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

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:

No load conditions (no wind/ice load)
Heavy loading district: Ice thickness = 12.5 mm, Wind load = 190 Pa, t = -20 °C
Maximum Conductor Temperature: t_max = 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)

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

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

$\tiny L =S + (X_{A}^{3} + X_{B}^{3} )\left ( \frac{w^{2}}{6H^{2}} \right )$

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.

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

Where:

S – horizontal distance between conductor supports

h – vertical distance between conductor supports

S_L – 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 S_L

L – total initial conductor length

D_A / D_B – sag measured between lowest point and respective conductor supports

X_A / X_B – horizontal distance of the conductor supports from the lowest point.

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

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):

$\tiny S_{L} = \sqrt{h^{2} + S^{2}} = \sqrt{32^{2} + 474^{2}} = 475$

$\tiny \dpi{200} \fn_phv \tiny D = \frac{H_{1}}{W_{1}}\left ( cosh\left ( \frac{S_{L}}{2 H_{1}/W_{1}} \right ) - 1\right ) = \frac{W_{1}S_{L}^{2}}{8H_{1}}=\frac{10.89*475^{2}}{8*20,450}=15\: \: meters$

Solving the horizontal distance:

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

$\tiny X_{B} = S - X_{A} = 474 - 332.38 = 141.62$

Total conductor length:

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

Initial Sag:

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

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

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)

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

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

b. Calculate wind load

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

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

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

c. Total unit weight

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

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

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:

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

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

Simplification to cubic equation:

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

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:

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

Solving the horizontal distance:

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

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

Final conductor Sag:

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

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

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

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

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

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.

DOWNLOAD THE DETAILED COMPUTATIONS 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:

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

By substitution of values and simplication:

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

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)

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

Solving the horizontal distance:

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

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

Final conductor Sag:

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

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

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.

DOWNLOAD THE DETAILED COMPUTATIONS FROM EXCEL HERE.

THEORY:

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

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)

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