Design: 150 kVA 11000/440 V transformer with tubes

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: clearance between core and LV = 5 mm (not given), $\delta_{HV} = 1.05\times2.5 = 2.625$ A/mm².

Core and window

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

$$\begin{aligned} E_t &= 0.45\sqrt{150} = 5.511\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{5.511}{4.44\times50} = 0.02483\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.02483}{1.35} = 0.01839\ \text{m}^2 = 183.9\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.60}} = \sqrt{\frac{0.01839}{0.60}} = 0.1751\ \text{m} = 175.1\ \text{mm}\\ a &= 0.90d = 157.6\ \text{mm}\\ b &= 0.70d = 122.5\ \text{mm},\quad c = 0.42d = 73.5\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.02043$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{150}{3.33\times50\times1.35\times2.5\times10^6\times0.25\times0.01839\times10^{-3}}\\ &= 0.05806\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.3}} = 158.9\ \text{mm},\quad H_w = 2.3W_w = 365.4\ \text{mm}\\ D &= d + W_w = 334.0\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke (20% more than gross core area):

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

$$\begin{aligned} A_y &= 1.2A_i = 0.02207\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.02452\ \text{m}^2\\ D_y &= a = 157.6\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.02452}{0.1576} = 155.6\ \text{mm} \end{aligned}$$

(i) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 365.4 + 2(155.6) = 676.7\ \text{mm}\\ W &= 2D + a = 2(334.0) + 157.6 = 825.5\ \text{mm}\\ \text{Depth} &= D_y = a = 157.6\ \text{mm} \end{aligned}$$

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

(ii) Per unit voltage regulation at 0.85 pf lagging

Turns and conductors

Phase voltages: LV (star) $V_{LV} = 254.0$ V; HV (delta) $V_{HV} = 11000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 196.82$ A; $I_{HV} = 4.545$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{254.0}{5.511} = 46.09 \approx 46\\ E_t\ (\text{revised}) &= \frac{254.0}{46} = 5.522\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1991.9 \approx 1992\\ a_{LV} &= \frac{I_{LV}}{\delta} = 78.73\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 1.732\ \text{mm}^2 \end{aligned}$$

Winding build

Radial build (assumed clearance between core and LV = 5 mm):

Radial depths: $b_1 = 20$ mm (LV), $b_2 = 25$ mm (HV), duct $a = 15$ mm. Axial length $L_c = 231.25$ mm (given mean coil height).

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{185.1 + 225.1}{2} = 0.6442\ \text{m}\\ L_{mt,HV} &= \pi\frac{255.1 + 305.1}{2} = 0.8799\ \text{m}\\ L_{mt} &= \pi\frac{185.1 + 305.1}{2} = 0.7699\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: HV outside diameter 305 mm vs $D$ = 334 mm. Adjacent HV coils have about 29 mm clearance, so the coils fit.

Resistance

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1992\times0.8799}{1.732} = 21.256\ \Omega\\ r_{LV} &= \frac{0.021\times46\times0.6442}{78.73} = 0.0079048\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 21.256 + 0.0079048\left(\frac{1992}{46}\right)^2 = 36.079\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{4.545\times36.079}{11000.0} = 0.01491\ \text{p.u.} \end{aligned}$$

Reactance

Leakage reactance (per unit), with $AT$ per limb $= 9055$ and axial winding length $L_c = 231.2$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{9055}{0.2313}\cdot\frac{0.7699}{5.522}\left(15 + \frac{20+25}{3}\right)\times10^{-3}\\ &= 0.0647\ \text{p.u.} \end{aligned}$$

Regulation

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.01491(0.85) + 0.0647(0.527) = 0.0467\ \text{p.u.} = 4.67\% \end{aligned}$$

(iii) Number of cooling tubes

Total loss $P$ = 2850 W, tank 1.15 m × 0.425 m × 1.05 m (H × W × L), mean oil rise 35 °C, tubes 50 mm diameter, 1.05 m long.

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

$$\begin{aligned} S_t &= 2(W_t + L_t)H_t = 3.3925\ \text{m}^2\\ \theta_{plain} &= \frac{2850}{12.5S_t} = 67.21\ ^\circ\text{C} \end{aligned}$$

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

$$\begin{aligned} A_{tubes} &= \frac{1}{8.8}\left(\frac{P}{\theta} - 12.5S_t\right) = \frac{1}{8.8}\left(\frac{2850}{35} - 12.5\times3.3925\right) = 4.434\ \text{m}^2\\ \text{Area of one tube} &= \pi d_tl_t = \pi\times0.05\times1.05 = 0.1649\ \text{m}^2\\ n_t &= \frac{4.434}{0.1649} = 26.89 \approx 27\ \text{tubes} \end{aligned}$$

Answer: frame 677 × 825 × 158 mm; regulation ≈ 4.67% ($\varepsilon_r$ = 0.0149, $\varepsilon_x$ = 0.0647); 27 tubes.