Shocked from the plumbing pipe

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wirenut1980 said:
But as Don said it is not just current, one also needs voltage present to make it dangerous. One also needs to take a voltage reading between the pipe and reference ground.
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It's current with the pipe intact, and voltage with it severed. Of course, you can't have a current without a driving voltage.

The problem with your idea is that the metallic water-pipe, especially if it qualifies as an electrode, usually IS the reference.
 
wirenut1980 said:
But as Don said it is not just current, one also needs voltage present to make it dangerous. One also needs to take a voltage reading between the pipe and reference ground.
Click to expand...
That is not exactly what I said...with the pipe intact and carrying current, the only voltage to create a shock hazard would be the voltage drop on the water pipe and the parallel grounded conductor. If you open the pipe, now you have an open circuit and if the grounded conductor is in good condition, then the voltage across the open pipe will be the voltage drop of the grounded conductor. In most installations, this would not be a shock hazard. However if there is a problem with the grounded conductor and you open the pipe there will be a serious shock hazard. In this case it would be the same as getting across an open a neutral...as that is exactly what you are doing.
 
081215-2023 EST

GilbeSpark:

Here is my suggestion on how to troubleshoot this problem.

Use a high impedance AC meter like a Fluke 27. Make a probe with an insulated handle, could be a screwdriver. You may need to be careful in the following experiments because you do not really know what voltages you may encounter. You could somewhat protect yourself with a 33,000 to 100,000 ohm resistor in series with the connection at the reference point. Get an adequately long insulated wire. I use the ground conductor in an extension cord, or cords plugged together. You could just use some #24 to #16 wire but it is usually more of a mess.

For this experiment use the ground rod at the power company transformer, or put another short rod in the ground near the transformer ground rod as your reference. This could be a 10" long screwdriver. Connect your wire to this reference point. The other end of the wire goes to one input terminal of the voltmeter. Connect your probing probe to the other input terminal. The probing probe could be another screwdriver.

I am initially assuming that the transformer pole ground rod is actually connected to the transformer center tap. That is why this is a useful reference point. The assumption may be wrong, but it is a good starting point. If the pole transformer is in another yard, then put your reference point in the corner of your yard closest to the pole transformer. Note: there could be some large voltages.

Pull the main disconnect in the house. This means that there should not be any current flowing in the yard resulting from load in the house.

Record your measurements and their locations.

Probe the yard in various places and see if there is any voltage greater than 1 V. If there is, then the line of attack will be different. In particular probe at the service entrance to the house. Also look for a ground rod in the service entrance area. Apply load from the house, especially on one phase. Does this cause any substantial change in the voltage? If so, then you probably have neutral current flowing thru the earth, and probably an open or high impedance neutral.

Continue into the house and probe both the neutral and ground buses in the main panel for no house load and the heavy load on one phase.

Do the same test to the water pipe where the shock occurred.

Report back.

.
 
LarryFine said:
It's current with the pipe intact, and voltage with it severed. Of course, you can't have a current without a driving voltage.

The problem with your idea is that the metallic water-pipe, especially if it qualifies as an electrode, usually IS the reference.
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My comment was based on a continuous water pipe in spirit with the original poster's situation. Getting between two sections of broken water pipe carrying current will definately be a shock hazard.

When I say reference ground, I mean stick a probe into the ground that is not carrying a whole lot of current. Reference ground is meant to simulate an area where no measurable current is flowing.

