# Power loses estimation

#### Pyrex

Joined Feb 16, 2022
310
Hi,
the situation is like this- a company, let's say- company No1 pay the bills of the electricity according to the readings of their's electricity meter. An other company- let's say -company No2 - buying electricity from the company No1. Companys No2 building is located 100 meter away from the company No1 and their's electricity meter is installed inside the building.
The problem is that power loses in the cable, fuses, etc. will be estimated by the company's No1 electricity meter only. How to estimate those loses in order to add them to the expenses of company No2 ? Power compsumption is quite uneven -about 1kW at night time and over 100kW at daytime.

#### DickCappels

Joined Aug 21, 2008
10,246
Compare the readings on the electricity usage meters at the two sites and take the ratio.

#### Irving

Joined Jan 30, 2016
4,065
Why not fit a power monitor on the #2 feed at #1 premises?

Failing that, monitor line voltage at #1 and #2 simultaneously, and using power meter at #2 then losses are (volts@#1/volts@#2 - 1) * power@#2

#### Irving

Joined Jan 30, 2016
4,065
Compare the readings on the electricity usage meters at the two sites and take the ratio.
How does that work without a third independent variable?

#### BobTPH

Joined Jun 5, 2013
9,284
If I understand you correctly, the losses are already included in the metering done by co1. The reading at the source end of the cable is for the current and voltage before the voltage drop across the cable. But co2 would see less power since it would see the same current but lower voltage.

#### LowQCab

Joined Nov 6, 2012
4,309
The losses are simply Electrical-Power that the #2 Company pays for, but never receives.
There's virtually zero measurable losses for Company #1.
Everything is accounted for.

The amount of Power lost is very minor in almost any case.
If it is not minor,
because Company #2 has seriously increased their Electrical usage
to the point that the original "Service-Entrance-Conductors" have become over-loaded,
then the "Service-Entrance-Conductors" may need to be up-graded,
but this is not worth the expense,
and it may take ~5 or ~10 years to see a monetary return for the upgrade.

If there is no Fire-Hazard, or problematic Voltage-Sag,
don't up-grade, it's not worth the expense.

This sounds like somebody isn't working hard enough to expand their business,
but is instead, wasting huge amounts of valuable time squabbling over "Pinching-Pennies".

Install Phase-Compensation-Equipment in both buildings if You want to save a few Dollars,
and then get to work on promoting your business, and delivering goods and services.
The ROI may still be several years.
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#### MrChips

Joined Oct 2, 2009
31,088
Real power or reactive power?

#### Pyrex

Joined Feb 16, 2022
310

#### Pyrex

Joined Feb 16, 2022
310
The amount of Power lost is very minor in almost any case.

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The point is to include those losses to the bill of Company#2 . There's no reason for Company N#1 to pay foreign expenses. Installed power is 200 kW, it's not a flat or a garden house

#### BobTPH

Joined Jun 5, 2013
9,284
Why do you think co1’s meter reading does not include the losses?

Let me explain. At co1’s meter it is reading 240V at 100A or 24KW.

At co2’s distribution box it is reading 230V at 100A or 23KW but co2 is being billed for 24KW. So who is paying for the power loss?

#### Pyrex

Joined Feb 16, 2022
310
Why do you think co1’s meter reading does not include the losses?

Let me explain. At co1’s meter it is reading 240V at 100A or 24KW.

At co2’s distribution box it is reading 230V at 100A or 23KW but co2 is being billed for 24KW. So who is paying for the power loss?
Co2 pay according to their electricity meter, which is located in their building.Voltage in the building is 230V. So , their bill is for 23kW. Their el. meter is in their building, so loses in the cable not estimated

#### BobTPH

Joined Jun 5, 2013
9,284
An other company- let's say -company No2 - buying electricity from the company No1. Companys No2 building is located 100 meter away from the company No1 and their's electricity meter is installed inside the building.
Ah, this is ambiguous. I read it to mean the meter was in co1’s building.

So, there is your solution, move the meter.

#### LowQCab

Joined Nov 6, 2012
4,309
It sounds like the Tread-Starter doesn't want to hear an explanation of
how this situation is actually a "non-issue".
Rather it sounds like ....... I want more Money, and I think I've found a way to get it.
( And we're talking about "Pennies" )

Company "A" is not paying anything "extra" for any supposed "Cable-Losses" ........... end of story.

