--- title: "Design: 125 kVA 6600/400 V three phase transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:design-125kva-three-phase-windings" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-125kva-three-phase-windings site: BCT Engineering --- # Design: 125 kVA 6600/400 V three phase transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2075 Chaitra, 16 marks - Video: [Three Phase Transformer complete design core, window, winding, space factor, specific loading, turns](https://www.youtube.com/watch?v=O7FdTj5wCV0) — ShreeDevi Classes - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-125kva-three-phase-windings - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Design a 125 kVA, 50 Hz, 6600/400 V, 3-phase, Δ/Y, core type oil immersed natural cooled distribution transformer. (Assume suitable data if necessary) Maximum flux density in core = 1.35 Wb/m², core type = cruciform; Current density = 2.75 A/mm²; Window space factor = 0.4; Hw/Ww = 2.5; Take hot rolled steel sheet and area of yoke = 1.2 × area of core; Axial depth of L.V. = 268 mm; Axial depth of H.V. = 276 mm; Radial depth of L.V. = 14 mm; Radial depth of H.V. = 18 mm; Width of insulation between L.V. and H.V. = 11 mm; Outside diameter of L.V. = 293 mm, Inside diameter of L.V. = 255 mm; Inside diameter of H.V. = 314 mm, Outside diameter of H.V. = 351 mm. Calculate: i) Dimension of core, window and yoke ii) Overall dimension of frame iii) Leakage reactance of transformer iv) Draw overall dimension of frame ## Short answer Core diameter 173.1 mm, window width 187.9 mm, height 469.7 mm, yoke depth 147.2 mm, height 152.1 mm. Overall frame 773.9 × 869.2 × 147.2 mm. Leakage reactance computed from winding dimensions and turns (value not completed in excerpt). ## Answer Standard method (A.K. Sawhney, *A Course in Electrical Machine Design*): $E_t = K\sqrt{Q}$, net core area ratio for the core type, output equation for the window, yoke depth = width of largest stamping. $K$ = 0.45 assumed (3-phase distribution transformer), $\rho = 0.021\ \Omega$-mm²/m. Winding data are given: LV 255/293 mm (inside/outside diameter), HV 314/351 mm, radial depths $b_1$ = 14 mm, $b_2$ = 18 mm, gap $a$ = 11 mm. Axial length taken as the mean of 268 and 276 mm, $L_c$ = 272 mm. #### i) Dimensions of core, window and yoke Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA, total for 3 phases): $$ \begin{aligned} E_t &= 0.45\sqrt{125} = 5.031\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{5.031}{4.44\times50} = 0.02266\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.02266}{1.35} = 0.01679\ \text{m}^2 = 167.9\ \text{cm}^2 \end{aligned} $$ For a cruciform (two-stepped) core with stacking factor 0.9, $A_i = 0.56d^2$ (standard ratio, Sawhney): $$ \begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01679}{0.56}} = 0.1731\ \text{m} = 173.1\ \text{mm}\\ a &= 0.85d = 147.2\ \text{mm}\\ b &= 0.53d = 91.8\ \text{mm} \end{aligned} $$ Gross core area $A_{gi} = A_i/0.9 = 0.01865$ m². Output equation (three phase): $Q = 3.33fB_m\delta K_wA_wA_i\times10^{-3}$ $$ \begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{125}{3.33\times50\times1.35\times2.75\times10^6\times0.4\times0.01679\times10^{-3}}\\ &= 0.03012\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 109.8\ \text{mm},\quad H_w = 2.5W_w = 274.4\ \text{mm}\\ D &= d + W_w = 282.9\ \text{mm (distance between adjacent limb centres)} \end{aligned} $$ Check with the given windings: $d$ = 173 mm fits inside the LV (255 mm), but the HV outside diameter 351 mm > $D$ = 283 mm and the coil height 276 mm > $H_w$ = 274 mm. So the window is enlarged to suit the given coils (10 mm between adjacent HV coils), and its height is increased to keep the given ratio $H_w/W_w = 2.5$: $$ \begin{aligned} D &= 351 + 10 = 361\ \text{mm},\quad W_w = D - d = 187.9\ \text{mm}\\ H_w &= 2.5W_w = 469.7\ \text{mm} \end{aligned} $$ Yoke: Yoke area 20% more than core area (hot-rolled steel, to reduce yoke loss and magnetising current). Yoke depth = width of largest stamping: $$ \begin{aligned} A_y &= 1.2A_i = 0.02014\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02238\ \text{m}^2\\ D_y &= a = 147.2\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.02238}{0.1472} = 152.1\ \text{mm} \end{aligned} $$ #### ii) Overall dimensions of frame $$ \begin{aligned} H &= H_w + 2H_y = 469.7 + 2(152.1) = 773.9\ \text{mm}\\ W &= 2D + a = 2(361.0) + 147.2 = 869.2\ \text{mm}\\ \text{Depth} &= D_y = a = 147.2\ \text{mm} \end{aligned} $$ #### iii) Leakage reactance Phase voltages: LV (star) $V_{LV} = 230.9$ V; HV (delta) $V_{HV} = 6600.0$ V. Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 180.42$ A; $I_{HV} = 6.313$ A. $$ \begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{230.9}{5.031} = 45.90 \approx 46\\ E_t\ (\text{revised}) &= \frac{230.9}{46} = 5.020\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1314.6 \approx 1315\\ a_{LV} &= \frac{I_{LV}}{\delta} = 65.61\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 2.296\ \text{mm}^2 \end{aligned} $$ Radial depths: $b_1 = 14$ mm (LV), $b_2 = 18$ mm (HV), duct $a = 11$ mm. Mean lengths of turn: $$ \begin{aligned} L_{mt,LV} &= \pi\frac{255.0 + 293.0}{2} = 0.8608\ \text{m}\\ L_{mt,HV} &= \pi\frac{314.0 + 351.0}{2} = 1.0446\ \text{m}\\ L_{mt} &= \pi\frac{255.0 + 351.0}{2} = 0.9519\ \text{m (both windings, for reactance)} \end{aligned} $$ Leakage reactance (per unit), with $AT$ per limb $= 8302$ and axial winding length $L_c = 272.0$ mm: $$ \begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{8302}{0.2720}\cdot\frac{0.9519}{5.020}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\ &= 0.0495\ \text{p.u.}\\ X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0495\times\frac{6600.0}{6.313} = 51.75\ \Omega\ \text{(referred to HV)}\\ X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 0.06332\ \Omega \end{aligned} $$ #### iv) Overall dimensions of frame (sketch) ```text +-----------------------------+ --- | yoke, Hy = 152 mm | ^ +-----+-----+-----+-----+-----+ | |limb | |limb | |limb | | | R | Ww | Y | Ww | B | H = 774 mm | |188 | | | | | +-----+-----+-----+-----+-----+ | | yoke | v +-----------------------------+ --- |<------------ W ------------>| W = 869 mm, D = 361 mm (limb centres) Depth of core and yoke = 147 mm Window height Hw = 470 mm ``` **Answer: $d$ = 173 mm; window 188 × 470 mm; yoke 152 × 147 mm; frame 774 × 869 × 147 mm; leakage reactance ≈ 51.7 Ω/phase referred to HV (4.95%).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the sequence used to find the core area for the 125 kVA design? A. E_t = 0.45 sqrt(125), phi_m = E_t/(4.44 f), A_i = phi_m/B_m B. E_t = 0.45 x 125, phi_m = E_t x 4.44 f, A_i = phi_m x B_m C. A_i = Q/(3.33 f B_m delta) D. A_i = 0.56 d^2 first, then E_t from A_i Answer: A. E_t = 5.031 V, phi_m = 0.02266 Wb and A_i = 0.02266/1.35 = 0.01679 m^2. (Key point: Core area from flux) 2. For the cruciform core, after d = 173.1 mm is found, what are the stepped dimensions a and b? A. a = 0.90d, b = 0.70d B. a = 0.85d, b = 0.53d C. a = 0.53d, b = 0.85d D. a = d, b = 0.5d Answer: B. A two-stepped cruciform core has widths a = 0.85d and b = 0.53d, giving 147.2 mm and 91.8 mm. (Key point: Cruciform core steps) 3. The output equation gives D = 283 mm and H_w = 274 mm. What is done and why? A. They are accepted because the HV coil fits in the window B. D is reduced to 173 mm so the core and window are compact C. The window is enlarged to D = 361 mm and H_w = 316 mm because the 351 mm HV coil and 276 mm coil height do not fit D. Only the yoke is enlarged, the window stays unchanged Answer: C. HV outside diameter 351 mm > D and coil height 276 mm > H_w, so clearances are added: D = 351 + 10 and H_w = 276 + 2(20). (Key point: Window enlargement) 4. With D = 361 mm, a = 147.2 mm, H_w = 316 mm and H_y = 152.1 mm, what is the frame height H and width W for the three-limb core? A. H = 620 mm, W = 869 mm B. H = 468 mm, W = 508 mm C. H = 620 mm, W = 508 mm D. H = 316 mm, W = 869 mm Answer: A. H = H_w + 2H_y = 316 + 304.2 = 620.2 mm and W = 2D + a = 722 + 147.2 = 869.2 mm. 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