--- title: "Distribution transformer efficiency and number of tubes" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:distribution-efficiency-cooling-tubes" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/distribution-efficiency-cooling-tubes site: BCT Engineering --- # Distribution transformer efficiency and number of tubes Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2078 Kartik, 3+3 marks; 2070 Chaitra, 4+4 marks (asked 2 times) - Video: [5EE4-05 Lecture 5 - Unit 2 - EMD - Design for minimum losses or maximum efficiency - Gauri Shankar](https://www.youtube.com/watch?v=6Rigbbw2ah8) — MITRC LECTURE - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/distribution-efficiency-cooling-tubes - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Why are distribution transformers designed to have maximum efficiency at loads much lower than full load? Derive the expression for calculating the number of tubes to be provided in a transformer tank. ## Short answer Distribution transformers are designed for maximum all-day efficiency because they are energized 24 h with varying load; iron loss is constant, copper loss varies with load square. Placing maximum efficiency at 50–70% load reduces total energy losses. Required cooling tubes: n_t = (1/(8.8πd_t l_t))[(P_i+P_c)/θ – 12.5S_t] where d_t,l_t are tube diameter and length, S_t plain wall area, θ permissible oil temperature rise. ## Answer #### Why maximum efficiency at less than full load A distribution transformer stays energised for 24 hours but its load changes through the day and is often light (night hours). So the important figure is **all-day (energy) efficiency**: $$ \eta_{all\ day} = \frac{\text{kWh output in 24 h}}{\text{kWh output} + \text{kWh losses in 24 h}} $$ - Iron loss $P_i$ occurs for all 24 hours, whatever the load. - Copper loss $P_c$ occurs only in proportion to (load)², and is large only for a few peak hours. Maximum efficiency occurs when copper loss = iron loss, i.e. at load fraction $x = \sqrt{P_i/P_{c,FL}}$. By keeping $P_i$ small (lower $B_m$, good-quality steel) the maximum efficiency is placed at about 50–70% of full load, where the transformer runs most of the time. This minimises total energy loss in a day. #### Number of cooling tubes Let - $P_i + P_c$ = total full-load loss (W) - $S_t$ = dissipating surface of plain tank walls (m²) - $\theta$ = permissible mean temperature rise of oil (°C) A plain tank wall dissipates by radiation ≈ 6 W/m²°C and by convection ≈ 6.5 W/m²°C, total 12.5 W/m²°C: $$ \theta = \frac{P_i + P_c}{12.5\,S_t} $$ If this $\theta$ is too high, tubes are added. Let the tube area be $xS_t$. Tubes do not add to radiation much (they screen each other), but they improve convection by about 35%: $$ 6.5 \times 1.35 \approx 8.8\ \text{W/m}^2{}^\circ\text{C} $$ Heat dissipated with tubes: $$ \begin{aligned} P_i + P_c &= \theta\left[12.5\,S_t + 8.8\,xS_t\right]\\ xS_t &= \frac{1}{8.8}\left(\frac{P_i + P_c}{\theta} - 12.5\,S_t\right) \end{aligned} $$ If each tube has diameter $d_t$ and length $l_t$, its surface is $\pi d_tl_t$, so $$ n_t = \frac{xS_t}{\pi d_tl_t} = \frac{1}{8.8\,\pi d_tl_t}\left(\frac{P_i + P_c}{\theta} - 12.5\,S_t\right) $$ Round $n_t$ up to the next whole number. Usually $\theta$ = 35 °C (mean oil rise), tube diameter 50 mm and spacing 75 mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is maximum efficiency of a distribution transformer placed at below full load? A. Copper loss is zero at night B. Full-load efficiency cannot be calculated C. Iron loss exists all 24 h while load is mostly light, so iron loss is kept small D. Regulation is better at low load Answer: C. All-day efficiency is improved by low iron loss, which is present all day. (Key point: Why low-load maximum efficiency) 2. Maximum efficiency occurs at load fraction x equal to: A. Pc,FL / Pi B. sqrt(Pi / Pc,FL) C. Pi x Pc,FL D. (Pi / Pc,FL)^2 Answer: B. Copper loss x^2 Pc,FL equals iron loss Pi when x = sqrt(Pi/Pc,FL). (Key point: Condition for maximum efficiency) 3. How do tubes mainly improve heat dissipation in the derivation? A. They raise radiation by 35% B. They improve convection by about 35% (6.5 x 1.35 = 8.8 W/m2 C) C. They remove the tank walls D. They reduce iron loss Answer: B. Tubes screen each other so radiation gain is negligible, but convection improves about 35%. (Key point: Tube effect on dissipation) 4. The number of tubes is: A. nt = (Pi+Pc)/(12.5 St theta) B. nt = 8.8 pi dt lt/(Pi+Pc) C. nt = theta St/(Pi+Pc) D. nt = (1/8.8 pi dt lt)((Pi+Pc)/theta - 12.5 St) Answer: D. Tube area from the heat balance divided by the area of one tube, pi dt lt. 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