Design: 125 kVA single phase 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. In a single-phase core type transformer half of each winding is on each limb, so the ampere-turns per limb are $I_{HV}T_{HV}/2$. $K$ = 0.75 assumed (single-phase core type), $\rho = 0.021\ \Omega$-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):

$$\begin{aligned} E_t &= 0.75\sqrt{125} = 8.385\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{8.385}{4.44\times50} = 0.03777\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.03777}{1.35} = 0.02798\ \text{m}^2 = 279.8\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.02798}{0.56}} = 0.2235\ \text{m} = 223.5\ \text{mm}\\ a &= 0.85d = 190.0\ \text{mm}\\ b &= 0.53d = 118.5\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.03109$ m².

Output equation (single phase): $Q = 2.22fB_m\delta K_wA_wA_i\times10^{-3}$

$$\begin{aligned} A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{125}{2.22\times50\times1.35\times2.75\times10^6\times0.3\times0.02798\times10^{-3}}\\ &= 0.03614\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 120.2\ \text{mm},\quad H_w = 2.5W_w = 300.6\ \text{mm}\\ D &= d + W_w = 343.8\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Check: $d$ = 224 mm fits inside the LV (255 mm) and coil height 276 mm < $H_w$. But HV outside diameter 351 mm > $D$ = 344 mm, so the window is widened (10 mm between the HV coils on the two limbs):

$$\begin{aligned} D &= 351 + 10 = 361\ \text{mm},\quad W_w = D - d = 137.5\ \text{mm},\quad H_w = 300.6\ \text{mm} \end{aligned}$$

Yoke:

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

$$\begin{aligned} A_y &= 1.2A_i = 0.03357\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.03731\ \text{m}^2\\ D_y &= a = 190.0\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.03731}{0.1900} = 196.3\ \text{mm} \end{aligned}$$

ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 300.6 + 2(196.3) = 693.3\ \text{mm}\\ W &= D + a = 361.0 + 190.0 = 551.0\ \text{mm}\\ \text{Depth} &= D_y = a = 190.0\ \text{mm} \end{aligned}$$

 +-----------------------+  ---
 |       top yoke        |   ^
 +-----+-----------+-----+   |
 |limb |  window   |limb |   H
 |     |  Ww x Hw  |     |   |
 +-----+-----------+-----+   |
 |      bottom yoke      |   v
 +-----------------------+  ---
 |<--------- W --------->|
    |<----- D ----->|

iii) Leakage reactance

Currents: $I_{LV} = \frac{Q}{V_{LV}} = 312.50$ A; $I_{HV} = 18.939$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{400.0}{8.385} = 47.70 \approx 48\\ E_t\ (\text{revised}) &= \frac{400.0}{48} = 8.333\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 792.0 \approx 792\\ a_{LV} &= \frac{I_{LV}}{\delta} = 113.64\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 6.887\ \text{mm}^2 \end{aligned}$$

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 $= 7500$ and axial winding length $L_c = 272.0$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{7500}{0.2720}\cdot\frac{0.9519}{8.333}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\ &= 0.0269\ \text{p.u.}\\ X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0269\times\frac{6600.0}{18.94} = 9.388\ \Omega\ \text{(referred to HV)}\\ X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 0.03448\ \Omega \end{aligned}$$

Answer: $d$ = 224 mm; window 137 × 301 mm; yoke 196 × 190 mm; frame 693 × 551 × 190 mm; leakage reactance ≈ 9.39 Ω referred to HV (2.69%).