--- title: "Kelvin's law and its limitations" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:kelvins-law-limitations" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/kelvins-law-limitations site: BCT Engineering --- # Kelvin's law and its limitations Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2078 Chaitra, 8 marks; 2071 Bhadra, 6 marks (asked 2 times) - Video: [PS08 Kelvins Law to Calculate Economic Size of Conductor](https://www.youtube.com/watch?v=03hjI2C6zGs) — Lectures in Electrical Engineering - 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/kelvins-law-limitations - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question State and prove Kelvin's law of most economical conductor size and explain its limitation. ## Short answer The most economical conductor size is when annual cost of energy loss equals annual interest and depreciation on conductor cost. Its limitations include uncertainty in costs and load, neglect of technical limits like voltage regulation and thermal capacity, and assumptions of linear continuous cost. ## Answer **Kelvin's law:** the most economical conductor cross-section is the one for which the **annual cost of energy lost** in the conductor equals the **annual interest and depreciation on the part of the capital cost that is proportional to the cross-section**. #### Proof Let $a$ be the conductor cross-sectional area. 1. Capital cost of the line = $P_1 + P_2 a$. - $P_1$ is the fixed part: towers, insulators, erection. - $P_2 a$ is the part proportional to area: conductor material and the extra tower strength. 2. With a rate $r$ of annual interest plus depreciation, the annual capital charge is $$ C_1 = r(P_1 + P_2 a) = P_1' + P_2' a $$ 3. Energy loss: resistance $R = \rho l/a$. With an equivalent loss current $I$ over a year of $t$ hours, the energy lost is $3I^2\rho l t / a$. At a cost of $k$ per kWh: $$ C_2 = \frac{P_3}{a} $$ 4. Total annual cost: $$ C = P_1' + P_2' a + \frac{P_3}{a} $$ 5. For minimum cost: $$ \begin{aligned} \frac{dC}{da} &= P_2' - \frac{P_3}{a^2} = 0 \\ \Rightarrow P_2' a &= \frac{P_3}{a} \end{aligned} $$ Also $d^2C/da^2 = 2P_3/a^3 > 0$, so this is a minimum. So at the most economical area, the variable annual capital charge equals the annual cost of energy loss, which proves the law. The economical area is $$ a = \sqrt{\frac{P_3}{P_2'}} $$ and the economical current density is $J = I/a$. ```text cost |\ / total | \ energy loss / | \ (P3/a) ___/__ | \ _-- / | `-.__ / capital (P2'a) | ``--/--.____ +-------------+---------> area a a_economical ``` #### Limitations 1. Interest and depreciation are hard to estimate exactly. Costs, interest rates and energy prices change over the life of the line. 2. The load varies, so the loss (equivalent) current is uncertain. Load growth is not considered. 3. It considers economy only. It ignores the technical limits of **voltage regulation, corona, thermal ampacity, stability and mechanical strength**. The economic size may be too small to meet these limits. 4. The conductor cost is not exactly proportional to area. Standard sizes are discrete, and the installation cost is not linear. 5. It does not include the cost of the extra generation capacity needed to supply the losses. 6. The capital cost of towers does not depend linearly on conductor area, especially when wind and ice loading govern the design. For these reasons Kelvin's law is used only as a first guide. The final conductor is chosen by technical checks plus a present-worth comparison of standard conductors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Kelvin's law states that the most economical area is when: A. Annual cost of energy loss equals annual interest and depreciation on the capital proportional to area B. Total capital cost is minimum C. Energy loss is zero D. Annual cost of energy loss equals the fixed capital charge Answer: A. Only the area-proportional part of the capital charge is balanced against the loss cost. (Key point: Statement of law) 2. In the proof, the annual energy loss cost varies with area a as: A. P3 a B. P3 / a C. P3 / a^2 D. P3 sqrt(a) Answer: B. Resistance is proportional to 1/a, so loss cost is P3/a. (Key point: Cost model) 3. Differentiating C = P1' + P2'a + P3/a and setting dC/da = 0 gives: A. a = P3 P2' B. a = P3/P2' C. a = sqrt(P3/P2') D. a = (P3/P2')^2 Answer: C. P2' = P3/a^2 gives a = sqrt(P3/P2'). (Key point: Minimisation) 4. Which is a main limitation of Kelvin's law? A. It ignores energy loss B. It considers only economy and ignores voltage regulation, corona, ampacity, stability and mechanical limits C. It only applies to cables D. It assumes zero interest Answer: B. The economic size may fail technical constraints, so it is only a first guide. 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