integration postporcessing is done on the meshed entities)

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Good luck
Ivar