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ELEC9712-2008: Introduction p. 1/62

ELEC9712 High Voltage Systems

Common Background Material for Electrical Equipment Design and Operation

HEAT GENERATION EFFECTS Temperature Constraints Heat loss mechanisms in equipment Rating of equipment Transient thermal heating Flammability of materials

FIELD PROPERTIES AND CALCULATIONS Equipotentials and flux lines Field mapping Effect of boundaries and change of material Macroscopic properties: Potential and Flux Microscopic properties: Potential gradient and Flux

density


MECHANICAL EFFECTS Electrodynamic forces Tensile strength – tension Material Compatibility Arcing Containment Electrical Contacts

ELECTRIC FIELDS Capacitance and voltage Breakdown strength Discharge inception levels Insulating materials: solids, gases and liquids Dielectric heating of solid insulators Induced effects – capacitive coupling Transient effects and field distribution Shielding against electric fields

MAGNETIC FIELDS (e.g. Transformer Core) Inductance and magnetic potential Magnetic materials Magnetic circuits


Hysteresis losses and eddy current losses Saturation effects in magnetic materials Shielding against magnetic fields Compatibility of equipment with magnetic fields

ELECTRICAL CONTACTS Contact resistance Contact materials Static contacts Dynamic contacts Arcing contacts

TESTING TECHNIQUES Performance tests Routine tests and Type tests Abnormal operation tests Acceptance tests Maintenance tests On-line tests


Thermal Characteristics and Ratings All electrical equipment has some constraint on its operation imposed by limitations in the temperature of the equipment. This constraint results in a maximum thermal rating which must be determined for the equipment. This is done by consideration of the heat generation during operation and the heat loss during operation. The rating determination requires the following information:

1. The environment and its effects on heat dissipation 2. Heat generation mechanisms in the equipment 3. Heat dissipation mechanisms 4. Temperature limitations of the materials 5. Short circuit currents 6. Material properties of the equipment

There are a variety of different ratings that must be considered. They are used for different operational conditions and for different items of equipment. The different forms of thermal rating are:

1. Continuous rating 2. Cyclic ratings (daily) 3. Emergency ratings: continuous, cyclic 4. Restricted cyclic (1-3 days) 5. Short duration rating (1-2 hours) 6. Short circuit rating (1-3 seconds)


1. Heat Generation Mechanisms Ohmic heat loss (direct and indirect) Magnetic hysteresis in magnetic materials Eddy currents in magnetic materials and in conductors Dielectric loss in insulating materials

1.1. Ohmic Heating Generation Direct ohmic heating results from the effects of current flow in the main conductors. For calculation of this heat loss we need an accurate determination of the resistance R. Indirect ohmic heating results from the effects of eddy currents induced by exposure of conductors to alternating magnetic fields. This is much more difficult to quantify. For direct ohmic heating, the simple calculation uses 2I R for the heat loss. We need to know R accurately. R depends on the following parameters:

Conductor material Temperature: ( ) ( ) ( )20 1 20R T R Tα = + − Impurities in the conductor The conductor geometry Frequency of the current flow – skin effect Effects of external fields (i.e. indirect heating)


[AS3768-1990: Guide to the effects of temperature on electrical equipment]


AS3768-1990 Appendix B


Resistance dependence on temperature For pure metals, resistivity is due to scattering of conduction electrons by thermal vibrations of the atoms in the lattice. The resistivity increases linearly with temperature. For metallic alloys, resistivity is only weakly temperature dependent. Here, resistivity also arises from scattering of electrons from the impurities which is not temperature dependent. The material resistivity ρ is determined by electron scaterring from the lattice:

1en

ρµ

=

where: e = electron charge n = number of electrons µ = mobility As lattice atoms vibrate, effective cross section increases and causes more scattering. So:

ρ ∝ temperature i.e. A Tρ = × But impurities will also cause scattering:

1i

ienρ

µ=

Density of impurities is independent of temperature, i.e.


