Transformer Design

The output equation relates the kVA rating of a transformer to its main dimensions (net core area $A_i$ and window area $A_w$) through the chosen magnetic loading $B_m$ and electric loading $\delta$. It is the starting point of transformer design.

Notation

• $f$ = frequency (Hz), $B_m$ = maximum flux density (Wb/m²), $\phi_m = B_mA_i$
• $A_i$ = net iron area of limb (m²), $A_w$ = window area (m²)
• $K_w$ = window space factor = copper area / window area
• $\delta$ = current density (A/m²), $T$ = turns, $I$ = current
• Subscripts 1, 2 for primary and secondary

Step 1: Voltage

EMF per turn and terminal voltage (neglecting drops):

$$E_t = 4.44 f\phi_m = 4.44fB_mA_i, \qquad V \approx E = 4.44fB_mA_iT$$

Step 2: Window (copper) area

In a single-phase core type transformer the window holds one primary and one secondary winding (half of each on each limb, but a full set passes through the window).

$$\begin{aligned} a_1 &= \frac{I_1}{\delta},\quad a_2 = \frac{I_2}{\delta}\\ \text{Copper area in window} &= T_1a_1 + T_2a_2 = \frac{T_1I_1 + T_2I_2}{\delta} = \frac{2AT}{\delta} \end{aligned}$$

since $T_1I_1 = T_2I_2 = AT$. This copper area is also $K_wA_w$, so

$$K_wA_w = \frac{2AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{2}$$

Step 3: Output

$$\begin{aligned} Q &= V_1I_1\times10^{-3} = 4.44fB_mA_iT_1I_1\times10^{-3}\\ &= 4.44fB_mA_i \cdot \frac{K_wA_w\delta}{2}\times10^{-3}\\ Q &= 2.22\,f\,B_m\,\delta\,K_w\,A_w\,A_i\times10^{-3}\ \text{kVA} \end{aligned}$$

Remarks

• The same equation applies to a single-phase shell type transformer, because there too the window carries one full set of windings.
• For a given rating, higher $B_m$ or $\delta$ gives a smaller core and window, but increases iron or copper losses and temperature rise.
• In design, $E_t = K\sqrt{Q}$ is used first to get $A_i$; the output equation then gives $A_w$.

In a core type transformer the windings surround the limbs of the core; in a shell type transformer the core surrounds the windings (the windings sit on the central limb, and the flux divides into two outer limbs).

   Core type (1-ph)          Shell type (1-ph)
   +--------------+          +---+------+---+
   | ##        ## |          |   |##  ##|   |
   |[##]      [##]|          |   |##  ##|   |
   |[##]      [##]|          |   |##  ##|   |
   | ##        ## |          |   |##  ##|   |
   +--------------+          +---+------+---+
  windings on 2 limbs     windings on centre limb

In practice the core type is preferred for high voltages because insulating concentric cylindrical windings from the core is easier.

A power transformer is used in generating stations and transmission substations to step voltages up or down at high power, and runs near full load most of the time. A distribution transformer feeds consumers at the final voltage (for example 11 kV/400 V) and its load varies widely through the day.

So a distribution transformer is designed with smaller iron loss and better regulation, while a power transformer is designed for high full-load efficiency and to withstand large fault levels.

Leakage reactance arises from the flux that links only the primary or only the secondary winding. For a core type transformer with concentric cylindrical windings it can be estimated as follows.

Assumptions

1. Primary and secondary windings have the same axial length $L_c$.
2. Leakage flux paths in the windings and the duct are parallel to the axis of the core.
3. The reluctance of the leakage path is that of the axial path of length $L_c$ only; the iron and the return path have negligible reluctance.
4. Ampere-turns of the two windings are equal ($I_1T_1 = I_2T_2 = AT$), and the mmf rises linearly across each winding.
5. Half the leakage flux in the duct links each winding.
6. Magnetising current is neglected.

 core | LV (b1) | duct (a) |  HV (b2)  |
      |<------->|<-------->|<--------->|
 mmf  0  /      AT  -----  AT  \      0
        /                       \

Derivation

Let $L_{mt}$ be the mean length of turn of both windings.

