Design data: 1000 kVA 66/11 kV transformer

The list of quantities is missing from the paper, so the usual set is worked out: core, window and yoke; overall frame; per unit resistance and reactance; regulation at 0.8 pf lagging. 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 10 mm, end clearance 50 mm (66 kV), $\rho = 0.021\ \Omega$-mm²/m.

Core and window

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

$$\begin{aligned} E_t &= 0.6\sqrt{1000} = 18.974\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{18.974}{4.44\times50} = 0.08547\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.08547}{1.45} = 0.05894\ \text{m}^2 = 589.4\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.05894}{0.56}} = 0.3244\ \text{m} = 324.4\ \text{mm}\\ a &= 0.85d = 275.8\ \text{mm}\\ b &= 0.53d = 171.9\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.06549$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{1000}{3.33\times50\times1.45\times2.75\times10^6\times0.13\times0.05894\times10^{-3}}\\ &= 0.19657\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.75}} = 267.4\ \text{mm},\quad H_w = 2.75W_w = 735.2\ \text{mm}\\ D &= d + W_w = 591.8\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke

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

$$\begin{aligned} A_y &= 1.2A_i = 0.07073\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.07859\ \text{m}^2\\ D_y &= a = 275.8\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.07859}{0.2758} = 285.0\ \text{mm} \end{aligned}$$

Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 735.2 + 2(285.0) = 1305.2\ \text{mm}\\ W &= 2D + a = 2(591.8) + 275.8 = 1459.3\ \text{mm}\\ \text{Depth} &= D_y = a = 275.8\ \text{mm} \end{aligned}$$

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

Turns and conductors (delta/delta)

Phase voltages: LV (delta) $V_{LV} = 11000.0$ V; HV (delta) $V_{HV} = 66000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 30.30$ A; $I_{HV} = 5.051$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{11000.0}{18.974} = 579.75 \approx 580\\ E_t\ (\text{revised}) &= \frac{11000.0}{580} = 18.966\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 3480.0 \approx 3480\\ a_{LV} &= \frac{I_{LV}}{\delta} = 11.02\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.837\ \text{mm}^2 \end{aligned}$$

Winding dimensions

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

Radial depths: $b_1 = 40$ mm (LV), $b_2 = 50$ mm (HV), duct $a = 20$ mm. Axial length $L_c = H_w - 2(50) = 635.2$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{344.4 + 424.4}{2} = 1.2077\ \text{m}\\ L_{mt,HV} &= \pi\frac{464.4 + 564.4}{2} = 1.6161\ \text{m}\\ L_{mt} &= \pi\frac{344.4 + 564.4}{2} = 1.4276\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: HV outside diameter 564 mm vs $D$ = 592 mm. Adjacent HV coils have about 27 mm clearance, so the coils fit.

Per unit resistance

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times3480\times1.6161}{1.837} = 64.309\ \Omega\\ r_{LV} &= \frac{0.021\times580\times1.2077}{11.02} = 1.3349\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 64.309 + 1.3349\left(\frac{3480}{580}\right)^2 = 112.37\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{5.051\times112.37}{66000.0} = 0.00860\ \text{p.u.} \end{aligned}$$

Per unit leakage reactance

Leakage reactance (per unit), with $AT$ per limb $= 17576$ and axial winding length $L_c = 635.2$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{17576}{0.6352}\cdot\frac{1.4276}{18.966}\left(20 + \frac{40+50}{3}\right)\times10^{-3}\\ &= 0.0411\ \text{p.u.} \end{aligned}$$

Regulation at 0.8 pf lagging

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.00860(0.8) + 0.0411(0.600) = 0.0315\ \text{p.u.} = 3.15\% \end{aligned}$$

Answer: $d$ = 324 mm; window 267 × 735 mm; frame 1305 × 1459 × 276 mm; $\varepsilon_r$ = 0.0086, $\varepsilon_x$ = 0.0411; regulation ≈ 3.15%.