iρ = constant Hence: total i iATρ ρ ρ ρ= + = +

or: AT Bρ = + (Matthiessen’s rule) Define temperature coefficient of resistivity (i.e. fractional change in resistivity per unit temperature rise at the reference temperature):

0

00

1

T TTδρα

ρ δ =

=

Usually, we choose 0 273T = 0K or 2930K. If we assume 0α = constant over a small range:

( ) ( )0 0 01T T Tρ ρ α = + − Note that 0α is dependent on 0ρ and hence on T0. If we use ATρ =

then 0 0 0 0 0 0AT T Tρ ρ α ρ α= + −

and 0 0A ρ α=

Thus: 0 0 0 0 0Tρ ρ α− =

Hence: 00

1 1273T

α = = for 00 0T C=


The effect of impurities and their lack of dependence on temperature is useful in resistance alloys such as nichrome, kanthal, etc, which can operate over wide ranges of T with little variation in R. Impurities in steel are important in reducing eddy currents in iron cores for transformers and machines. Resistance and conductor geometry

bVaV

aSbS

E

dl

ds bVaV

aSbS

E

dl

ds

We require calculation of R, the total resistance between the two equi-potential surfaces (Sa and Sb), which have potentials Va and Vb. I is the total current flow (flux) bounded by Sa and Sb. In the most general case, we use the expression:

a bV VRI−

= (Ohm’s law)


where: b

ab b aa

V V V E dl= − = ⋅∫

a aS S

I j ds E dsσ= ⋅ = ⋅∫∫ ∫∫ (or use Sb)

If the cross section is not uniform or able to be represented analytically, we need to:

solve Laplace’s equation to determine the electric field ( ), ,E r zθ .

perform a field mapping exercise.

use some known functional relationship for E and j (current density).

Example: Insulation resistance of a coaxial system

arbr

j

1m

arbr

j

1m

1Er

∝

Hence: ( ) 0arj r jr

= ⋅

( ) ( )1E r E rσρ

= =

where j0 = current density at ar r=


( )

( ). .

b

aab

j r drR

j r rd dx

ρ

θ

⋅=∫

∫∫

0

2

00 0

1 . .

b

aa

x

a

drr jr

r j rd dxr

π

ρ

θ=

∫

∫ ∫

ln2

bab

a

rRx r

ρπ

=

ohms

i.e. ln2

bab

a

rRr

ρπ

=

ohms/metre

Alternative method:

br

ar rdr

xbr

ar rdr

x

2

drdRrx

ρπ

=

ln2 2

b bb

abaa a

rdrR dRx r x r

ρ ρπ π

= = ⋅ =

∫ ∫ ohms


SKIN EFFECT

In conductors and magnetic materials, the effect of an AC (external) magnetic field is to induce eddy currents which alter the B-field distribution, and the current distribution. We can analyse the situation by solving the wave equation in the situation shown below. Consider a semi-infinite slab of conductor, with B field and current density as shown.

yB

x

yz

zJ( )zE

( )yH

yB

x

yz

zJ( )zE

( )yH

We have: ( )0,0, zJ J= ( )0, ,0yB B= J and B vary only with x and we assume sinusoidal variation

j te ω . Maxwell’s curl equations are:

BEt

∂∇× = −

∂

DH Jt

∂∇× = +

∂


But B Hµ= and if we neglect D t∂ ∂ at low frequencies then: B Jµ∇× = Also, we have Ohm’s law:

EJ Eσρ

= =

Substitution gives (note that 0y z∂ ∂≡ ≡

∂ ∂):

zy

J j Bx

ρ ω∂= −

∂

yz

BJ

xµ

∂= −

∂

which gives (denote yB B≡ , zJ J≡ ):

2

2 0B j Bx

µωρ

∂− =

∂

2

2 0J j Jx

µωρ

∂− =

∂

Thus, J and B have the same x dependence. The solution is:

( ) 1 2exp expx xJ x J j J jd d

= + −

0 exp xJ jd

= −

(for finite J)


where: d ρµω

=

But: ( )1 12

j j= +

Hence: ( ) 0 exp expx xJ x J jδ δ− = −

where: 22d ρδµω

= = = skin depth

Similarly: ( ) 0 exp expx xB x B jδ δ− = −

x

0J

x δ=

0 01 0.37J Je

=

0 x

0J

x δ=

0 01 0.37J Je

=

0 At x δ= below the surface:

B and J are 37% of surface values

their phase angles are 1 radian behind the phase of 0B and 0J .


