How to get the impedance of part of a conductor loop?
How to get the impedance of part of a conductor loop?
For example, in the steady AC magnetic application, a wire carries high frequency AC current in the centre. Then the induced eddy current flows in the copper ring.
However the 1/4 ring (shown in the light brown) is of interest. how to get the impedance of the 1/4 ring?
I set up the circuit like above with solid conductors. But don't know how to assign the terminals to the geometry.
Answers
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Hi Wenjun,
what you are doing is correct. But there is an issue is that you can't consider the ring as two solid conductor region as you can't assign external terminal since the ring is considered as closed coil by Flux and you can only assign internal terminal.
What I think you should do is to delete the left portion of the circuit (related to the ring) and assign each volume (portion) to a different volume region of solid conductor type. you don't need the circuit since the current is induced. In this case you don't have access to the voltage and current directly, but I am thinking you can use the active and reactive power to perform the computation, those can be computed on each region separately. you can use a sensor of integration type to compute the total current on a region if needed.
Also the problem seem to be symmetrical from the picture above. if this the case you can compute for the full ring and just divide by 4.
Thanks,
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Mohammed Elamin_22169 said:
Hi Wenjun,
what you are doing is correct. But there is an issue is that you can't consider the ring as two solid conductor region as you can't assign external terminal since the ring is considered as closed coil by Flux and you can only assign internal terminal.
What I think you should do is to delete the left portion of the circuit (related to the ring) and assign each volume (portion) to a different volume region of solid conductor type. you don't need the circuit since the current is induced. In this case you don't have access to the voltage and current directly, but I am thinking you can use the active and reactive power to perform the computation, those can be computed on each region separately. you can use a sensor of integration type to compute the total current on a region if needed.
Also the problem seem to be symmetrical from the picture above. if this the case you can compute for the full ring and just divide by 4.
Thanks,
Thanks, Elamin.
when I perform the integration of the current on a region.
the results have 3 components. could you explain more on that? why 3 components?
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Wenjun Zhu said:
Thanks, Elamin.
when I perform the integration of the current on a region.
the results have 3 components. could you explain more on that? why 3 components?
Hi Wenjun,
The current density (J) is a vector quantity, so the three parts are X,Y and Z components of that vector. I think you should look at the magnitude (Modulus of the vector) as Mod(J).
Thanks,
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Mohammed Elamin_22169 said:
Hi Wenjun,
The current density (J) is a vector quantity, so the three parts are X,Y and Z components of that vector. I think you should look at the magnitude (Modulus of the vector) as Mod(J).
Thanks,
to get the current in the copper ring, you mean using the integration on face (like the cut-section of the ring) of the Mod(J)? But it is weird that different results on the two cut-sections.
Consider this new model below. a copper wire carries AC current in the centre. Rectangle shaped steel forms a closed loop which the induced current flows in.
Questions:
1. the reactive power on the centre copper wire from the compute on region is 18.4 Var, but from the circuit solid conductor is 732.9Var. Huge gap there.
2. the integral on face of Mod(J) for different cut-sections varies. why?
on face No.6 highlight in the right pic.
on face No.20 highlight in the right pic.
if I can send you the Flux project to make it more clear?
Thanks.
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Wenjun Zhu said:
to get the current in the copper ring, you mean using the integration on face (like the cut-section of the ring) of the Mod(J)? But it is weird that different results on the two cut-sections.
Consider this new model below. a copper wire carries AC current in the centre. Rectangle shaped steel forms a closed loop which the induced current flows in.
Questions:
1. the reactive power on the centre copper wire from the compute on region is 18.4 Var, but from the circuit solid conductor is 732.9Var. Huge gap there.
2. the integral on face of Mod(J) for different cut-sections varies. why?
on face No.6 highlight in the right pic.
on face No.20 highlight in the right pic.
if I can send you the Flux project to make it more clear?
Thanks.
Hi Wenjun,
1- This is normal. The reactive power measured by the region it is only for the wire. while the one measured by the solid conductor not only include the reactive power for the wire but the one leaked to the air and the one induced on the steel loop. As the solid conductor is connected to the source and it has to respect the power balance. You can double check this by computing the reactive power on the entire domain (or on other region) and check if it balance out.
2. As the current on the loop is induced, I don' expect it to have a uniform density and direction along the loop (like the case when it supplied with current using source). Hence, the two faces having different values might be normal. I think it would make sense to do the integration over the volume of interest instead of the faces.
I hope this helps,
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Mohammed Elamin_22169 said:
Hi Wenjun,
1- This is normal. The reactive power measured by the region it is only for the wire. while the one measured by the solid conductor not only include the reactive power for the wire but the one leaked to the air and the one induced on the steel loop. As the solid conductor is connected to the source and it has to respect the power balance. You can double check this by computing the reactive power on the entire domain (or on other region) and check if it balance out.
2. As the current on the loop is induced, I don' expect it to have a uniform density and direction along the loop (like the case when it supplied with current using source). Hence, the two faces having different values might be normal. I think it would make sense to do the integration over the volume of interest instead of the faces.
I hope this helps,
Thanks for your patient! one more thing is the unit for J or Mod(J) would be A/m2. but taking the integration over volume (m3) , ends up in A*m. that's not the current, right?
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