Inside the LV winding: at distance $x$ from its inner edge, mmf $= AT\,x/b_1$. Flux in strip $dx$:

$$d\phi_x = \mu_0\frac{AT\,x}{b_1}\cdot\frac{L_{mt}\,dx}{L_c}$$

This flux links $T_1x/b_1$ turns, so the linkage is

$$\psi_1 = \int_0^{b_1}\mu_0\frac{AT\,x}{b_1}\frac{L_{mt}}{L_c}\cdot\frac{T_1x}{b_1}\,dx = \mu_0\,AT\,T_1\frac{L_{mt}}{L_c}\cdot\frac{b_1}{3}$$

In the duct (width $a$): mmf $= AT$, flux $= \mu_0AT\,L_{mt}a/L_c$; half of it links the primary, giving $\mu_0AT\,T_1\frac{L_{mt}}{L_c}\frac{a}{2}$.

Total primary leakage linkage and reactance:

$$\begin{aligned} \psi_1 &= \mu_0AT\,T_1\frac{L_{mt}}{L_c}\left(\frac{b_1}{3} + \frac{a}{2}\right)\\ x_1 &= 2\pi f\mu_0T_1^2\frac{L_{mt}}{L_c}\left(\frac{b_1}{3} + \frac{a}{2}\right) \end{aligned}$$

Similarly, referred to the primary, $x_2' = 2\pi f\mu_0T_1^2\frac{L_{mt}}{L_c}\left(\frac{b_2}{3} + \frac{a}{2}\right)$.

Total leakage reactance referred to primary:

$$X_p = 2\pi f\mu_0T_p^2\frac{L_{mt}}{L_c}\left(a + \frac{b_1 + b_2}{3}\right)$$

Per-unit form

With $I_pT_p = AT$ and $V_p = E_tT_p$:

$$\varepsilon_x = \frac{I_pX_p}{V_p} = 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1 + b_2}{3}\right)$$

Conclusions

• Reactance rises with $T^2$, $L_{mt}$, duct width and winding thickness.
• It falls with a longer winding ($L_c$), so tall narrow windows give low reactance.
• It can be reduced by sandwiching (splitting) windings.

The output equation gives the kVA rating of a three-phase core type transformer in terms of the net core area $A_i$, window area $A_w$, flux density $B_m$ and current density $\delta$.

   +-------+-------+-------+
   |       |       |       |
  [L|H] W1 [H|L|H] W2 [H|L]
   |  R    |  Y    |  B    |
   +-------+-------+-------+
  3 limbs, 2 windows; each window holds
  LV + HV of two different phases

Step 1: Voltage per phase

$$E_t = 4.44f\phi_m = 4.44fB_mA_i, \qquad V_p \approx 4.44fB_mA_iT_p$$

Step 2: Copper in a window

A three-phase core type transformer has three limbs and two windows. Each window contains half the windings of two phases, that is, one complete primary and one complete secondary winding of two limbs facing each other:

$$\text{Copper area in a window} = 2(a_pT_p + a_sT_s)$$

With $a_p = I_p/\delta$, $a_s = I_s/\delta$ and $I_pT_p = I_sT_s = AT$:

$$K_wA_w = 2\left(\frac{AT}{\delta} + \frac{AT}{\delta}\right) = \frac{4AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{4}$$

Step 3: Output

$$\begin{aligned} Q &= 3V_pI_p\times10^{-3} = 3\times4.44fB_mA_iT_pI_p\times10^{-3}\\ &= 3\times4.44fB_mA_i\times\frac{K_wA_w\delta}{4}\times10^{-3}\\ Q &= 3.33\,f\,B_m\,\delta\,K_w\,A_w\,A_i\times10^{-3}\ \text{kVA} \end{aligned}$$

where $Q$ is the total three-phase rating in kVA.

