Transmission line sag and slack

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

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    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

    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
    The effective deflection would presumably be determined from the catenary relationship:

    D=\frac{T_h}{W} \[ \cosh{\( \frac{SW}{2T_h} \)}-1 \]


    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
    Thank TNK I will continue to have a look at this and post my answer.

    Appreciate your help...