A grounding electrode carrying current (which would not simulate your everyday section of earth), with enough voltage behind it will show up in a voltage measurement to reference ground as dangerous to touch even if the pipe is unbroken.
 
wirenut1980 said:
But as Don said it is not just current, one also needs voltage present to make it dangerous. One also needs to take a voltage reading between the pipe and reference ground.
Click to expand...
I've always heard it's not the voltage that kills it's the current?
 
zappy said:
I've always heard it's not the voltage that kills it's the current?
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While it is true that current will get you, it is not the whole story. I was troubleshooting a job where a telephone company was measuring 5 amps on a ground bar in a commercial building's communication room and they were afraid to touch it or work near it. The ground bar was connecting the grounds from CATV, telephone, and power. After extensive testing I determined that the current was neutral current from the medium voltage system outside the building. Performing a voltage measurement from the ground bar to reference ground outside resulted in around 3 V. Inserting a 500 ohm resistor resulted in the voltage dropping to nothing. I then safely touched the ground bar with my bare hand without receiving a shock.
 
zappy said:
I've always heard it's not the voltage that kills it's the current?
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You have to have a potential difference between two points or surfaces to drive a current through a conductive pathway. For a given voltage, the lower the impedance of the pathway, the higher the resultant current.

If the conductive pathway is of very low impedance (as is desireable in any intentional pathway), such as an intact water pipe, the voltage between the two points of contact will be extremely low, and the current high.

If you create a break in the pathway, the full potential difference now falls across the air gap, and the current drops to zero. That's why a bonding jumper is required around insulative interruptions in the piping.

The benefit is that the parts of the electrical system that should be at earth potential are at least close to that. The disadvantage is that a current is forced through the piping, but it's openings that create a hazard.

The reason an intact water pipe might still create a shock hazard is due to impedance-caused voltage drop between the pipe and the earth (or more accurately, the utility's GES), not between parts of the pipe.

An equipotential bonding system does not assure zero volts to earth, merely zero volts between various parts of the bonded equipment and materials. Zero volts to earth is desireable, but not guaranteed.

The above presumes you understand that the greatest potential between points in a current-carrying circuit is across the part of the circuit with the highest impedance (which simply speaking means AC resistance.)

If there's current traveling through a waterpipe system, it's from an unwanted impedance in the service's neutral conductive pathway between the transformer's neutral terminal and the panel's neutral terminal.

The voltage causes a current, which causes a voltage across the impedance, which causes a voltage drop which allows a voltage difference. Eliminate the impedance, and you eliminate the voltage difference.

A shock in your body is a result of current that the voltage difference causes to appear across your body's impedance. A good bonding system is supposed to minimize the voltage across points you can contact.
 
081216-1424 EST

zappy:

At zero volts there is zero current thru a finite non-zero resistor.

Many years ago with a 6 V era car and a hot 6 V to chassis horn ring I felt an electrical tingle between the horn ring and the car body when I had very sweaty hands. In those days a wire from the horn relay went up the inside of the steering column, no slip rings, to the horn ring. Pushing on the horn ring caused a short to ground (chassis) energizing the horn relay and horn. Open circuit there was 6 V on the horn ring.

My resistance was probably in the range of 6000 ohms under these conditions.

Most of this resistance is at or near the surface of the skin. Internally the body has relatively low resistance.

Normally I have about 500,000 ohms resistance between the surface of a pair of fingers gripping a conductor on one hand with a similar connection on the other hand. Today on a Fluke 27 in resistance mode I read about 3 megohms (about 1 V source voltage), and separately using the same meter in microamp mode I get 2 microamps with 6 V source, 3 megohms.

Different subject: with a vacuum tube voltmeter I read 0.4 V between my two hands today. Got near zero voltage reading on the Fluke.

.
 
LarryFine said:
You have to have a potential difference between two points or surfaces to drive a current through a conductive pathway. For a given voltage, the lower the impedance of the pathway, the higher the resultant current.

If the conductive pathway is of very low impedance (as is desireable in any intentional pathway), such as an intact water pipe, the voltage between the two points of contact will be extremely low, and the current high.

If you create a break in the pathway, the full potential difference now falls across the air gap, and the current drops to zero. That's why a bonding jumper is required around insulative interruptions in the piping.

The benefit is that the parts of the electrical system that should be at earth potential are at least close to that. The disadvantage is that a current is forced through the piping, but it's openings that create a hazard.