Use in design

• First, $E_t = K\sqrt{Q}$ (with $K \approx 0.45$ for distribution and 0.6–0.7 for power transformers) gives $\phi_m$ and $A_i = \phi_m/B_m$.
• Then the output equation gives $A_w$, from which window width and height are found with a chosen $H_w/W_w$ ratio (2 to 4).

Why maximum efficiency at less than full load

A distribution transformer stays energised for 24 hours but its load changes through the day and is often light (night hours). So the important figure is all-day (energy) efficiency:

$$\eta_{all\ day} = \frac{\text{kWh output in 24 h}}{\text{kWh output} + \text{kWh losses in 24 h}}$$

• Iron loss $P_i$ occurs for all 24 hours, whatever the load.
• Copper loss $P_c$ occurs only in proportion to (load)², and is large only for a few peak hours.

Maximum efficiency occurs when copper loss = iron loss, i.e. at load fraction $x = \sqrt{P_i/P_{c,FL}}$. By keeping $P_i$ small (lower $B_m$, good-quality steel) the maximum efficiency is placed at about 50–70% of full load, where the transformer runs most of the time. This minimises total energy loss in a day.

Number of cooling tubes

Let

• $P_i + P_c$ = total full-load loss (W)
• $S_t$ = dissipating surface of plain tank walls (m²)
• $\theta$ = permissible mean temperature rise of oil (°C)

A plain tank wall dissipates by radiation ≈ 6 W/m²°C and by convection ≈ 6.5 W/m²°C, total 12.5 W/m²°C:

$$\theta = \frac{P_i + P_c}{12.5\,S_t}$$

If this $\theta$ is too high, tubes are added. Let the tube area be $xS_t$. Tubes do not add to radiation much (they screen each other), but they improve convection by about 35%:

$$6.5 \times 1.35 \approx 8.8\ \text{W/m}^2{}^\circ\text{C}$$

Heat dissipated with tubes:

$$\begin{aligned} P_i + P_c &= \theta\left[12.5\,S_t + 8.8\,xS_t\right]\\ xS_t &= \frac{1}{8.8}\left(\frac{P_i + P_c}{\theta} - 12.5\,S_t\right) \end{aligned}$$

If each tube has diameter $d_t$ and length $l_t$, its surface is $\pi d_tl_t$, so

$$n_t = \frac{xS_t}{\pi d_tl_t} = \frac{1}{8.8\,\pi d_tl_t}\left(\frac{P_i + P_c}{\theta} - 12.5\,S_t\right)$$

Round $n_t$ up to the next whole number. Usually $\theta$ = 35 °C (mean oil rise), tube diameter 50 mm and spacing 75 mm.

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).

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%.

Output equation of a 3-phase core type transformer

Let $A_i$ = net core area, $A_w$ = window area, $B_m$ = maximum flux density, $\delta$ = current density, $K_w$ = window space factor, $T$ = turns per phase, $I$ = phase current.

EMF per phase:

$$V \approx E = 4.44fB_mA_iT$$

There are three limbs and two windows. Each window holds one primary and one secondary of two adjacent limbs, so the copper area in one window is

$$K_wA_w = 2\left(\frac{I_pT_p}{\delta} + \frac{I_sT_s}{\delta}\right) = \frac{4AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{4}$$

Output:

$$\begin{aligned} Q &= 3VI\times10^{-3} = 3\times4.44fB_mA_i\,(IT)\times10^{-3}\\ &= 3\times4.44fB_mA_i\times\frac{K_wA_w\delta}{4}\times10^{-3}\\ Q &= 3.33\,fB_m\delta K_wA_wA_i\times10^{-3}\ \text{kVA} \end{aligned}$$

Why windings are made circular

1. Short-circuit forces: under a short circuit the radial forces act outward on the HV and inward on the LV winding. A circular coil takes them as uniform hoop stress and keeps its shape; a rectangular coil would bend at the flat sides.
2. Least copper: for a given enclosed area, a circle has the smallest perimeter, so the mean length of turn, copper weight, cost and $I^2R$ loss are least.
3. Ease of winding: circular coils are easily wound on formers on a lathe, with uniform tension.
4. Uniform insulation and cooling: clearances between core, LV and HV are uniform all round, and oil ducts are easy to provide.

