Design: 150 kVA 6600/400 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):

$$\begin{aligned} E_t &= 0.45\sqrt{150} = 5.511\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{5.511}{4.44\times50} = 0.02483\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.02483}{1.35} = 0.01839\ \text{m}^2 = 183.9\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01839}{0.56}} = 0.1812\ \text{m} = 181.2\ \text{mm}\\ a &= 0.85d = 154.0\ \text{mm}\\ b &= 0.53d = 96.0\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.02043$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{150}{3.33\times50\times1.35\times2.75\times10^6\times0.27\times0.01839\times10^{-3}}\\ &= 0.04887\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 139.8\ \text{mm},\quad H_w = 2.5W_w = 349.5\ \text{mm}\\ D &= d + W_w = 321.0\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke:

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

$$\begin{aligned} A_y &= 1.2A_i = 0.02207\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02452\ \text{m}^2\\ D_y &= a = 154.0\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.02452}{0.1540} = 159.2\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 349.5 + 2(159.2) = 667.9\ \text{mm}\\ W &= 2D + a = 2(321.0) + 154.0 = 796.1\ \text{mm}\\ \text{Depth} &= D_y = a = 154.0\ \text{mm} \end{aligned}$$

 +-----------------------------+  ---
 |          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

Turns and conductors

Phase voltages: LV (star) $V_{LV} = 230.9$ V; HV (delta) $V_{HV} = 6600.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 216.51$ A; $I_{HV} = 7.576$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{230.9}{5.511} = 41.90 \approx 42\\ E_t\ (\text{revised}) &= \frac{230.9}{42} = 5.499\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1200.3 \approx 1200\\ a_{LV} &= \frac{I_{LV}}{\delta} = 78.73\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 2.755\ \text{mm}^2 \end{aligned}$$

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 $L_c = H_w - 2(20) = 309.5$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{191.2 + 231.2}{2} = 0.6635\ \text{m}\\ L_{mt,HV} &= \pi\frac{261.2 + 311.2}{2} = 0.8992\ \text{m}\\ L_{mt} &= \pi\frac{191.2 + 311.2}{2} = 0.7892\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: HV outside diameter 311 mm vs $D$ = 321 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):

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1200\times0.8992}{2.755} = 8.2252\ \Omega\\ r_{LV} &= \frac{0.021\times42\times0.6635}{78.73} = 0.0074337\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 8.2252 + 0.0074337\left(\frac{1200}{42}\right)^2 = 14.294\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{7.576\times14.294}{6600.0} = 0.01641\ \text{p.u.} \end{aligned}$$

Leakage reactance (per unit), with $AT$ per limb $= 9091$ and axial winding length $L_c = 309.5$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{9091}{0.3095}\cdot\frac{0.7892}{5.499}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\ &= 0.0499\ \text{p.u.} \end{aligned}$$

(iv) Per unit regulation at 0.85 pf lagging

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.01641(0.85) + 0.0499(0.527) = 0.0402\ \text{p.u.} = 4.02\% \end{aligned}$$

Answer: $d$ = 181 mm, window 140 × 350 mm, yoke 159 × 154 mm; frame 668 × 796 × 154 mm; $\varepsilon_r$ = 0.0164, $\varepsilon_x$ = 0.0499; regulation ≈ 4.02%.