# Sample Calculation of Sag and Tension in Transmission Line – Uneven Elevation

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In the previous post, we calculated the sag and tension of a transmission line given that the conductors supports are at the same elevation.In this example we will calculate the sag and tension if the conductor supports are at different elevation. Creep is not considered as a factor in final sag in this calculation. Also, loading of the conductors are based on the National Electrical Safety Code 2017.

Problem:

A transmission line conductor was strung between two towers (Structure 1 and 2), 474 meters apart and different elevation as shown in the figure above. During the time of installation, the conditions were t =15°C and initial horizontal tension = 25% of Rated Tensile Strength (RTS). Line 3 is 12 meters above the ground and the distance between phases is 3 meters.

Note: The elevation profile between the towers were derived from google earth. This is good for graphical presentation and initial planning.

Calculate the sag and tension at:

2. Heavy loading district: Ice thickness = 12.5 mm, Wind load = 190 Pa, t = -20 °C
3. Maximum Conductor Temperature: tmax = 90, no wind, no ice
 Conductor Name 403-A1-37 Conductor Common Name 403 mm2 37 AAC (Arbutus) Span Length 300 m Outside Diameter 26.1 mm Conductor Unit Weight 10.89 N/m Rated Tensile Strength 61.8 kN Modulus of Elasticity 58.9 GPa Coefficient of Thermal Expansion 23 x 10-6 /°C Total Conductor Area 402.9 mm2

#### THEORY:

The equations derived for the even support elevation can be used to analyze the uneven conductor support heights.

Below are the derived equations: (see the figure)

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;X_{A}&space;=&space;\frac{S}{2}&space;\left&space;[&space;1&space;+&space;\frac{h}{4D}&space;\right&space;]$                       $\dpi{200}&space;\fn_phv&space;\tiny&space;D_{A}&space;=\frac{wX_{A}^{2}}{2H}$                     $\dpi{200}&space;\fn_phv&space;\tiny&space;D_{A}&space;=&space;D&space;\left&space;(&space;1&space;+&space;\frac{h}{4D}&space;\right&space;)^{2}$                          $\dpi{200}&space;\fn_phv&space;\tiny&space;T_{A}&space;=&space;H&space;+&space;wD_{A}$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\fn_phv&space;X_{B}&space;=&space;\frac{S}{2}&space;\left&space;[&space;1&space;-&space;\frac{h}{4D}&space;\right&space;]$                    $\dpi{200}&space;\fn_phv&space;\tiny&space;D_{B}&space;=\frac{wX_{B}^{2}}{2H}$                      $\dpi{200}&space;\fn_phv&space;\tiny&space;D_{B}&space;=&space;D&space;\left&space;(&space;1&space;-&space;\frac{h}{4D}&space;\right&space;)^{2}$                       $\dpi{200}&space;\fn_phv&space;\tiny&space;T_{B}&space;=&space;H&space;+&space;wD_{B}$

$\dpi{200}&space;\fn_phv&space;\tiny&space;L&space;=S&space;+&space;(X_{A}^{3}&space;+&space;X_{B}^{3}&space;)\left&space;(&space;\frac{w^{2}}{6H^{2}}&space;\right&space;)$

The horizontal component of tension are the same at all point in the conductor. However, the vertical component of tension at the upper support is greater, hence the total tension at that point is also greater.

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;T_{upper}&space;=&space;T_{lower}&space;+&space;wh$

Where:

S – horizontal distance between conductor supports

h – vertical distance between conductor supports

SL – straight line distance between conductor supports

D – sag measured vertically through the points of conductor         supports to a line tangent to the conductor

– also called the midpoint sag

– approximately equal to a sag in even elevation with span of  SL

L – total initial conductor length

D / DB – sag measured between lowest point and respective conductor supports

XA / XB – horizontal distance of the conductor supports from the lowest point.

t1 = 15°C
H1 = 25% RTS = 61,800 * 0.25 = 15,450 N
A = 0.0004029 m2
S = 474 m
W1 = 10.89 N/m
E = 58.9 x 109 Pa

h = 32 m

Solving the conductor sag D (see figure above):

$\dpi{200}&space;\fn_phv&space;\tiny&space;S_{L}&space;=&space;\sqrt{h^{2}&space;+&space;S^{2}}&space;=&space;\sqrt{32^{2}&space;+&space;474^{2}}&space;=&space;475$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D&space;=&space;\frac{H_{1}}{W_{1}}\left&space;(&space;cosh\left&space;(&space;\frac{S_{L}}{2&space;H_{1}/W_{1}}&space;\right&space;)&space;-&space;1\right&space;)&space;=&space;\frac{W_{1}S_{L}^{2}}{8H_{1}}=\frac{10.89*475^{2}}{8*15,450}=19.879\:&space;\:&space;meters$

