Please discuss volts/mil vs electrical stress

[electrofelon]

I am still a little fuzzy on the precise effect of shielding in MV power cables. OF course the textbook definition of the purpose of shielding is to evenly distribute the electrical stress in the dielectric. I think what has been confusing me is that I have assumed electrical stress would be the same for a given volts/mil. So is this not the case? Please see the attached drawing. Lets say this is a typical 15KV cable with an insulation thickness of 175 mils, so 85 volts/mil. Now I dont see how we could EVER have more than 85 volts/mil, even on the bottom of the cable where it is resting on the grounded conduit. The potential difference is what it is. But is there more electrical stress on the bottom?

 

[gar]

220216-0943 EST

electrofelon:

If you assume all the conductors in the cable are at the same potential, then you can view them as a single conductor.

Next place an insulator of 175 mils around that conductor, and make its dielectric constant the same as air, 1 . This insulated wire is now placed in a conductive tube much larger than the insulated wire so it is located way off center of the tube. The air in the tube is also a dielectric constant of 1. Making both dielectric constants the same is simply to make it easier to visualize the electric field.

Now you need to understand how to hand sketch electric field flux lines, and equipotential lines. This you do by drawing curvilinear squares to map the field. Manually it is a trial and error process. The flux lines in that drawing are not well drawn. But to some extent illustrate what is taking place. The flux lines will enter the conductors perpendicular to the conductive surface.

Equipotential lines will be much closer together near the bottom of the drawing, and therefore the electric field gradient much greater than at the top of the internal conductor. It is about 65 years since I last did any such sketches.

Having two different dielectric constants makes sketching more difficult, but the same concept applies.

.

 

[electrofelon]

220216-0943 EST

electrofelon:

If you assume all the conductors in the cable are at the same potential, then you can view them as a single conductor.

Next place an insulator of 175 mils around that conductor, and make its dielectric constant the same as air, 1 . This insulated wire is now placed in a conductive tube much larger than the insulated wire so it is located way off center of the tube. The air in the tube is also a dielectric constant of 1. Making both dielectric constants the same is simply to make it easier to visualize the electric field.

Now you need to understand how to hand sketch electric field flux lines, and equipotential lines. This you do by drawing curvilinear squares to map the field. Manually it is a trial and error process. The flux lines in that drawing are not well drawn. But to some extent illustrate what is taking place. The flux lines will enter the conductors perpendicular to the conductive surface.

Equipotential lines will be much closer together near the bottom of the drawing, and therefore the electric field gradient much greater than at the top of the internal conductor. It is about 65 years since I last did any such sketches.

Having two different dielectric constants makes sketching more difficult, but the same concept applies.

.

Gar,

Yes, just so we have it to look at, attached is what we have in a shielded conductor.

The part in red from your quote: So am I understanding correctly that without shielding, like in the first picture, for a given path through the dielectric will not have a uniform volts/mil? Say it could be say 120 volts per mil for the first 87.5 mils and then 51 volts per mil for the outer 87.5 mils? IS that what can happen?

 

[gar]

220216-1113 EST

electrofelon:

In a shielded conductor as drawn you have symmetry radially between the center wire and the coaxial outer shield. Thus, there is a uniformity of flux and field lines radially. Move the center conduct off center and this distorts the uniform field.

Do something different. Split the coax cable down the middle vertically, and use two different dielectric constant insulators. Now you develop nonuniform fields even though the conductors are coaxial.

.

 

[gar]

220216-1124 EST

electrofelon:

Looking back I did not answer your question. With the coaxial cable, and assuming a uniform dielectric constant, the field intensity is not a constant as you move radially outward from the inner conductor. This is because as the radius gets greater, then as you move outward the curvilinear squares become larger. Thus, the voltage gradient becomes less.

.

 

[wwhitney]

The part in red from your quote: So am I understanding correctly that without shielding, like in the first picture, for a given path through the dielectric will not have a uniform volts/mil? Say it could be say 120 volts per mil for the first 87.5 mils and then 51 volts per mil for the outer 87.5 mils? IS that what can happen?

I may be wrong, but my understanding of the first picture you posted is that the surface of the cable insulation need not all be at 0V relative to ground. Where the cable insulation is in contact with the grounded conduit, it will be (or close). But on the opposite side where there is a large air gap, air has a nonzero dielectric constant, and so the voltage at the insulation surface may be higher than 0V, with the voltage reducing along the air gap to reach 0V at the grounded conduit.

