Losses on Power Transmission Lines

In summary, the conversation discusses a power station that transmits 1008 MWh of power per year to a facility but only receives 900 MWh due to power loss. The transmission line is 15 km long and uses a FLAMINGO (ACSR) cable with a voltage of 132kV and a diameter of 25mm. The theoretical average power loss is calculated to be 7.2 MWh per km, but the exact calculation requires more information about the load conditions. Approximate calculations can be made using the formula Power = I^2 x R, and sanity checks can be used to verify the accuracy of the results.
  • #1
Barry Fends
5
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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
 
Last edited:
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  • #2
Barry Fends said:
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.
 
  • #3
I ap
Drakkith said:
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
 
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  • #4
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
Drakkith said:
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.
 
  • #6
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
jim hardy said:
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?
 
  • #8
Barry Fends said:
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?
 
  • #9
jim hardy said:
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
 
  • #10
Barry Fends said:
None at all
I'd use 1.0 then.
 
  • #11
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].
 
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1. What causes losses on power transmission lines?

There are several factors that contribute to losses on power transmission lines, including resistance in the conductors, capacitance between the conductors, and inductance from the magnetic fields around the conductors. These losses are known as line losses or I2R losses.

2. How do line losses affect power transmission?

Line losses have a direct impact on the efficiency of power transmission. As energy is lost through resistance, less power is delivered to the end user. This results in higher electricity costs for consumers and can also lead to power outages if the losses are too high.

3. What are the different types of line losses?

There are two main types of line losses: AC losses and DC losses. AC losses occur when power is transmitted using alternating current, while DC losses occur when power is transmitted using direct current. AC losses are typically higher due to the skin effect, which causes current to flow more on the surface of the conductor rather than through the entire cross-section.

4. How can line losses be reduced?

One way to reduce line losses is by using materials with lower resistance for the conductors. This can include using larger conductors or using materials with higher conductivity, such as copper rather than aluminum. Another method is to increase the voltage of the transmission, which reduces the current and therefore the losses. Additionally, implementing more efficient transmission technologies, such as High Voltage Direct Current (HVDC) systems, can also help reduce losses.

5. Can line losses be completely eliminated?

No, it is not possible to completely eliminate line losses. However, through advancements in technology and improvements in transmission infrastructure, losses can be minimized to a certain extent. The goal is to find a balance between cost and efficiency, as reducing losses typically requires higher upfront costs for materials and technology.

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