Design: 100 kVA 11/132 kV transformer with winding data

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. $K$ is not given; $K$ = 0.45 is taken (3-phase distribution transformer). Also assumed $\rho = 0.021\ \Omega$-mm²/m. Winding data are given: LV 255/293 mm (inside/outside diameter), HV 314/351 mm, radial depths $b_1$ = 14 mm, $b_2$ = 18 mm, gap $a$ = 11 mm. Axial length taken as the mean of 268 and 276 mm, $L_c$ = 272 mm.

(i) Dimensions of core, window and yoke

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

$$\begin{aligned} E_t &= 0.45\sqrt{100} = 4.500\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{4.500}{4.44\times50} = 0.02027\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.02027}{1.35} = 0.01502\ \text{m}^2 = 150.2\ \text{cm}^2 \end{aligned}$$

For a three-stepped core with stacking factor 0.9, $A_i = 0.60d^2$ (standard ratio, Sawhney):

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.60}} = \sqrt{\frac{0.01502}{0.60}} = 0.1582\ \text{m} = 158.2\ \text{mm}\\ a &= 0.90d = 142.4\ \text{mm}\\ b &= 0.70d = 110.7\ \text{mm},\quad c = 0.42d = 66.4\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.01668$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{100}{3.33\times50\times1.35\times2.75\times10^6\times0.4\times0.01502\times10^{-3}}\\ &= 0.02694\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{3}} = 94.8\ \text{mm},\quad H_w = 3W_w = 284.3\ \text{mm}\\ D &= d + W_w = 252.9\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Check with the given windings: core $d$ = 158 mm fits inside the LV (255 mm). But the HV outside diameter (351 mm) exceeds $D$ = 253 mm, and the coil (276 mm) is almost as tall as the window (284 mm). So the window is enlarged to suit the given coils, with 20 mm between adjacent HV coils, and its height is increased to keep the given ratio $H_w/W_w = 3$:

$$\begin{aligned} D &= 351 + 20 = 371\ \text{mm},\quad W_w = D - d = 212.8\ \text{mm}\\ H_w &= 3W_w = 638.4\ \text{mm} \end{aligned}$$

(A real 132 kV winding would need far larger clearances; the given figures are used as stated.)

Yoke:

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

$$\begin{aligned} A_y &= 1.2A_i = 0.01802\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02002\ \text{m}^2\\ D_y &= a = 142.4\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.02002}{0.1424} = 140.6\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 638.4 + 2(140.6) = 919.7\ \text{mm}\\ W &= 2D + a = 2(371.0) + 142.4 = 884.4\ \text{mm}\\ \text{Depth} &= D_y = a = 142.4\ \text{mm} \end{aligned}$$

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

(iii) Leakage reactance

Phase voltages: LV (delta) $V_{LV} = 11000.0$ V; HV (star) $V_{HV} = 76210.2$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 3.03$ A; $I_{HV} = 0.437$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{11000.0}{4.500} = 2444.44 \approx 2444\\ E_t\ (\text{revised}) &= \frac{11000.0}{2444} = 4.501\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 16932.5 \approx 16933\\ a_{LV} &= \frac{I_{LV}}{\delta} = 1.10\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 0.159\ \text{mm}^2 \end{aligned}$$

Radial depths: $b_1 = 14$ mm (LV), $b_2 = 18$ mm (HV), duct $a = 11$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{255.0 + 293.0}{2} = 0.8608\ \text{m}\\ L_{mt,HV} &= \pi\frac{314.0 + 351.0}{2} = 1.0446\ \text{m}\\ L_{mt} &= \pi\frac{255.0 + 351.0}{2} = 0.9519\ \text{m (both windings, for reactance)} \end{aligned}$$

Leakage reactance (per unit), with $AT$ per limb $= 7406$ and axial winding length $L_c = 272.0$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{7406}{0.2720}\cdot\frac{0.9519}{4.501}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\ &= 0.0493\ \text{p.u.}\\ X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0493\times\frac{76210.2}{0.4374} = 8583\ \Omega\ \text{(referred to HV)}\\ X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 178.8\ \Omega \end{aligned}$$

(iv) Regulation at 0.8 pf lagging, full load

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times16933\times1.0446}{0.159} = 2335.4\ \Omega\\ r_{LV} &= \frac{0.021\times2444\times0.8608}{1.10} = 40.093\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 2335.4 + 40.093\left(\frac{16933}{2444}\right)^2 = 4260\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{0.4374\times4260}{76210.2} = 0.02445\ \text{p.u.} \end{aligned}$$ $$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.02445(0.8) + 0.0493(0.600) = 0.0491\ \text{p.u.} = 4.91\% \end{aligned}$$

Answer: $d$ = 158 mm; window 213 × 638 mm; yoke 141 × 142 mm; frame 920 × 884 × 142 mm; $X$ ≈ 8583 Ω/phase on HV (4.93%); regulation ≈ 4.91%.