# Losses on Power Transmission Lines

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1. May 21, 2017

### Barry Fends

Say a power station transmits 1008 MWh of power per year to a facility however they only receive 900 MWh say the transmission line is 15 km long how would you calculate average power loss per km, now I know that theoretically it will just be 7.2 MWh per km but what calculations would you make to get there?
Cable Type: FLAMINGO (ACSR)
Voltage:132kV
Cable Diameter:25mm
Resistance: 0.315 Ohms per MFT

Last edited: May 21, 2017
2. May 21, 2017

### Drakkith

Staff Emeritus
A couple of things:

1. Are you sure your numbers are correct? 10008 - 900 = 9108, but 15km * 7.2 MW/km = 108 MW, and the two results don't add up to the total.

2. I'm assuming you mean MWh and not MW. Watt-hours are units of energy, but watts are units of power. Electric power usage is usually a measure of energy usage, not power, despite everyone using electric "power". So using some amount of MWh over the course of a year makes sense and you can convert that into Joules, the SI unit of energy. However, saying that you used some amount of MW over the course of a year makes no sense because a megawatt is already energy/time and dividing it by time again makes it meaningless.

3. May 21, 2017

### Barry Fends

I ap
I apologise I have edited the post to make more sense it was 1008 not 10008 and the units were mixed up

Last edited: May 21, 2017
4. May 21, 2017

### Drakkith

Staff Emeritus
Well, I'm not sure I can help you. I'm not even sure there is a way to calculate the power loss in the conductor without knowing the exact load conditions. But perhaps someone else here knows better than I.

5. May 21, 2017

### Barry Fends

I think it is just assumed there is no load conditions.

6. May 21, 2017

### jim hardy

This seems to me an exercise designed to:
make you keep units straight,
make you attack the problem in small logical steps
drive home the difference between power and energy.
I rather like it as a teaching tool..

so you want to check the measured MWH numbers?
You don't have enough information for an exact calculation
but you can make an approximate one as follows

Power = I2 x R

you can find resistance R per conductor, that's just changing from ohms per Mft to ohms per KM
and you can find approximate current I from power and voltage leaving the station, but you'll have to either neglect or guess at the power factor just use 1.0 for first approximation
I2R gives you watts which you'll have to convert to Mwh/year.

How does the resulting I2R loss compare with your 7.2 MWH/Km over a year ?
I call such approximate comparison Sanity Checks" . They're quite useful for finding errors of logic, incorrect unit conversion or mis-keyed calculator entries.

7. May 21, 2017

### Barry Fends

Thanks for the reply what do you mean by power factor?

8. May 21, 2017

### jim hardy

Inductive loads like transformers and motors cause a teeny extra bit of current to flow,
so that power isn't quite Volts X Amps it's Volts X Amps X Power Factor.
Power factor is can only be 1.0 or less. A factory will get charged for power factor less than 1.0 so they try to keep it real close to 1.0.

So the "approximate calculation" i described will be in error by the power factor's difference from 1.0. Assuming power factor is not less than 0.95, that's just a few per cent.

Did teacher give any clue as to that factory's power factor?

9. May 21, 2017

### Barry Fends

None at all

10. May 21, 2017

### jim hardy

I'd use 1.0 then.

11. May 27, 2017

Correction: ACSR Flamingo AC resistance 75oC it is only 0.0314 ohm/1000 ft.=0.103 ohm/km

If we shall take 365 days/year 24 h the losses will be 12.3 kW that means 51.5 A/phase.

(1008-900)*1000/365/24=12.32 kW

[Ploss=3*0.103*15*51.5^2=12.29 kW].

If we shall take full only 8 h/day the losses will be 37 kW that means 89.3 A/phase

It is a very low load for this 132 kV transmission line [the rated is about 800 A].