Design: 500 kVA 66/11 kV power 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), end clearance 50 mm (66 kV), $\rho = 0.021\ \Omega$-mm²/m, regulation at 0.8 pf lagging (pf not given).

(i) Overall dimensions of core

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

$$\begin{aligned} E_t &= 0.6\sqrt{500} = 13.416\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{13.416}{4.44\times50} = 0.06043\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.06043}{1.6} = 0.03777\ \text{m}^2 = 377.7\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.03777}{0.56}} = 0.2597\ \text{m} = 259.7\ \text{mm}\\ a &= 0.85d = 220.8\ \text{mm}\\ b &= 0.53d = 137.6\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.04197$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{500}{3.33\times50\times1.6\times2.5\times10^6\times0.22\times0.03777\times10^{-3}}\\ &= 0.09035\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.75}} = 181.3\ \text{mm},\quad H_w = 2.75W_w = 498.4\ \text{mm}\\ D &= d + W_w = 441.0\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Winding build (LV next to core):

Radial build (given clearance between core and LV = 10 mm):

Radial depths: $b_1 = 50$ mm (LV), $b_2 = 60$ mm (HV), duct $a = 20$ mm. Axial length of coil, from the output-equation window height: $L_c = H_w - 2(50) = 398.4$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{279.7 + 379.7}{2} = 1.0358\ \text{m}\\ L_{mt,HV} &= \pi\frac{419.7 + 539.7}{2} = 1.5071\ \text{m}\\ L_{mt} &= \pi\frac{279.7 + 539.7}{2} = 1.2871\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: the HV outside diameter is 539.7 mm but $D$ from the output equation is only 441.0 mm, so the coils of adjacent limbs would overlap. The window is widened so that there is 30 mm clearance between adjacent HV coils (66 kV), and its height is increased to keep the given ratio $H_w/W_w = 2.75$. The coil height stays at $L_c$ = 398.4 mm.

$$\begin{aligned} D &= d_{o,HV} + 30 = 569.7\ \text{mm}\\ W_w &= D - d = 310.0\ \text{mm},\quad H_w = 2.75W_w = 852.5\ \text{mm} \end{aligned}$$

Yoke:

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

$$\begin{aligned} A_y &= 1.2A_i = 0.04533\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.05036\ \text{m}^2\\ D_y &= a = 220.8\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.05036}{0.2208} = 228.1\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 852.5 + 2(228.1) = 1308.8\ \text{mm}\\ W &= 2D + a = 2(569.7) + 220.8 = 1360.2\ \text{mm}\\ \text{Depth} &= D_y = a = 220.8\ \text{mm} \end{aligned}$$

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

(iii) Per unit resistance and leakage reactance drop

Phase voltages: LV (star) $V_{LV} = 6350.9$ V; HV (delta) $V_{HV} = 66000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 26.24$ A; $I_{HV} = 2.525$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{6350.9}{13.416} = 473.36 \approx 473\\ E_t\ (\text{revised}) &= \frac{6350.9}{473} = 13.427\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 4915.6 \approx 4916\\ a_{LV} &= \frac{I_{LV}}{\delta} = 10.50\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.010\ \text{mm}^2 \end{aligned}$$

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times4916\times1.5071}{1.010} = 154.03\ \Omega\\ r_{LV} &= \frac{0.021\times473\times1.0358}{10.50} = 0.98013\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 154.03 + 0.98013\left(\frac{4916}{473}\right)^2 = 259.9\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{2.525\times259.9}{66000.0} = 0.00994\ \text{p.u.} \end{aligned}$$

Leakage reactance (per unit), with $AT$ per limb $= 12414$ and axial winding length $L_c = 398.4$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{12414}{0.3984}\cdot\frac{1.2871}{13.427}\left(20 + \frac{50+60}{3}\right)\times10^{-3}\\ &= 0.0668\ \text{p.u.} \end{aligned}$$

(iv) Per unit voltage regulation (0.8 pf lagging)

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.00994(0.8) + 0.0668(0.600) = 0.0480\ \text{p.u.} = 4.80\% \end{aligned}$$

Answer: $d$ = 260 mm; window 310 × 853 mm (after widening); frame 1309 × 1360 × 221 mm; $\varepsilon_r$ = 0.0099, $\varepsilon_x$ = 0.0668; regulation ≈ 4.80%.