electrocomp final

8/3/2019 ElectroComp Final

1/28

FET Tuzla

Numerical Calculation of Magnetic DissipationNumerical Calculation of Magnetic Dissipationat Power Transformersat Power Transformers

Prof. Vlado MadareviProf. Alija Muharemovi

Dr. Amir NuhanoviMr. Hidajet Salki

Florida 16 18 March 2005.

ElectroComp - Orlando,


2/28

FET Tuzla

Motivation: Transformer Design

Better efficiencyLess dissipation = increased efficiency

Reliability Less heat dissipation Increased device duration

Quality of service Better voltage stability of secondary winding

Cost

Problem formulation: find dissipative inductance fora given transformer


3/28

FET Tuzla

Possible Approaches To The Problem

Analytical Hard

Experimental Expensive

Numerical is our choice Input - transformer geometry [+ regime] Output

Magnetic field at each point in space

Dissipative inductance, energy


4/28

FET Tuzla

Conceptual View of Dissipation

F F =i g-

produced - useful = dissipative

F g

F i


5/28

FET Tuzla

Core

PrimarySecondary

Winding

UsefulDissipated

Flux

Dissipation in PracticalTransformers

Insulation


6/28

FET Tuzla

Short Circuit Setup

Core

Insulation

Secondary current setup so that useful flux = 0.

Enables measurement of dissipative flux


7/28

FET Tuzla

Energy Method - Outline

Given prim. and sec. currents

Find cross section current density [J]

Find magnetic vector potential [A]

Find energy stored in dissipative field [ ]m

W

Find dissipative inductance [L]

Step 1

Step 2

Step 3


8/28

FET Tuzla

Energy Stored in Dissipative Magnetic Field

Energy formula

Current density [ ] Magnetic vector potential [ ]

Magnetic field strength [ ] Magnetic induction [ ]

ArJ

r

Br H

r

( )rot A = Br r

( )D

rot H = J +t

rr r

[layered transformer design]

m

(V)

1W = H B dV

2 r r

m

(V)

1W = J A dV

2

rr


9/28

FET TuzlaEnergy Method for Computing Dissipative

Magnetic Field Energy

m(V)

1

W = J A dV2

rr

m(S)

1

W ' = J A dS2

Solve using

method

of finite moments

Problem is

2 - dimensional

( )D

rot H = J +t

rr r

( )rot A = Hr r

Step2: Using A and J perform numeric integration.

Step1: Using current density [J] find the magnetic vector potential [A].

Step3: use formulam

2

2W 'L =

I

( )( )-1rot rot A = Jr r


10/28

FET TuzlaStep 1: Method of Finite Elements - Outline

=

Ferromag.ConductorInsulator

Construct local equation for each element

Solve the set of equations globally

Use iterative procedures because of complexity [FEM 2D]

( )( )-1rot rot A = Jr r


11/28

FET Tuzla

Step 2: Numerical Integration

m

(S)

1W ' = J A dS

2 Need to Find

There are 4 possible values of J Inside ferromagnetic material = 0 Inside insulator = 0 Inside primary windingInside secondary winding

z

p

i= ii,a i,b i,c

m i

i= i

A + A + AJW ' = S

2 3

For each value of J compute

iS

i,cA

i,bAi,aA

element inside

primary winding


12/28

FET Tuzla

Method of Linked Fluxes - Outline

Given prim. and sec. currents

Find cross section current density [J]

Find magnetic induction [B]

Find dissipative flux [ ]

Find dissipative inductance [L]

Step 1

Step 2

Step 3


13/28

FET TuzlaLinked Fluxes

1F

2F

3F

[ ]1 1 2 2 3 3-d n + n + n e =

dt


14/28

FET Tuzla

Step1: Using current density [J] find the magnetic induction [B]

( )D

rot H = J +t

rr r

B =H

r r ( )-1rot B = Jr r Solve using

method

of finite moments

Step2: Using B, calculate dissipative flux [ ] by numeric integration

Step3: use formula

nL =

I

Method of Linked Fluxes

1 1 2 2 3 3

n + n + n =

n

gF


15/28

FET Tuzla

Practical Example

Manufacturer data Nominal power 300 VA Model OKT10 Make 1f Primary voltage 200V Primary windings 550 Primary wire diameter 0.7mm Secondary voltage 24V Secondary windings 60 Secondary wire diameter 0.7mm

Core EI-VIII Body 90x108 mm


16/28

FET Tuzla

a

b

g

k l

f e d c

Practical Example [cont.]

a = 90 mm b = 108 mm c = 2 mm d = 9 mm e = 3 mm f = 2.5 mm g = 18 mm k = 50 mm l = 54 mm


17/28

FET Tuzla

0.0

0.5

1.0

1.5

2.0

2.5

0 5000 10000 15000 20000 25000 30000 35000 40000 45000 50000

FeSi MD200

H (A/m)

40

160

250

400

1000

2000

5000

10000

50000

B (T)

0.125

0.86

1.03

1.17

1.34

1.45

1.57

1.70

2.01

Magnetization Characteristic of the Core

H [A/m]

B [T]


18/28

FET Tuzla

Core

Insulator

Primary

Secondary

1 - 1884

1885 - 3054

3055 - 3726

3727 - 4398

Concrete Discretization for Method of Finite Elements


19/28

FET Tuzla

Concrete Magnetic Vector Potential


20/28

FET Tuzla

Concrete Magnetic Induction


21/28

FET TuzlaResults of the Two Proposed Methods

Winding Energy Method Method of Linked Fluxes

Prim.L

P= 1.761 (mH) L

P= 1.719 (mH)

Sec.L

S= 2.061 (mH) L

S= 2.011 (mH)

Total

L = Lp+Ls = 3.821 (mH) L = Lp+Ls = 3.73 (mH)


22/28

FET Tuzla

Error Analysis

TransformerMakeOKT10

L = Lp+ L

s(mH)

EM MUF Experimental

NN energytransformer

3.821 3.730 3.895


23/28

FET Tuzla

Conclusion

Both methods give satisfactory results Useful for transformer design Take into account non-linearity in the core

Energy method is more reliable

Energy method is simpler for implementation Method of linked fluxes can be used in any regime


24/28

FET Tuzla

Backup Slides


25/28

FET Tuzla

1F

2F

3F

4F


26/28

FET Tuzla

Core

Insulation


27/28

FET TuzlaDissipation in Practical

Transformers

Core

Insulation

PrimarySecondary

Winding

UsefulDissipated

Flux


28/28

FET TuzlaDissipation in Practical

Transformers

Core

Insulation

PrimarySecondary

Winding

UsefulDissipated

Flux

Only

CurrentDensity[J] isgiven

electrocomp final

Documents