Transmission line sag and slack

Discussion in 'Homework Help' started by Kayne, Sep 27, 2012.

1. Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Hi all,

I am haven trouble with a question to do with solving the Sag and Slack of a transmission line, I have been starting at this for a while now without any luck.The question is

Using a level-span catenary model, calculate the sag and slack for a 335-m span of ACSR cable having a cable mass per unit length of 1.63 kg / m and a diameter of 2.86 cm.
The load on the cable consists of (i) ice with a uniformly distributed radial thickness of 0.965 cm and (ii) a constant horizontal wind pressure of 383 Pa of total projected area. Take the breaking tension for the cable as 133 kN and use a safety factor of 3.
Take the density of ice as $915 kg.m^-3$ and gravitational acceleration constant as $9.81 m.s^-2$
Neglect thermal and elongation effects.

Span $S = 335m$
Mass $M = 1.63 kg/m$
diameter $d = 2.86cm$
ice radius thickness $irt = 0.965 cm$
horizontal wind pressure $hwp = 383Pa$
breaking tension $br = 133kN$
density of ice $id = 915 kg.m^-3$
gravity $g = 9.81 m.s^-2$

So I have found that the effects of Ice and Wind loading is

Xsectional area of ice
$Ai = pi*irt(d+irt) = 1.16*10^-3 m^3$

Weight of ice per unit lenght of cable
$Wi = Ai*id*g =10.409 kg$

projected area per unit lenght of ice covered cable
$ai=d+2*irt=0.048$

Horizontal wind force of ice covered cable
$Fw = hwp*ai=18.346$

Resultant force due to wind and ice
$Fwi = (\sqrt(Fw^2)+\sqrt(M*g+Wi)^2)=32.148$

Now this is where I am stuck as I need to calculate Sag and slack. Am I on the right path or is there another way I should be solving this??

Any guidance would be a great help

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The effective deflection would presumably be determined from the catenary relationship:

$D=\frac{T_h}{W} $\cosh{$$\frac{SW}{2T_h}$$}-1$$

Where:

$T_h=\frac{133 \ kN}{3} = 44.333 \ kN$

Applying a loading safety factor of 3

$W=32.148 \ N$

as you have calculated as the effective per meter loading

$S=335 \ m$

being the cable span

This effective deflection D may then be resolved into its vertical and horizontal components per the relationship of vertical [ice + cable weight] and horizontal [wind load] force components.

Kayne likes this.
3. Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Thank TNK I will continue to have a look at this and post my answer.