calculating step potentials 2

[strongone]

ab - parallel to x-axis ab = 1m
cd - parallel to y-axis cd = 1m

Now if i want to determine the step potential at x
Vstep^2 = (Va-Vb)^2 + (Vc-Vd)^2 where Va is the potential at a etc.
Is this correct???
c
|
a - x - b
|
d

 

[jbartos]

Comment: If a portion of a region containing a two-dimensional potential field, divided into squares of side h is considered, the potential Vo at the center is approximately equal to the average of the potentials at the four neighborhood points.
Solution of:
d^2V/dx^2 + d^2V/dy^2~(V1+V2+V3+V4-4Vo)/h^2=0
or
Vo~(V1+V2+V3+V4)/4
~=approximately equal
Specifically,
Vx~(Va+Vb+Vc+Vd)/4

 

[strongone]

Heppe R.J., 1979, Computation of potential at surface above an energized grid or other electrode, allowing for non-uniform current distribution", IEEE Trans. on Power App. and Sys., 98, 1978-88.

Heppe gives the step voltage as
Vstep = sqrt((dV/dx)^2 + (dV/dy)^2)
where Vstep is also known as the potential gradient
therefore i dont understand what you mean by this d^2V/dx^2 + d^2V/dy^2

 

[jbartos]

Comment: Those are second partial derivatives, one by x the other by y. This is an approximate solution of the Laplace partial differential equation. See for example:
William H. Hayt, Jr., Engineering Electromagnetics, McGraw-Hill Book Company, 1967,
Experimental Mapping Methods
The Iteration Method
Figure 6.8, page 168

 

[jbartos]

Comment: Please notice that Heppe mentioned relationship is not applicable to the calculation of Vx given Va, Vb, Vc and Vd as indicated.
Vx could be expressed as for example:
Vx=sqrt[(Va-Vo)^2+(Vd-Vo)^2)
where:
Va---Vx
| |
| |
Vo---Vd
Heppe writes:
Step voltage was computed by finding voltage differences between points 1 m apart in the x and y directions. This was done at 1 m intervals. These differences are called dv/dx and dv/dy. Step voltage was calculated as sqrt[(dv/dx)^2+(dv/dy)^2]. The maximum step voltage was 1622 V/m, located at the edge of the grid, 8 m from a corner.

 

[jbartos]

Correction: I beg your pardon.
Vx=sqrt[(Va-Vo)^2+(Vd-Vo)^2)]

 

[strongone]

Well it seems that i have interpreted Heppe completely wrong. Could you give me a demonstration of Heppe's method, for some reason i just cant visualize it.

i would like to determine step potentials throughout a grid. the potentials at every 1m is know. how do i go about calculating the step potentials??

should i go about it this way
Vx=sqrt[(Va-Vo)^2+(Vd-Vo)^2)
where:
Va---Vx
| |
| |
Vo---Vd

or simply
Va---Vx----Vb
Vx = abs(Va-Vb)

 

[jbartos]

Comment: I believe that Heppe used some mesh or grid consisting of squares with the edge 1 meter long. Ultimately, Vo=0Volt may be at one square.

 

[strongone]

Vo = 0 volts, that doesnt make sense to me,
how would you calcuate the step voltage through, supposing that you knew the potentail at each point of the grid??

 

[Alex68]

Dear strongone
you can simplify your calculation in this way:

usually a man (not a woman!) walks along a straight line, so the most popular prgs calculate only a step voltage PROFILE along a line!

My suggestion is to calculate the surface potential in a area, the touch potential in the same area (GPR - Vsurface) but only profiles of step potential.

 

[GordS]

I don't understand how Vstep = sqrt( dV/dx^2 + dV/dy^2). Consider a "grid" consisting of one rod. Assume under fault that the GPR=10V. The equipotential lines around the rod would, by symetry, be concentric circles. Assume that 1 m from the rod, the surface potential is 5V. Due to the concentric equipotential lines, Vstep is 5V (place one foot on the rod - all points 1 m away are at 5V).

The referenced equation for Vstep would calculate dV/dx=dV/dy = 5 V/m and Vstep= 7V = sqrt(5^2 + 5^2) which seems incorrect.

 

[strongone]

a,b,c,d are all 1m from x

the true step potential at x
VmaxX = max(abs(Va-Vx) and abs(Vb-Vx))
VmaxY = max(abs(Vc-Vx) and abs(Vd-Vx))

Vstep at x = sqrt(VmaxY^2 + VmaxX^2)
where Va is the potential at a etc.
Is this correct???
c
|
a - x - b
|
d

Heppe writes:
Step voltage was computed by finding voltage differences between points 1 m apart in the x and y directions. This was done at 1 m intervals. These differences are called dv/dx and dv/dy. Step voltage was calculated as sqrt[(dv/dx)^2+(dv/dy)^2].

does this correspond to Heppe???

 

[strongone]

c
i | i
a - x - b
i | i
d
a,b,c,d and i are all 1m from x

I think the reason why Heppe did this was to take into account , the voltage at the diagonals
Vstep = sqrt[(dv/dx)^2+(dv/dy)^2].

However is there anywrong with
Vstep = max(abs(Va-Vx) and abs(Vb-Vx) and abs(Vc-Vx) and abs(Vd-Vx)

 

[jghrist]

I think Heppe's formula is a reasonable approach to approximate the maximum voltage gradient (V/m) in any direction where measurements are calculated one meter apart only in an even grid. It is not the voltage gradient in a diagonal direction necessarily, but the maximum voltage gradient that is needed. It is an approximation unless a potential is calculated one meter away from x in all directions.

In the case of concentric circular equipotential lines around a single rod, this gives a gradient that is too high, as GordS points out. In this case, Vstep=dv/dx=dv/dy.

If there were a single conductor of infinite length parallel to the x-axis, then you would have straight equipotential lines parallel to the x-axis. Va=Vx=Vb, and dv/dx=0 and sqrt[(dv/dx)²+(dv/dy)²]=dv/dy. In this case, the maximum voltage gradient is in the y direction. The voltage gradient in a diagonal direction would be less than dv/dy, and thus would not be Vstep.

In practical terms, I think if you calculate surface potentials every meter, then it is adequate to take Vstep at x to be equal to the maximum of |Vx-Va|, |Vx-Vb|, |Vx-Vc|, or |Vx-Vd|.

 

[jbartos]

Comment on strongone (Electrical) Mar 24, 2004 marked ///\\Vo = 0 volts, that doesnt make sense to me,
///It may happen that at some starting point Vo=0V. It depends on the voltage boundary conditions.\\\
how would you calcuate the step voltage through, supposing that you knew the potentail at each point of the grid??
///E.g., the way Heppe calculates it in its paper, or via solution of the Laplace partial differential equation. Remember, that solving equipotential lines is not only in the substation grounding area step potentials, it is also needed in insulation systems, capacitor dielectrics, etc.\\\

 

[jghrist]

In terms of substation surface potentials during a fault, V=0V would only occur at remote ground by definition. This voltage would not occur within the substation.

 

[jbartos]

Comment: Heppe approach, using the vector gradient magnitude, has an important prerequisite, namely, each point must have the gradient vector existing, i.e.:
gradFI=DelFI or NablaFI
where
Del or Nabla operator =ixd/dx+jd/dy+kd/dz, in x,y,z space or
Del or Nabla operator =ixd/dx+jd/dy, in x,y plane
d/dx,d/dy,d/dz are partial derivatives