Design: 25 kVA transformer frame, regulation, tubes

Design follows the standard method (A.K. Sawhney): $E_t = K\sqrt{Q}$, output equation for $A_w$, yoke 20% larger than core (hot-rolled steel assumed), $\rho = 0.021\ \Omega$-mm²/m.

Core and window

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

$$\begin{aligned} E_t &= 0.45\sqrt{25} = 2.250\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{2.250}{4.44\times50} = 0.01014\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.01014}{1} = 0.01014\ \text{m}^2 = 101.4\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.01014}{0.56}} = 0.1345\ \text{m} = 134.5\ \text{mm}\\ a &= 0.85d = 114.4\ \text{mm}\\ b &= 0.53d = 71.3\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.01126$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{25}{3.33\times50\times1\times2.3\times10^6\times0.18\times0.01014\times10^{-3}}\\ &= 0.03578\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 119.6\ \text{mm},\quad H_w = 2.5W_w = 299.1\ \text{mm}\\ D &= d + W_w = 254.2\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Check: HV outside diameter (239 mm) < $D$ (254.2 mm), so adjacent HV coils have about 15 mm clearance. LV inside diameter 138 mm > $d$, so the windings fit.

Yoke

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

$$\begin{aligned} A_y &= 1.2A_i = 0.01216\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.01351\ \text{m}^2\\ D_y &= a = 114.4\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.01351}{0.1144} = 118.2\ \text{mm} \end{aligned}$$

i) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 299.1 + 2(118.2) = 535.5\ \text{mm}\\ W &= 2D + a = 2(254.2) + 114.4 = 622.7\ \text{mm}\\ \text{Depth} &= D_y = a = 114.4\ \text{mm} \end{aligned}$$

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

ii) Per unit regulation at full load, 0.8 pf lagging

Turns and currents

Phase voltages: LV (star) $V_{LV} = 250.0$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 33.33$ A; $I_{HV} = 0.758$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{250.0}{2.250} = 111.11 \approx 111\\ E_t\ (\text{revised}) &= \frac{250.0}{111} = 2.252\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 4884.1 \approx 4884\\ a_{LV} &= \frac{I_{LV}}{\delta} = 14.49\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 0.329\ \text{mm}^2 \end{aligned}$$

(The given conductor areas, 14.9 mm² and 0.312 mm², match these values.)

Resistance

Radial depths: $b_1 = 9.1$ mm (LV), $b_2 = 26.4$ mm (HV), duct $a = 15$ mm. Axial length $L_c = 253$ mm (given).

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{138.0 + 156.2}{2} = 0.4621\ \text{m}\\ L_{mt,HV} &= \pi\frac{186.2 + 239.0}{2} = 0.6679\ \text{m}\\ L_{mt} &= \pi\frac{138.0 + 239.0}{2} = 0.5922\ \text{m (both windings, for reactance)} \end{aligned}$$

Using the given conductor areas:

$$\begin{aligned} r_{HV} &= \frac{0.021\times4884\times0.6679}{0.312} = 219.6\ \Omega\\ r_{LV} &= \frac{0.021\times111\times0.4621}{14.9} = 0.0723\ \Omega\\ R_p &= 219.6 + 0.0723\left(\frac{4884}{111}\right)^2 = 359.5\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{0.7576\times359.5}{11000} = 0.0248 \end{aligned}$$

Reactance

Leakage reactance (per unit), with $AT$ per limb $= 3700$ and axial winding length $L_c = 253.0$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{3700}{0.2530}\cdot\frac{0.5922}{2.252}\left(15 + \frac{9.1+26.4}{3}\right)\times10^{-3}\\ &= 0.0407\ \text{p.u.} \end{aligned}$$

Regulation

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi = 0.0248(0.8) + 0.0407(0.6)\\ &= 0.0442\ \text{p.u.} = 4.42\% \end{aligned}$$

iii) Number of cooling tubes

Total loss $P = 1200$ W, allowed mean oil rise $\theta = 35\ ^\circ$C, tube diameter 50 mm, length 1.35 m.

Tank walls (top and bottom neglected), dissipation 12.5 W/m²°C (6 radiation + 6.5 convection):

$$\begin{aligned} S_t &= 2(W_t + L_t)H_t = 2.2610\ \text{m}^2\\ \theta_{plain} &= \frac{1200}{12.5S_t} = 42.46\ ^\circ\text{C} \end{aligned}$$

This exceeds 35 °C, so tubes are needed. Tubes improve convection by 35%: $6.5\times1.35 \approx 8.8$ W/m²°C.

$$\begin{aligned} A_{tubes} &= \frac{1}{8.8}\left(\frac{P}{\theta} - 12.5S_t\right) = \frac{1}{8.8}\left(\frac{1200}{35} - 12.5\times2.2610\right) = 0.684\ \text{m}^2\\ \text{Area of one tube} &= \pi d_tl_t = \pi\times0.05\times1.35 = 0.2121\ \text{m}^2\\ n_t &= \frac{0.684}{0.2121} = 3.23 \approx 4\ \text{tubes} \end{aligned}$$

Answer: frame 535 mm high × 623 mm wide × 114 mm deep; regulation ≈ 4.42%; minimum 4 tubes (50 mm dia, 1.35 m).