The reason an intact water pipe might still create a shock hazard is due to impedance-caused voltage drop between the pipe and the earth (or more accurately, the utility's GES), not between parts of the pipe.

An equipotential bonding system does not assure zero volts to earth, merely zero volts between various parts of the bonded equipment and materials. Zero volts to earth is desireable, but not guaranteed.

The above presumes you understand that the greatest potential between points in a current-carrying circuit is across the part of the circuit with the highest impedance (which simply speaking means AC resistance.)

If there's current traveling through a waterpipe system, it's from an unwanted impedance in the service's neutral conductive pathway between the transformer's neutral terminal and the panel's neutral terminal.

The voltage causes a current, which causes a voltage across the impedance, which causes a voltage drop which allows a voltage difference. Eliminate the impedance, and you eliminate the voltage difference.

A shock in your body is a result of current that the voltage difference causes to appear across your body's impedance. A good bonding system is supposed to minimize the voltage across points you can contact.
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So it's all about resistence?if your touching a pipe that has voltage on it but your body's resistence is higher then the pipe's resistence it won't shock you?But if it's lower then the pipe's resistence it will shock you?Is it always true electricity always follows the path of least resistence?
 
zappy said:
So it's all about resistence?if your touching a pipe that has voltage on it but your body's resistence is higher then the pipe's resistence it won't shock you?But if it's lower then the pipe's resistence it will shock you?Is it always true electricity always follows the path of least resistence?
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Electricity will follow all of the pathways it can, not just the lowest resistance. Just like water on top of a mountain, the widest stream has the heaviest flow, but they all flow something.

The relative resistances control the relative division of current; the lower the resistance, the higher the current. If another pathway becomes available, the new current adds to the total.

Think about the various circuits in your house. They're all in parallel. They all flow current, not just the one with the lowest resistance. The total current is the sum of that of the individual loads.

The current through each circuit depends on two things: the source voltage and the load (and circuit) resistance. Only with resistances in series does one affect another, but that's another lesson.

Added: Now, if you touch two different points on an intact pipe carrying a current, you would only see the voltage drop between the two points, which is dependent on the current and the resistance.
 
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Pierre C Belarge said:
It could also be a neighbor's service problem, such as their neutral connection has been compromised, and the next best path is through the neighbor's service via the GEC connection to the water pipe.
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I agree with you Pierre I have seen in city water systems where a neighbor loses a neutral and his service is bonded to the same piping system as his neighbors the return path to the transformer is back thru the piping system in the neighbors house with the good neutral connection. The piping system will carry the unbalanced load from the neighbor without a neutral and you can get shocked from the piping system because the neutral is searching for any and all paths back to the transformer .
 
wirenut1980 said:
My comment was based on a continuous water pipe in spirit with the original poster's situation. Getting between two sections of broken water pipe carrying current will definately be a shock hazard.
Click to expand...
Not unless there is an excessive voltage drop on the grounded conductor. If the grounded conductor is in good condition the water pipe is in parallel with the grounded conductor and will be carrying current however when you open the pipe the only voltage between the two sections of pipe is the voltage drop on the grounded conductor. As long as the neutral is in good shape and of the correct size the voltage between the two sections of pipe will not be a shock hazard.
 
081216-2029 EST

zappy:

You are not applying the basic electrical theory of parallel circuits in your statement. A high resistance and low resistance in parallel with a voltage across the parallel resistors has current flowing thru each resistance. i = e/R. If one resistor is 100 times larger than the other, then the current thru it will be 1/100 of the current thru the low resistance.

If my finger to finger body resistance is 24000 ohms, then across a 120 V 60 Hz voltage source the RMS current is 5 MA, the threshold for a GFCI. I will get a very substantial shock. At 240,000 ohms I will get a shock. At 2.4 megohms probably a tingle. I am not experimenting to verify these statements.

I have heard that some electricans measure voltage and troubleshoot with their fingers on one hand. I would not do this nor suggest that anyone even experiment with this technique.

.
 
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