Edgar J. Kaiser
Certified Consultant
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Posted:
1 decade ago
7 mai 2013, 05:34 UTC−4
Hi,
the field is the gradient of the potential. So your request is not physical. A uniform potential gives zero field. A linear potential distribution gives a uniform field.
Cheers
Edgar
--
Edgar J. Kaiser
www.emphys.com
Hi,
the field is the gradient of the potential. So your request is not physical. A uniform potential gives zero field. A linear potential distribution gives a uniform field.
Cheers
Edgar
--
Edgar J. Kaiser
http://www.emphys.com
Please login with a confirmed email address before reporting spam
Posted:
1 decade ago
7 mai 2013, 12:57 UTC−4
Thank u sir
I apply 33Kv voltage at HV end. I got result, in this result potential along the insulator surface is linear distribution, but not uniform field distribution along the surface and also field at HV (33KV) terminal as same as ground terminal, why sir ?
i attach my model image.
Thank u sir
I apply 33Kv voltage at HV end. I got result, in this result potential along the insulator surface is linear distribution, but not uniform field distribution along the surface and also field at HV (33KV) terminal as same as ground terminal, why sir ?
i attach my model image.
Edgar J. Kaiser
Certified Consultant
Please login with a confirmed email address before reporting spam
Posted:
1 decade ago
7 mai 2013, 15:08 UTC−4
I would expect a higher field close to the terminals and the geometry is the same on both ends.
--
Edgar J. Kaiser
www.emphys.com
I would expect a higher field close to the terminals and the geometry is the same on both ends.
--
Edgar J. Kaiser
http://www.emphys.com
Please login with a confirmed email address before reporting spam
Posted:
1 decade ago
8 mai 2013, 09:02 UTC−4
Hi
In HV end : field is high bcoz 33Kv applied, but same field will occur in ground end (zero potential), why sir?
Hi
In HV end : field is high bcoz 33Kv applied, but same field will occur in ground end (zero potential), why sir?
Edgar J. Kaiser
Certified Consultant
Please login with a confirmed email address before reporting spam
Posted:
1 decade ago
8 mai 2013, 10:09 UTC−4
Sure, your geometry is symmetric. It doesn't matter where HV and ground are connected.
But looking at your pictures, I think you are making a very basic mistake. You don't have an air volume around your device. So you confine the model into the interior of the device. That is probably not what you want.
Another hint: The device has axial symmetry. You can save tremendous memory and computing time by setting up this in 2Daxial geometry. And the precision of the results can be much better.
Cheers
Edgar
--
Edgar J. Kaiser
www.emphys.com
Sure, your geometry is symmetric. It doesn't matter where HV and ground are connected.
But looking at your pictures, I think you are making a very basic mistake. You don't have an air volume around your device. So you confine the model into the interior of the device. That is probably not what you want.
Another hint: The device has axial symmetry. You can save tremendous memory and computing time by setting up this in 2Daxial geometry. And the precision of the results can be much better.
Cheers
Edgar
--
Edgar J. Kaiser
http://www.emphys.com