calculating step potentials
calculating step potentials
(OP)
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
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






RE: calculating step potentials
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
RE: calculating step potentials
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
RE: calculating step potentials
William H. Hayt, Jr., Engineering Electromagnetics, McGraw-Hill Book Company, 1967,
Experimental Mapping Methods
The Iteration Method
Figure 6.8, page 168
RE: calculating step potentials
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.
RE: calculating step potentials
Vx=sqrt[(Va-Vo)^2+(Vd-Vo)^2)]
RE: calculating step potentials
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)
RE: calculating step potentials
RE: calculating step potentials
how would you calcuate the step voltage through, supposing that you knew the potentail at each point of the grid??
RE: calculating step potentials
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.
RE: calculating step potentials
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.
RE: calculating step potentials
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???
RE: calculating step potentials
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)
RE: calculating step potentials
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|.
RE: calculating step potentials
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.\\\
RE: calculating step potentials
RE: calculating step potentials
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