Voltage drop calculation (finding distance of existing circuit)

[Dsg319]

I know this is quite simple but have never really done it before until today, and while it might not be spot on I believe it should be relatively close. Any thoughts?

So anyhow bored on a Friday no work got a cold or allergies can’t go around anyone for respect with all the stuff going around. But would someone care to double check me and tell me if this could actually work.

What I have done was used a hair dryer for a load and measured voltage before on and while in use and used those numbers in my Calculation. Although my clamp meter is not here I had to use name plate specs to get the current. I figure should be close.

 

[gar]

210312-1637 EST

Dsg319:

Here is my guesstimate --- Copper wire #12 is about 1.6 ohms/1000ft. If your voltage drop and current values are approximately correct, then loop resistance is about 5/15 = 0.33 0hms. This is about a loop distance of 1000*0.33/1.6 = 206 ft, or one way distance of 103 ft. You are in the ballpark. if you allowed the wire to heat some, then distance would be a little less. Also you might need more accurate measurements. Also your voltage drop might be somewhat less because of voltage drop of your source under load.

.

 

[Dsg319]

210312-1637 EST

Dsg319:

Here is my guesstimate --- Copper wire #12 is about 1.6 ohms/1000ft. If your voltage drop and current values are approximately correct, then loop resistance is about 5/15 = 0.33 0hms. This is about a loop distance of 1000*0.33/1.6 = 206 ft, or one way distance of 103 ft. You are in the ballpark. if you allowed the wire to heat some, then distance would be a little less. Also you might need more accurate measurements. Also your voltage drop might be somewhat less because of voltage drop of your source under load.

.

Thanks. Also This didn’t exactly make sense to me. Circuit before load was 125v. While under load measured 120volts from phase conductor to neutral. I figured I should have still measured 125volts from phase to neutral, and than be able to measure 5volts from neutral to EGC.

But when I measured from neutral to EGC while under load I was around 2.5volts. I expected to see the full voltage drop.

 

[retirede]

Thanks. Also This didn’t exactly make sense to me. Circuit before load was 125v. While under load measured 120volts from phase conductor to neutral. I figured I should have still measured 125volts from phase to neutral, and than be able to measure 5volts from neutral to EGC.

But when I measured from neutral to EGC while under load I was around 2.5volts. I expected to see the full voltage drop.

Draw out the equivalent circuit and see if it makes sense. If you measured 125 at the load L-N, the voltage drop would be zero.

 

[don_resqcapt19]

Thanks. Also This didn’t exactly make sense to me. Circuit before load was 125v. While under load measured 120volts from phase conductor to neutral. I figured I should have still measured 125volts from phase to neutral, and than be able to measure 5volts from neutral to EGC.

But when I measured from neutral to EGC while under load I was around 2.5volts. I expected to see the full voltage drop.

With equal length conductors half of the voltage drop is on the ungrounded conductor, so 2.5 volts from the neutral to the EGC is correct.

 

[Dsg319]

With equal length conductors half of the voltage drop is on the ungrounded conductor, so 2.5 volts from the neutral to the EGC is correct.

Ahh there lies the answer. It kind of had me baffled. But do I have this right in thinking like you mentioned (as in most cases) of both grounded and ungrounded conductors are equal length. So say voltage is 120.
After the load I have a VD of 4volts,

So line to neutral voltage should be 116volts. Line to EGC still 120volts (or would it be 118volts since half the VD is on the ungrounded conductor)
Neutral to EGC will be half of the VD SO 2volts.
Is that correct ?

 

[Dsg319]

Draw out the equivalent circuit and see if it makes sense. If you measured 125 at the load L-N, the voltage drop would be zero.

125v was before the circuit was underload. 120volts after from line to neutral.

 

[oldsparky52]

Thanks. Also This didn’t exactly make sense to me. Circuit before load was 125v. While under load measured 120volts from phase conductor to neutral. I figured I should have still measured 125volts from phase to neutral, and than be able to measure 5volts from neutral to EGC.

But when I measured from neutral to EGC while under load I was around 2.5volts. I expected to see the full voltage drop.

