---
title: "Design: 150 kVA 11000/440 V transformer with tubes"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:design-150kva-11000-440-tubes"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-150kva-11000-440-tubes
site: BCT Engineering
---

# Design: 150 kVA 11000/440 V transformer with tubes

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2081 Bhadra, 7+7+4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-150kva-11000-440-tubes
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

The design parameters for a 150 kVA, 50 Hz, 11000/440 V, 3-phase, delta/star, core type, oil immersed natural cooled distribution transformer are given below. Constant for output voltage per turn = 0.45; Maximum flux density in the core = 1.35 Wb/m²; Current density in conductor = 2.5 A/mm²; Take 5% more current density in HV winding; Core type = three stepped; Window space factor = 0.25; Stacking factor = 0.9; Ratio of window height to width = 2.3; Take hot rolled steel and area of yoke is 20% greater than gross area of core; Width of LV winding = 20 mm; Width of HV winding = 25 mm; Width of duct between HV and LV winding = 15 mm; Mean height of coil = 231.25 mm; Resistivity of copper = 0.021 Ω-mm²/m; Height of tank = 1150 mm; Width of tank = 425 mm; Length of tank = 1050 mm; Total losses = 2850 W. Assuming all other required parameters, calculate: (i) Overall dimension of frame. (ii) Per unit voltage regulation at 0.85 pf. (iii) The minimum number of tubes of diameter 50 mm with average length of 1.05 m required for maintaining the mean temperature within the permissible limit. The mean temperature of the oil should not exceed 35°C. The rate of heat dissipation from plain wall is 6.5 and 6 W/m²-°C for convection and radiation respectively. The provision of tubes improves the rate of heat dissipation by 35%.

## Short answer

Frame: H=676.7 mm, W=825.5 mm, depth=157.6 mm. Core d=175.1 mm, three-stepped; window 158.9×365.4 mm; yoke depth 157.6 mm height 155.6 mm. Regulation and tube count not fully provided.

## Answer

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

```text
 +-----------------------------+  ---
 |          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):

| Item | Inside dia (mm) | Outside dia (mm) |
|---|---|---|
| LV winding | 185.1 | 225.1 |
| HV winding | 255.1 | 305.1 |

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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For this 150 kVA three-phase transformer, which relations size the core and window?
   A. E_t = K sqrt(Q) = 5.51 V; Q = 2.22 f B_m delta K_w A_w A_i x 10^-3
   B. E_t = K sqrt(Q) = 5.51 V; A_i = phi_m/B_m with phi_m = E_t/(4.44 f); Q = 3.33 f B_m delta K_w A_w A_i x 10^-3
   C. E_t = K Q = 67.5 V; Q = 3.33 f B_m delta K_w A_w A_i
   D. E_t = K sqrt(Q) with Q per phase; Q = 2.22 f B_m delta K_w A_w A_i
   Answer: B. E_t = 0.45 sqrt(150) = 5.511 V gives phi_m = 0.02483 Wb and A_i = 0.01839 m^2. The three phase output equation with 3.33 then gives the window area. (Key point: Three-phase core and window sizing)

2. Why is T_LV = 254/5.511 about 46 rather than 440/5.511?
   A. The LV winding is delta, so the phase voltage equals 440 V
   B. Turns are always based on 440/sqrt(2)
   C. The LV winding is star connected, so its phase voltage is 440/sqrt(3) = 254 V; HV delta phase voltage is the full 11000 V
   D. Because the core has three steps
   Answer: C. Turns are counted per phase winding. A star winding has phase voltage line/sqrt(3) = 254 V, whereas the delta HV winding has phase voltage 11000 V. (Key point: Phase voltages and turns for delta/star)

3. How is the per unit regulation of 4.67% at 0.85 pf obtained?
   A. er cos(phi) + ex sin(phi) = 0.01491(0.85) + 0.0647(0.527)
   B. er sin(phi) + ex cos(phi) = 0.01491(0.527) + 0.0647(0.85)
   C. er + ex = 0.01491 + 0.0647
   D. ex cos(phi) - er sin(phi)
   Answer: A. Regulation = er cos(phi) + ex sin(phi) with cos = 0.85 and sin = 0.527. This gives 0.0127 + 0.0341 = 0.0467, i.e. 4.67%. (Key point: Regulation formula and its components)

4. How is the number of tubes found?
   A. Divide the loss by 35 and by the tube area
   B. Tubes carry all the loss: n = P/(8.8 x 35 x tube area)
   C. Tank walls dissipate 12.5 S_t x 35; the remaining loss P - 12.5 S_t theta is dissipated by tubes at 8.8 W/m^2 C. Tube area = 4.434 m^2, so n = 4.434/0.1649 = 27
   D. The tank alone gives a rise of 67 C, so no tubes are needed
   Answer: C. Plain tank would reach 67.2 C, above 35 C. The balance of heat 2850/35 - 12.5 x 3.3925 is removed by tubes at 8.8 W/m^2 C, giving 4.434 m^2, or about 27 tubes of 0.1649 m^2 each. (Key point: Number of cooling tubes)

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