# ground potential rise

Im trying to get a good qualatative sense of ground potential rise (both during a fault and lighting strike). Can someone be of
assistance? Every reference on the subject always quotes ohms law. But, that would refer to voltage drop across a resistance (in this case, the ground path or ground conductor path back to source). Is GPR simply the potential at the point and the gradient drop of potential in the circuit from where it is faulted back to the source (ie the OC potential of the system - potential drop across the resistor, in this case the return path resistance= 0V). and in alot of cases the gpr is termed so because someone walking along the return path can possibly become part of the return path....
J007
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Ground is often simply the negative terminal on the battery.
Ground, earth, zero potential rail, negative terminal, neutral pin on socket, all the same really.
There is always a potential differendce between atmosphere and earth, if you construct a little truck of wood, with wooden wheel, put thousands of windings around it and head it off rolling along a correct direction, right angles to the field, it will light a low voltage lamp you hook to it. I foget what it is called.
-- Billy H
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Billy H wrote:

Im not confused about what ground in terms charge distribution or electric potential. I am looking for a qualatative feel gpr during a fault. As above, Im a little bit confused on how and why most references always quote ohms law for voltage drop across a resistor when referring to gpr - the OC voltage of the souce (essentially - infinite bus) / ground impedance = fault current. Is the ground potential rise referring to the ground now part of the circuit path, under the influence of the E field, where current can flow through back to the source, and the voltage drop across the ground as an impedance....if someone was to stand far enough apart along the E field influence they could become a part of the circuit (a large enough potential difference b/t legs), a small parallel path for current to flow?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On 11/13/06 9:33 AM, in article snipped-for-privacy@i42g2000cwa.googlegroups.com, "jack007"

When talking about such things, you should try to be more specific. There are various levels of detail in the phenomena associated with grounding. For example, the cabinet potential reached a fault from a hot line conductor to the case of a dishwasher will depend on how the case is grounded. Is it a long run of grounding conductor through the garage and house to a grounding panel, or a bad connection to a water pipe? And there are many more variations on that theme.
Bill -- Fermez le Bush
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

I used to do ground fault potential rise calculations for substations when we applied for telecom service. That was about 18 years ago! I no longer remember the exact methods used. I suggest you muck around in some telecom websites. I think that is where the standards exist.
Charles Perry P.E.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

http://shop.ieee.org/ieeestore/Product.aspx?product_no=SS94491
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
operator jay wrote:

Okay to be more specific, lets do three cases of phase to ground fault: 1. solidly grounded: neutral stays at G potential, other lines remain at l-n(g) voltage r 2. resistance grounded: G floats slightly from neutral, other lines l-n(g) voltage rises slightly 3. no grounding: G floats to the leg of the faulted phase. other phase to ground voltages become line to line
in each case the only difference is the magnitude of the return fault current. now, can someone explain a bit about the gpr in each situation...
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Wrong. In a real system, the faulted phase goes to zero at the point of the fault. The other two phases can actually experience a voltage rise. The amount of this rise is related to how far the fault is from the source transformer. On long distribution lines, this can be a real problem. Oh, and the neutral does not stay at ground potential. That is only true at the bonding point.
Charles Perry P.E.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Charles Perry wrote:

Agreed. I was using conceptual ideal situations to describe things. Lets use the case where there is a phase to ground fault and some impedance in the line on a solidly grounded source. The faulted point goes to zero (reference) potential. the other two phases experience a line to ground voltage rise (similar to what i described in the resistance grounded source). In addition there is some unintentional resistance or impedance between the bonding point and ground - adding to the neutral drifting from reference ground potential and the voltage rise on the other phases. back to the phase to ground fault point... There is impedance in the ground path back to the source and current flowing (the ohms law part - oc source voltage/impedance, neglecting the angles for now). I need a step by step of what this ground potential rise is with regards to this type of fault (as explained above)....
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

GPR is about the potential of the earth and ground systems around the location at which current is being injected. Step potentials are about people stepping and getting hurt.
What is it you need this for?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
operator jay wrote:

I need this for my general knowledge/understanding. People throw around the term and I wanted a complete understanding of it. So, it is as i thought, basically the potential of earth and ground system when a phase is faulted. ie the potential drop from the point of the fault through the ground impedance back to the source. if someone steps in the influence of the E field and there is a potential difference between the two feet, they can become part of the path (same as a bird on power line, if they had long legs they could be electrocuted). and, correct me if im wrong, it is the oc voltage-line impedance vdrop up to the faulted phase to ground point? this is where i am a bit confused, how does the point where the phase faults to ground go to zero? ground essentially is uncharged (neutral), but so is a conductor downstream of any switch in a designed circuit. so closing a switch on the latter system is essentially the same effect as the line to ground fault with regards to potentials upstream and downstream of the switch and closing the switch(i dont mean the consequences related to impedance and fault current, i mean E field and potentials)
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
remove the X to answer ----------------------------

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

## Site Timeline

• ### Call for Tutorials @ The 11th International Conference on Computational Creativity ...

• Share To

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.