Losses on Power Transmission Lines

[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?
Additional Info:
Cable Type: FLAMINGO (ACSR)
Voltage:132kV
Cable Diameter:25mm
Resistance: 0.315 Ohms per MFT

 

[Drakkith]

ay a power station transmits 10008 MW of power per year to a facility however they only receive 900 MW 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 MW per km but what calculations would you make to get there?

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.

 

[Barry Fends]

I ap

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.

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

 

[Drakkith]

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.

 

[Barry Fends]

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.

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

 

[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.

 

[Barry Fends]

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.

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

 

[jim hardy]

what do you mean by power factor?

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?

 

[Barry Fends]

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?

None at all

 

[jim hardy]

None at all

I'd use 1.0 then.

 

[Babadag]

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].