# Sample Calculation of Sag and Tension of Transmission Line – Even Elevation Posted by

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.

## 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:

1. No Load Conditions (no wind / no ice)
2. Heavy Loading District: Ice Thickness = 12.5 mm, Wind Pressure = 190 Pa, t = -20 deg C
3. 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

• 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}

\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}

### 3. Maximum Conductor Temperature: tmax = 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}