Maxwell 3D (ANSOFT)

emredemir

Fuxue;

i want to know that. i must be done transient analyses,arent i?

so can i apply ac current ? also i applied in windings and current value is a parametre(current1= 100*sin(w*t) ) is it true or false. Also i applied 500 Volt dc voltage in Winding part and 5 ohm resistance value. Is it ok?

Fuxue Jin

yes, you need to use transient solution in order to get losses.

Excitation 100*sin(W*t) is OK (I guess that should be 100*sin(2*pi*60*time), if frequency is 60Hz, for example)

But I don't know if you can overlap a dc voltage on top of the current excitation. I have not done that so far.

Fuxue Jin

Also, when using voltage excitation, and your winding is specified as stranded, then Maxwell doesn't solve the resistance of the winding coil. The resistance you entered is the winding resistance of the coil, so it will be the same as you calculate the current V/R, and copper loss I^2R by hand. There will be no need to use Maxwell.

emredemir

so you say that dont use maxwell but i can see lots of map in power loss . For example stranded loss copper loss core loss they are avaible i think . in addition What do think about calculate power loss on coil is equal to stranded loss or copper lose.

Fuxue Jin

Sorry, but I don't understand your question very well.

In general,

If your copper winding is specified as stranded, and you use Voltage excitation, then there is no need to use Maxwell to calculate the copper loss, which is simply V/R = I, and loss is I^2R. Here V and R are specified by you as input, and you already know ahead.

If your copper winding is specified as stranded, and you use current excitation, then you will get strandedloss from Maxwell for your copper winding. However, if you can calculate the resistance of your winding using the length and cross section of your copper, then you will get the copper loss by I^2R since your current is known factor. This I^2R loss will be the same as you will get in Maxwell (the strandedloss).

If your copper winding is specified as solid, then you will get solidloss in Maxwell. In this case, the Maxwell will simulate the conduction path resistance and give out copper loss. When you specify your copper winding as solid, then you have to draw the real winding. Meaning if you have 10 turns, then you have to drawing real 10 turns. If you have 100 turns, you have to draw 100 turns. But it is not possible for Maxwell to simulate a 100 turns winding unless you have a very powerful machine with lots of memory.

So you have to decide how you want to specify your winding (solid or stranded) in order to know if using Maxwell makes sense to you or not.

pourmoosa

Dear all
I'm a new user and I need help about maxwell 2D transient.
i want to simulate a linear induction motor using maxwell 14.
when i define a band for transient state and then run maxwell,an error occures.
error is:"there might be an isolated hole in the band"

how i can remove this error?
sorry for my English;
thank you;

emredemir

Fuxue i'm now understand that everything you told me. i use copper windings stranded and voltage excitations on coil. So that is very senseless to calculate power . Because i would give the value of the voltage and resistance but i want to explain that i gave this excitations and i could saw the losses do you understand. for example i would give v=100 V and R= 5 ohm so İ = 20 A but if i gave V= 100*sin(W*t) i could see the losses because of the AC variables. Do you understand?

if i have wrong things you can say me... but what do you suggest me to solve?
Because this is my thesis so important for me and i'm grateful to you very very thank you Fuxue

Fuxue Jin

I have not done anything like this and may not be that helpful. But based on my understanding, a band object needs to be a sheet in 2D or a region in 3D without any interruption. The band object should not interfere with anything else. It is between the moving objects and the stationary ones.

The error sounds like the band object is discontinuous due to some interruption.

Fuxue Jin

Let's see the copper ohmic loss;

For DC, V=100V, R =5 ohm, we have I=20A, and power loss is 2000Watts.

For AC = 100* Sin(W*t), in this case, 100 is peak current, you need take the rms value for power loss, so Vrms = 70.7. For R = 5ohm, the power loss is , 999.70Watts.

In order to have the same equivalent DC current 100A, the AC peak current need to be 141.4A.

For AC current, if the wire is multi-stranded, then the current density is uniform and it will have the same power loss if the AC rms value is equal to the DC current value. It doesn't matter it is AC or DC because the current density is uniform.

Now, for AC input, if you want to consider skin effect/eddy current at high frequency (say a few MHz), and want to show the different power loss between DC and AC, you need to specify the copper as solid, not stranded, in Maxwell. As I said, in order to specify the copper as solid, you have to draw your winding turns. If it is 10 turns or so, then it will be OK. Otherwise, it is not practical to draw a few hundred turns in Maxwell.

gerbi

According to definitions, real power is calculated as average per cycle not rms.
From my expierence, as you've said, maxwell avg power and hand calcs are pretty close in value.

emredemir

Dear Fuxue;

now i'm understand and i want to see eddy current/skin effect. So power loss may be
different from the when we use stranded AC current values. So you suggest me to give Copper windings are solid current value,arent it? so i must draw real turns.

but i want to send you subject of my thesis :http://www.maglev.ir/eng/documents/p...2006_T5_11.pdf

and i use a document from ansoft ,name was; Ansoft Maxwell 11 3D UserGuide you can see how did i draw coil.(page 149 150 151).

gerbi

The way to draw a coil taken from Maxwell User Guide is good for stranded currents.
Yes, you need to create every turn to observe eddy effect. How many turns do you have ?

emredemir

Gerbi;


i have lots of turns but firstly i dont know how to draw this turns. draw by a helix or spiral? i realy dont know but i made turns by helix a month ago. Gerbi my problem
is power loss and frequency. I am working in nearly 20 KHz. So What do suggest me to see power loss on coil. Fuxue said that If i wanted to see Eddy current and differ
between AC to DC power losses ,i must be draw turns. He said that me you can see.

İndeed ,The first problem for me power loss. so What do think, i have to draw solid turns or i m continue to Stranded Current windings

gerbi

To observe eddy current losses in AC do as Fuxue Jin said. Draw solid turns of copper.

Lots of turns you say. I see no way to analyze this device in full geometry.
You need to reduce number of turns to pretty low number (it depends on computer CPU and RAM memory) or you won't be able to create proper mesh to calculate this accurately.

emredemir

but i want to know that how can i draw this turns . draw helix is usefull or not.
i think 1 turn even enough for me but HOW? in addition gerbi can i send you my maxwell project by mail or other ways. may be it is very easy to see what i do

gerbi

Tricky job. Coil will be made of rectangular or round conductor ? It's wounded on rectangular core, right ?

If this is rectangular wire then just draw rectangles as part of conductor, merge all and fillet the edges.

If this is round wire: draw a cylinder of required lenght and diameter of straight part of wire. Then duplicate it around axis (90 deg). You'll have two cylinders perpedicular with one common corner (there is a hole but it can be filled with spherical element). Merge all and you have one element ready.

I don't know how good are you in drawing in maxwell. There are pretty many options there (merging, substracting, splitting.. - planes can be changed) so take your time.

I'm sorry but currently I'm pretty busy with some other projects.

emredemir

Ok Gerbi;


i really understand you are bussy but when you have free time Can you turn me for this topic.
also i can say my coil made by rectangular so i have to use rectangular for turns ,arent i? So i draw lots os rectangular and merge them . but i think draw one rectangular also useful for me,isnt it? because i will be take same results from programme.

For example one turn with current is 500 A. Other type is 5 turn with 100 A. they are the same things F= N*I . so Fuxue said and show me that i have to give in windings part Current= 100*sin(2*pi*20000*time) and solid . so it will solve eddy current and skin effect i think so...