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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,
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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
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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
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Conceptual View of Dissipation
F F =i g-
produced - useful = dissipative
F g
F i
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Core
PrimarySecondary
Winding
UsefulDissipated
Flux
Dissipation in PracticalTransformers
Insulation
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Short Circuit Setup
Core
Insulation
Secondary current setup so that useful flux = 0.
Enables measurement of dissipative flux
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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
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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
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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
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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
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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
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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
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FET TuzlaLinked Fluxes
1F
2F
3F
[ ]1 1 2 2 3 3-d n + n + n e =
dt
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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
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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
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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
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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]
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Core
Insulator
Primary
Secondary
1 - 1884
1885 - 3054
3055 - 3726
3727 - 4398
Concrete Discretization for Method of Finite Elements
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Concrete Magnetic Vector Potential
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Concrete Magnetic Induction
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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)
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Error Analysis
TransformerMakeOKT10
L = Lp+ L
s(mH)
EM MUF Experimental
NN energytransformer
3.821 3.730 3.895
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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
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Backup Slides
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1F
2F
3F
4F
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Core
Insulation
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FET TuzlaDissipation in Practical
Transformers
Core
Insulation
PrimarySecondary
Winding
UsefulDissipated
Flux
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FET TuzlaDissipation in Practical
Transformers
Core
Insulation
PrimarySecondary
Winding
UsefulDissipated
Flux
Only
CurrentDensity[J] isgiven