Solving the horizontal distance:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;X_{A}&space;=&space;\frac{S}{2}&space;\left&space;[&space;1&space;+&space;\frac{h}{4D}&space;\right&space;]&space;=\frac{474}{2}&space;\left&space;[&space;1&space;+&space;\frac{32}{4*19.879}&space;\right&space;]&space;=&space;332.38$

$\dpi{200}&space;\fn_phv&space;\tiny&space;X_{B}&space;=&space;S&space;-&space;X_{A}&space;=&space;474&space;-&space;332.38&space;=&space;141.62$

Total conductor length:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;L&space;=S&space;+&space;(X_{A}^{3}&space;+&space;X_{B}^{3}&space;)\left&space;(&space;\frac{w^{2}}{6H^{2}}&space;\right&space;)=&space;474&space;+&space;(332.38^{3}&space;+&space;141.62^{3}&space;)\left&space;(&space;\frac{10.89^{2}}{6*15,450^{2}}&space;\right&space;)=&space;477.28$

Initial Sag:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{A}&space;=\frac{wX_{A}^{2}}{2H}&space;=&space;\frac{10.89*332.38^{2}}{2*15,450}=38.9&space;\,&space;\:&space;meters$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{B}&space;=\frac{wX_{B}^{2}}{2H}&space;=&space;\frac{10.89*141.62^{2}}{2*15,450}=7.07&space;\,&space;\:&space;meters$

We can see the lowest point of the conductor is closer to the lower conductor support. Also, the difference between the horizontal span S and total conductor length is just 3.28 m but has great effect on the sag.

#### 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)

$\fn_jvn&space;\small&space;W_{ice}&space;=\rho&space;_{ice}&space;*\pi&space;t_{ice}(D&space;+t_{ice})&space;=&space;915*\pi&space;*0.0125&space;\;&space;(0.0261&space;+&space;0.0125)$

$\fn_jvn&space;\small&space;\fn_jvn&space;\small&space;W_{ice}&space;=1.386\;&space;kg/m&space;\;&space;or&space;\;13.6&space;\;N/m$

$\fn_jvn&space;\small&space;Total\;Diameter&space;=&space;D&space;+&space;2t_{ice}&space;=&space;0.0261&space;+2*0.0125&space;=&space;0.0511\:m$

$\fn_jvn&space;\small&space;\fn_jvn&space;\small&space;W_{wind}&space;=&space;Wind\;pressure&space;*&space;Conductor\;diameter&space;(including&space;\;ice)$

$\fn_jvn&space;\small&space;\fn_jvn&space;\small&space;W_{wind}&space;=&space;190&space;*&space;0.0511&space;=&space;9.71\;N/m$

c. Total unit weight

$\fn_jvn&space;\small&space;W_{total}&space;=&space;\sqrt{W_{wind}^{2}+(W_{1}+W_{ice})^{2}}&space;=\sqrt{9.71^{2}+(10.89+13.6)^{2}}$

$\fn_jvn&space;\small&space;W_{total}&space;=&space;26.34\;N/m$

d. Final Conductor tension, H2

 Initial Condition Final Condition t1 = 15°C H1 = = 15,450 N A = 0.0004029 m2 S = 474 m W1 = 10.89 N/m E = 58.9 x 109 Pa α =23 x 10-6 /°C t2 = -20°C H2 = _______N ? A = 0.0004029 m2 S = 474 m W2 = 26.34 N/m E = 58.9 x 109 Pa

From the conductor state change equation:

$\fn_jvn&space;\small&space;\fn_jvn&space;\small&space;\fn_jvn&space;\fn_phv&space;\fn_phv&space;H_2^{3}&space;+H_2^{2}&space;\left&space;(&space;\frac{(W_{1}S)^{2}AE}{24H_{1}^{2}}&space;-&space;H_{1}&space;+&space;(t_{2}-t_{1})&space;\alpha&space;AE\right&space;)&space;-&space;\frac{(W_{2}S)^{2}AE}{24}&space;=&space;0$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;H_2^{3}&space;+H_2^{2}&space;\left&space;(&space;\frac{(10.89*474)^{2}&space;40.29x10^{-5}*58.9&space;x10^{9}}{24*15,450^{2}}&space;-&space;15,450&space;+&space;(-20-15)*&space;23x10^{-6}*40.29x10^{-5}*58.9&space;x10^{9}\right&space;)&space;-&space;\frac{(26.34*474)^{2}40.29x10^{-5}*58.9&space;x10^{9}}{24}&space;=&space;0$

Simplification to cubic equation:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;H_{2}^{3}&space;+&space;75818.16\;H_{2}^{2}&space;-1.541\:x\:10^{14}=0$

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

H2 = 36,964 Newtons (63% of RTS)

The design engineer must make sure that the final horizontal tension should not exceed the required limit. If the tension is exceeded, the designer could change the type of conductor and the initial horizontal tension, etc. to meet the requirement.