The voltage drop is on the grounded and ungrounded conductor, 2.5 volts on each. So when you read (underload) from the two circuit conductors you would read the 5-volt drop, hence the 120V instead of 125V.

Since each conductor drops 2.5 volts and the ground has no current, then the drop on the neutral would show up when reading from neutral to ground and should read 2.5 volts. The hot to ground (under load) should read 125-2.5 or 122.5V.

 

[Dsg319]

The voltage drop is on the grounded and ungrounded conductor, 2.5 volts on each. So when you read (underload) from the two circuit conductors you would read the 5-volt drop, hence the 120V instead of 125V.

Since each conductor drops 2.5 volts and the ground has no current, then the drop on the neutral would show up when reading from neutral to ground and should read 2.5 volts. The hot to ground (under load) should read 125-2.5 or 122.5V.

Thank you! Just answered my question. Not sure if you had already read it or not lol

 

[oldsparky52]

Thank you! Just answered my question. Not sure if you had already read it or not lol

I answer some posts before I read the whole thread. I should stop that.

 

[Dsg319]

I answer some posts before I read the whole thread. I should stop that.

It needed answered anyways haha

 

[gar]

210312-2023 EST

Dsg319:

Just to make it very clear on how you should define your problem I will make the following points:

1. Take a 100 ft length of #12 copper wire and assume it is at room temperature, you momentarily apply 15 A of low frequency current thru that wire, then there is little temperature rise in the wire because of the short time, and its resistance will be close to 1.6/10 = 0.16 ohms. Thus, voltage drop along the 100 ft of wire is about 15*0.16 = 2.4 V.

2. Next add an additional piece of wire of the same length in close proximity to the first wire, and connect the wires at one end. Keep the current the same. The voltage at the input end is 2*2.4 = 4.8 V. Thus, the loop voltage drop is 4.8 V.

3. Next insert a 5 ohm resistor (can be more wire) at the input end of the loop, and still keep the current at 15 A. The voltage drop of the 200 ft loop, 100 ft each way, is still 4.8 V. But the input voltage to the 5 ohm plus 0.32 ohm resistance is 15*(5+0.32) = 79.8 V.

4. Next remove the shorted end of the loop and move the 5 ohm resistor to replace the short end as a load. Now apply 120 V to the 5.32 ohm loop. The current thru the wire is 120/5.32 = 22.56 A. The input voltage to the circuit is 120 V, and the load voltage is 22.56*5.32= 112.8 V.

5. Now add more resistance at the input end to obtain 15 A form 120 V. This is a total resistance of 120/15 = 8.0 ohms, or an added resistance of 8 - 5.32 = 2.68 ohms. But the voltage drop across the 200 ft of wire is still 4.8 V. In other words measure the input voltage to the start of the 100 ft section (79.8 V) and the voltage at the end of the 100 ft distance (5*15 = 75), and 79.8 -75 = 4.8 V.

If I made no mistakes you should be able to draw this out and get comparable results.

.

 

[xptpcrewx]

I know this is quite simple but have never really done it before until today, and while it might not be spot on I believe it should be relatively close. Any thoughts?

So anyhow bored on a Friday no work got a cold or allergies can’t go around anyone for respect with all the stuff going around. But would someone care to double check me and tell me if this could actually work.

What I have done was used a hair dryer for a load and measured voltage before on and while in use and used those numbers in my Calculation. Although my clamp meter is not here I had to use name plate specs to get the current. I figure should be close.View attachment 2555734

I think it’s great you are experimenting with this!

Based on my experience, back solving for distance is not accurately possible because of calculation assumptions, variance associated with the impedance of a wire/cable and other contributions generally neglected ; i.e. measurement errors, readings not taken simultaneously, temperature variation along the length of the line, inability to distinguish impedance contributions, assuming total impedance correlates with wire/cable length, ignoring load power factor, assuming the load is linear, assuming pure sine waves for voltage/current, assuming the mechanics of voltage drop is just a simple algebraic difference between sending and receiving end voltages, etc.

There are two methods generally used for calculating voltage drop: (I) the oversimplified method used in this thread, (ii) the correct method (which considers cable resistance and reactance, load power factor and complex angles). DM me if you want a one-on-one explanation on this. I wouldn’t mind doing a screen share with some examples.


