MECHANICAL DESIGN of TRANSMISSION LINE

MECHANICAL DESIGN of TRANSMISSION LINE

Introduction The mechanical design of this transmission line includes pole structure, wind load pressure, sag and the tensions on the conductor and also the guy wires. Some details on the line insulators and the line accessories are further discussed in this part.

Determination of Conductor Sag and Tension in Overhead Lines: While erecting an overhead line, it is very important that conductors are safe tension. If the conductors are too much stretched between supports in a bid to save conductor material, the stress in the conductor may reach unsafe value and in certain cases the conductor may break due to excessive tension. In order to permit safe tension in the conductors, they are not fully stretched but are allowed to have a dip or sag. The difference in level between points of supports and the lowest point on the conductor is called sag.

Conductor Sag and Tension: This is an important consideration in the mechanical design of overhead lines. The conductor sag should be kept to a minimum in order to reduce the conductor material required and to avoid extra pole height for sufficient clearance above ground level. It is also desirable that tension in the conductor should be low to avoid mechanical failure of conductor and to permit the use of less strong supports. However, low conductor tension and minimum sag are not possible. It is because low sag means a tight wire and high tension, whereas a low means a loose wire and increased sag. Therefore, in actual pratise, a compromise in made between the two.

Loadings:

The strength to be designed into a transmission line depends to a large extent on wind and ice loads that may be imposed on the conductor, overhead ground wire and supporting structure. When selecting appropriate design loads, the engineer should evaluate climatic conditions, previous line operation experience and the importance of the line to the system. Conservative load assumptions should be made for a transmission line which is the only tie to important load centers.

Wind load and wind pressure affecting the tension of the line:

Formula: Wind load = p x [D/12]

Where: p = wind pressure lb/ft p = 0.00256 (V)2 KzGRFCf

For Basic Wind Speed 200 km /hr considering the Transmission Line is at zone II *the value for basic wind speed came from the “New Wind Load Provisions in Philippine Structural Code” shown in the image below.

Where

Kz = velocity pressure exposure coefficient Kz = 2.01(h/900)(2/9.5)

Where h = height of the wire at the structure and is between 33 feet and 900 feet h = 30 + 0.3 (KV) = 30 +0.3(345kv) = 133.5 ft

Kz = 2.01 x

= 1.34

V = basic wind speed V = 124.27 miles/hr

GRF = gust response factor GRF= 0.70 (for 750 to 1000 span length in ft.)

Cf = the force coefficient = 1.0 for stranded wires

D = diameter of wire * The constants found here are all based on the Electrical Engineering Handbook tables for corresponding data.

Solution: P = 0.00256 (124.27)2(1.34)(0.70)(1.0) P = 37.08 lb/ft2

Wind load

Conductor Load

From Table, The conductor ORTOLAN w/ a cross sectional area of 1033 Kcmil, the nominal weight is 1165/1000 ft.

/ft

Effective Weight of Conductor Considering the Wind Load

√

√

Calculation of Sag and Tension In an overhead line, the sag should be adjusted that in the conductors is within safe limits. The tension is governed by conductor weight, effects of wind, ice loading and temperature variations. It is standard practice to keep conductor tension less than 50 % of its ultimate tensile strength i.e. minimum factor of safety in respect to conductor should be 2. We shall now calculate sag and tension of a conductor when (i) supports are at equal level (ii) supports are at unequal levels.

(i)

When supports are at equal level

Consider a conductor between two equilevel supports A and B with O as the lowest point is shown in the Fig. It can be proved that the lowest point will be at the mid span.

L = Length of span W = Weight per unit length of conductor T = tension in the conductor

Sag, S =

The sag is as result of the tensioning of the line and must not be too low otherwise the safety clearances may not be met. Also, the sag had to be such that it caters for ice loading in the winter of temperate climates. If the sag is large, and the line becomes heavily loaded, then the sag will further increase and branch the safety clearances. Similarity, if the sag is low, then

when the line contracts in the winter, low sag will indicate a high tension, and as a result of this contraction, the line may snap.

Required Clearances: 1. Clearance of conductors passing by buildings 2. Minimum clearances of conductors above ground or rails 3. Crossing clearances of wires carried of wires carried on different supports 4. Horizontal clearances at support between line conductors based on sag.

Working Tension

Given: Conductor Name = ORTOLAN Ultimate Strength = 27,700 lb Safety Factor = 2 L = 250 m  820.21 ft. span length

T=

T=

= 13850lb

Sag of Conductor a. Consider conductor load only

For bundling Sag = 7.07 ft x 2 Sag = 14.14 ft. or 4.3 m

b. Consider both conductor and wind load

This is the slant sag in a direction making an angle θ with the vertical, where value of θ is given by: θ = tan-1

= tan-1

= 72.74o

(ii)

When supports are unequal levels

In hilly areas, we generally come across conductors suspended between supports at unequal levels. Fig. shows a conductor suspended between supports A and B which at different levels. The lowest point of the conductor is O. Let l = Span Length h = Difference in levels between two supports x1 = Distance of supports at lower level x2 = Distance of supports at higher level T = Tension S1 = Sag using distance X1 S2 = Sag using distance X2

Calculations: Span 820.12 ft X1 + X2 = 820.12; X1 = 820.12 - X2 For h = 20 ft (to assume that there’s a distance difference level of the 2 towers)

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LINE SUPPORTS The supporting structure for overhead line conductors are various type of pole towers called line supports. In general, the line support should have the following properties: a. High mechanical strength to withstand the weight of conductor and wind loads etc. b. Light in weight without the loss of mechanical strength c. Cheap in cost and economical to maintain. d. Longer life. e. Easy accessibility of conductors for maintenance.