If that's correct, then the volts/mil is higher at the bottom of the cable than at the top, as the voltage difference is higher.

Cheers, Wayne

 

[Hv&Lv]

In the first picture, think “path of least resistance”.
The V only has to go through the insulation to get to the grounded conduit in an It isn’t a solidly grounded path, as there is some resistance there.
This allows corona discharge to develop and eat away at the insulation, and anything else around that’s affected by nitric acid and ultraviolet light.
Add in a tiny bit of moisture, dust, dirt, etc, and tracking direct to ground leading to a cable blowout follows very quickly.

The top side of your cable in the picture the field has to travel through the insulation AND the air to get to ground.

MV cable, XLPE in particular, is bad to “tree” when stressed. The shield around the cable distributes all these forces equally (or real close) rather than stress one particular area.
Remember, voltage is “pressure”.

 

[electrofelon]

Ok so I get the concept of equally distributing the electrical stress and that shielding does that. BUT I'm still unclear what is happening to the volts per mil if we unshield. Is it going up? Is the electrical stress going up despite the volts/mil remaining the same? Walk with me thru my thought process here and tell me where I am going wrong:

Take our hypothetical 15kv shielded conductor with 175 mil insulation, so 86 volts/mil. No let's unshield it and bring a grounded point object, or unequal proximity to a grounded object like in picture number one. Intuitively it makes sense that we have a concentration of electrical stress but the potential is still 15kv so there is still 86 volts/mil there right? So what's going on?

 

[Hv&Lv]

Ok so I get the concept of equally distributing the electrical stress and that shielding does that. BUT I'm still unclear what is happening to the volts per mil if we unshield. Is it going up? Is the electrical stress going up despite the volts/mil remaining the same? Walk with me thru my thought process here and tell me where I am going wrong:

Take our hypothetical 15kv shielded conductor with 175 mil insulation, so 86 volts/mil. No let's unshield it and bring a grounded point object, or unequal proximity to a grounded object like in picture number one. Intuitively it makes sense that we have a concentration of electrical stress but the potential is still 15kv so there is still 86 volts/mil there right? So what's going on?

I’m going to take a stab at it.

In a MV cable the voltages are such that you have to protect the insulation from what it’s insulating.

let’s look at a cable.
There is the conductor, like any other conductor there are air voids. Next is a semi conductive layer (the black carbon containing layer next to the conductor), then the insulation. XLPE is somewhere around 600-800 volts per mil.
Then there is another semiconductive layer (black semi-con) over the insulation, then the copper tape or concentric neutral cable.

With these voltages the air dielectric will break down (50-70 volts per mil for air) with corona discharges.
This creates Nitric acid, UV light, and some heat.
This breaks down the insulation, leading to “treeing” which causes blowouts.

The first black semiconductive layer around the conductor itself basically provides an equal potential area within the actual conductors themselves. There is no real good place for the voltage to go when it’s wrapped within this first cocoon. No corona, or partial discharge, no insulation degradation.

the insulation itself does its job at its Volt per mil
Level, then the semiconductive layer on the outside does basically the same thing with the ground, returning fault currents.

you ever see a Tesla ball? Ever run your hand over it and see what happens with all the “sparks”?
They concentrate where your finger is located and “follow” it.
In your scenario, think of your unshielded cable as a Tesla ball with little sparks going everywhere. When you bring the grounded object closer you concentrate the corona discharge.
with the ball, it’s no big deal and cool to watch. With MV cable, it’s detrimental to the cable.

 

[gar]

220216-1529 EST

electrofelon:

Lets start with two parallel plates, and a uniform dielectric material between the parallel plates. Also lets ignore fringing effects near the edges of the plates. Then the electric flux density lines run straight and perpendicular to the plates. And the electric field intensity lines are parallel to the plates, and are equally spaced between the plates.

When we change this to a circular coaxial structure, then the flux lines are radial and perpendicular to the inner and outer coaxial conductors. These lines start closer together at the inner conductor, and spread out at the outer conductor. The equipotential increment lines which are circles centered on the inner conductor have to grow in incremental radius as they move away from the center of the cable in order to still create curvilinear squares. This means the voltage gradient reduces as you move outward from the center.

