Design: 4000 kVA 66/11 kV power transformer

Standard method (A.K. Sawhney). Assumptions: hot-rolled steel, yoke area = 1.2 × core area, yoke depth = largest stamping width, $\rho = 0.021\ \Omega$-mm²/m, end clearance 50 mm at each end of the 66 kV winding.

(i) Overall core dimensions

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

$$\begin{aligned} E_t &= 0.6\sqrt{4000} = 37.947\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{37.947}{4.44\times50} = 0.17093\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.17093}{1.6} = 0.10683\ \text{m}^2 = 1068.3\ \text{cm}^2 \end{aligned}$$

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

$$\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.10683}{0.56}} = 0.4368\ \text{m} = 436.8\ \text{mm}\\ a &= 0.85d = 371.3\ \text{mm}\\ b &= 0.53d = 231.5\ \text{mm} \end{aligned}$$

Gross core area $A_{gi} = A_i/0.9 = 0.11870$ m².

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

$$\begin{aligned} A_w &= \frac{Q}{3.33fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{4000}{3.33\times50\times1.6\times2.5\times10^6\times0.22\times0.10683\times10^{-3}}\\ &= 0.25554\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.75}} = 304.8\ \text{mm},\quad H_w = 2.75W_w = 838.3\ \text{mm}\\ D &= d + W_w = 741.6\ \text{mm (distance between adjacent limb centres)} \end{aligned}$$

Yoke

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

$$\begin{aligned} A_y &= 1.2A_i = 0.12820\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.14244\ \text{m}^2\\ D_y &= a = 371.3\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.14244}{0.3713} = 383.7\ \text{mm} \end{aligned}$$

(ii) Overall dimensions of frame

$$\begin{aligned} H &= H_w + 2H_y = 838.3 + 2(383.7) = 1605.6\ \text{mm}\\ W &= 2D + a = 2(741.6) + 371.3 = 1854.5\ \text{mm}\\ \text{Depth} &= D_y = a = 371.3\ \text{mm} \end{aligned}$$

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

(iii) Per unit resistance and leakage reactance drop

Turns and conductors (both windings delta):

Phase voltages: LV (delta) $V_{LV} = 11000.0$ V; HV (delta) $V_{HV} = 66000.0$ V.

Phase currents: $I_{LV} = \frac{Q}{3V_{LV}} = 121.21$ A; $I_{HV} = 20.202$ A.

$$\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{11000.0}{37.947} = 289.88 \approx 290\\ E_t\ (\text{revised}) &= \frac{11000.0}{290} = 37.931\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 1740.0 \approx 1740\\ a_{LV} &= \frac{I_{LV}}{\delta} = 48.48\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 8.081\ \text{mm}^2 \end{aligned}$$

Winding dimensions

Radial build (given clearance between core and LV = 10 mm):

Radial depths: $b_1 = 50$ mm (LV), $b_2 = 50$ mm (HV), duct $a = 20$ mm. Axial length $L_c = H_w - 2(50) = 738.3$ mm.

Mean lengths of turn:

$$\begin{aligned} L_{mt,LV} &= \pi\frac{456.8 + 556.8}{2} = 1.5921\ \text{m}\\ L_{mt,HV} &= \pi\frac{596.8 + 696.8}{2} = 2.0319\ \text{m}\\ L_{mt} &= \pi\frac{456.8 + 696.8}{2} = 1.8120\ \text{m (both windings, for reactance)} \end{aligned}$$

Check: HV outside diameter 697 mm < $D$ = 742 mm, leaving about 45 mm between adjacent HV coils.

Per unit resistance

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

$$\begin{aligned} r_{HV} &= \frac{\rho T_{HV}L_{mt,HV}}{a_{HV}} = \frac{0.021\times1740\times2.0319}{8.081} = 9.1879\ \Omega\\ r_{LV} &= \frac{0.021\times290\times1.5921}{48.48} = 0.19998\ \Omega\\ R_p &= r_{HV} + r_{LV}\left(\frac{T_{HV}}{T_{LV}}\right)^2 = 9.1879 + 0.19998\left(\frac{1740}{290}\right)^2 = 16.387\ \Omega\\ \varepsilon_r &= \frac{I_{HV}R_p}{V_{HV}} = \frac{20.2\times16.387}{66000.0} = 0.00502\ \text{p.u.} \end{aligned}$$

Per unit leakage reactance

Leakage reactance (per unit), with $AT$ per limb $= 35152$ and axial winding length $L_c = 738.3$ mm:

$$\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{35152}{0.7383}\cdot\frac{1.8120}{37.931}\left(20 + \frac{50+50}{3}\right)\times10^{-3}\\ &= 0.0479\ \text{p.u.} \end{aligned}$$

(iv) Per unit regulation at 0.8 pf lagging

$$\begin{aligned} \varepsilon &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi\\ &= 0.00502(0.8) + 0.0479(0.600) = 0.0327\ \text{p.u.} = 3.27\% \end{aligned}$$

Answer: core $d$ = 437 mm; window 305 × 838 mm; frame 1606 × 1854 × 371 mm; $\varepsilon_r$ = 0.0050, $\varepsilon_x$ = 0.0479 p.u.; regulation ≈ 3.27%.