e. Calculate final sag:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D&space;=&space;\frac{H_{2}}{W_{2}}\left&space;(&space;cosh\left&space;(&space;\frac{S_{L}}{2&space;H_{2}/W_{2}}&space;\right&space;)&space;-&space;1\right&space;)&space;=&space;\frac{W_{2}S_{L}^{2}}{8H_{2}}=\frac{26.34*475^{2}}{8*36964}=20.1\:&space;\:&space;meters$

Solving the horizontal distance:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;X_{A}&space;=&space;\frac{S}{2}&space;\left&space;[&space;1&space;+&space;\frac{h}{4D}&space;\right&space;]&space;=\frac{474}{2}&space;\left&space;[&space;1&space;+&space;\frac{32}{4*20.1}&space;\right&space;]&space;=&space;331.3&space;\:&space;meters$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;X_{B}&space;=&space;S&space;-&space;X_{A}&space;=&space;474&space;-&space;331.3&space;=&space;142.7\:&space;meters$

Final conductor Sag:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{A}&space;=\frac{w_{2}X_{A}^{2}}{2H_{2}}&space;=&space;\frac{26.34*331.3^{2}}{2*36964}=39.1&space;\,&space;\:&space;meters$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{B}&space;=\frac{w_{2}X_{B}^{2}}{2H_{2}}&space;=&space;\frac{26.34*142.7^{2}}{2*36964}=&space;7.3&space;\,&space;\:&space;meters$

$\fn_jvn&space;\small&space;\fn_jvn&space;\small&space;\theta&space;=&space;\tan^{-1}{\left&space;(&space;\frac{W_{wind}}{W_{ice}+W}&space;\right&space;)}&space;=\tan^{-1}\left&space;(&space;\frac{9.71}{13.6&space;+10.89}&space;\right&space;)=21.63\degree$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{A-vertical}&space;=D_{A}\cos{\theta&space;}&space;=&space;39.1&space;\cos{21.63}&space;=&space;36.3\:meters$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{B-vertical}&space;=D_{B}\cos{\theta&space;}&space;=&space;7.3&space;\cos{21.63}&space;=6.8\:meters$

Since the sag of the conductor at this loading was already identified, the designer must factor this in the selection of the height of the structures or poles to install. This is to make sure that the conductor meets the minimum line to ground clearance as per applicable standards.

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

a. Calculate Final Horizontal Tension

From the conductor state change equation:

$\fn_jvn&space;\small&space;\fn_phv&space;H_2^{3}&space;+H_2^{2}&space;\left&space;(&space;\frac{(W_{1}S)^{2}AE}{24H_{1}^{2}}&space;-&space;H_{1}&space;+&space;(t_{2}-t_{1})&space;\alpha&space;AE\right&space;)&space;-&space;\frac{(W_{1}S)^{2}AE}{24}&space;=&space;0$

By substitution of values and simplication:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;H_{2}^{3}&space;+135857.11\;H_{2}^{2}&space;-2.635\:x\:10^{13}=0$

Using excel or pocket calculator or manual trial and error:

H2 = 13292 Newtons (23% of RTS)

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

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D&space;=&space;\frac{H_{2}}{W_{2}}\left&space;(&space;cosh\left&space;(&space;\frac{S_{L}}{2&space;H_{2}/W_{2}}&space;\right&space;)&space;-&space;1\right&space;)&space;=&space;\frac{W_{2}S_{L}^{2}}{8H_{2}}=\frac{10.89*475^{2}}{8*13292}=23.1\:&space;\:&space;meters$

Solving the horizontal distance:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;X_{A}&space;=&space;\frac{S}{2}&space;\left&space;[&space;1&space;+&space;\frac{h}{4D}&space;\right&space;]&space;=\frac{474}{2}&space;\left&space;[&space;1&space;+&space;\frac{32}{4*23.1}&space;\right&space;]&space;=&space;319.1&space;\:&space;meters$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;X_{B}&space;=&space;S&space;-&space;X_{A}&space;=&space;474&space;-&space;319.1&space;=&space;155\:&space;meters$

Final conductor Sag:

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{A}&space;=\frac{w_{2}X_{A}^{2}}{2H_{2}}&space;=&space;\frac{10.89*319.1^{2}}{2*13292}=41.71&space;\,&space;\:&space;meters$

$\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;\dpi{200}&space;\fn_phv&space;\tiny&space;D_{B}&space;=\frac{w_{2}X_{B}^{2}}{2H_{2}}&space;=&space;\frac{10.89*155^{2}}{2*13292}=&space;9.84&space;\,&space;\:&space;meters$

Since there is no wind, these values are the final conductor sag at this loading. It can be observed that when the conductor elongates because if increase in temperature, the value of sag increases while the tension decreases.