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Air clearances, 220 kV line for 200 MW

Short answer

Cross arm length 4.20 m, insulator string 2.34 m (14 discs), vertical separation 4.88 m, horizontal separation 9.89 m, height of earth wire above top conductor 8.57 m. Based on 220 kV and a minimum air clearance of 2.54 m.

For a transmission line to deliver 200 MW of power over 120 km, it has been decided to design a 220 kV transmission line of single circuit. Find the following air clearances with justifications. i) Cross arm length ii) Insulator string length iii) Horizontal and vertical separation of conductor iv) Height of earth wire from the top most conductor.

The air clearances fix the tower head dimensions. They depend on the voltage, not on the power or the length; those decide the conductor. The 200 MW, 120 km data are used later for conductor selection.

Data and assumptions: 220 kV nominal, maximum system voltage 245 kV (phase rms 141.45 kV); 254 × 146 mm discs; maximum string swing 45°; minimum air clearance 6.5 inch per 10 kV (phase rms) plus 8 inch safety; tower body width 1.5 m; single circuit in a triangular (offset vertical) layout, with two phases on one side and one on the other.

Minimum air clearance

a=6.5×141.4510+8=99.94 in=2.54 ma = 6.5\times\frac{141.45}{10} + 8 = 99.94\ \text{in} = 2.54\ \text{m}

ii) Insulator string length

  • 14 discs, the standard 220 kV suspension string. The table check (1-min wet withstand 395 kV and impulse 900 kV) needs 10 discs; extra discs are added for switching surges, pollution and one defective disc.
  • Length:
l=14×0.146+0.3=2.34 ml = 14\times0.146 + 0.3 = 2.34\ \text{m}

i) Cross arm length

With the string swung 45°, the conductor must still be at least aa from the tower body:

X=a+lsin⁡45∘=2.54+2.344×0.707=4.20 mX = a + l\sin45^\circ = 2.54 + 2.344\times0.707 = 4.20\ \text{m}

iii) Separation of conductors

  • Vertical: the conductor below one cross arm must clear the next cross arm by aa:
Y=l+a=2.344+2.539=4.88 mY = l + a = 2.344 + 2.539 = 4.88\ \text{m}
  • Horizontal (between the phases on opposite sides of the tower):
Sh=2X+w=2(4.196)+1.5=9.89 mS_h = 2X + w = 2(4.196) + 1.5 = 9.89\ \text{m}
  • The resulting phase distances are 9.892+4.882=11.03\sqrt{9.89^2 + 4.88^2} = 11.03 m between adjacent phases and 2Y=9.772Y = 9.77 m between the top and bottom phases. The GMD is 10.59 m.

iv) Height of earth wire above the top conductor

Use one earth wire at the peak with a shielding angle of at most 30°:

he=X+w/2tan⁡30∘=4.9460.5774=8.57 mh_e = \frac{X + w/2}{\tan30^\circ} = \frac{4.946}{0.5774} = 8.57\ \text{m}

This is about 6.2 m above the top cross arm.

            E   (1 earth wire)
            |\  30 deg
            | \
       o----+--*  top phase (A)
  (B)  *----+     Y = 4.88 m
            +--*  bottom phase (C)
            |  X = 4.2 m
   <------ 9.89 m ------>
ItemValue
Cross arm length4.20 m
Insulator string length2.34 m (14 discs)
Vertical separation4.88 m
Horizontal separation9.89 m
Earth wire above top conductor8.57 m

✓ Answer checked by Claude Opus 5.5 (high effort)

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