--- title: "Ratio of vertical separation, 132 vs 220 kV" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:vertical-separation-ratio-132-220kv" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/vertical-separation-ratio-132-220kv site: BCT Engineering --- # Ratio of vertical separation, 132 vs 220 kV Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2072 Asoj, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/vertical-separation-ratio-132-220kv - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question Compute ratio of vertical separation of conductors of 132 kV and 220 kV double circuit transmission line. Derive any expressions used. ## Short answer Vertical separation Y = string length l + air clearance a. For 132 kV: l=1.614 m, a=1.585 m, Y=3.20 m. For 220 kV: l=2.344 m, a=2.539 m, Y=4.88 m. Ratio = 3.20/4.88 ≈ 0.66. ## Answer The vertical separation between conductors is fixed by the insulator string length $l$ plus the minimum air clearance $a$. #### Derivation of the expressions **1. Minimum air clearance.** As given with IOE papers, allow 6.5 inch per 10 kV of the maximum phase voltage (rms), plus 8 inch as a factor of safety: $$ a = \left(6.5\times\frac{V_{max}/\sqrt3}{10} + 8\right)\ \text{inch} $$ **2. String length.** $l = n\times 146$ mm plus about 0.3 m of hardware, where $n$ is the number of 254 × 146 mm discs. **3. Vertical spacing.** Take a double-circuit tower with conductors arranged vertically. The upper conductor hangs a distance $l$ below its cross arm, and it lies directly above the tip of the next cross arm, which is $Y$ below. ```text ======o upper cross arm | l (string) * upper conductor : >= a (air clearance) ======o lower cross arm | l * lower conductor ``` The gap between the upper conductor and the lower cross arm is $Y - l$, and it must be at least $a$: $$ Y - l \ge a \;\Rightarrow\; Y = l + a $$ When the string swings by θ, the conductor rises: its vertical distance from the cross arm above becomes $l\cos\theta$. That only increases the gap, so the unswung position governs. #### 132 kV double circuit - Maximum system voltage 145 kV, so the phase voltage is $145/\sqrt3 = 83.72$ kV. - $a = 6.5\times8.372+8 = 62.42$ inch $= 1.585$ m. - Discs: 9 discs, the standard practice for 132 kV. This covers the table check (1-min wet withstand 230 kV needs 6 discs; impulse 550 kV needs 7 discs) plus a margin for switching surges, pollution and one defective disc. - $l = 9\times0.146+0.3 = 1.614$ m. - $Y_{132} = 1.614+1.585 = 3.199$ m. #### 220 kV double circuit - Maximum system voltage 245 kV, so the phase voltage is 141.45 kV. - $a = 6.5\times14.145+8 = 99.94$ inch $= 2.539$ m. - Discs: 14, the standard 220 kV suspension string. - $l = 14\times0.146+0.3 = 2.344$ m. - $Y_{220} = 2.344+2.539 = 4.883$ m. #### Ratio $$ \frac{Y_{132}}{Y_{220}} = \frac{3.199}{4.883} = 0.655 $$ **Answer: the vertical separation is about 3.2 m for 132 kV and 4.9 m for 220 kV, a ratio of about 0.66 : 1. The 132 kV spacing is roughly two-thirds of the 220 kV spacing, because both the clearance and the string length grow with voltage.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What fixes the vertical separation Y between conductors in this answer? A. Insulator string length plus minimum air clearance (Y = l + a) B. Twice the insulator string length (Y = 2l) C. Minimum air clearance only (Y = a) D. String length times cos of the swing angle Answer: A. The conductor hangs l below its arm and must stay a from the arm below, so Y = l + a. (Key point: Vertical separation formula) 2. How is the minimum air clearance a obtained for 132 kV? A. 6.5 inch per 10 kV of line-to-line voltage, with no safety addition B. 8 inch per 10 kV of phase voltage, plus 6.5 inch safety C. 6.5 inch per 10 kV of maximum phase rms voltage, plus 8 inch safety D. 6.5 inch per 10 kV of nominal voltage, plus 8 m safety Answer: C. a = 6.5 x (Vmax/sqrt3)/10 + 8 inch, using the phase rms voltage. (Key point: Air clearance formula) 3. What is the ratio Y132 : Y220 and why is it below 1? A. About 1.5, because 132 kV needs a longer string B. About 0.66, because only the air clearance changes with voltage C. About 0.5, because string length is proportional to voltage squared D. About 0.66, because both clearance and string length grow with voltage Answer: D. Y132 = 3.199 m and Y220 = 4.883 m, ratio 0.655; both a and l increase with voltage. 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