Designing a 1600kVA/35kVA, 50Hz Distribution Oil Transformer

General Information

Technical Specification

Input voltage 3 x 35000/20230V, star sine wave
Transformer output voltage

3 x 690/400V, star

Line output current

3 x 1340A, continuous operating mode

Frequency 50Hz
Average oil temperature 55°C
Max. temperature rise and/or max. Cu-winding losses at 75°C 25°K 18000W => 1.125%
Short-circuit voltage 6.5%
°Short-circuit voltage 6.5%
Max. core losses 3200W => 0.2%
Max. no-load current 1.3%
Test Voltage at 50Hz, 1 minute Primary 85kV, outside Secondary 4kV, inside
Steel & Core Assembly M5, annealed, strips for alternated stacking (4x45°+3x90° per shape), "round" cross section with 8 steps
Core Size Optimized for minimal material price for: Cu_Price/Fe_Price = 2 with Cu-winding

Creating Input

4 input screens are used to set the input parameters for designing a transformer:

  • Winding parameters per limb
  • Core
  • Environment
  • Other

and 3 screens for selection and set up of material:

  • wires
  • steels
  • cores.

Criteria and Parameters of Design

The design of a distribution transformer is always framed by 5 criteria which have to be put into effect simultaneously:

  • Short-circuit voltage
  • Winding losses at 75 °C
  • Winding temperature rise
  • Core losses
  • No-load output current

Under this condition the first step is the optimizing the core size to match the above mentioned prescribed design criteria for the optimal material price using some additional parameters such as:

  • Cooling media
  • Testing voltages
  • Steel quality and core assembly
  • Winding type and wire type & material
  • Cu/Al and Fe price relationship

Normally the user of this software will create an optimized core family for a typical design criteria and parameters and select a desired core per click.

In order to demonstrate the procedure for core optimization, note that the following parameters of optimization are a summery of 5-6 versions:

  • Max. winding losses at 75 °C = 18000W
  • Inductive short-circuit voltage = 6.4 %
  • Max. temperature rise 25 °K

    For 18000W @ 25 °K you need a very big cooling surface using the vertical and horizontal cooling channels in both windings.

    The optimal windings construction is presented in the next picture.

    Note that the secondary winding can be realized by using foil with 4 cooling channels within the winding and approx. 40% more Cu material for the outside primary winding.

  • Max. core losses = 3200 W
  • Max, no-load current = 1.3%

    These 2 criteria of design can be easy realized with annealed strips of M111 (M6) grain oriented steel at the induction 1.6T with the following shape and 8 steps "round" cross-section:

  • For 85kV, 50Hz, 1 minute and the power 1600kVA test voltage the following min. spacing is recommended:

Windings parameters per limb


The primary is created in star connection. The sine wave input voltage is 20230V.

There are no voltage harmonics and there is no duty cycle operation mode.

The primary will be manufactured with Cu-flat wire in disc winding technology (view picture above) with the horizontal cooling channel of h=5mm.

The advantage of the disc windings is low voltage per turn without any partial discharging problems.

In order to suppress the high line voltage discharge the turns of the first and last disc can easily add stronger insulation.

The following picture describes the manufacturing of a continuous disc winding:



The secondary winding is set inside. It is wound with 2 parallel connected "bifilar screw" strands (view picture above).

Between each turn there are horizontal cooling channels. h=5mm.

In order to avoid the circulating currents in parallel connected wires per strand you have to use the transposition through the rotation of the wire position in the strand in accordance with some rules:


The sine wave output voltage is 399V.

The rms output current is 1336Arms. There are no current harmonics:

Also, there is no duty cycle operation mode on the secondary.

With the eddy current losses factor (RacRdc) 1.4 the number of parallel connected flat wires per strand will be limited.

Note that at this point of the design you cannot prescribe the wire size.

You can select only the wire or family which the program must use in order to select the suitable wires for your application.



On this input screen you can:

  • select and manipulate the selected steel M97, 030mm (M5l)
  • set the operating induction (1.6T) and the frequency (50Hz)
  • select the core assembly
  • and prescribe the core selection out of an input file. This option will not be used because the core size has to be optimized.