Because the winding is circular, the core is made stepped (cruciform or multi-stepped) to fill the circle as fully as possible.

(i) Maximum efficiency of a distribution transformer at loads much smaller than full load

• A distribution transformer is connected to the supply for all 24 hours, so its iron loss is present all the time.
• Its load varies widely: high only for a few peak hours, light at night and during much of the day. The average load is often only 50–70% of the rating.
• Copper loss varies with the square of load, so it is large only during the short peak periods.
• Efficiency is maximum when copper loss equals iron loss, at load fraction

$$x = \sqrt{\frac{P_i}{P_{c,FL}}}$$

• The useful measure is all-day efficiency (kWh output / kWh input over 24 h). To make it high, the iron loss is kept small by using a lower flux density and good grain-oriented steel, while a higher copper loss at full load is accepted.
• So the designer makes $P_i$ about one-third to one-half of $P_{c,FL}$, which puts maximum efficiency at roughly 50–70% of full load, matching the actual load pattern. Energy loss and running cost over the day are then minimum.

Power transformers, in contrast, run near full load and are designed for maximum efficiency near full load.

(ii) LV winding near the core and HV winding outside

  core | ins | LV | duct | HV |
       |thin |    | thick|    |

1. Insulation saving: the core is at earth potential. Placing LV next to it needs only thin insulation between core and LV. If HV were placed next to the core, thick insulation for full HV would be needed there, increasing the core-winding gap, the mean turn length and cost.
2. Less leakage and copper: with the smaller gap, the core diameter and winding diameters stay small, giving shorter mean turns, less copper and lower leakage reactance.
3. Tappings and repair: tappings for voltage control are taken on the HV winding. On the outside they are easy to bring out, and the HV winding is easy to inspect and repair.
4. Lower current: HV winding carries smaller current with thin conductors, which are easier to wind on the outside on a large diameter.

In a core type transformer the LV winding is placed next to the core, and the HV winding is wound concentrically over it, mainly to save insulation.

  core | thin ins | LV | duct | HV

Reasons

1. Insulation: the core is earthed. The insulation needed between a winding and the core depends on the winding voltage. LV next to the core needs only thin insulation. If HV were next to the core, a thick insulation barrier for the full HV would be needed there as well as between the windings.
2. Smaller size and cost: less insulation near the core keeps the winding diameters small. Mean length of turn, copper weight, $I^2R$ loss and cost are reduced.
3. Lower leakage reactance: a smaller radial gap between core and windings means less leakage flux.
4. Tappings and maintenance: tappings for voltage control are put on the HV winding (it carries less current). With HV outside, tappings are easy to bring out and the winding is easy to reach for repair.

So the order core → LV → HV gives the most economical and practical design.

Why windings are circular

• Under short circuit, radial forces on a circular coil produce uniform hoop stress, so the coil keeps its shape. Rectangular coils would deform.
• For a given area enclosed, a circle has the smallest perimeter, so the mean length of turn, copper weight and copper loss are minimum.
• Circular coils are easy to wind on formers and give uniform clearances and oil ducts.

Advantages of stepped cores

A square core inside a circular coil leaves much space unused. A stepped core fills the circle better.