Effect of “Skin Effect” and “Proximity Effect” on electrical conductor impedance

When a current-carrying conductor is subject to an alternating magnetic field, eddy currents will be generated within the conductor and these eddy currents will then interact with the intrinsic current in the conductor and change the effective current distribution. The net effect of this change in current will be an effective increase in the conductor resistance and thus an effective increase in the ohmic losses in the conductor. When the interacting magnetic field is the AC field of the intrinsic current in the conductor itself, the result is the “skin effect”. When the interacting magnetic field is an external AC magnetic field, the result is the “proximity effect”.

(i) Resistance The current density when modified by skin effect is given by:

( ) 0 exp expr jrj r jδ δ− − =


where: 2 skin depthρδµω

= =

The redistribution of the current density j(r) causes an effective increase in the resistance

of the conductor. The new AC resistance is designated:

( )1AC DC sR R Y= +

where RDC is the resistance of the conductor assuming uniform current distribution with radius and Ys is a multiplier factor dependent on the conductor geometry and properties. (ii) Inductance In addition to the change in current distribution, the magnetic field distribution within the conductors is also changed because of the skin effect, becoming:

( ) exp expor jrB r Bδ δ− − =

This change in magnetic field and the change in the coupling with the current thus affects the internal inductance of the conductor and this will change the internal inductive reactance of the conductor. The end result will be an effective decrease in the internal inductance

and the internal inductive reactance.


[The external inductance will remain unchanged and this is the predominant inductive component in most conductors, with the exception of steel wires].

Conductor size limits

The increase in the electrical resistance of conductors with skin effect puts an effective uppper limit on size for the carrying of AC currents. This upper limit is nominally a diameter of about twice the skin depth of the material at the AC frequency: 2d δ≈ When δ is known, the effective cross-section area is, for a cylindrical conductor

:

( )22A a aπ π δ= − − 2 for a aπ δ δ Hence:

2 2 2AC

AC

RA a a

ρµωρ ρπ δ π

= = =

Observe: 2

AC

DC

R aR δ

= [cylindrical conductor]

( )2

aba b δ

=+

[rectangular bar a x b]

Note that: 1/ 2 1/ 2 ACR fω∝ ∝ 1/ 2 1/2

r µ µ∝ ∝ 1/ 2 ρ∝


At 50Hz, the skin effect can be significant in low voltage cables in particular, and also in large transformer windings and motors. In large transformers and motors, some segmentation of the conductor is necessary (Roebell construction) to counteract the skin effect.

Determination of AC resistance at 50Hz

The usual method for cable conductors is simply to use existing tables for given geometries. For example: ( )1AC DC sR R Y= + where:

( )4 4192 0.8s s sY x x= + and 2 78 10sDC

fxRπ −= ×

Thus saxδ

∝


Alternatively, graphs of RAC/RDC are available for use, as shown in the figures.

[Note that 1000 DC

f aR δ

∝ (ratio of radius to skin depth)]


Effect of external AC magnetic fields on losses in current-carrying conductors Whenever any electrical conductor is exposed to an external AC magnetic field, the field penetrates into the conductor and the resulting effect is eddy current generation in the conductor (in fact this is also the effect of the intrinsic internal magnetic field that causes skin effect in a current-carrying conductor)] In an isolated conductor material, these eddy currents will generate eddy current loss due to ohmic heating and in current-carrying conductors the eddy currents will interact with the internal current flow and alter the current distribution and hence increase the resistance and thus also increase heat loss. This will occur in and affect the following:

Transformers and large rotating machines Any electrical conductors Cable sheaths and armour (particularly steel armour) Metallic busbar trunking material Transformer tanks

The eddy current losses in a metal slab are given by:

2 2

3 [W/m ]mE

kf BPρ

=

where: k = a material constant mB = peak magnetic flux density


f = AC frequency ρ = material electrical resistivity Note that: 2 EP f∝ 2 mB∝ 1 ρ∝ Note that if we laminate the slab of material by using N insulated layers, the total loss is reduced to:

* 32 [W/m ]E

EPPN

=

Proximity effect When a current-carrying conductor is subject to influence by an external magnetic field, the modification of the current distribution and the resulting increase in resistance and heating is termed the “proximity effect”. The resistance is then designated:

( )1AC DC s pR R Y Y= + + where Ys is the skin effect multiplier and Yp is the proximity effect multiplier. RDC is the geometric resistance of the conductor, as before. Note that Ys will be zero if the current is DC. Minimising the skin and proximity effects on resistance


For large conductors the skin effect can be reduced by dividing the conductor into segments which are insulated (by an enamel coating) from each other, and then the proximity effect can be reduced by regular transposition of the conductor segments.


1.2. LOSSES IN MAGNETIC MATERIALS (when subject to AC magnetic fields) Magnetic material losses are a major cause of heating in large electrical equipment items and they must be limited as much as possible. There are two components of magnetic material loss:

Hysteresis loss Eddy current loss

1.3. Eddy current loss The eddy current loss is due to the same mechanism as just previously discussed. However the heating effect is greatly magnified in magnetic materials because of the high magnetic permeability of such materials which will magnify the applied B field and thus cause Bm to be very high. Lamination is the traditional way of reducing such eddy current losses in magnetic material cores. Hysteresis loss Hysteresis losses in magnetic materials are generated by frictional effects of the magnetic domain movement in the materials as they try to follow the changing direction of the applied alternating field. When the material is subject to the field, the B-H relationship is the well-known hysteresis loop of magnetic materials, as shown below:


There is an associated energy storage in the magnetic field in the medium:

212

W Hµ=

and the hysteresis loss per cycle can be obtained by integrating this energy around the B-H loop over one cycle of the alternating field. For example, for a toroid with N turns carrying a current of I amps, we have:

mmf NI H= = I I dI→ + H H dH→ + B B dB→ +

If in a time dt, dV Ndtφ

= − is generated.

We need to increase supply voltage:

W VIdt NIdφ∆ = − =


But: BA d AdBφ φ= ⇒ = From Ampere’s law (in the toroid case):

2NIH

rπ=

Hence: 2 volume W rA HdB HdBπ∆ = ⋅ = ×

Hence: max

0

B

W v HdB= ∫

Net loss per cycle is:

J/cycleW v HdB= ∫

Power loss wattsWf= wattsvf HdB= ∫ Thus, hysteresis power density is:

3 watts/mhp f HdB= ∫ For the evaluation of ph, we need the appropriate B-H loop and then we can calculate the energy from the area of the loop. This area calculation can be done in any of a number of ways:

Linear approximation [approximate the area] Draw a grid and count squares within the B-H loop Use the manufacturer’s data on losses Use an appropriate empirical formula for the losses.


The most common formula used for magnetic losses is that due to Steinmetz:

3max J/m /cyclen

hw Bλ= 3

max watts/mnhW B fλ= [f is frequency]

[Note: the above is an empirical formula only and does not cover the full range of possible value of B.] Usually, 1.5 < n < 2.5 with a most common value of about n=1.6. Often a value of 2 is used to give consistency with the eddy current losses when determining the total magnetic material losses [eddy + hysteresis] and their dependence on the magnetic field level. The value of λ depends on the material used. Typical values of losses The following typical values of λ and n for various materials are for values of Bmax in the range 1.0 – 1.6 tesla.

Cast steel: 2500λ = 1.6n = [ 5 32.65 10 W/mhW = × at 1.6 tesla] Silicon steel: 130λ = 1.58n = [ 4 31.37 10 W/mhW = × at 1.6 tesla]


Ni-Fe high µ: 25λ = 1.75n = [ 3 32.85 10 W/mhW = × at 1.6 tesla]

Typically, Wh = 2 – 3 watts/kg for normal transformer core materials at Bmax ~ 1.6 tesla for good low-loss steel. A Bmax of 1.6 tesla is a typical operating flux density in a transformer core material.


