--- title: "Design: 500 kVA 66/11 kV power transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:design-500kva-66-11kv" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-500kva-66-11kv site: BCT Engineering --- # Design: 500 kVA 66/11 kV power transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2078 Kartik, 18 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-500kva-66-11kv - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question For a 500 kVA, 3-phase, 50 Hz, 66/11 kV, delta/star, core type, oil immersed natural cooled power transformer the design data are: Maximum flux density in the core = 1.6 Wb/m²; Constant for output voltage per turn = 0.6; Core type = cruciform; Current density in the conductor = 2.5 A/mm²; Window space factor = 0.22; Stacking factor = 0.9; Ratio of window height to width = 2.75. Width of duct between LV and core, LV winding, HV winding and duct between HV and LV are 10 mm, 50 mm, 60 mm, 20 mm respectively. Assuming all other required parameters, calculate: (i) Overall dimension of core (ii) Overall dimension of frame (iii) Per unit resistance and leakage reactance drop (iv) Per unit voltage regulation ## Short answer Cruciform core d=260 mm, stampings 221×138 mm; window 853 mm high × 310 mm wide; yoke depth 221 mm, height 228 mm; overall frame 1309×1360×221 mm. Per unit resistance drop 0.010, leakage reactance drop 0.065, voltage regulation at 0.8 pf lagging 0.0472 (4.72%). ## 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. Assumed: yoke area 1.2 × core area (hot-rolled steel), end clearance 50 mm (66 kV), $\rho = 0.021\ \Omega$-mm²/m, regulation at 0.8 pf lagging (pf not given). #### (i) Overall dimensions of core Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA, total for 3 phases): $$ \begin{aligned} E_t &= 0.6\sqrt{500} = 13.416\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{13.416}{4.44\times50} = 0.06043\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.06043}{1.6} = 0.03777\ \text{m}^2 = 377.7\ \text{cm}^2 \end{aligned} $$ For a cruciform (two-stepped) core with stacking factor 0.9, $A_i = 0.56d^2$ (standard ratio, Sawhney): $$ \begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.03777}{0.56}} = 0.2597\ \text{m} = 259.7\ \text{mm}\\ a &= 0.85d = 220.8\ \text{mm}\\ b &= 0.53d = 137.6\ \text{mm} \end{aligned} $$ Gross core area $A_{gi} = A_i/0.9 = 0.04197$ m². Output equation (three phase): $Q = 3.33fB_m\delta K_wA_wA_i\times10^{-3}$ $$ \begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{500}{3.33\times50\times1.6\times2.5\times10^6\times0.22\times0.03777\times10^{-3}}\\ &= 0.09035\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.75}} = 181.3\ \text{mm},\quad H_w = 2.75W_w = 498.4\ \text{mm}\\ D &= d + W_w = 441.0\ \text{mm (distance between adjacent limb centres)} \end{aligned} $$ **Winding build** (LV next to core): Radial build (given clearance between core and LV = 10 mm): | Item | Inside dia (mm) | Outside dia (mm) | |---|---|---| | LV winding | 279.7 | 379.7 | | HV winding | 419.7 | 539.7 | Radial depths: $b_1 = 50$ mm (LV), $b_2 = 60$ mm (HV), duct $a = 20$ mm. Axial length of coil, from the output-equation window height: $L_c = H_w - 2(50) = 398.4$ mm. Mean lengths of turn: $$ \begin{aligned} L_{mt,LV} &= \pi\frac{279.7 + 379.7}{2} = 1.0358\ \text{m}\\ L_{mt,HV} &= \pi\frac{419.7 + 539.7}{2} = 1.5071\ \text{m}\\ L_{mt} &= \pi\frac{279.7 + 539.7}{2} = 1.2871\ \text{m (both windings, for reactance)} \end{aligned} $$ Check: the HV outside diameter is 539.7 mm but $D$ from the output equation is only 441.0 mm, so the coils of adjacent limbs would overlap. The window is widened so that there is 30 mm clearance between adjacent HV coils (66 kV), and its height is increased to keep the given ratio $H_w/W_w = 2.75$. The coil height stays at $L_c$ = 398.4 mm. $$ \begin{aligned} D &= d_{o,HV} + 30 = 569.7\ \text{mm}\\ W_w &= D - d = 310.0\ \text{mm},\quad H_w = 2.75W_w = 852.5\ \text{mm} \end{aligned} $$ Yoke: Yoke area 20% more than core area (hot-rolled steel, to reduce yoke loss and magnetising current). Yoke depth = width of largest stamping: $$ \begin{aligned} A_y &= 1.2A_i = 0.04533\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.05036\ \text{m}^2\\ D_y &= a = 220.8\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.05036}{0.2208} = 228.1\ \text{mm} \end{aligned} $$ #### (ii) Overall dimensions of frame $$ \begin{aligned} H &= H_w + 2H_y = 852.5 + 2(228.1) = 1308.8\ \text{mm}\\ W &= 2D + a = 2(569.7) + 220.8 = 1360.2\ \text{mm}\\ \text{Depth} &= D_y = a = 220.8\ \text{mm} \end{aligned} $$ ```text +-----------------------------+ --- | top yoke | ^ +-----+-----+-----+-----+-----+ | |limb | |limb | |limb | | | R | Ww | Y | Ww | B | H | | | | | | | +-----+-----+-----+-----+-----+ | | bottom yoke | v +-----------------------------+ --- |<------------ W ------------>| |<--- D --->|<--- D --->| ``` #### (iii) Per unit resistance and leakage reactance drop Phase voltages: LV (star) $V_{LV} = 6350.9$ V; HV (delta) $V_{HV} = 66000.0$ V. Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 26.24$ A; $I_{HV} = 2.525$ A. $$ \begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{6350.9}{13.416} = 473.36 \approx 473\\ E_t\ (\text{revised}) &= \frac{6350.9}{473} = 13.427\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 4915.6 \approx 4916\\ a_{LV} &= \frac{I_{LV}}{\delta} = 10.50\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.010\ \text{mm}^2 \end{aligned} $$ Resistance referred to HV, with $\rho = 0.021\ \Omega\text{-mm}^2/\text{m}$ (copper at about 75 °C): $$ \begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times4916\times1.5071}{1.010} = 154.03\ \Omega\\ r_{LV} &= \frac{0.021\times473\times1.0358}{10.50} = 0.98013\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 154.03 + 0.98013\left(\frac{4916}{473}\right)^2 = 259.9\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{2.525\times259.9}{66000.0} = 0.00994\ \text{p.u.} \end{aligned} $$ Leakage reactance (per unit), with $AT$ per limb $= 12414$ and axial winding length $L_c = 398.4$ mm: $$ \begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{12414}{0.3984}\cdot\frac{1.2871}{13.427}\left(20 + \frac{50+60}{3}\right)\times10^{-3}\\ &= 0.0668\ \text{p.u.} \end{aligned} $$ #### (iv) Per unit voltage regulation (0.8 pf lagging) $$ \begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.00994(0.8) + 0.0668(0.600) = 0.0480\ \text{p.u.} = 4.80\% \end{aligned} $$ **Answer: $d$ = 260 mm; window 310 × 853 mm (after widening); frame 1309 × 1360 × 221 mm; $\varepsilon_r$ = 0.0099, $\varepsilon_x$ = 0.0668; regulation ≈ 4.80%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this design, how is the net core area A_i obtained? A. A_i = Q/(3.33 f B_m) directly from the kVA rating B. From E_t = K sqrt(Q), then phi_m = E_t/(4.44 f), then A_i = phi_m/B_m C. From the window area using A_i = A_w x K_w D. From the current density and the phase current: A_i = I/delta Answer: B. The voltage per turn E_t = 0.6 sqrt(500) = 13.416 V gives phi_m = E_t/(4.44 f) = 0.0604 Wb, and dividing by B_m = 1.6 gives A_i = 0.0378 m^2. (Key point: Voltage per turn and core area) 2. For a cruciform (two-stepped) core with stacking factor 0.9, which relation gives the circumscribing circle diameter d from the net area? A. A_i = 0.45 d^2 B. A_i = 0.60 d^2 C. A_i = 0.56 d^2 D. A_i = 0.785 d^2 Answer: C. The standard ratio for a cruciform core is A_i = 0.56 d^2, so d = sqrt(0.03777/0.56) = 259.7 mm, with a = 0.85d and b = 0.53d. (Key point: Cruciform core dimensions) 3. Why was the window width widened after the first calculation? A. The output equation gave a negative window area B. The HV outside diameter (539.7 mm) exceeded the limb centre distance D (441 mm), so adjacent coils would overlap C. The yoke area had to be increased by 20% D. The LV winding was too thin to carry the current Answer: B. Coils on adjacent limbs must not overlap, so D was increased to the HV outside diameter plus 30 mm clearance, giving D = 569.7 mm and W_w = 310 mm. (Key point: Window widening check) 4. How are the yoke dimensions and the frame width obtained? A. Yoke area = 1.2 x core area, yoke depth = width of the largest stamping, and W = 2D + a B. Yoke area = 0.8 x core area, yoke depth = d, and W = D + a C. Yoke area = core area, yoke depth = b, and W = 2D D. Yoke area = 1.2 x core area, yoke depth = d, and W = D + d Answer: A. A hot-rolled steel yoke is made 20% larger in area to lower yoke loss and magnetising current. The depth equals the largest stamping width a, and the three-limb frame width is W = 2D + a. 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