Design: 100 kVA 2200/480 V 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 $I_{HV}T_{HV}/2$. Assumed: yoke area 1.2 × core area (hot-rolled steel), $\rho = 0.021\ \Omega$-mm²/m, end clearance 20 mm at each end of the windings.

(i) Overall dimensions of core

Voltage per turn is given: $E_t = 7.50$ V.

$$\begin{aligned} \phi_m &= \frac{E_t}{4.44f} = \frac{7.500}{4.44\times50} = 0.03378\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.03378}{1.2} = 0.02815\ \text{m}^2 = 281.5\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.02815}{0.56}} = 0.2242\ \text{m} = 224.2\ \text{mm}\\ a &= 0.85d = 190.6\ \text{mm}\\ b &= 0.53d = 118.8\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.03128$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{100}{2.22\times50\times1.2\times2.5\times10^6\times0.28\times0.02815\times10^{-3}}\\ &= 0.03810\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2}} = 138.0\ \text{mm},\quad H_w = 2W_w = 276.0\ \text{mm}\\ D &= d + W_w = 362.2\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Winding build (LV next to core):

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

Radial depths: $b_1 = 25$ mm (LV), $b_2 = 30$ mm (HV), duct $a = 10$ mm. Axial length of coil, from the output-equation window height: $L_c = H_w - 2(20) = 236.0$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{234.2 + 284.2}{2} = 0.8144\ \text{m}\\ L_{mt,HV} &= \pi\frac{304.2 + 364.2}{2} = 1.0500\ \text{m}\\ L_{mt} &= \pi\frac{234.2 + 364.2}{2} = 0.9400\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: the HV outside diameter is 364.2 mm but $D$ from the output equation is only 362.2 mm, so the coils of adjacent limbs would overlap. The window is widened so that there is 10 mm clearance between adjacent HV coils, and its height is increased to keep the given ratio $H_w/W_w = 2$. The coil height stays at $L_c$ = 236.0 mm.

$$\begin{aligned} D &= d_{o,HV} + 10 = 374.2\ \text{mm}\\ W_w &= D - d = 150.0\ \text{mm},\quad H_w = 2W_w = 300.0\ \text{mm} \end{aligned}$$

Yoke:

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

$$\begin{aligned} A_y &= 1.2A_i = 0.03378\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.03754\ \text{m}^2\\ D_y &= a = 190.6\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.03754}{0.1906} = 197.0\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 300.0 + 2(197.0) = 693.9\ \text{mm}\\ W &= D + a = 374.2 + 190.6 = 564.8\ \text{mm}\\ \text{Depth} &= D_y = a = 190.6\ \text{mm} \end{aligned}$$

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

(iii) Per unit resistance and leakage reactance drop

Currents: $I_{LV} = \frac{Q}{V_{LV}} = 208.33$ A; $I_{HV} = 45.455$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{480.0}{7.500} = 64.00 \approx 64\\ E_t\ (\text{revised}) &= \frac{480.0}{64} = 7.500\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 293.3 \approx 293\\ a_{LV} &= \frac{I_{LV}}{\delta} = 83.33\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 18.182\ \text{mm}^2 \end{aligned}$$

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times293\times1.0500}{18.182} = 0.35533\ \Omega\\ r_{LV} &= \frac{0.021\times64\times0.8144}{83.33} = 0.013134\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 0.35533 + 0.013134\left(\frac{293}{64}\right)^2 = 0.6306\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{45.45\times0.6306}{2200.0} = 0.01303\ \text{p.u.} \end{aligned}$$

Leakage reactance (per unit), with $AT$ per limb $= 6659$ and axial winding length $L_c = 236.0$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{6659}{0.2360}\cdot\frac{0.9400}{7.500}\left(10 + \frac{25+30}{3}\right)\times10^{-3}\\ &= 0.0396\ \text{p.u.} \end{aligned}$$

(iv) Per unit regulation at 0.8 pf lagging

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.01303(0.8) + 0.0396(0.600) = 0.0342\ \text{p.u.} = 3.42\% \end{aligned}$$

Answer: $d$ = 224 mm, $A_i$ = 281.5 cm²; window 150 × 300 mm; frame 694 × 565 × 191 mm; $\varepsilon_r$ = 0.0130, $\varepsilon_x$ = 0.0396; regulation ≈ 3.42%.