The cooling medium is oil with the average temperature 55°C. The cooling surface of the core is increased by using 4 L-brackets on the core.

The minimum distance between the primary windings of 2 phases is 30mm.

There is no flange but both windings have to be fixed in order to suppress the axial forces during the short circuit operation mode.

There is no air in the transformer!




The selected criteria of the design and core optimization are the winding losses (18000W => 1.125%) at 75 °C and the inductive short-circuit voltage 6.4%.

If you prescribe also the temperature rise then the program has to use the criterion which is more critical: either the winding losses or the temperature rise with the prescribed short circuit voltage.

The core losses and the no-load input current can be manipulated only with steel quality, core assembly and induction

Core optimization

After you have set all input screens you need to select a core family and a core as template: 3 phase core family with 8 steps "round" cross section.

Core optimization

Click Core to open the input screen for reading the parameters of the selected core.

Core optimization

Click Optimize to optimize the core.

Core optimization

The yellow output fields are optimal results. Both other columns have a higher material price for 2%.

Here you can round off the core diameter (260mm instead 261.1) and click Create. This is the optimized core after the setting X = Y = 1050 (in order to use only 3 strip sizes per shape).

Core optimization


The first step of the presentation of the output screen is DIAGNOSIS: it is the summary of the most important calculated parameters of your transformer.


Note that the program uses the numerical calculation of the magnetic fields and the temperature rises.

Due to this technology the calculations of the eddy current losses, the steel losses, the short-circuit voltage, the circulating current and the transposition are very powerful.

The following picture shows the magnetic field outside&inside the core window.


Note that the criterion of design is the winding losses. With this criterion, the program optimizes the relationship of the primary and secondary losses.

Due to the higher eddy current losses in the secondary winding and better cooling of the primary winding the temperature rise of the secondary is higher than the temperature rise of the primary winding.

A very important detail is the max. oil temperature in the cooling channel (points 2 ,5, 7 and 10).


Finally here are 4 printed pages showing the design results.






The secondary winding (2 x Scr) is wound with 2 parallel strands. Each strand has 6 parallel flat wires.

The transposition (rotation) of the wires in these 2 strands has to be done after 1., 3., 5., 7., 9., 11., 13., 15., 17., 19., 21. and 23. turn. The horizontal cooling channel between these 2 strands is 5 mm.

The calculated number of the discs of the primary winding is 64 discs. In order to set the -5.0%, -2.5%, +2.5% and +5.0% taps for voltage regulation, the primary winding is normally cut in the middle.

At this point there should be a horizontal cooling channel 12-15 mm instead 5 mm.


Due to high voltage line discharge each turn in the 2 first and the 2 last discs have to be additionally insulated with approx 0.75 - 1.00mm one-side insulation.

For these two reasons, the number of the discs should be set to 62, wound as follows:

  • Discs 1&2 &61&62 => 10
  • 28&29&30&31&32&33&34&35 => 15 turns
  • Other =>20&21 turns

Nominal operating mode

Nominal operating mode

On this page you can check the prescribed parameter:

  • winding losses at 75 °C :0.99%<1.125%
  • short voltage:6.49% (instead of 6.4%)
  • core losses:2920W < 3200W
  • No-load current : 1.2% < 1.3%
  • Max temperature rise :24.8 °K < 25 °K
  • Max.radial tension in short-circuit: 18.22N/mm^2 < 60 N/mm^2
  • Max temp. rises during 4s in short-circuit:59.99°K

Test Mode

If you are not satisfied with the solution made by the program you can switch into the Test Mode and change your transformer by hand:

  • Turns 24.8
  • Wire size
  • Material (Cu or Al)
  • Number parallel connected wires and their order in strand
  • Cooling channels and insulations
  • Margin
  • Steel
  • Technology parameter (impregnation, gaps,...)

and then you can set it under an operation mode changing:

  • Input voltage
  • Frequency
  • Loads and their K-factors
  • Duty cycle of each winding
  • Ambient temperature
  • Air flow
Test Mode

In order to optimize the material costs you need to reduce the very high eddy current losses.