Eddy current heat loss

Assume:

(i) B is sinusoidal ( ) ( )exp 2mB B j t fω ω π= = (ii) b

(iii) B is not modified by induced current

B is normal to face. ρ is material resistivity. Calculate current through the element xδ . Flux linkage of the area enclosed: max 2m B xφ = ×


Or: max2 cosx B tφ ω=

(Faraday's law)dVdtφ

= −

2 sinmx B tω ω= Eddy current loop resistance:

2'1

Rxρδ

=× (c.f.

Aρ )

Hence, eddy current loss in loop:

2 2

' 2V V xdwR

δρ

= =

( )221

2 2mx B xω δρ

= ×

[Note: ½ to give RMS value]

2 2 2 2

mx Bdw xω δρ

=

Hence, total loss is:

2 2 2 2

0

bmx BW dxω

ρ= ⋅∫


2 3 2

3mb Bω

ρ=

watts/metre depth

or:

2 3 2

'3

mb BW ωρ

= watts/unit area ( )1m=

Hence: ' W ∝ (i) 2ω (i.e. 2f ) (ii) 3b (iii) 2

mB (iv) 1 ρ Replace by N laminations, each of thickness ( )2b N

( )32 2/

3mb N B

W Nω

ρ= ⋅

2

1 'WN

= i.e loss reduced by a factor of N2

Note: laminations must be insulated from each other. Total core losses: Core loss = hysteresis loss + eddy current loss

2 21 2

nm mP k fB k f B= +

where k1 and k2 are constants. Often, we use the following equation for total losses:

( )2 21 2 mP k f k f B= +


DIELECTRIC HEATING (1.4. Dielectric loss in insulating materials) This phenomenon is an analogue of magnetic hysteresis (i.e. hysteresis due to an AC electric field). Used to assess the efficiency of insulators.

RC

IC IR

I

V

V

IIC

IR

δ

φRC

IC IR

I

V

V

IIC

IR

δ

φ

V.IR represents the dielectric losses.

Typically: 90 is oδ φ φ= −

tan R

C

I PI Q

δ = =

tanδ is the loss factor (also called dielectric dissipation factor) and is a measurable quantity.

tanR CP VI VI δ= =

But: CI CVω= Hence:

2 tanP CVω δ= 2 tanP CVω δ=


If: ACdε

=

Then: 2 tanAVP

dωε δ

=

( )2

tanVAdd

ωε δ =

or: ( )2 3tan power density W/mP EAd

ωε δ= =

i.e. 2 3tan (W/m )o rp Eωε ε δ= The dielectric dissipation factor (tanδ ) is measured and used as an insulation quality factor. In so far as heating is concerned, it only becomes significant at V > 50 – 60 kV and then only in cables. Equivalent circuit for dielectric loss There are 2 possible equivalent circuits (a) Parallel circuit


RpCp

IC IR

I

V

V

IIC

IR

δ

φRpCp

IC IR

I

V

V

IIC

IR

δ

φ

2

2 tanpp

VP C VR

ω δ= =

as derived before. Thus:

tan pR

C p

V RII C V

δω

= =

1

p pR Cω=

(b) Series circuit

Cs

IVCω

=Rs

Cs

I

V

V

I

δ

φ

R sV IR=

Cs

IVCω

=Rs

Cs

I

V

V

I

δ

φ

R sV IR=

Here: tan Rs s

C

V C RV

δ ω= =


coss C sI C V C Vω ω δ= = Hence: 2 sin cosR sP IV C Vω δ δ= =

22

tan1 tansC V δω

δ=

+

2 tan for small sC Vω δ δ Both circuits are expected to give the same answer. But:

1tan s sp p

C RC R

δ ωω

= =

So Rs, Rp, and Cs, Cp must differ. Using:

2

2 tanpp

VP C VR

ω δ= =

2 22

tan1 tans sC V I Rδω

δ= =

+

We have:

21 tans pC C δ = +

and: 2

11tanp sR R

δ = +

[show how ??]

