# Calculating Watt loss in a line

• posted

Hi,

I am trying to figure the watt loss on a 3 phase power line. The line is 30 miles long with .015 resistance per mile. There is 900 amps across the line.

• posted

Assuming that the phase loads are balanced and there is no neutral current or ground current and that the resistance is that per conductor:

Then each line has 900 amps flowing through a total resistance of 0.015 x 30 ohm.

Thus each line loss is I^2(R) = 900x 900 x 0.015 x30

The total loss is three times that, ie 900 x900 x0.015 x30 x 3 watts.

-- Sue

• posted

Ohm's law is quite useful here. power = current squares x resistance.

Power = 900 x 900 x 0.015 x 30 which my calculator gives as 364.5kW (& probably 3 times that if there is 900 amps flowing in each line)

• posted

Thanks Charles & Sue! I appreciate your quick response. I thought it was this simple but I wanted to check with you guys first.

• posted

0.15 ohms per mile comes to 0.15/5280 x 1000 =3D 0.0284 ohms per 1000 ft. This corresponds to 400 Kcmil at 75 degrees C copper using table 8 of the NEC. The ampacity of 90 degree C 400 kcmil is 380 amperes. Utility overhead lines are run at 120 degrees and assume a 40 degree C ambient. Even at this higher temperature, it appears without further calculations that 900 amperes is much too high of a current for this conductor.
• posted

hmmm, now we're thinking!

local nuke here (Dominion) wants nrc special dispensation to run the rods higher to increase plant output beyond originally spec rating.

engineering says orginal specs were conservatively written to ensure original permitting was accepeted without severe review.

safety measures were reviewed to assure any unusual incident occurence could be met within nrc standard.

ok, so now somebody in newsgroups wants to know if some xmiss line can handle more than it was designed for....

am i paranoid, or is the element of pure profit causing bending of rules and compromise of safety??

ok, now lets find out if the BWR really can huff and puff its way over the hil.......... :-))

0.15 ohms per mile comes to 0.15/5280 x 1000 = 0.0284 ohms per 1000 ft. This corresponds to 400 Kcmil at 75 degrees C copper using table 8 of the NEC. The ampacity of 90 degree C 400 kcmil is 380 amperes. Utility overhead lines are run at 120 degrees and assume a 40 degree C ambient. Even at this higher temperature, it appears without further calculations that 900 amperes is much too high of a current for this conductor.
• posted

Disregard my previous post. I used 0.15 ohms but should have used

• posted

---------------------------- Except that "power = current squares x resistance. is not Ohm's Law even though true.

• posted

0.15 ohms per mile comes to 0.15/5280 x 1000 = 0.0284 ohms per 1000 ft. This corresponds to 400 Kcmil at 75 degrees C copper using table 8 of the NEC. The ampacity of 90 degree C 400 kcmil is 380 amperes. Utility overhead lines are run at 120 degrees and assume a 40 degree C ambient. Even at this higher temperature, it appears without further calculations that 900 amperes is much too high of a current for this conductor.

------------------ I would suggest that the person who set this problem just extracted some numbers out of the air. numbers together without considering that they were not realistic. However they suffice to deal with the concept at a beginner stage.

Also, I am not too sure that I would want to run ACSR (as opposed to copper which no utility uses) at 120C. There will be an irreversible deterioration in conductor life which will happen every time the conductor reaches this temperature.

• posted

You are reading too much into this question. Odds are that its someone's homework.

• posted

And with such temperatures isn't there also a problem with sagging in the line? ISTR that running very hot lines can lead to the expansion of the line and sagging to dangerous levels.

But the OP had said 0.015 ohms per mile, so it must be larger than 400 kcmil...

daestrom

• posted

IIRC, the big outage back in '03 began with a heavily loaded transmission line (on a hot day) sagging into some trees.

'Dangerous levels' is somewhat ambiguous. All power lines sag more at higher temperatures. Their design must take this into account when allowing for clearances, insulator string swings, etc. Then, its up to the maintenance folks to keep trees clear of the lines and operations not to overload them beyond their thermal limits.

• posted

In addition, with ACSR, there can be a problem with "birdcaging" due to differential expansion of the steel core and the aluminum conductor. I don't think that this is completely reversible- correct me if I am wrong. I am going on memory of a long ago look into the effects of overloads (particularly repeated ones) on overhead conductors.

You reminded me of a situation many years ago (over 50) in Quebec when, on a hot day with well above normal loading due to a combination of things there were repeated mysterious outages on a 240KV line. Mysterious until a local farmer called in and complained that the wires were hitting his barn roof.

This is an exercise in some circuits course where the intention was to find out whether the student could apply I^2R to a 3 phase line- as opposed to a single phase line. The realism of the actual numbers given isn't important at this stage (sure, it would be nice to use data pertaining to a real conductor just to say there is an real application for this) - the concept (which Troy apparently hadn't grasped) is the point.

The side track is more interesting than the original question.

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