Graphically you can sketch these field maps with pencil, eraser, paper, straight edge, compass, and circle guide. Much erasing occurs.

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[wwhitney]

Take our hypothetical 15kv shielded conductor with 175 mil insulation, so 86 volts/mil.

Two uninformed questions:

Per my earlier comment, how do you know that the outside of the insulation is at 0V? With the shield you know that, but I would think without it different parts of the insulation perimeter could be at different voltages. That would make the volts/mil different along different radii.

Is it just the magnitude of volts/mil that matters, or does the spatial distribution of how volts/mil is changing along nearby radii also matter?

Cheers, Wayne

 

[gar]

220216-1909 EST

wwhitney:

If you put an insulated wire, with a uniform insulator, in an electric field that is radially uniform pointing to the wire center ( a coaxial cable ), then the potential ( electric field vector ) around the surface of that concentric insulator is a constant, and points to the wire center. If you change to a nonuniform perpendicular vector to the surface of the insulator, then you will have a varying potential on the surface of the insulator as you change the angular position around the insulator.

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[Hv&Lv]

220216-1909 EST

wwhitney:

If you put an insulated wire, with a uniform insulator, in an electric field that is radially uniform pointing to the wire center ( a coaxial cable ), then the potential ( electric field vector ) around the surface of that concentric insulator is a constant, and points to the wire center. If you change to a nonuniform perpendicular vector to the surface of the insulator, then you will have a varying potential on the surface of the insulator as you change the angular position around the insulator.

.

Been a while, but that sounds like Gauss‘s law in differential form.

 

[electrofelon]

I’m going to take a stab at it.

In a MV cable the voltages are such that you have to protect the insulation from what it’s insulating.

let’s look at a cable.
There is the conductor, like any other conductor there are air voids. Next is a semi conductive layer (the black carbon containing layer next to the conductor), then the insulation. XLPE is somewhere around 600-800 volts per mil.
Then there is another semiconductive layer (black semi-con) over the insulation, then the copper tape or concentric neutral cable.

With these voltages the air dielectric will break down (50-70 volts per mil for air) with corona discharges.
This creates Nitric acid, UV light, and some heat.
This breaks down the insulation, leading to “treeing” which causes blowouts.

The first black semiconductive layer around the conductor itself basically provides an equal potential area within the actual conductors themselves. There is no real good place for the voltage to go when it’s wrapped within this first cocoon. No corona, or partial discharge, no insulation degradation.

the insulation itself does its job at its Volt per mil
Level, then the semiconductive layer on the outside does basically the same thing with the ground, returning fault currents.

you ever see a Tesla ball? Ever run your hand over it and see what happens with all the “sparks”?
They concentrate where your finger is located and “follow” it.
In your scenario, think of your unshielded cable as a Tesla ball with little sparks going everywhere. When you bring the grounded object closer you concentrate the corona discharge.
with the ball, it’s no big deal and cool to watch. With MV cable, it’s detrimental to the cable.

Great explanation HV. I think the Tesla ball analogy does it for me.

 

[electrofelon]

Two uninformed questions:

Per my earlier comment, {color=red]how do you know that the outside of the insulation is at 0V? [/color]With the shield you know that, but I would think without it different parts of the insulation perimeter could be at different voltages. That would make the volts/mil different along different radii.

Is it just the magnitude of volts/mil that matters, or does the spatial distribution of how volts/mil is changing along nearby radii also matter?

Cheers, Wayne

Wayne, I believe the spatial distribution is part of it too. For example see the following diagram. The presenter states that "the equipotential lines and flux lines tend to concentrate near the end of the semicon so you get a high electrical stress a that point."

 

[gar]

220217-0847 EST

electrofelon:

Consider two flat parallel conductive plates spaced X apart. Fill half that X space adjacent to one of the conductive plates with a material of dielectric constant greater than 1 and the remaining space with air, which is a dielectric constant very close to 1.

Thus, two capacitors in a series circuit. And in turn this means exactly the same current flows thru each capacitor. The potential at the interface between the dielectric material and the air gap relative to the conductor next to the dielectric material will be less than 1/2 half the potential between the two conductive plates.

The current flow vectors will be the same on both sides of the air and dielectric material interface. The voltage gradient will be lower in the dielectric material than the air gap.

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