From a material costs point of view, here is a better version with secondary 2 x 12 flat wires 8mm x 2mm and primary wires 1mm x 8mm in only 48 discs.

Test Mode

Designing Water Cooled Inverter Filter Choke, 0.5mH, 600Arms, 60Hz

General about inverter filter choke

Fig.1 illustrates the main circuit diagram of a three-phase inverter.

The 3-phase mains Uin supplies the controlled rectifier R through the 3-phase commutation choke CC.

The DC voltage Udc is regulated by the rectifier and smoothed with the capacitor C.

The 3-phase AC voltage Uout is produced at the inverter outputs. The amplitude, frequency and form of this 3-phase AC voltage are regulated with the inverter and rectifier.

Designing Water Cooled Inverter Filter Choke, 0.5mH, 600Arms, 60Hz

The typical form of the inverter output voltage per phase is illustrated in Fig.2.

Designing Water Cooled Inverter Filter Choke, 0.5mH, 600Arms, 60Hz

At an inverter modulation frequency N*f, the output voltage Uout essentially consists of three components:

  • Fundamental frequency U with the frequency f (50Hz or 60Hz)
  • First harmonic U1=0.45*U with the frequency (N-1)*f
  • Second harmonic U2=0.45*U with the frequency (N+1)*f.

Accordingly, the current through the inverter filter choke Iout essentially comprises of 3 components:

  • Fundamental frequency I with the frequency f. This current is "impressed" and its amplitude depends on the inverter power
  • First harmonic I1 with the frequency (N-1) * f. The amplitude of this current depends on the voltage U1, the modulation frequency N*f and the inductance L of the inverter filter choke: I1=U1/(2* ¶*f*(N-1)*L)
  • Second harmonic I2 with the frequency (N+1) * f. The amplitude of this current depends on the voltage U2, the modulation frequency N*f and the inductance L of the inverter filter choke: I2=U2/(2* ¶*f*(N+1)*L).

About input parameters

Inductance at 1000Hz and 900Apeak 0.5mH
Inverter frequency


Modulation frequency (N=18)


RMS current at main frequency 600Arms at 60Hz
RMS currents at 1020Hz and 1140Hz 30Arms
Maximal total losses (warm) 10kW
Insulation clas H
Indirect water cooling

20°C incoming

55°C, maximal outcoming

1.5at max. pressure drop

1.5 m/s, max. speed

Test voltage winding-cooler, winding-core 4kV, 1minute, 60Hz

About indirect water cooling

The indirect water cooling of the three phase inverter filter choke will be ralized with accordance to the construction in the Fig.3.

About indirect water cooling

Note that the Small Chokes Program does not support water cooling. I selected this design example in order to transfer the know-how for water cooling design to my users.

Power per cooler in W and kcal/s

  1. Pc = K*Ptot/Nc
  2. Qc = Kd*Ptot/4180/Nc
  • Kd => Factor of the losses distribution between cooler and air (0.8-0.9)
  • Ptot => Choke total losses in W
  • Nc => Number of coolers
  • Pc => Power per cooler in W
  • Qc => Power per cooler in kcal/s

Amount of water in l/s

q = Qc/dTw

  • Qc => Power per cooler in kcal/s
  • dTw => maximal temperature rise of water in °K
  • q => Amount of water in l/s

Water speed in m/s

v = 10q/Apcs

  • q => Amount of water in l/s
  • Apcs => Cross section of pipe in cm2
  • v => Water speed in m/s (It must not be higher tan 1.5 m/s)

Equivalent pipe diameter

If the pipe cross section is rectangular then the equivalent pipe diameter can be calculated

Dk = 2ab/(a+b)

  • a, b => Sides of the rectangular pipe
  • Dk => Equivalent pipe diameter

Factor of the convection between water and pipe in W/°K/cm^2

α = 0.313v0.87Dk-0.13

  • V => Water speed in m/s
  • Dk => pipe diameter in m
  • Al => Factor of the convection in W/°K/cm2

