Design: 100 kVA 4000/433 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: core-to-LV clearance 5 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):

$$\begin{aligned} E_t &= 0.45\sqrt{100} = 4.500\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{4.500}{4.44\times50} = 0.02027\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.02027}{1.3} = 0.01559\ \text{m}^2 = 155.9\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01559}{0.56}} = 0.1669\ \text{m} = 166.9\ \text{mm}\\ a &= 0.85d = 141.8\ \text{mm}\\ b &= 0.53d = 88.4\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.01733$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{100}{3.33\times50\times1.3\times2.5\times10^6\times0.25\times0.01559\times10^{-3}}\\ &= 0.04741\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 137.7\ \text{mm},\quad H_w = 2.5W_w = 344.3\ \text{mm}\\ D &= d + W_w = 304.6\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke:

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

$$\begin{aligned} A_y &= 1.2A_i = 0.01871\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02079\ \text{m}^2\\ D_y &= a = 141.8\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.02079}{0.1418} = 146.6\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 344.3 + 2(146.6) = 637.4\ \text{mm}\\ W &= 2D + a = 2(304.6) + 141.8 = 751.0\ \text{mm}\\ \text{Depth} &= D_y = a = 141.8\ \text{mm} \end{aligned}$$

 +-----------------------------+  ---
 |          top yoke           |   ^
 +-----+-----+-----+-----+-----+   |
 |limb |     |limb |     |limb |   |
 |  R  | Ww  |  Y  | Ww  |  B  |   H
 |     |     |     |     |     |   |
 +-----+-----+-----+-----+-----+   |
 |         bottom yoke         |   v
 +-----------------------------+  ---
 |<------------ W ------------>|
    |<--- D --->|<--- D --->|

(iii) Per unit voltage regulation at 0.8 pf lagging

Turns and conductors

Phase voltages: LV (star) $V_{LV} = 250.0$ V; HV (delta) $V_{HV} = 4000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 133.34$ A; $I_{HV} = 8.333$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{250.0}{4.500} = 55.55 \approx 56\\ E_t\ (\text{revised}) &= \frac{250.0}{56} = 4.464\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 896.0 \approx 896\\ a_{LV} &= \frac{I_{LV}}{\delta} = 53.33\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 3.333\ \text{mm}^2 \end{aligned}$$

Winding build

Radial build (assumed clearance between core and LV = 5 mm):

Radial depths: $b_1 = 20$ mm (LV), $b_2 = 25$ mm (HV), duct $a = 15$ mm. Axial length $L_c = H_w - 2(20) = 304.3$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{176.9 + 216.9}{2} = 0.6185\ \text{m}\\ L_{mt,HV} &= \pi\frac{246.9 + 296.9}{2} = 0.8541\ \text{m}\\ L_{mt} &= \pi\frac{176.9 + 296.9}{2} = 0.7441\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: HV outside diameter 296.9 mm < $D$ = 304.6 mm (about 8 mm between adjacent coils; just adequate at 4 kV).

Resistance

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times896\times0.8541}{3.333} = 4.8212\ \Omega\\ r_{LV} &= \frac{0.021\times56\times0.6185}{53.33} = 0.013637\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 4.8212 + 0.013637\left(\frac{896}{56}\right)^2 = 8.3122\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{8.333\times8.3122}{4000.0} = 0.01732\ \text{p.u.} \end{aligned}$$

Reactance

Leakage reactance (per unit), with $AT$ per limb $= 7467$ and axial winding length $L_c = 304.3$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{7467}{0.3043}\cdot\frac{0.7441}{4.464}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\ &= 0.0484\ \text{p.u.} \end{aligned}$$

Regulation

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.01732(0.8) + 0.0484(0.600) = 0.0429\ \text{p.u.} = 4.29\% \end{aligned}$$

Answer: $d$ = 167 mm, window 138 × 344 mm, yoke 147 × 142 mm; frame 637 × 751 × 142 mm; regulation ≈ 4.29%.