Sag & Tension of Distribution lines

[engratcorner]

Hello all,

I have had this question from long time and I couldn't find an accurate answer to this anywhere!

Well, you may be familiar with Sag of a Transmission or Distribution line and it is defined as
Sag = [(weight in lbs/ft)*(span distance in ft)^2]/[8 * (Horizontal Tension)]

All the parameters in the above equation looks obvious but the "Horizontal Tension"!

What does "Horizontal tension" actually mean? (I guess not the tension of the conductor but the tension of conductor at lowest point??)

Can anyone please help me in understanding this?

Thanks,

 

[jghrist]

It is the tension at the low point. The total tension in the conductor is in the direction of the conductor at any point in the span. The tension can be resolved into a vertical and a horizontal component. At the low point, the conductor is horizontal, and the total tension is in the horizontal direction. The horizontal component of the tension is equal at all points in the span.

 

[engratcorner]

Informative @jghrist.

Moving forward on this discussion, my understanding from the above is, the horizontal tension is the tension at which the conductor was strung - since only horizontal component is present at low point and it is equal to the total tension at which the conductor was strung. Correct me if I am wrong.

 

[cswilson]

Tangentially related (forgive the possible pun) question:

In the US, long distance transmission lines are referrerd to as "high-tension" lines. I had always assumed this meant high mechanical tension. But then I learned that the French word for "voltage" is "tension", so it occurred to me that it might refer to high voltage instead.

Does anyone know?

 

[davidbeach]

I'm pretty sure it is related to voltage. It would seem that the same type of conductor, for the same ruling span, would be tensioned similarly regardless of voltage.

 

[cuky2000]

although the word tension is sinonimum of voltage is also sinonimum of voltage. For sag and TENSION application the word tension = force. Sag and tension is basically the mechanical modeling of cable catenary and no much related with the electrical characteristics.

The word Voltage= electrical potential difference, electric tension or electric pressure. Also Tension = force as well in Other languages.

The classification of long, medium and short overhead transmission line is associated with the approximation of modeling techniques considering how the line capacitance is taking into account (ABCD) parameters.

Prior computer became popular, a great deal of simplification was convenient to use T. Lines modeling simplification. So short line < 50 mi approx. Medium: 50 to 150 mi and large T. Line >150mi approx.

 

[stevenal]

See [URL unfurl="true"]http://www.rurdev.usda.gov/SupportDocuments/UEP_Bulletin_1724E-200.pdf [/url]section 9.12.

 

[jghrist]

Moving forward on this discussion, my understanding from the above is, the horizontal tension is the tension at which the conductor was strung - since only horizontal component is present at low point and it is equal to the total tension at which the conductor was strung. Correct me if I am wrong.

It doesn't refer to the tension at which the conductor was strung (stringing tension). It refers to the direction of the tension or component of tension. Conductor tension is directional and can be resolved into a vertical and a horizontal component. The total tension is the vector sum = sqrt(H^2 + V^2), where H is the horizontal component and V is the vertical component.

 

[tinfoil]

{Assuming that no motion is occurring in the wires}

At the low point of the span (midspan if the two terminals are at equal height-above-ground), the vertical component of the tension VECTOR is zero, so the magnitude of the tension and the horizontal tension component are the same*.

As you look towards the terminals, the horizontal component remains the same (or else the wire would accelerating as F= m * a ALWAYS applies), but a vertical component begins to appear.

At the pole/tower connection points, you have some noticeable slope on the wire. the horizontal force is STILL the same, and you can work out the vertical component, and the wire tension using trig*.
(The force vector in a conductor at rest MUST lie in the direction of that conductor, or else the wire would be under acceleration).

* Note that if there is wind blowing on the wire, there can also be a second lateral horizontal component contributing to the total tension, perpendicular to the 'longitudinal' horizontal path from tower to tower.

The horizontal tension therefore is a component of the total tension that that particular span of wire has under that particular set of conditions.

 

[stuj]

Hi

you can use my free line design software to calculate and disign lines. the tension is horizontal (mid span zero slope = horizontal tension) vertical component at the attachment point is weight span.

http://www.wazazulu.com/lview/lview.htm

have a look at this reference, this references the ESAA guideline cb1 that is very useful.

https://www.ergon.com.au/__data/assets/pdf_file/0004/64228/Section-10-Design-Programs.pdf

 

[racookpe1978]

Can you make the approximation that the vertical component at the tower mounting points is only weight?

If a tower is on a high point, and its neighbor in the valley, doesn't the load carry sideways? Or not enough to matter?

 

[stuj]

The weightspan is the vertical component. In equlibrium there will be no sideways load. However, as conditions change temp and wind will be some sidways movement and the insulator will swing. Best to do a FEA on big height differences and significant adjacent span sizes.

weightspan is based on the projected vertex and if there is significant height difference the vertex can be outside the span. You can model this on lineview and turn on the vertex to see where it is (vertex number can be displayed). use this software - lineview...

http://www.wazazulu.com/lview/lview.htm

download the tutorial...its