The line supports used for transmission and distribution of electric power are of various types including wooden poles, steel poles, RCC poles, and lattice steel towers. The choice od supporting structure for a particular case depends upon the line span, cross-sectional area, line voltage, cost and local conditions.

Steel Poles. The steel poles are often used as a substitute for wooden poles. They possess greater mechanical strength, longer life and permit longer spans to be used. Such poles are generally used for distribution purposes in the cities. This type of supports need galvanized or painted in order to prolong its life. The steel poles are three types., (i)rail poles (ii) tubular poles and (iii)rolled steel joints.

Spacing and Clearance

1. Vertical Clearance - NESCode 232 Vertical Clearance (Above Ground, Roadway, Rail or Water Suface)

Ground clearance required for a 345KV transmission voltage is 48 ft. 2. Spacing of Phase Conductors S = 27.3 ft

3. Space between OHGW and Top Phase Conductor Standard 1m for Neutral to top phase conductor = 3.28ft

Pole Structure:

Pole Height: H = 30 + 0.30(Kv) H = 30 + 0.30(345) H = 133.5 ft. At Tension = T = 13,850 lbs

Where: H1 = height starting from the ground to the 1st phase conductor H2 = height starting from the ground to the 2nd phase conductor H3 = height starting from the ground to the 3rd phase conductor H4 = height starting from the ground to the OHGW

H1 = 133.5 – 27.3 – 27.3 – 3.28 = 75.62 ft H2 = 75.62 ft + 27.3 = 102.92 ft H3 = 75.62 ft + 54.6 = 130.22 ft H4 = 75.62 ft + 57.68 = 133.5 ft

Mass: Where: M = T1H1 + T2H2 + T3H3 + [(10% x T)(H4)] M = [(13850) (75.62+ 102.92 + 130.22)] + (0.1) (13,850)(133.5) M = 4,465,378 lb-ft = 53,584,542 lb-in

Circumference (c) bottom: Where s = ultimate fiber stress using 46,350.36 lb/ in2 √

√

Butt Diameter:

Butt Diameter = 54.54 in x

x

= 1.38 m

Top Diameter: Ratio of the top to bottom is 0.60 Top Diameter = 0.6 x 54.54 in. Top Diameter= 32.72 in Top Diameter = 32.72 in x

x

= 0.83 m

Pole Circumference, Top: C = top diameter x C = 32.72 x C = 102.8 in. Taper

Projected Area: ( (

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POLE SETTING DEPTHS It is indicated that the 10% of the total length of the pole above the ground is depth of the portion of the pole to be on the ground.

The total length of pole above the ground is 133.5 ft. so the 10% of the 133.5 ft pole is 13.35 ft pole.

The additional 13.35 ft will be the depth of the pole to be under the ground. These giving the total length of the pole to:

H = 133.5 + 13.35 H = 146.85 ft. CORNER POLE: The Angle 30˚ & 40˚ of two cables Since T1 = T2 = 13,850 Kg is already computed ∑

Vertical Force: Fv = T1 Sin 30 – T2 Sin 40 Fv=13,850 Sin 30 – 13,850 Sin 40 Fv= -1977.60

Horizontal Force: Fh = T1 Cos 30 – T2 Cos 40 Fh = 13,850 Cos 30 - 13,850 Cos 40 Fh = 1384.74

Resultant Force: √ √(

) 0

Angle:

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GUY WIRES The various grades of guy strand are almost universally furnished in accordance with ASTM specifications. The ultimate strength for each size and grade is given. The so-called double galvanized is commonly used. In transmission construction a factor of safety of 2 is general for guys, although this may be somewhat reduced. GUY REQUIREMENT: IEEE C2-1997 (EE Handbook pp(18-66))

Distance dig from the pole not less than ¼ or more than 1 to 1 ½ of the height of the guy attachment.

GUY WIRE TENSIONS D = 0.551 x H D = 0.551 (146.85) D= 80.91 ft

Length of guy

√ √

√ √

√ √

√ √

L1  T 110.88  1385   1898.02lb D 80.91 L  T 130.99  13850 T2  2   22,422.58lb D 80.91 L  T 153.30  13850 T3  31   26,241.56lb D 80.91 L  T 156.10  13850 T4  4   26,720.86lb D 80.91 T1 

DESIGN DATA FOR GUYS

According to the guy requirements, for dead ends, the allowable stressed must be less than 66.67% of the ultimate strength of the guy used.

Guy at Phase Conductors Siemens-Martin grade Guy Plow Steel Diameter of 3/4 inch 26,200 lbs Nominal Breaking Strength

Guy at Static Wire Siemens-Martin grade Guy Mild Plow Steel Diameter of 3/4 35,900 lbs Nominal Breaking Strength