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 mm^{2} 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 | mm^{2} |
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)
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.
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)
t_{1} = 15°C
H_{1} = 25% RTS = 61,800 * 0.25 = 15,450 N
A = 0.0004029 m^{2}
S = 474 m
W_{1} = 10.89 N/m
E = 58.9 x 10^{9} Pa
h = 32 m
Solving the conductor sag D (see figure above):
Solving the horizontal distance:
Total conductor length:
Initial Sag:
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/m^{3})
b. Calculate wind load
c. Total unit weight
d. Final Conductor tension, H_{2}
Initial Condition | Final Condition |
t_{1} = 15°C α =23 x 10^{-6} /°C |
t_{2} = -20°C H2 = _______N ? A = 0.0004029 m^{2} S = 474 m W2 = 26.34 N/m E = 58.9 x 10^{9} Pa |
From the conductor state change equation:
Simplification to cubic equation:
By trial and error method or goalseek in excel or modern pocket calculator:
H_{2} = 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:
Solving the horizontal distance:
Final conductor Sag:
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:
By substitution of values and simplication:
Using excel or pocket calculator or manual trial and error:
H_{2} = 13292 Newtons (23% of RTS)
b. Calculate final sag (note that there is no change in conductor weight)
Solving the horizontal distance:
Final conductor Sag:
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.