Question on line sag - September 12 Frisco outage

I read the report on the incident on Sept. 12 that shut down about 1 M people's lives. One of the things in the report was that there was no power from an adjacent area (North, maybe?) because due to the load, the line dropped 25 feet and contacted a tree, causing a line loss.

I am curious about this. I think the outage only lasted a couple of hours; and so I'm wondering how much current there would have to be in a line to lengthen a span by 50 feet (25 feet down; then another 25' back up. Let's ignore any catenary characteristics for now.)

First, how long of spans would you expect to see here, anyway? And secondly, I'm still struggling with the vision of a 25' drop in line clearance. Where can I learn more about this?

I have heard of this happening at other times, but I have no frame of reference for this. I'd appreciate any help, please.

HR

Reply to
Rowbotth
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Major high voltage, long distance transmission lines get warm when under heavy load. A warmer metal becomes longer - droops. These transmission lines have a maximum summer and a maximum winter current rating. Since these lines can remain cooler in winter, their current rating is higher.

To appreciate the concept, learn how First Energy of Ohio all but intentionally created the NE blackout. Among the violations included raising the summer and winter spec ratings of some transmission line so that the summer current rating was even as high as winter's (this reeks oF MBAs changing specifications). Letting trees grow too high so that the sagging Hanna Jupiter power line shorted to those trees (see testimony from employees of Robert's Tree Service who saw it and fortunately were not under that line when the fault occurred). First Energy apparently even chopped down that tree below ground level hoping that investigators would not discover this (and other) violations.

Due to a large NE blackout created by numerous divisions of First Energy, industry standards were changed. I believe laws were created as to what electric companies must do because of why and how these electric lines sag when under heavy currents

- especially in hot weather.

Pictures showing those lines and how lines sag are included in NERC reports at

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including Gerry Cauley's Presentation and the final report of

5 April 2004.

Rowbotth wrote:

Reply to
w_tom

A similar problem was contributory to the large outage in the US last year(?). The overloaded lines coincided with a lack of tree pruning under them, so there was less clearance than there should have been to start with.

Reply to
Andrew Gabriel

I do understand the concept of line sag, and I've done a few calcs myself. I am just really curious about how you'd get an extra 50' of conductor between spans. (Of course, all we use is Drake or Linnet, but still...)

Thanks for the link to the pictures, but I was actually more looking for some info on the physics of so much expansion in the conductors. What might these folks have been using - a 4/0 conductor? This sort of information...

HR.

Reply to
Rowbotth

Before an intelligent conversation can start, the voltage of the line needs to be known. Voltage will and terrain will dedicate span lengths.

I seriously doubt that a single span lengthen 25 feet with out a cable failure.

I have not strung any lines in some time. The last towers I worked on were

345kv lines and the towers were ~800 feet apart. I have run some 15kv over a mesa in Arizona where the poles were ~350 feet apart.

I have seen lines "reverse sag" in winter storms. Which is pretty scary.

Lines are tensioned between supports. Some times the tension can be measured in tons.

Reply to
SQLit

I don't remember the line voltage, but I'd bet at least 240kV, considering the population and the area. I'm used to 300 m line lengths (300/0.3048 = feet) and that is sort of why I am wondering about the quoted line sag.

Hence the posting....

HR.

Reply to
Rowbotth

Child's play. In eastern Kentucky, I know of 15kV spans over 1000 feet in length! Transmission spans over 1500 feet are not unhead of.

Charles Perry P.E.

Reply to
Charles Perry

First, what is reverse sag.

Secondly, it does not take much increased wire length to cause a considerable sag.

As a rough estimate, consider a span of 2a consisting of two bars pivoted to each other and to the extremes of the span, each of length c. The sag b for such a combination is easily estimated from the Pythagorean theorem,

c^2=a^2+b^2.

Assuming that a is constant, take the differential of the equation.

2*c*dc = 2*b*db or db = (c/b)*dc,

Thus, it is easy to see that for small sags small differences in conductor length make for large changes in sag.

Take the numbers given above. Assume there is a 25 foot sag to begin with. Increase the length of each member by one foot. The increase in sag for a one foot increase in c is approximately

db = (400/25)*1 = 16 feet.

Bill

Reply to
<salmonegg

Thanks Bill;

Its good to see there are some engineers that still understand the question and provide a direct answer.....versus...the rhetoric of the previous responses.

"SQLit"

Reply to
SERTEC

Feather Mesa in AZ is not all that steep but there were no roads, just bloody rugged. Holes for the poles were blasted into the granite. Then we assembled the poles in a field and had them flown in with a helicopter. Only thing that could get to the locations was a crawler, which I refused to ride on. Looked like it was going to tip over all of the time.

I did a transmission line over the edge of one of the mines near Globe, AZ. Span was just short of 3400 feet. Took us almost 6 months to build both sides of that one. Flew the wire over with helo.

Reply to
SQLit

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I have seen wind take the wire from its natural state, drooping toward the earth. To the same position vertically. Gets pretty scary when the lines are not in sync. Arcy sparky time, with the poles or towers rockin and reelin.

snipped

Reply to
SQLit

Well, that's where you went awry. The conductors don't have to grow by anything near 25' to get a 25' sag. When they are strung 'tightly' between towers, a relatively modest growth of just a couple of inches can make for a noticable sag. The tighter the original line, the less change is needed to get a given amount of sag.

Think of it in reverse, how tightly do you have to tension a line to get it perfectly straight? (impossible) How much tension to get the line 'pretty straight' versus 'droopy'. Now, how much expansion in the length of conductor to 'ease' that tension from the 'pretty straight' value to the 'droopy' tension? Not very much.

To get into all of it you need sinh and cosh functions, but 'salmonegg's approach is not too bad an approximation.

Not intuitive, I know. Kind of like how many feet do you add to a ring encircling the world in order to expand it to encircle the world at a height of 3.14 feet (answer, 1 foot).

daestrom

Reply to
daestrom

Reverse those? Still neat, though.

j
Reply to
operator jay

Big deal. It happens every earthquake around here. :=)

Bill

Reply to
<salmonegg

Oops! Do the math again: PI(D+(2*3.14)) - PI(D) = PI(2*3.14)

At a height of 1 foot: PI(D+2) - PI(D) = 2PI = ~6.28 feet

-- VWW

Reply to
VWWall

Yeah, you guys are right.

Like I said, 'not intuitive' ;-)

daestrom

Reply to
daestrom

Reply to
w_tom

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