Design: 150 kVA 11000/420 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: yoke area 1.2 × core area (hot-rolled steel), core-to-LV clearance 5 mm, end clearance 20 mm, $\rho = 0.021\ \Omega$-mm²/m.

(i) Overall dimensions of core

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

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

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): $Q = 3.33fB_m\delta K_wA_wA_i\times10^{-3}$

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{150}{3.33\times50\times1.35\times2.5\times10^6\times0.25\times0.01839\times10^{-3}}\\ &= 0.05806\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 152.4\ \text{mm},\quad H_w = 2.5W_w = 381.0\ \text{mm}\\ D &= d + W_w = 333.6\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke:

Yoke area is 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

$$\begin{aligned} H &= H_w + 2H_y = 381.0 + 2(159.2) = 699.4\ \text{mm}\\ W &= 2D + a = 2(333.6) + 154.0 = 821.3\ \text{mm}\\ \text{Depth} &= D_y = a = 154.0\ \text{mm} \end{aligned}$$

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

Phase voltages: LV (star) $V_{LV} = 242.5$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 206.20$ A; $I_{HV} = 4.545$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{242.5}{5.511} = 44.00 \approx 44\\ E_t\ (\text{revised}) &= \frac{242.5}{44} = 5.511\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1996.0 \approx 1996\\ a_{LV} &= \frac{I_{LV}}{\delta} = 82.48\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.818\ \text{mm}^2 \end{aligned}$$

Radial build (assumed clearance between core and LV = 5 mm). Each radial gap or depth appears on both sides of the diameter, so it is added twice:

• LV inside diameter $= 181.2 + 2(5) = 191.2$ mm
• LV outside diameter $= 191.2 + 2(20) = 231.2$ mm
• HV inside diameter $= 231.2 + 2(15) = 261.2$ mm
• HV outside diameter $= 261.2 + 2(25) = 311.2$ mm

Radial depths: $b_1 = 20$ mm (LV), $b_2 = 25$ mm (HV), duct $a = 15$ mm. Axial length $L_c = H_w - 2(20) = 341.0$ 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 $D$ = 334 mm. Adjacent HV coils have about 22 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):

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1996\times0.8992}{1.818} = 20.729\ \Omega\\ r_{LV} &= \frac{0.021\times44\times0.6635}{82.48} = 0.0074337\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 20.729 + 0.0074337\left(\frac{1996}{44}\right)^2 = 36.027\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{4.545\times36.027}{11000.0} = 0.01489\ \text{p.u.} \end{aligned}$$

Leakage reactance (per unit), with $AT$ per limb $= 9073$ and axial winding length $L_c = 341.0$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{9073}{0.3410}\cdot\frac{0.7892}{5.511}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\ &= 0.0451\ \text{p.u.} \end{aligned}$$

(iv) Per unit regulation at 0.85 pf lagging

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.01489(0.85) + 0.0451(0.527) = 0.0364\ \text{p.u.} = 3.64\% \end{aligned}$$

Answer: $d$ = 181 mm, window 152 × 381 mm; frame 699 × 821 × 154 mm; $\varepsilon_r$ = 0.0149, $\varepsilon_x$ = 0.0451; regulation ≈ 3.64%.