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HVDC short circuit force on a conductor bending at 90 degrees.

HVDC short circuit force on a conductor bending at 90 degrees.

HVDC short circuit force on a conductor bending at 90 degrees.

(OP)
Hi,

I have got to work out the effect of the DC short circuit force on a conductor which bends on an angle almost 90 degrees. The level of fault current is 28kA. I have attached a sketch to show what I mean.

There is only one conductor as shown in the link. What I would like to know is how to work out what sort of force is experienced by the conductor as the direction of the current changes due to the bend.

Any help will be appreciated. If require more info then do let me know.

Thanks

RE: HVDC short circuit force on a conductor bending at 90 degrees.

You need to show more details of the return path.  If the return flow of current is in the opposite direction as you show, you'll get a compressive force that will try to pull the conductors.  If the right angle bend is perpendicular to the return current, it will not have any force on it.

RE: HVDC short circuit force on a conductor bending at 90 degrees.

As magoo says, you need to know where the return current is flowing.  The forces will be between the current and the return current.  Since the current in the two sections shown is perpendicular, there will be no forces between the two sections.
 

RE: HVDC short circuit force on a conductor bending at 90 degrees.

I don't want to think about the numerical solution, but there will be a magnetic field around each section of the conductor and where the fields interact at and near the corner there will be a force tending to straighten the corner.
Pete may want to help with the mathematics.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: HVDC short circuit force on a conductor bending at 90 degrees.

(OP)
Thanks for replying guys.

Yes I need help with maths in forming an equation to solve this problem.

There is a typical problem that I came across in the books is where you have A square loop of wire with side length a carries a current I_1. The center of the loop is located a distance d from an infinite wire carrying a current I_2 . The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular.
What is the magnitude, , of the net force on the loop?

See the attached photo for description.

I'm trying to use this problem as the basis for solving mine. all the answers I came across so far say that the perpendicular part of the square to the infinite wire cancel out cause of force being in the opposite direction.

What I'm trying to find is how to work out the magnitude and direction of the force in the top perpendicular section of the square.

This an answer I found but can't seem to figure out how this equation was created. If somebody can help me by explaining what is happening in the equation will be great.

the origin (0,0) is taken on the infinite wire so that the vector of current J2=j*|J2|, while center of the loop is C=(d, 0);
at distance x from infinite wire magnetic field is a vector
B2(x) = -k*μ0* |J2|/(2pi*x);

♦ take a part of the loop where current J1 is directed from J2; then force on this part is
F3= ∫ |J1|*[dx×B] = j*|J1|*∫B(x)*dx =
= j*μ0* |J1|*|J2| ∫dx /(2pi*x) {x=(d-a/2) to (d+a/2)};
♦ take a part of the loop where current J1 is directed to J2; then force on this part is
F4= -F3;


 

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