Estimation Of Transmission Lines, Materials Required For 66 KV

Hello everyone, in this article I will discuss the estimation of transmission lines, estimation of materials required for 66 KV single circuit transmission lines, the code of practice related to transmission lines, and materials required for the transmission lines.

If you want an article on some other topics then comment us below in the comment section. You can also catch me @ Instagram – Chetan Shidling.

Also read – Electrical Estimation And Costing Questions And Answers. and now I will discuss estimation of transmission lines.

Estimation Of Transmission lines

  • 1. The minimum clearance of conductors from buildings where extra – high voltage line passes above or adjacent, as recommended by IE rule NO. 79 under the worst conditions. i.e., maximum temperature and wind pressure not less than 3.66 meters + 0.305 meters for every additional 33 KV or part thereof.
  • 2. For extra-high voltage over the headline the minimum vertical clearance between ground and conductors, as recommended by vied IE rule 77 shall not be less than 5.2 meters + 0.305 meters for every 33 KV or part thereof. The minimum clearance along/across any public street shall not be less than 6.1 meters.
  • 3. For extra-high voltage horizontal clearance between any part of the building and nearest conductors as recommended vide IE rule no. 79 shall not be less than 1.83 meters and 0.305 meters for an additional 33 KV or part thereof.
  • 4. A Guarding should be provided whenever the line crosses any road, railways line, building, telecommunication line or electrical distribution lines, etc.,
  • 5. Phase plates, Danger plates, and anti-climbing should be provided for each support.
  • 6. Continuous earth wire should run on the top of towers to protect the transmission line against lightning discharges.

Problem On Transmission Lines

Q: Prepare the schedule of materials required for running 100 KM single circuit of 66 KV transmission line across jungle terrain using a four-legged fabricated steel structure. Assume span length of 300 m.

Solution : 

Materials required :

(1) Towers :

Span = (100*1000) / 300

         = 333.33 or say 333

(a) No of supports = 333 + 1 = 334

(b) No of ‘C’ type towers for anchoring and deviation of 15 degrees to 30 degree

      = 334 / 4 = 83.5 or say 84.

(c) No of ‘B’ type towers for deviation of 2 degrees to 15 degree

      = 334 / 20 = 16.7 or say 17

(d) No of ‘A’ [tangent towers] tower

      = 334 – (84 + 17) = 233

Weight of towers :

  • (a) Weight of ‘A’ tower = 233 * 2.00 = 466 M Ton
  • (b) Weight of ‘B’ tower = 17 * 2.5 = 42.5 M Ton
  • (c) Weight of ‘C’ tower = 84 * 3.5 = 294 M. Ton

Total weight of tower required in M Tones

 = 466 + 42.5 + 294 = 802.5 M Ton.

(02) Extension pieces :

(a) No of 6-meter extension pieces

       = Length of line / 20 = 100 / 20 = 5

(b) No of 3 m extension pieces

       = length of line / 10 = 100 / 10 = 10

weight of extension pieces = 5 * 1.33 * 10 + 1

                                           = 16.65 M Ton.

(03) Length of transmission conductor of a coyote with 1.5 % extra for sag.

= 3 * 100 + [(300 * 1.5) / 100 ]

= 304.5 KM

(04) Conductor accessories :

(a) No of suspension clamps i.e., for ‘A’ type tower

      = (No of clamps / tower ) * total ‘A’ tower

      = 3 * 233 = 699

(b) No of tension clamps i.e., for ‘B’ and ‘C’ type towers

      = (No of clamps / tower) * No of B and C towers

      = 6 * 101 = 606

(c) No of mid-span compression joints = one per 10 KM length of conductor approximately

     = 304.5 KM / 10 KM = 30.45 or say 31

(d) No of armored rods

      = No of suspension clamps

      = 3 * 233 = 699

(e) No of vibration clampers

     = ‘A’ type towers * 1 + (length of the transmission line) * 2

     = (233 * 1) + (100*2)

     = 433

(f) No of repair sleeves = 304.5 / 5 = 60.9 or say 61

(05) Disc insulators :

(a) In suspension clamps

     = ( No of suspension clamps) * 4

     = 699 * 4

     = 2796

(b) In Tension Clamps

      = (No of Tension clamps ) * 5

      = 606 * 5

      = 3030

Total no of Disc insulators with 1 % for breakage

= ( 2796 + 3030) + (5826 * 1 / 100)

= 5884.26 or say 5885

(06) Ground wire (7/8 SWG wire) with 1.5% sag

= 100 * 1.015 = 101.5 KM

(07) Ground wire accessories :

(a) No of suspension clamps

= ( No of ‘A’ type tower ) * 1

= one each for ‘A’ type tower

= 233

(b) No of strain clamps

= ( No of B and C type towers ) * 2

= 101 * 2 = 202

(c) No of midspan compression joints = one per 10 KM

= 100 / 10 = 10

(d) No of repair sleeves = one per 5 KM

= 100 / 5 = 20

(08) Earthing material :

(a) No of pipe type earthing = 334 (No of towers)

(b) Earth strip = 334 * 5 = 1670 M (5 M for each tower)

(c) No of lugs with bolts and nuts

= 334 * 2(two for each tower)

= 668

(09) Cement concrete :

(a) For ‘A’ type towers = 25 * 233 = 5825 cmt

(b) For ‘B’ type towers = 36 * 17 = 612 cmt

(c) For ‘C’ type towers = 45 * 84 = 3780 cmt

Total = 5825 + 612 + 3780 = 10217 cmt.

(10) Anti-climbing devices :

(a) Weight of barbed wired = 334 * 3 Kg = 1002 Kg.

(b) Weight of Danger boards = 334 * 5 Kg = 1670 Kg.

(11) Boards :

(a) No of phase plates = 334 * 3 = 1002

(b) No of Danger boards = 334 * 1 = 334

(12) Miscellaneous materials such as nuts, bolts, screws, etc.

I hope this may help you…….

Tags: Estimation Of Materials Required For 66 KV Single Circuit Transmission Line, Estimation Of Transmission Lines.

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Chetan Shidling

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