In the United-States this "Power-sharing" arrangement is illegal.
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#### Irving

Joined Jan 30, 2016
4,065
It sounds like the Tread-Starter doesn't want to hear an explanation of
how this situation is actually a "non-issue".
Rather it sounds like ....... I want more Money, and I think I've found a way to get it.
( And we're talking about "Pennies" )

Company "A" is not paying anything "extra" for any supposed "Cable-Losses" ........... end of story.

In the United-States this "Power-sharing" arrangement is illegal.
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Umm, though we don't have evidence, I'd hazard a guess they are not in the US. In which case 'illegal' may or may not be relevant, nor Codes relating to installation. Also 100kW @ 240v or 440v is going to be lossy even over 100m.

#### WBahn

Joined Mar 31, 2012
30,303
Hi,
the situation is like this- a company, let's say- company No1 pay the bills of the electricity according to the readings of their's electricity meter. An other company- let's say -company No2 - buying electricity from the company No1. Companys No2 building is located 100 meter away from the company No1 and their's electricity meter is installed inside the building.
The problem is that power loses in the cable, fuses, etc. will be estimated by the company's No1 electricity meter only. How to estimate those loses in order to add them to the expenses of company No2 ? Power compsumption is quite uneven -about 1kW at night time and over 100kW at daytime.
A very simple model can be put together based on a single-phase equivalent.

Let's say that the power is at a line voltage of Vs and the wire between Company A and Company B has a total resistance of Rw.

If Company A is getting billed for an amount of power Pa and Company B is getting billed for an amount of power Pb, with the difference being the power, Pw, lost in the wire, then the total current leaving Company A is

Ia = Pa/Vs

The power lost in the wire is

Pw = Ia²·Rw

The power at Company B's meter is then

Pb = Pa - Pw

What you are interested in is the ratio Pa/Pb, as this is the surcharge you need to multiply the meter reading at Company B to get what the power delivered by Company A would be.

Pw = (Pa/Vs)²·Rw = Pa² / (Vs²/Rw)

Pa/Pb = Pa/(Pa - Pw)

Pa/Pb = 1/(1 - Pw/Pa)

Pa/Pb = 1/{1 - [Pa / (Vs²/Rw)]}

Notice that

P' = Vs²/Rw

is the power that would be delivered to the wire if the lines were shorted at Company B, which is going to be a very large amount compared to the power measured at Company A, so this fraction is going to be very small and the ratio of Pa/Pb is going to be very close to unity.

Pa/Pb = 1/(1 - (Pa/P'))

You didn't say what the voltage was that was being sent from Company A to Company B. If the power is 200 kW max, it seems unlikely that it is at 440 V as that would be over 500 A. I don't know what it might be between that and, say, 13.8 kV at 15 A.

Let's run both, with the smallest rated wire, and see what we get.

At 440 V, you would need two runs of 4/0 which would have a total wire resistance of about 16 mΩ for 100 m. So the

P' = (440 V)² / 16 mΩ = 12.1 MW

So

Pa/Pb = 1.017

So you would need to charge a 1.7% surcharge when they are actually consuming 200 kW. At 100 kW the surcharge would be 0.8 % and at 1 kW is would only be less that 0.01%.

If the distribution is at 13.8 kV, then even if you used #10 AWG (and I don't know what the smallest that would be allowed at that voltage, regardless of current) then the resistance could be about 650 mΩ, making

P' = (13800 V)² / 650 mΩ = 293 MW

Now, even at full draw of 200 kW, the ratio is only

Pa/Pb = 1.0007

making the surcharge less than 0.1%

Now, these are fast numbers that don't properly take into account scaling for a single-phase equivalent, but the results are likely within a factor of two or so.

Now, if they are consuming 100 kW for 12 hours a day, five days a week, then their total monthly metered consumption would be about 24 MWh. At $0.10/kWh, that would be$2400/mo and if the average line loss came out to be 0.1% that would be a surcharge of just \$2.40/mo.

Ask them to treat you to a few dozen donuts a couple times a year and call it a day.