Temperature drop between cooler and water in °K

dTcw = Pc/α/Apw

  • Pc = 1000Ptot/Nc => Losses per cooler in W
  • α => Factor of convection in W/°K/cm2
  • Apw => Contact surface cooler-water in cm2
  • dTcw => Temperature drop between the cooler and water in the pipe in °K

Temperature drop between the pipe and the cooler surface in °K

dTcc = 2 Pc Lcp/(Ac+Apw)/λc

  • Pc = => Losses per cooler in W
  • Lcp => Equivalent distance cooler surface-pipe in cm
  • Apw => Contact surface cooler-water in cm2
  • Ac => Surface between the cooler and winding in cm2
  • λc=> Thermal conductivity of cooler material (normally Al) 1.8 W/°K/cm
  • dTcc = Temperature drop between the pipe and cooler surface in °K

Temperature drop within the cooler insualtion in °K

dTci = Pc Lci/Ac/λci

  • Pc => Losses per cooler in W
  • Lci => Thickness of the cooler insulation cm
  • Ac => Surface between the cooler and winding in cm2
  • λci => Thermal conductivity of cooler insulation in W/°K/cm
  • dTci = Temperature drop within the cooler insulation in °K

Temperature drop within the wire insualtion in °K

dTwi = Pc Lwi/Ac/λwi

  • Pc => Losses per cooler in W
  • Lwi => Thickness of the wire insulation cm
  • Ac => Surface between the cooler and winding in cm2
  • λwi => Thermal conductivity of wire insulation in W/°K/cm
  • dTwi = Temperature drop within the wire insulation in °K

Temperature drop within the winding in °K

dTw = Pc Lw/Aw/λw/16

  • Pc = => Losses per cooler in W
  • Lw => Average turn length in cm
  • Aw => Winding cross section (all turns) in cm2
  • λw=> Thermal conductivity of winding material (normally Cu) 3.5 W/°K/cm
  • dTw = Temperature drop within the winding in °K

Water pressure drop in the pipe in at

dp = 0.01 K v2 Lt/Dk

  • Lp => Pipe length in cm
  • Dk => Pipe diameter in cm
  • K => Factor
Dk (m) 0.0.004 0.006 0.006 0.007 0.008 0.009 0.010 0.011 0.012
K 0.094 0.087 0.083 0.080 0.075 0.073 0.070 0.068 0.066

Provisional "core selection"

Before you start coke design you need approximately to know how big has to be the cooler.

Due to the fact that the cooler fits tightly over the leg along the window height its cooling surface to the winding is approximately equal with the leg surface along the window height.

Pc = Kd Ptot/Nc = 0.9x10000/6 = 1500 W

Qc = Pc/4180 = 0.358 kcal

dTw = 3.58°K

q = Qc/dTw = 0.358/3.58 = 0.1 l/s

Provisional core selection

Winding teperature rise will be 2+41+40+15 +9.2 + 42 =149.2°K. Due to the fact that the winding losses are lower than 3000W (view the design results) the temperature rise will be approx. 20% lower ( 125°K instead 149°K)

This provisional thermal design of the cooler shows that the core with the leg sizes 3" x 12" is, from thermal point of view, big enough.

General design rules

  • The cooler size is H=12", W=3" T=1.2". In order to avoid any gaps between the cooler and the winding the contact form has to be oval (see the Fig3). The cooler has to be coated with 1.5mm resin from Huntsman XB 2710+XB2711. This resign has very high thermal conductivity (0.015W/°K/cm) and relative high break down voltage (>20kV/mm).
  • Due to the fact that the water cooled are small the amperturns are very high and the gap size very big. For a "good" gap of approx <0.1" the choke has to be made by more than 10 gaps per leg and with the high induction.
  • The criterion of the design is Q-factor:at the max. temperature 165°C: Qf = Lω/Irms2/Ptot/Kd/(Nc/2) = 0.0005x376/6002/10000/3/0.9 >24 =2
  • The insulation in the gaps and around the gaps has to be class H.
  • From thermal point of view the optimal construction is one layer winding with Cu rectangular wire
  • Design page 1

    Design page 1

    Design page 2

    Design page 2

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