single-phase transformer design

8/2/2019 Single-Phase Transformer Design

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3/9/20

Single-phaseTransformer DesignEE 181 Maintainability Engineering

1

Outline

Common criteria for design

Heating issues

Current density and heat transfer

Electrical aspect of design: E-I transformer

example

2

Real transformer criteria

Core and winding losses

Economy

Performance

Reliability

Environmental impact

Location of use: (V/Hz ratio)

3

Examples of Insulations

Organic: up to 105o C

Cotton

Wax paper

Polyester film

Varnish

Inorganic

Mica

Asbestos

Fiberglass

Silicon Varnish (most used) 4

Insulation Classes

Class Maximum allowable

temperature (C)

Maximum allowable

temperature (F)

A 105 221

B 130 266

F 155 311

H 180 356

5

Source: Standard NEMA Classifications (retrieved from engineeringtoolbox.com)

Heating effect on windings

Increase in heat = reduction of flux (because of

thermal motion interfering with alignment ofmagnetic moments)

Reduction of flux = loss in magnetism

Loss of magnetism = loss of torque (rotation)

and of course, electrical production

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Curie temperature

Also known as curie point

The temperature at which a ferromagnetic

material becomes paramagnetic; that is, loses

ALL magnetic ability (important for ferrite

core use)

7

Material Curie Temperature (C)

Iron 770

Cobalt 1130

Nickel 358

Oxidation

If metals insulation is insufficient, excessive

increase in temperature may cause oxidation

Oxidation = chemical reaction, breaking up of

molecular chains

End result: Rust

8

Managing heating effects

Electrically, most of the temperature drop is in

the form ofi2R losses in the iron core and

conductors

Dissipated heat must be transferred to the

cooling medium

Commonly used transformer cooling

mechanisms: air (natural cooling)

For larger units: Transformer oil (askarel)

H+ cooling (gaseous hydrogen), water9

Heat transfer coefficient

The amount of heat that can be removed from a

surface of areaA with a temperature difference T

Denoted by the symbol h , unit W/m2C

Typical range: 20 < h < 80 (air cooling)

For oil or water cooled (i.e. forced cooling), maximum

level ofh is around 250 W/m2C.

10

Current density (JC) and heat

transfer

Define currentdensity asampere/square meter

d


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By definition of current density (A/m2)

Substituting conductor area considered:

Therefore, power loss equation becomes:

13

AJI C

SC KdyJI

S

SCKdy

lKdyJP

2

Cancelling common terms leaves us with:

Or, using conductor volume:

Design considerations:

we want low resistivity/high conductivity

wires

l,y,d dependent on wirings/number of turns

14

SC KdylJP 2

CC VJP 2

Current density and surface area

Surface area, in m2:

Taking the power dissipated per surface area:

Simplifying:

15

ylAS 2

ly

KdylJ

A

P SC

S

L

2

2

22watts/m,

2SC

S

L KdJ

A

P

Equate this to heat transfer coefficient by

Heat Balance Equation:

LHS: heat as dissipated from copper wiring

RHS: heat that the cooling medium cantransfer16

match)(units22

m

WxC

Cm

WTh

ThKdJ SC 2

2

Wire ampacity

Wire ampacity can be chosen by rearranging

the heat balance equation and isolatingJC:

17

S

CKd

ThJ

2

Example:

Consider the following section of transformer where

the ambient temperature is 40 degrees C and the coiltemperature is currently 135 degrees C. The coils are

insulated by a material which provides a 15 C drop.

The resistivity of the coil material is 2.5 x 10-8-m.

The heat transfer coefficient is 25 W/m2C. If the coil

parameters are given to be 10 mm with a spacing

factor of 0.5, solve for the required ampacity of the

wire.

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Solution

Using the formula:

19

S

C Kd

Th

J

2

)5.0)(01.0)(105.2(

)80)(/25)(2(8

2

mmx

CCmWJC

80)1540(135 T

2A/m249.854,656,5CJ

2A/mm656.5CJ

Circular Mil and current

density 1 CM = area of a circular cross section with diameter =

1 mil.

1000 mil = 1inch

1 CM = (0.001)2/4 inches2

1 CM = (/4) x 10-6 in2

20

Converting Jc unit

Example: JC = 2.52 A/mm2

A higher value means greater area for the samecurrent, and thus less power loss.

21

AmmJC

/3952.01 2

2622

22

10)4/(

1*

)4.25(

in1*3952.0

inx

CM

mmA

mm

AmpCM/783

Design of 1 Transformer:

Electrical Aspect Faradays law:

Amperes law:

Also,

22

B

A

B

A

V

V

N

N

BBAA ININ

BA

By Faradays law of EM induction:

Assuming the flux density, B, to be sinusoidal

(since we assume the source is sinusoidal),

23

dt

dNV AA

dt

dBANV iAA

)sin()( max tBtB

Therefore,

The above expression for VA is in terms of peak

values, we take RMS equivalent as

We can now derive an expression for the core

area, Ai :

24

tBdt

tdBcos

)(max

tBANV iAA cosmax

2

maxBANV iAARMS

max

2

BN

VA

A

A

iRMS


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Laminates

E-I transformer

utilizes laminations;

thin sheets of E-Isections are stacked,

with each sheet

insulated.

