Initial line design for 200 MW over 120 km

Given: $P = 200$ MW, $L = 120$ km. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$, factors FWR = 1.15, NACF = 1.1, FOS = 1.1 (usual design values), disc spacing 154 mm. Maximum system voltage of the 220 kV class = 255 kV (Table A-2).

a. Economical voltage, capability, standard voltage and circuits

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi\,N_c \times 150}}$$

$N_c = 1$: $V = 5.5\sqrt{75 + 1481.48} = 5.5 \times 39.452 = 216.99$ kV $N_c = 2$: $V = 5.5\sqrt{75 + 740.74} = 5.5 \times 28.561 = 157.09$ kV

m.f. at 120 km $= 2.75 - \frac{40}{80}(0.5) = 2.50$.

$V_{econ}$ for S/C (217 kV) almost equals 220 kV. Selected: 220 kV, single circuit (vertical/triangular formation, one earth wire).

b. Number of insulator discs

$k = 1.15 \times 1.1 \times 1.1 = 1.3915$

Switching: $V_{ss} = 2.8 \times \sqrt{2} \times 255/\sqrt{3} = 582.98$ kV peak; $\times k = 811.21$ kV peak $= 573.62$ kV rms. This exceeds the 14-disc wet FOV (555 kV), so 15 discs are needed.

Number of discs = 15 per suspension string (switching surge governs).

c. Air clearances

Minimum air clearance (6.5 in per 10 kV of maximum phase rms voltage + 8 in):

$$\begin{aligned} V_{ph} &= \frac{255}{\sqrt{3}} = 147.22\ \text{kV} \\ a &= 6.5 \times \frac{147.22}{10} + 8 = 103.70\ \text{in} = 2.63\ \text{m} \end{aligned}$$

ii. Insulator string length (15 discs × 154 mm, fittings neglected):

$$l = 15 \times 0.154 = 2.31\ \text{m}$$

i. Cross-arm length. With a 45° string swing, the conductor moves $l\sin45°$ toward the tower and must still be $a$ away from the tower body:

$$CA = l\sin45° + a = 2.31 \times 0.7071 + 2.634 = 1.633 + 2.634 = 4.27\ \text{m}$$

iii. Separation of conductors

• Vertical separation between cross-arms: the conductor hangs $l$ below its cross-arm and must be $a$ above the cross-arm below:

$$d_v = l + a = 2.31 + 2.634 = 4.94\ \text{m}$$

• Horizontal separation between conductors on opposite sides of the tower (tower body width neglected):

$$d_h = 2 \times CA = 2 \times 4.267 = 8.53\ \text{m}$$

iv. Height of earth wire above the top conductor. With a single earth wire on the tower axis and a shielding angle of 30°:

$$h = \frac{CA}{\tan30°} = 4.267 \times 1.732 = 7.39\ \text{m}$$

          o  earth wire
          |\
          | \ 30 deg
     7.39 m  \
          |   \
   -------+----o  top phase
   4.27 m |    | string 2.31 m
          |
   4.94 m |
   o------+------o  lower phases
     8.53 m apart
          |
       tower

(Also: minimum ground clearance $= (255-33)/33 + 17 = 23.73$ ft $= 7.23$ m.)

(If the IEC value $V_{max} = 245$ kV were used instead, the switching requirement would be 551.12 kV rms, 14 discs would just suffice, and the clearances would be slightly smaller: $a = 2.54$ m, $l = 2.16$ m, $CA = 4.06$ m.)

Summary: 220 kV S/C; 15 discs; string length 2.31 m; cross-arm 4.27 m; vertical spacing 4.94 m; horizontal spacing 8.53 m; earth wire 7.39 m above the top conductor.