Besoker blender

[T.M.Haja Sahib]

I took the following paras from
http://forums.mikeholt.com/showthread.php?t=145581 (post#103)

You are no doubt conversant with the definition of the Amp?re as the constant current that willproduce an attractive force of 2 ? 10^ -7 newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in a vacuum.

That gives us a starting point. Let’s assume that the fault current is 10kA. We have something under that from the supply in our test bay so maybe not an unreasonable assumption. How far apart are the MOVs? Say 20mm (about ? of an inch). Chuck those numbers into the Besoeker blender and you come up with a force of 5N.

The result of 5N from Besoeker blender appears to be too small to be correct.

 

[charlie b]

Looks too high to me. I am presuming that the force varies linearly with the distance between the conductors. So if the force is 2E-7 at one meter with one amp of current, then at 20 mm (1/50 meters), with 10,000 amps of current, the force would be (2E-7) x 50 x 10,000, or 0.1N. What am I missing? :?

 

[Besoeker]

I took the following paras from
http://forums.mikeholt.com/showthread.php?t=145581 (post#103)

The result of 5N from Besoeker blender appears to be too small to be correct.

Then perhaps you'd be good enough the present your calculations.

 

[Besoeker]

Looks too high to me. I am presuming that the force varies linearly with the distance between the conductors.

It's generally taken as being an inverse square law rather than linear.

 

[Smart $]

It's generally taken as being an inverse square law rather than linear.

The inverse square law is based on force emanating from a single point. The ampere definition uses parallel conductors, so the emanation of force is cylindrical rather than spherical (i.e. think of it as an infinite series of overlapping spheres along each conductor as the equal force lines).

 

[T.M.Haja Sahib]

Force=2x10^-7xIXIXL/D, where I=Current, L=length of current element, D=distance between them. Assuming I=10000A, L=0.08m, D=0.02m, the force
F= 2x10^-7x10000x10000x0.08/0.02=2x10x4=80N

 

[charlie b]

It's generally taken as being an inverse square law rather than linear.

Things that spread out from a single point tend to follow an inverse square law. Things that sit between infinite parallel planes tend to be uniform. I wasn't sure about infinite parallel lines. But if you are right about that relationship, it would make the force (2E-7) x 50 x 50 x 10,000, or 5N. So I confirm your calculation.

 

[charlie b]

Force=2x10^-7 x I x I x L/D

I think it is the distance between conductors that gets squared, not the current.

 

[T.M.Haja Sahib]

But I referred to manuals before posting.

 

[Besoeker]

I think it is the distance between conductors that gets squared, not the current.

It is the current. He is correct.
Still. 80N isn't so much.

 

[Smart $]

... 80N isn't so much.

No, it isn't...

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html#c2

 

[pfalcon]

Dissipation due to distance is derived from the area formula of the expanding wave.

From a point: Spherical Area = 4(pi)r²
From a line: Cylindrical Area = 2(pi)rh
From a plane: Planar Area = wh

A point wave ratio of 1m (1000mm) to 20mm would be (1000/20)²=50²=2500:1
A line wave ratio of 1m (1000mm) to 20mm would be (1000/20)=50:1
A plane wave ratio of 1m (1000mm) to 20mm would be 1:1

 

[T.M.Haja Sahib]

Still. 80N isn't so much.

One reason is you limited current to 10kA, while surge current of even 100kA may flow.

 

[Besoeker]

One reason is you limited current to 10kA, while surge current of even 100kA may flow.

I limited it to that because it's fairly typical of what I come across on the LV supply.

 

[T.M.Haja Sahib]

I limited it to that because it's fairly typical of what I come across on the LV supply.

The surge voltage in the building wiring system does not exceed 10kV. But the surge current can have much higher magnitude. Arresters with rating of even 160kA are available.

 

[Besoeker]

The surge voltage in the building wiring system does not exceed 10kV. But the surge current can have much higher magnitude. Arresters with rating of even 160kA are available.

I'm not sure that's entirely relevant.
The (closed) thread from which you took my post was about surge suppression devices on a 600V (i.e LV) system.

 

[T.M.Haja Sahib]

The web link below shows a surge arrester that can be used in LV system for a surge current of even 160 kA.
http://www.directindustry.com/prod/apt-inc/surge-arresters-type-2-68051-603384.html

 

[Besoeker]

The web link below shows a surge arrester that can be used in LV system for a surge current of even 160 kA.

And??

 

[T.M.Haja Sahib]

And??

I think you now agree that a surge arrester with parallel elements can also be damaged by mechanical force of surge current passing through them and not only by heat energy of the currents.

 

[Besoeker]

I think you now agree that a surge arrester with parallel elements can also be damaged by mechanical force of surge current passing through them and not only by heat energy of the currents.

Now agree???
I don't believe I ever excluded that as a possibility.