• For the same circumscribing diameter $d$, the net iron area is larger (square $0.45d^2$, cruciform $0.56d^2$, 3-step $0.60d^2$).
• For the same iron area, the diameter $d$ and so the mean turn of the windings are smaller, saving copper and copper loss.
• Less space wasted, so a smaller, cheaper transformer with lower leakage reactance.

Most economical dimensions of a two-stepped (cruciform) core

          <---- a ---->
          +-----------+
       +--+-----------+--+  ^
       |  |           |  |  b
       +--+-----------+--+  v
          +-----------+
     circle of diameter d

Let $a$ = width of the larger stamping, $b$ = width of the smaller one, and $\theta$ the angle that the diagonal makes with the side $a$. Then

$$a = d\cos\theta,\qquad b = d\sin\theta$$

Gross core area (two rectangles $a\times b$, minus the common square $b^2$):

$$\begin{aligned} A_{gi} &= 2ab - b^2 = 2d^2\sin\theta\cos\theta - d^2\sin^2\theta\\ &= d^2\left(\sin2\theta - \sin^2\theta\right) \end{aligned}$$

For maximum area:

$$\begin{aligned} \frac{dA_{gi}}{d\theta} &= d^2(2\cos2\theta - \sin2\theta) = 0\\ \tan2\theta &= 2 \;\Rightarrow\; \theta = 31.72^\circ \end{aligned}$$

So

$$\begin{aligned} a &= d\cos31.72^\circ = 0.85\,d\\ b &= d\sin31.72^\circ = 0.53\,d\\ A_{gi,max} &= d^2(\sin63.43^\circ - \sin^2 31.72^\circ) = 0.618\,d^2 \end{aligned}$$

With stacking factor 0.9, net iron area $A_i = 0.9 \times 0.618d^2 \approx 0.56\,d^2$. The cruciform core uses 0.618/0.785 ≈ 79% of the circle area, against 64% for a square core.

A transformer has minimum total cost of active material (iron + copper) when the cost of iron equals the cost of copper.

Derivation

Let

• $\phi_m$ = maximum flux, $AT$ = ampere-turns per limb
• $c_i$, $c_c$ = cost per kg of iron and copper
• $\rho_i$, $\rho_c$ = densities; $l_i$ = mean length of flux path; $L_{mt}$ = mean length of turn
• $B_m$, $\delta$ = flux density and current density (fixed by losses and heating)

For a single-phase transformer:

$$Q = 4.44f\phi_m\,(IT)\times10^{-3} = 4.44f\,\phi_m\,AT\times10^{-3}$$

so for a given rating the product $\phi_m \cdot AT$ is constant:

$$\phi_m\cdot AT = \frac{Q\times10^3}{4.44f} = \text{constant} = C$$

Weight of iron $G_i = \rho_il_iA_i = \rho_il_i\dfrac{\phi_m}{B_m}$.

Copper area (both windings) $= \dfrac{2AT}{\delta}$, so weight of copper $G_c = \rho_cL_{mt}\dfrac{2AT}{\delta}$.

Total cost:

$$C_t = c_iG_i + c_cG_c = \underbrace{\frac{c_i\rho_il_i}{B_m}}_{k_1}\phi_m + \underbrace{\frac{2c_c\rho_cL_{mt}}{\delta}}_{k_2}AT$$

Put $AT = C/\phi_m$:

$$C_t = k_1\phi_m + \frac{k_2C}{\phi_m}$$

For minimum cost:

$$\begin{aligned} \frac{dC_t}{d\phi_m} &= k_1 - \frac{k_2C}{\phi_m^2} = 0\\ k_1\phi_m &= \frac{k_2C}{\phi_m} = k_2\,AT \end{aligned}$$

that is,

$$c_iG_i = c_cG_c \quad\Longrightarrow\quad \text{cost of iron} = \text{cost of copper}$$

Optimum ratio of magnetic to electric loading

From $k_1\phi_m = k_2AT$:

$$r = \frac{\phi_m}{AT} = \frac{k_2}{k_1} = \frac{2c_c\rho_cL_{mt}B_m}{c_i\rho_il_i\delta}$$

This ratio fixes the voltage per turn, $E_t = K\sqrt{Q}$ with $K = \sqrt{4.44fr\times10^3}$.

