White paper: measuring 3-wire pump resistances with unknowns

[Reach4]

I am not saying information is often needed . It may be of interest to somebody.

This is a somewhat technical discussion. In a well, you may not know the wire resistance of a pump that is in place. With (assumption #1) that the 3 current carrying wires have equal resistance for practical purposes, we can measure both the resistance of the pump windings and the resistance of the wires.

Edit: Note the simpler method at the bottom of reply #7 below just uses ohmmeter and math. A 4-wire ohmmeter with Kelvin clips makes low resistance readings more accurately, but a good 2-lead ohmmeter can do pretty well. The current source method just below would be more accurate, because the probe tip resistance does not matter, the ohmmeter-only method in #7 is simpler.​

This uses Ohm's law, a multimeter, and a current source of some sort. Note that most of us don't have a calibrated "current source" on the bench. You can make one with a battery and a resistor. In the drawing 1 amp is shown, but that is arbitrary for simple math. 600 milliamps or 2 amps or anything in between is fine. You want something big enough to make the voltages high enough to measure easily, and low enough to not overheat the windings. Feed a battery through a power resistor, and you can know the current for other calculations. Put the resistor into line. Measure the voltage across the resistor. The current (I) is the voltage divided by the resistance. Measure for each hook-up, and use the result. Alternatively measure the current with an ammeter in series. Use the actual measured current wherever 1 amp is referenced below. Also note that (Vc-Vb) means the voltage between Vc and Vb. The resistor will need to be a power resistor, such as a 10 ohm 20 watt resistor. The power consumed in the resistor is the voltage across the resistor times the current in amps.

For each measurement, we pass current through the common (yellow) wire and one other wire. A key is that the other wire is not carrying current (assumption #2). So in the drawing shown, we know V2 = V3 = Vc. Thus the resistance of the yellow wire is (Vc-Vb)/1 amp. Due to assumption #1, we now know the resistance of each of the 3 wires.

We then know that the resistance of
(Rblack+Rrun+Ryellow)= (Va-Vb)/ (1 amp)

So subract 2*Ryellow, from the total, and you have Rrun.

Change things around running the current through Rred instead of Rblack. Then remeasure+compute the new current, and you can compute the Rstart resistance.

 

[efinley]

As an electrical engineer I understand what you're trying to do, but no the why. Are you simply trying to create a higher current or higher voltage ohm meter?

 

[Sponsor]

 

[Reach4]

As an electrical engineer I understand what you're trying to do, but no the why. Are you simply trying to create a higher current or higher voltage ohm meter?

Trying to measure the resistance of the motor windings and or wire resistance if the wire length or wire characteristics are unknown. This wire could include the drop wire, it could include the wire between the control box and the pump. This could be handier than unsplicing the wires at the well for those who have the control box mounted in the house. If you know the resistance of the wires, you can just use an ohmmeter and subtract the round-trip wire resistance as you did.

Suppose you just bought a house, and the pump was at an unknown depth. The pump winding resistance is unknown. But maybe you know the wire is #12 copper. You could measure the wire resistance and compute how about deep the pump is set ( although there could be some wire slack to add a bit of uncertanty).

I was at one point thinking this could apply to you, but re-thought that it might be useful to somebody with a different situation.

 

[efinley]

I don't see how you are isolating the resistance of the wire from the pump. can you explain?

 

[Reach4]

I don't see how you are isolating the resistance of the wire from the pump. can you explain?

The key is that we know the voltages at both ends of the yellow wire (Vb and V2), because we know that V2=Vc. And we know the current through the yellow wire.

 

[efinley]

How do you know v2?

 

[Reach4]

How do you know v2?

Measure Vc. Because there is no current through the red wire or the start winding, we know that voltage is the same as V2.

I think this should be solvable with 3 resistance readings. It could be done in an iterative spreadsheet solver, but it is doable symbolically. Here is the algebra problem:

Let Rw be the resistance of each wire.
Rac is the measured resistance between black and red up top.
Rab is the measured resistance between black and yellow.
Rbc is the measured resistance between yellow and red
Rrun is unknown but constant. Rstart is unknown but constant.

3 equations with 3 unknowns:
Rac=2*Rw+Rstart+Rrun
Rab=2*Rw+Rrun
Rbc=2*Rw+Rstart

Solve for Rw knowing Rac, Rab, and Rbc. Once Rw is known, Rrun and Rstart are easy.

Solution:
Rw= (Rab + Rbc -Rac) /2
Rrun=Rab -(2*Rw)
Rstart=Rbc-(2*Rw)

 

[efinley]

Just to clarify, you are assuming Rblack, Ryellow and Rred are all equal? If so I think this is a solvable system of equation, but if not then it is not. It seems reasonable that each wire would have the same resistance unless you have a wiring problem. This just becomes a problem to be solved with Kirchoff's current or voltage laws (KCL, KVL).

 

[Reach4]

Just to clarify, you are assuming Rblack, Ryellow and Rred are all equal? If so I think this is a solvable system of equation, but if not then it is not. It seems reasonable that each wire would have the same resistance unless you have a wiring problem. This just becomes a problem to be solved with Kirchoff's current or voltage laws (KCL, KVL).

Yes. Assumption #1 is that the 3 current carrying wires have equal resistance for practical purposes.

I tried the solution formulas with toy values in a spreadsheet, and the equations checked out.

 

[DonL]

Measuring a inductive load using resistance measurements is a stab in the dark.

Inductance of a AC motor winding should be made measuring the inductance at the AC operating frequency.

Have fun.