Introduce stacking

factor Ki(usually

0.9 or higher)

25

Estimate of core area, Ai

Taking into account the stacking factor,

Parameter observations

Ai estimated core area, for a certain voltage VArms,

Bmax how much flux can be produced at a certainfrequency ; also, hysteresis loss is dependent on

NA number of turns => resistance

For same core area, Bmax decreases as NA increases;hysteresis and core losses are minimized at the cost ofincreasing resistance

26

iA

A

i KBN

V

ARMS

max

2

Window Area

Total MMF:

Also,

27

Aw, JcAABBAA INININ 2

Ai

SwCenclosed KAJI

Combining the two equations yields the

estimate ofwindow area:

Current density JC is calculated depending on

coil thickness; resulting estimate may be

recalculated (i.e. iteration) depending on

result of design28

AASwC INKAJ 2

SC

AAw

KJ

INA

2

Area-window product

Product of Ai and Aw estimates:

Area-window product a function of magnetic

capability of material, stack and space factors,

frequency, and power (VI product)29

iA

A

SC

AAwi

KBN

V

KJ

INAA rms

max

22

siC

ArmsAwi

KKBJ

VIAA

max

22

Design Criteria

Energy conservation: requires high quality core for

peak flux density Thickness: chosen for lower core loss

Less conductor current density = less winding losses

If transformer is operated at rated load: max efficiency

equates winding (Cu) and core loss

30

PCu increases with turns

PCore core loss goes down with dB/dt

(where dB/dt decreases with NA)

NA

Plosses


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Example: Transformer Design Design a single-phase transformer with the following given

requirements:

Voltage: 115 V pri / 24 V sec; Ks = 0.4, Ki = 0.9

60 Hz, 50 W; JC = 4 A/mm2 = 493 CM/Ampere

Total core loss (eddy + hysteresis) : 4.2 W/kg @ Bmax = 1.5 T

Core density = 7600 kg/m3, wire resistivity 2 x 10 -8-m

EI dimensions:

31

a3a

3a

3a

a/2

a/2

a/2

a/2

a/2

a/2

Solution

Calculate area-window product (core property):

32

siC

ArmsAwi

KKBJVIAAmax

22

)4.0(9.0)60(25.1104

5022

2

6 Tm

Ax

WAA wi

46m101736.0

xAAwi

From the E-I lamination figure:

Estimating using a square core area assumption,

Solving for the value ofa,

33

25.1)2/)(3( aaaAw

62 101736.05.1 xaAAA iwi

2

2

6

5.1

101736.0a

a

xAi

mmmmma 1944.18or01844.0

Window area is calculated as:

Core area estimate:

Stacking depth (adjusting for the rounded a)

34

222 5.541)19(5.15.1 mmmmaAw

2

2

6

5.3205.541

101736.0mm

mm

xAi

mmmm

mm

mm

a

Ai1786.16

19

5.320 2

Assuming a standard laminate (individual sheet)

thickness, Lt, of0.35 mm

35

thicknesslaminate

depthstackpiecesof#

piecesI-E497.480.35mm

17mmpiecesof#

Number of primary turns, NA (or NP)

Number of secondary turns:

36

ii

A

AKBA

VN RMS

max

2

)9.0()/377()5.1()10320(2)115(

26

sradTmx

V

997turns03.997 AN

24115

997xV

V

NN S

P

PS

turns209208.6 SN


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Core loss: consider core material density and volume

We need to calculate volume using E-I dimensions and

calculated value ofa. First we take total area of each

laminate

37

CuVolumemasscore Cu

3a3a

3a

a/2

a/2

a

a/2

a/2

A1A2

A3

A4

A5

Area calculations:

A1 = (3a)(a/2)=1.5a2 A2 = (3a)(a/2)=1.5a2

A3 = (3a)(a/2)=1.5a2 A4 = (3a)(a) = 3a2

A5 = (3a)(a/2) = 1.5a2

ATOTAL = A1+A2+A3+A4+A5 = 9a2

Total Volume = ATx (stack depth) x (stacking factor)

Multiply this by volume density to obtain mass:

38

)9.0()017.0()019.0(9 2 mmVT

)9.0()017.0()019.0(97600 23

mmm

kgMass

kg0.377masstotal

Core loss computation:

For copper loss, consider the amount of winding

involved:

l = 4 quarter arcs (1 circle) + straight wire segments

= (2*19/4) + 2(19+17)l= 101.84 mm (mean length per turn) 39

WkgkgWPCore 586.1377.0/2.4

windings

r

19mm

17mm

ra/2

Define copper loss as a function of volume, current

density, and resistivity: (conductor volume is the

window area multiplied by the mean length)

lKaJVJPSCCCCu

22

)(

2

)( 5.1

mmmAxmxPCu 10184.04.0019.05.1/10410222268

WPCu

05.7

%25.8505.7586.150

50

WWW

W

Wire sizing

Different for primary and secondary windings

Calculated using the following formula:

In the example:

41

sec/

*1

(CM)sizewirepri

out

V

P

J

115V

W50*)

Amp

CM(493.38CMpri

CM/Amp38.4934

112

A

mm

J

24V

W50*)

Amp

CM(493.38CMsec

#27AWGCM5.214CMpri

#20AWGCM027.871CMsec

Adjustments/Approximations

Maximum efficiency happens when core = copper loss

(or at least as close as possible; i.e. around 5% diff) Current density was done on a coil dimension

estimate (may vary with the actual temperatures

involved)

Core loss factor (Watts/kg)

Note that core loss is composed of eddy+hysteresis;

hysteresis loss is frequency dependent

Copper loss is calculated on mean length per turn l,

may also vary depending on operating temperature

(i.e. resistance variation with T) 42

single-phase transformer design

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