How would you approach this customer complaint?

[gobblerhuntr]

Had a customer, I work for a POCO, ct. metered on the transformer bank (120/240) pole. At the rear of this plant was a local small town water tank and building which was caught off this same bank. Long story short, the plant's secondary was caught off on the load side of the plant's meter. I told the customer that I would figure up how much he has paid extra and reimburse him for it. He asked after I presented my number if I had included the loss in the wire going to the water tank. I told him that I had added a few percent but didn't calculate is out as the result would be small. He disagreed, he said he had been a plant manager and had to figure amperage rise based on how far the motor was from the source and insisted that a 20hp motor that was 150 feet from the source would pull a great deal more amps at the start of the 150 feet than it would at the end of the 150 feet. I asked him if he might have amperage and voltage turned around and he insured me that he didn't, he said he would look for the formula he used and get back with me. I haven't heard from him yet.

What are some of your thoughts? I had a few things I wanted to try and explain to him but I figured I would wait until he calmed down a little.

 

[stephenw22]

Amps won't change down a line. Voltage will drop, but amps will not drop. If the amps into a wire don't equal the amps coming out, it's called a fault.

 

[521AB]

Cable has a capacitance, and if long cable, the capacitance is big. This capacitance creates a so called "charging current" that goes from the cable to "ground". This means that at the beginning of the cable you can measure, let's say, 250 A, and at the end you can measure 230A, because the other 20 A go away through the cable capacitance.
I don't know if this is what you are meaning..

 

[jghrist]

Capacitive current on a 120/240 volt 150 foot overhead service conductor is negligible.

Even though the current stays the same, there will be real power lost in the service. Unless it's pretty small wire, thought, I agree that it won't be much. Less than the margin of error in estimating how long the water pump ran.

 

[521AB]

jghrist, I am sorry, I didn't understand it was a LV cable.. actually it is not even a cable.. Sorry.
Anyway the loss you have is the Joule effect on the cable. It can be calculated: Ploss = R * I ^2 watts where R = rho * l / S ohms is the resistance of the cable and where rho is the Cu resistivity in ohm * meter, l is the length of the cable in meters and S is the cross section of the cable, in square meters.
Multiply the power by the "seconds" and you get the energy (in Joule), which is probably quite small...
Leave the converion in kWh to your customer...

 

[apowerengr]

I would speculate the state you work in (assumes you're in the US) has regulations on how to handle rebates for metering errors, how to calculate the error and payment for interest, etc.

 

[rcw retired EE]

Here's my thinking, A 20 HP 240V motor will pull more amps on starting if it is close to the source than if it is further from the source. The inrush current on start is only limited by the motor and system impedances so with a longer wire feeder there will be more impedance and less inrush current.

Once it is running a motor is a torque-matching device. With a constant load there will be a constant power input to the motor. If the voltage at the motor terminals is less due to voltage drop in the feeder wire, the motor draws more amps to keep the power constant. P=VIxpf.

More amps flowing in the transformer and wires = more losses. But they won't be very large.

 

[gobblerhuntr]

Talked to the customer today and asked him to provide the formula he was using to calculate this and he said he would try and find it but it has been years since he has used it. We'll see..

 

[swn1]

You'll just embarrass your customer, he's wrong, so don't push it.

Instead, run a simulation using some decent EE software and give him a generous interpretation of the result. The "garbage in, gospell out" effect will dispell most doubt, and it wasn't his error.

 

[PHovnanian]

First of all, your customer is using a formula intended for a different purpose (the motor curent vs voltage calculation).

Second: I'm assuming that the second service which was erroneously tapped off the CT load side was itself metered and you are proposing to subtract those KWh from his bill. You are practically there. The customer is theoretically correct in that the CT metering is measuring the I^2R loss in the tapped service in addition to the metered load amount. In practice, this will be insignificant, but you can calculate it by finding the R of the service and the average current of the tapped service (making some p.f. assumptions). It may turn out that this energy correction is insignificant. If its not, the figure can be refined somewhat by determining a loss factor for the given load factor. But getting correct data for this may be time consuming and ignoring it will be in the customers favor anyway.

 

[tinfoil]

Seeing is believing:

Put an instantaneous power meter on his meter point. I am sure that your meter guys have an appropriate analyzer handy.

Observe the realtime power with the town water tank pump running. Leave the analyzer running, and stop the water tank pump.

Let your complainant observe the before and after. The difference is the incremental load at his meter point.

Reinstall the analyzer at the water tank. Measure the load. Assuming this load remains constant, you have now measured the line losses via an empirical method.

If your analyzer is calibrated and certified, your Regulatory Authority will be content as well.

HTH