Notes

• In the same way, minimum total loss (maximum efficiency) at a load occurs when iron loss = copper loss.
• In practice designers use the "minimum cost" ratio only as a guide, since loss capitalisation, regulation and cooling also matter.

The emf per turn is fixed by the ratio of magnetic loading to electric loading, which designers keep nearly constant for a given type of transformer.

Proof (single-phase)

Rating in kVA:

$$Q = VI\times10^{-3} = E_tTI\times10^{-3} = E_t\,AT\times10^{-3}$$

where $AT = IT$ (ampere-turns per winding) and $E_t = 4.44f\phi_m$.

Let the ratio of magnetic to electric loading be

$$r = \frac{\phi_m}{AT} \;\Rightarrow\; AT = \frac{\phi_m}{r}$$

Then

$$\begin{aligned} Q &= 4.44f\phi_m\cdot\frac{\phi_m}{r}\times10^{-3} = \frac{4.44f\phi_m^2}{r}\times10^{-3}\\ \phi_m &= \sqrt{\frac{rQ\times10^3}{4.44f}} \end{aligned}$$

Hence

$$\begin{aligned} E_t &= 4.44f\phi_m = 4.44f\sqrt{\frac{rQ\times10^3}{4.44f}} = \sqrt{4.44f\,r\times10^3}\,\sqrt{Q}\\ E_t &= K\sqrt{Q},\qquad K = \sqrt{4.44f\,r\times10^3} \end{aligned}$$

For a given frequency and a fixed ratio $r$ (chosen for minimum cost), $K$ is constant.

Typical values of K

The maximum flux density $B_m$ in the core is chosen as high as possible to reduce the size and cost of the core, but it is limited by iron loss, magnetising current, noise and the type of steel.

Factors deciding $B_m$

1. Core loss and efficiency: hysteresis loss ∝ $B_m^{1.6}$ and eddy loss ∝ $B_m^2$. Higher $B_m$ means more iron loss and lower all-day efficiency.
2. Magnetising current: near the knee of the B–H curve the mmf rises sharply, so $I_0$ and its harmonics increase.
3. Type of transformer: distribution transformers are energised 24 h, so a lower $B_m$ is used to keep iron loss low. Power transformers can use higher values.
4. Grade of steel: cold-rolled grain-oriented (CRGO) steel has low loss and high permeability along the grain, so it allows higher $B_m$ than hot-rolled steel.
5. Over-voltage and frequency: margin is kept so the core does not saturate at overvoltage or reduced frequency.
6. Noise: magnetostriction hum increases with $B_m$.

Usual values

So $B_m$ is a compromise between a small core (high $B_m$) and low iron loss, low no-load current and low noise (low $B_m$).

Per-unit resistance drop is the full-load resistive voltage drop expressed as a fraction of rated voltage, $\varepsilon_r = I_pR_p/V_p$. It can be found from winding dimensions as below.

Notation

• $T_p$, $T_s$ = primary and secondary turns; $I_p$, $I_s$ = currents
• $L_{mtp}$, $L_{mts}$ = mean lengths of turn
• $a_p$, $a_s$ = conductor areas; $\delta_p$, $\delta_s$ = current densities
• $\rho$ = resistivity of copper; $E_t$ = volts per turn

Derivation

Resistance of each winding:

$$r_p = \frac{\rho T_pL_{mtp}}{a_p},\qquad r_s = \frac{\rho T_sL_{mts}}{a_s}$$

Total resistance referred to primary:

$$R_p = r_p + r_s\left(\frac{T_p}{T_s}\right)^2$$

Per-unit resistance:

$$\varepsilon_r = \frac{I_pR_p}{V_p} = \frac{I_pr_p}{V_p} + \frac{I_pr_s(T_p/T_s)^2}{V_p}$$

