--- title: "Core, turns and conductors of 5 kVA transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:design-5kva-single-phase-core" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-5kva-single-phase-core site: BCT Engineering --- # Core, turns and conductors of 5 kVA transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2073 Shrawan, 12 marks - Video: [Transformer Main dimensions numerical problem 3 #electricalengineering #electricmotors](https://www.youtube.com/watch?v=bnBxtr7VY-Q) — EEE, SJBIT - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-5kva-single-phase-core - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Determine the main dimensions of the core, the number of turns and the cross sections of the conductors for a 5 kVA, 11000/400 V, 50 Hz, single phase core type distribution transformer. The net conductor area in the window is 0.6 times the net cross section of iron in the core. Assume a square cross-section for core, a flux density 1 Wb/m², a current density 1.4 A/mm², and window space factor 0.2. The height of window is 3 times its width. ## Short answer Core area 73.2 cm², square limb side 90.2 mm (diameter 127.6 mm), window 85.6×256.7 mm. HV winding: 6765 turns, 0.325 mm²; LV winding: 246 turns, 8.93 mm². ## Answer Data: $Q$ = 5 kVA, $f$ = 50 Hz, $B_m$ = 1 Wb/m², $\delta$ = 1.4 A/mm², $K_w$ = 0.2, $H_w = 3W_w$, square core, and copper area in window $K_wA_w = 0.6A_i$. #### Core area Single-phase output equation: $$ Q = 2.22fB_m\delta\,(K_wA_w)\,A_i\times10^{-3} $$ With $K_wA_w = 0.6A_i$: $$ \begin{aligned} Q &= 2.22fB_m\delta\,(0.6A_i)A_i\times10^{-3}\\ 5 &= 2.22\times50\times1\times1.4\times10^6\times0.6\times10^{-3}\,A_i^2 = 93240\,A_i^2\\ A_i &= \sqrt{\frac{5}{93240}} = 0.00732\ \text{m}^2 = 73.2\ \text{cm}^2 \end{aligned} $$ #### Core dimensions (square core) For a square core with stacking factor 0.9: $A_i = 0.9\times0.5d^2 = 0.45d^2$, side $a = d/\sqrt2 = 0.707d$. $$ \begin{aligned} d &= \sqrt{\frac{0.00732}{0.45}} = 127.6\ \text{mm}\\ a &= 0.707d = 90.2\ \text{mm (side of square limb)} \end{aligned} $$ #### Window $$ \begin{aligned} K_wA_w &= 0.6A_i = 0.00439\ \text{m}^2\\ A_w &= \frac{0.00439}{0.2} = 0.02197\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{3}} = 85.6\ \text{mm},\quad H_w = 3W_w = 256.7\ \text{mm}\\ D &= d + W_w = 213.1\ \text{mm (centre distance of limbs)} \end{aligned} $$ #### Turns $$ \begin{aligned} E_t &= 4.44fB_mA_i = 4.44\times50\times1\times0.00732 = 1.626\ \text{V}\\ T_{LV} &= \frac{400}{1.626} = 246.0 \approx 246\\ T_{HV} &= 246\times\frac{11000}{400} = 6765.0 \approx 6765 \end{aligned} $$ #### Conductor sections $$ \begin{aligned} I_{HV} &= \frac{5000}{11000} = 0.4545\ \text{A},\quad a_{HV} = \frac{0.4545}{1.4} = 0.325\ \text{mm}^2\\ I_{LV} &= \frac{5000}{400} = 12.5\ \text{A},\quad a_{LV} = \frac{12.5}{1.4} = 8.93\ \text{mm}^2 \end{aligned} $$ The HV conductor ≈ 0.33 mm² (about 0.65 mm diameter round wire); the LV conductor ≈ 8.9 mm² (round wire about 3.4 mm diameter, or a small strip). | Quantity | Value | |---|---| | Net core area $A_i$ | 73.2 cm² | | Core circle dia $d$ / side $a$ | 127.6 mm / 90.2 mm | | Window $W_w \times H_w$ | 85.6 × 256.7 mm | | Turns HV / LV | 6765 / 246 | | Conductor area HV / LV | 0.325 / 8.93 mm² | **Answer: $A_i$ ≈ 73.2 cm², square limb 90.2 mm (circle 127.6 mm), window 85.6 × 256.7 mm; HV 6765 turns of 0.325 mm², LV 246 turns of 8.93 mm².** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is A_i found using that the conductor area in the window is 0.6 times the core area? A. Substitute K_w A_w = 0.6 A_i into Q = 2.22 f B_m delta K_w A_w A_i x 10^-3, giving A_i^2 = Q/(...), A_i = 0.00732 m^2 B. A_i = Q/(4.44 f B_m) C. A_i = 0.6 x Q/delta D. A_i = sqrt(Q) / 100 Answer: A. 5 = 93240 A_i^2 so A_i = sqrt(5/93240) = 73.2 cm^2. (Key point: Core area from output equation) 2. For the square core with stacking factor 0.9, what relation gives d and the side a? A. A_i = 0.56 d^2, a = 0.85 d B. A_i = 0.9 x 0.5 d^2 = 0.45 d^2, a = 0.71 d C. A_i = 0.6 d^2, a = 0.9 d D. A_i = d^2, a = d Answer: B. A square inscribed in a circle of diameter d has area 0.5 d^2; with stacking factor 0.9, A_i = 0.45 d^2, so d = 127.6 mm and side 0.71d = 90.6 mm. (Key point: Square core size) 3. How are the window dimensions found? A. A_w = 0.6 A_i / 0.2 = 0.02197 m^2; W_w = sqrt(A_w/3) = 85.6 mm, H_w = 3 W_w = 256.7 mm B. A_w = 0.6 A_i x 0.2; H_w = W_w C. A_w = A_i / 0.2; W_w = A_w/3 D. A_w = Q / delta; W_w = d Answer: A. The copper area K_w A_w is 0.6 A_i, so A_w = 0.00439/0.2, and with H_w = 3 W_w, W_w = sqrt(A_w/3). (Key point: Window dimensions) 4. How are the turns and conductor areas determined? A. E_t = 4.44 f B_m A_i = 1.626 V, T_LV = 400/E_t = 246, T_HV = 246 x 11000/400 = 6765; conductor area = I/delta B. E_t = 4.44 f B_m d, T_LV = 11000/E_t; area = I x delta C. E_t = K sqrt(Q) only; area = I/B_m D. T_HV = T_LV x 400/11000; area = I/delta Answer: A. Turns come from voltage per turn, and conductor area is current divided by current density: a_HV = 0.4545/1.4 = 0.325 mm^2, a_LV = 12.5/1.4 = 8.93 mm^2. 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