Sample Calculation of Sag and Tension of Transmission Line – Even Elevation

This is an example calculation for sag and tension in the transmission line. You may opt to review first the fundamental principles and formulas from this post.

The foregoing calculation is based on the linear cable model on which plastic elongations are ignored. Also, the loading of the conductors is based on the National Electrical Safety Code 2017.

Table of Contents

Problem:

A transmission line conductor was strung between two towers, 300 meters apart and same elevation. During the time of installation, the conditions were t =15°C and initial horizontal tension = 25% of Rated Tensile Strength (RTS).

Calculate the sag and tension at:

No Load Conditions (no wind / no ice)
Heavy Loading District: Ice Thickness = 12.5 mm, Wind Pressure = 190 Pa, t = -20 deg C
Maximum Conductor Temperature : t max = 90 deg C, no wind, no ice

Solution:


Conductor Name	403-A1-37
Conductor Common Name	403 mm2 AAC (Arbutus)
Span Length	300 m
Outside Diameter	26.1 mm
Conductor Unit Weight	10.89 N/m
Rated Tensile Strength	81.8 kN
Modulus of Elasticity	58.9 GPa
Coefficient of Thermal Expansion	23 x 10-6 / deg C
Total Conductor Area	402.9 mm2

1. No load conditions (no wind/ice load)

t₁ = 15°C
H₁ = 25% RTS = 81, 800 * 0.25 = 20,450 N
A = 0.0004029 m
S = 300 m
W₁ = 10.89 N/m
E = 58.9 x 10⁹ Pa

Total Conductor Length:

\begin{align} L_{1} = S \left( 1 +\frac{S^{2}W_{1}^{2}}{24H_{1}^{2}} \right)= 300 \left( 1 +\frac{300^{2} * 10.89^{2}}{24*20,450^{2}} \right) = 300.32\ meters \end{align}

Initial Sag:

\begin{align} D=\frac{H_{1}}{W_{1}}\left(cosh\left(\frac{S}{2H_{1}/W_{1}}\right)-1\right)=\frac{W_{1}S^{2}}{8H_{1}}=\frac{10.89*300^{2}}{8*20,450}=5.99\ meters \\ \end{align}

\begin{align} \end{align}

2. Heavy loading district: Ice thickness = 12.5 mm, Wind load = 190 Pa, t = -20 °C

a. Calculate ice load (assume ice density is 915 kg/m³

\begin{align} W_{ice} &=\rho _{ice} *\pi t_{ice}(D +t_{ice}) = 915*\pi *0.0125 \; (0.0261 + 0.0125)\\ W_{ice} &= 1.386 \ kg/m \ or \ 13.6 \ N/m \end{align}

b. Calculate Wind Load

\begin{align} W_{wind} = Pressure * (D + 2t_{ice}) = 190 *(0.0261 + 2*0.0125) = 9.71 \ N/m \end{align}

c. Total Unit Weight

\begin{align} W_{total} &= \sqrt{(W_{cond} + W_{ice} )^{2} + W_{wind}^{2}} \\ W_{total} &= \sqrt{(10.89 + 13.6 )^{2} + 7.33^{2}} \\ W_{total} &= 26.34 \ N/m\\ \end{align}

Note: We will not include the “k” factor of NESC.

d. Final Conductor Tension, H₂

Initial Condition	Final Condition
t₁ = 15 °C	t₂ = -20 °C
H₁ = 20, 450 N	H₂ = ?
A = 0.0004029 m²	A = 0.0004029 m²
S = 300 m	S = 300 m
W₁ = 10.89 N/m	W₂ = 26.34 N/m
E = 58.9 x 10⁹ Pa	E = 58.9 x 10⁹ Pa
α = 23 x 10^-6/°C	α = 23 x 10^-6/°C

From the conductor state change equation:

\begin{align} \small H_2^{3} +H_2^{2} \left( \frac{(W_{1}S)^{2}AE}{24H_{1}^{2}} – H_{1} + (t_{2}-t_{1}) \alpha AE\right) – \frac{(W_{2}S)^{2}AE}{24} = 0 \end{align}

\begin{align} \small H_2^{3} +H_2^{2}\left(\frac{(10.89*300)^{2}0.0004029*58.9*10^{9}}{24*20,450^{2}}20,450+(-20-15)*23*10^{-6}*58.9*10^{9}\right)-\frac{(26.34*300)^{2}*0.0004029*58.9*10^{9}}{24}=0\\ \end{align}

Simplifying to a cubic equation:

\begin{align} H_{2}^{3} -14317.74\ H_{2}^{2} -6.176\ x\ 10^{13}=0 \end{align}

By trial and error method or goalseek in excel or modern pocket calculator:

H₂ = 44,921.94 Newtons (55% of RTS)

e. Calculate final sag and blowout angle

\begin{align} \theta&= \tan^{-1}{\left( \frac{W_{wind}}{W_{ice}+W} \right)} =\tan^{-1}\left( \frac{9.71}{13.6 +10.89} \right)=21.64^{\circ} \\ D_{slant} &= \frac{W_{2}S_{2}^{2}}{8H_{2}} = \frac{26.34*300^{2}}{8*44,922}= 6.60\:meters \\ D_{vertical} &=D\cos{\theta } = 6.60 \cos{21.63^{\circ} } = 6.14\:meters \\ D_{horizontal} &=D\sin{\theta } = 6.60 \sin{21.63^{\circ} } = 2.43\:meters \\ \end{align}

DOWNLOAD THE DETAILED COMPUTATION FROM EXCEL HERE.

3. Maximum Conductor Temperature: t_max= 90° (no wind/no ice)

a. Calculate Final Horizontal Tension

From the conductor state change equation:

\begin{align} \small H_2^{3} +H_2^{2} \left( \frac{(W_{1}S)^{2}AE}{24H_{1}^{2}} – H_{1} + (t_{2}-t_{1}) \alpha AE\right) – \frac{(W_{2}S)^{2}AE}{24} = 0 \end{align}

By substitution of values and simplification:

\begin{align} H_{2}^{3} +45721.21\;H_{2}^{2} -1.055\:x\:10^{13}=0 \end{align}

Using excel or pocket calculator or manual trial and error:

H₂ = 13,364.66 Newtons (16% of RTS)

b. Calculate final sag (note that there is no change in conductor weight)

\begin{align} D = \frac{W_{2}S^{2}}{8H_{2}}=\frac{10.89*300^{2}}{8*13,364.66}=9.18\: \: meters \end{align}