Design: 25 kVA 11000/433 V distribution transformer

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: clearance core-to-LV 2 mm, LV-to-HV duct 15 mm, end clearance 20 mm, $\rho = 0.021\ \Omega$-mm²/m.

i) Dimensions of core, window and yoke

Voltage per turn, $E_t = K\sqrt{Q}$ ($Q$ in kVA, total for 3 phases):

$$\begin{aligned} E_t &= 0.45\sqrt{25} = 2.250\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{2.250}{4.44\times50} = 0.01014\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.01014}{1} = 0.01014\ \text{m}^2 = 101.4\ \text{cm}^2 \end{aligned}$$

For a cruciform (two-stepped) core with stacking factor 0.9, $A_i = 0.56d^2$ (standard ratio, Sawhney):

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01014}{0.56}} = 0.1345\ \text{m} = 134.5\ \text{mm}\\ a &= 0.85d = 114.4\ \text{mm}\\ b &= 0.53d = 71.3\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.01126$ m².

Output equation (three phase): $Q = 3.33fB_m\delta K_wA_wA_i\times10^{-3}$

Window space factor $K_w = \dfrac{8}{30 + kV} = \dfrac{8}{30 + 11} = 0.1951$

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{25}{3.33\times50\times1\times2.3\times10^6\times0.195\times0.01014\times10^{-3}}\\ &= 0.03301\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 114.9\ \text{mm},\quad H_w = 2.5W_w = 287.3\ \text{mm}\\ D &= d + W_w = 249.4\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke:

Yoke area 20% more than core area (hot-rolled steel, to reduce yoke loss and magnetising current). Yoke depth = width of largest stamping:

$$\begin{aligned} A_y &= 1.2A_i = 0.01216\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.01351\ \text{m}^2\\ D_y &= a = 114.4\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.01351}{0.1144} = 118.2\ \text{mm} \end{aligned}$$

ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 287.3 + 2(118.2) = 523.6\ \text{mm}\\ W &= 2D + a = 2(249.4) + 114.4 = 613.2\ \text{mm}\\ \text{Depth} &= D_y = a = 114.4\ \text{mm} \end{aligned}$$

 +-----------------------------+  ---
 |          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) $V_{LV} = 250.0$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 33.33$ A; $I_{HV} = 0.758$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{250.0}{2.250} = 111.11 \approx 111\\ E_t\ (\text{revised}) &= \frac{250.0}{111} = 2.252\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 4884.1 \approx 4884\\ a_{LV} &= \frac{I_{LV}}{\delta} = 14.49\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 0.329\ \text{mm}^2 \end{aligned}$$

Radial build (assumed clearance between core and LV = 2 mm):

Radial depths: $b_1 = 9.1$ mm (LV), $b_2 = 26.22$ mm (HV), duct $a = 15$ mm. Axial length $L_c = H_w - 2(20) = 247.3$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{138.5 + 156.7}{2} = 0.4638\ \text{m}\\ L_{mt,HV} &= \pi\frac{186.7 + 239.2}{2} = 0.6690\ \text{m}\\ L_{mt} &= \pi\frac{138.5 + 239.2}{2} = 0.5933\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: HV outside diameter 239 mm vs $D$ = 249 mm. Adjacent HV coils have about 10 mm clearance, so the coils fit.

Resistance referred to HV, with $\rho = 0.021\ \Omega\text{-mm}^2/\text{m}$ (copper at about 75 °C):

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times4884\times0.6690}{0.329} = 208.32\ \Omega\\ r_{LV} &= \frac{0.021\times111\times0.4638}{14.49} = 0.074594\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 208.32 + 0.074594\left(\frac{4884}{111}\right)^2 = 352.73\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{0.7576\times352.73}{11000.0} = 0.02429\ \text{p.u.} \end{aligned}$$

Leakage reactance (per unit), with $AT$ per limb $= 3700$ and axial winding length $L_c = 247.3$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{3700}{0.2473}\cdot\frac{0.5933}{2.252}\left(15 + \frac{9.1+26.22}{3}\right)\times10^{-3}\\ &= 0.0417\ \text{p.u.} \end{aligned}$$

iv) Per unit regulation at 0.8 pf lagging

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.02429(0.8) + 0.0417(0.600) = 0.0444\ \text{p.u.} = 4.44\% \end{aligned}$$

v) Full-load efficiency at 0.8 pf

Output $= 25\times0.8 = 20$ kW; total full-load loss = 901 W.

$$\eta = \frac{20}{20 + 0.901}\times100 = 95.69\%$$

(Copper loss from the resistance above: $3I_{HV}^2R_p$ = 607 W, so the iron loss is about 294 W.)

Answer: $d$ = 134.5 mm, window 115 × 287 mm; frame 524 × 613 × 114 mm; $\varepsilon_r$ = 0.0243, $\varepsilon_x$ = 0.0417; regulation ≈ 4.44%; efficiency ≈ 95.69%.