Since $I_pT_p = I_sT_s$ and $V_p/T_p = V_s/T_s = E_t$, the second term equals $I_sr_s/V_s$. So

$$\varepsilon_r = \frac{I_pr_p}{V_p} + \frac{I_sr_s}{V_s}$$

Now $I_pr_p = I_p\dfrac{\rho T_pL_{mtp}}{a_p} = \rho\,\delta_p\,T_pL_{mtp}$ (as $I_p/a_p = \delta_p$), and $V_p = E_tT_p$:

$$\frac{I_pr_p}{V_p} = \frac{\rho\,\delta_pL_{mtp}}{E_t},\qquad \frac{I_sr_s}{V_s} = \frac{\rho\,\delta_sL_{mts}}{E_t}$$

Therefore

$$\varepsilon_r = \frac{\rho\left(\delta_pL_{mtp} + \delta_sL_{mts}\right)}{E_t}$$

If both windings have the same current density $\delta$ and a common mean turn $L_{mt}$:

$$\varepsilon_r = \frac{2\rho\,\delta\,L_{mt}}{E_t}$$

Other forms

Multiplying numerator and denominator by $I_pT_p$:

$$\varepsilon_r = \frac{I_p^2R_p}{V_pI_p} = \frac{\text{full-load copper loss (per phase)}}{\text{VA rating (per phase)}}$$

Example

With $\rho$ = 0.021 Ω-mm²/m, $\delta$ = 2.5 A/mm², $L_{mt}$ = 0.9 m and $E_t$ = 7.5 V:

$$\varepsilon_r = \frac{2\times0.021\times2.5\times0.9}{7.5} = 0.0126\ \text{p.u.} = 1.26\%$$

Conclusions

• $\varepsilon_r$ falls if $E_t$ is high (fewer turns) or $\delta$ is low.
• A short mean turn (stepped core, compact windings) reduces both copper loss and $\varepsilon_r$.

Core design fixes the net iron area, the shape of the limb section, the stampings, and the yoke, so that the flux $\phi_m$ is carried at the chosen flux density with low loss and low cost.

1. Net iron area

From $E_t = K\sqrt{Q}$:

$$\phi_m = \frac{E_t}{4.44f},\qquad A_i = \frac{\phi_m}{B_m}$$

$B_m$ is chosen according to the steel: about 1.1–1.45 Wb/m² for hot-rolled, 1.35–1.75 Wb/m² for CRGO.

2. Shape of limb section

Windings are circular, so the limb is made stepped to fill the circle of diameter $d$:

(stacking factor 0.9). More steps use the circle better and shorten the mean turn, but increase labour. Square and cruciform cores are used for small transformers; 3 to 7 steps for large ones.

3. Stampings and stacking

• Thin silicon-steel laminations (0.35 mm or less), insulated by varnish or oxide, to reduce eddy loss.
• Stacking factor (net/gross area) ≈ 0.9.
• Joints are staggered (overlapped) or mitred at 45° (for CRGO) to reduce reluctance and loss at the corners.
• Large cores have cooling ducts between packets.

4. Yoke

• With hot-rolled steel, yoke area is made 15–20% larger than limb area to lower yoke flux density, loss and magnetising current.
• Yoke is usually rectangular: depth $D_y = a$, height $H_y = A_{gy}/D_y$.
• With CRGO, yoke area is equal to limb area (stepped yoke).

5. Window

The output equation gives the window area, $A_w = Q/(2.22 \text{ or } 3.33\,fB_m\delta K_wA_i\times10^{-3})$, with $H_w/W_w$ = 2 to 4. The core frame then follows: $D = d + W_w$, $H = H_w + 2H_y$, $W = 2D + a$ (3-phase) or $D + a$ (1-phase).