How do I calculate the sag in power lines?
How do I calculate the sag in power lines?
(OP)
I need to calculate the sag in power lines and, also, I need to transform the wind velocity into pressure or force. How can I do it?
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How do I calculate the sag in power lines?
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How do I calculate the sag in power lines?How do I calculate the sag in power lines?(OP)
I need to calculate the sag in power lines and, also, I need to transform the wind velocity into pressure or force. How can I do it?
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RE: How do I calculate the sag in power lines?
There is more to it than that of course. Call your local Okonite cable rep and get a copy of EHB-90 "Engineering Data for Copper and Aluminum Conductor Electrical Cables". There is a very good writup on pages 45-49. If you can't get the book go to the Okonite Website and ask them to send you a copy of the book or those pages.http://www.okonite.com/
Southwire at http://www.southwire.com/ has a "Power Cable Manual" that they will send you that has a write up on sag calculations, They also have a form you can enter you parameter on and they will calculate sag it on their computer and fax ( maby Email by now )you back a solution .
RE: How do I calculate the sag in power lines?
Wind pressure in lbs/ft^2 is calculated using
Pw = 0.00256*(Vw)^2
Vw = Wind speed in miles per hour
Wind load per unit length is equal to the wind pressure multiplied by the conductor diameter.
Using the same units, Fw comes out in lbs/ft
Fw = Pw * (Dc + 2t)/12
Dc = conductor diameter (inches)
t = ice thickness, if you get ice in your area (inches)
RE: How do I calculate the sag in power lines?
RE: How do I calculate the sag in power lines?
1. Wadhwa C. L., "Electrical Power Systems," Second Edition, John Wiley & Sons, 1991
Reference 1 Chapter 7 Mechanical Design of Transmission Lines
includes a good treatment of sags, including wind, ice, and you could possibly consider conductor bundles. However, to leave it up to a specific software is often required.
RE: How do I calculate the sag in power lines?
Keep a stop watch on your left hand and a dry buzz stick on the right. Hit the conductor by the stick and start counting time. Let the stick rest on the conductor. When you receive the third wave return, stop the watch. The time noticed by the watch be squared and multiply it by 0.0372.
The answer is the sag in meters.
By using the sag tension formula, I know nobody who measures it. Sorry for this word.
Another way to measure is to put a mark on the tower at location approximately the sag value from the insulator position. Get the position on the other tower which is in line with the mark on the tower and the tangent to the sagging conductor. Get average of the two values. It is approximately the sag. This can be exactly calculated by considering the sagging conductor as a parabola.
RE: How do I calculate the sag in power lines?
The above sag is in fact in feet. It needs correction.
The correct formula in meter for:
First return wave sag = 1.006*T^2
Second return wave sag = 0.2515*T^2
Third return wave sag = 0.1118*T^2
T is time in seconds
RE: How do I calculate the sag in power lines?
the above formula provides sag in feet. The earlier formula is alright.
RE: How do I calculate the sag in power lines?
There are times when transmission lines are installed to very exacting initial tensions calculated specifically for a particular installation. In those cases, dynamometers are used to set the initial tension, and targets (or return wave) may be used to verify the correct sag.
I have seen Southwire's reference books, and I wholeheartedly agree with BJC that they are worth having.
RE: How do I calculate the sag in power lines?
Try
http://www.southwire.com
RE: How do I calculate the sag in power lines?
Reference 1 indicates that:
1. Ice loading produces:
wi=pi x rho x (2 x r x t + t**2), in kg/m
where
wi is weight of ice per unit length of conductor
r is radius of the conductor in meters
t is thickness of the ice in meters
rho is the density of ice, e.g. 910 kg/m**3
Please, notice that this does not account for icicles that can potentially become very heavy. Then safety factor has to account for them, e.g. 5 or more since the normal safety factor may be 2 only.
2. Wind loading produces:
wa=2 x (r + t) x p, in kg/m
where
wa is wind loading per unit length
p is the wind pressure in kg/m**2
t is ice coating thickness in meters
r is radius of conductor
Total loading W=sqrt[(w + wi)**2 + wa**2], in kg/m
where w is weight of conductor downward
Loading factor q=W/w