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[Carultch]

Based on my experience, back solving for distance is not accurately possible because of calculation assumptions, variance associated with the impedance of a wire/cable and other contributions generally neglected ; i.e. measurement errors, readings not taken simultaneously, temperature variation along the length of the line, inability to distinguish impedance contributions, assuming total impedance correlates with wire/cable length, ignoring load power factor, assuming the load is linear, assuming pure sine waves for voltage/current, assuming the mechanics of voltage drop is just a simple algebraic difference between sending and receiving end voltages, etc.

I would suggest that you de-energize the circuit completely, and use your multimeter's Ohm setting, so your sample voltage source is controlled. What your multimeter does in Ohm mode, is apply a sample known voltage, and measure the corresponding current, or vice versa. This also keeps the wire at its cold temperature, and allows you to have control over the voltage applied to test the circuit, rather than depending on grid voltage that we know is variable.

 

[ptonsparky]

I used VD calc on a 100 HP motor once to prove one set of the parallel conductors had failed. There was no room to clamp incoming lines at the pump panel so I had to have POCO run current checks at the padmount. Gophers had gotten to all four of one set.

Good use of downtime, OP.

 

[xptpcrewx]

I would suggest that you de-energize the circuit completely, and use your multimeter's Ohm setting, so your sample voltage source is controlled. What your multimeter does in Ohm mode, is apply a sample known voltage, and measure the corresponding current, or vice versa. This also keeps the wire at its cold temperature, and allows you to have control over the voltage applied to test the circuit, rather than depending on grid voltage that we know is variable.

Again, there is no adequate correlation between total impedance and circuit length (one loose terminal, the conductor lay, flexing the conductor and the raceway material type can throw everything off). A DMM only measures DC resistance, and there is more to the calculation than this. In 99% of the instances the DMM is outside its measurement range. Also you can try to zero out the DMM lead resistance but good luck getting a steady error free readying with moving those leads around. You’ll need a better instrument that uses a 4-wire kelvin technique, but at that point, why not just take a tape measure and measure the length of the line?

The OP is about back solving for distance, I maintain it’s not accurately possible. Less so if you use the simplified calculation method.


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[Carultch]

A DMM only measures DC resistance, and there is more to the calculation than this.

If anything, measuring DC resistance to infer a wire's length has an advantage over using AC, because it eliminates the influence of inductance and magnetism of the metal conduit impacting the measurement.

I do see your point about how most multi-meters are not sensitive enough to measure the resistance applicable to typical power circuits. It might work if the circuit has multiple full Ohms of resistance.

I assisted in solving a similar problem, where a co-worker measured the DC ohms of a data cable to try to figure out where it was routed. He gave me the ohms and gauge, and had me tell him how many feet it is.

 

[xptpcrewx]

If anything, measuring DC resistance to infer a wire's length has an advantage over using AC, because it eliminates the influence of inductance and magnetism of the metal conduit impacting the measurement.

I do see your point about how most multi-meters are not sensitive enough to measure the resistance applicable to typical power circuits. It might work if the circuit has multiple full Ohms of resistance.

I assisted in solving a similar problem, where a co-worker measured the DC ohms of a data cable to try to figure out where it was routed. He gave me the ohms and gauge, and had me tell him how many feet it is.

Now that I think about it, you might be able to get away with measuring DC resistance for tiny conductors where the resistance is much larger than the contributions I mentioned, but not with larger lower impedance power conductors or anything having parallel sets. Just to keep this within the realm of voltage drop and the OP, let’s stick with the calculation method and formula presented above.


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[oldsparky52]

The OP is about back solving for distance, I maintain it’s not accurately possible. Less so if you use the simplified calculation method.

So how much difference would you get in your answer if you use the oversimplified calculation? 1%, 5%, 25%?
Thanks

 

[xptpcrewx]

So how much difference would you get in your answer if you use the oversimplified calculation? 1%, 5%, 25%?
Thanks

Depends on the circuit. Give me some specifics and I’ll gladly calculate the difference.


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