For good insulators, δ is small, and tan2δ can be neglected. Thus: s pC C


and: 2tans pR R δ The series representation is generally used at low frequencies, such as in tanδ measurement at 50Hz.


FIELD ANALYSIS To solve a field problem, we need to obtain the flux or potential distribution. From this, we can obtain the required macroscopic and microscopic properties

e.g. R, C, L, RTH

, , , , etcJ D B E There are three methods of obtaining flux and potential distribution:

1. Solve Laplace’s equation ( )2 0φ∇ = or Poisson’s

equation ( )2φ ρ ε∇ = − analytically or numerically

2. Mannual field mapping 3. Numerical iteration

Field mapping (plotting) This technique uses known field properties to draw up a map of the field for a particular configuration. It can give surprisingly accurate results. Field properties:

(i) Flux lines and equipotentials are orthogonal (ii) Surfaces of flux sources are equipotentials (iii) Flux lines intersect sources and sinks orthogonally.


The aim is to divide the field structure into curvilinear squares (assuming there is symmetry in the third dimension)

If the element is part of current flow and the current density is J , and electric field E ( ) ( ) area = =1 mI J J x J x∆ = × × ∆ = ∆ V E y∆ = ∆ Hence, resistance of the elemental section is:

1V E y yRI J x xσ

∆ ∆ ∆∆ = = = ⋅

∆ ∆ ∆

1 if 1m and y xρσ

= = = ∆ = ∆


Thus, each curvilinear square of unit depth has resistance ρ. The total resistance can be obtained by counting squares in the full map. This holds for any field, the only difference being the property relative to the field type, e.g.

For an electric field, capacitance of the unit square ε= For a magnetic field, the unit square inductance µ= For a thermal field, the unit square conductance k=

e.g. for a current flow field above, if there are n equipotential drops and m flux tubes of current and the resistivity is ρ, then the total R is:

ohmsnRmρ

=

In the above: 44, 8 ohms8 2

n m R ρ ρ= = ⇒ = = (per m)


If a capacitance: and total 8Q Q m Q Qψ∆ = ∆ = ∆ = ∆ and total 4V V n V Vφ∆ = ∆ = ∆ = ∆ Hence:

Total 8 2 (F/m)4

Q QCV V

ε∆= = =

∆

If an inductance: Total flux 8ψ ψ= ∆ ( ) ( ) and total 4NI NI NIφ∆ ≡ ∆ = ∆ Hence:

Total ( )

8 24

LNI NIψ ψ µ∆

= = =∆

For flux density and potential gradient, use the appropriate single square and the applicable scale factor of the map:

Flux density1 xψ∆

=×∆

mx xψ ψ∆

= =∆ ∆

Pot. grad ny yφ φ∆

= =∆ ∆

Flux density1 xψ∆

=×∆

mx xψ ψ∆

= =∆ ∆

Pot. grad ny yφ φ∆

= =∆ ∆

Thus if we know total ψ and φ and get m and n, and

( ) =x y∆ ∆ from the map, we can find and yxψ φ∆ ∆ ∆ ∆ .


Example: A cable has an eccentric core and axial uniformity (see Figure). We want to find its capacitance and insulation resistance. There are 100 volts across the cable and 2.0rε = . The map divides total potential into 4 equal potential divisions, i.e. n = 4. Hence:

100100 25V4

φ φ= ⇒ ∆ = =

There are 2 14 28× = flux tubes, i.e. m = 28. (a) Capacitance: Each unit square (of unit depth 1= ) has a capacitance of ε farads, i.e. 2 oε= farads.

Total capacitance ( )28 2 14 F/m4 o oε ε= × =

124 pF/m= Note: for comparison, the exact formula for capacitance of this geometry is:

2 2 2

1

2

cosh2

a b

a b

Cr r c

r r

πε

−

= + −

This gives: 128 pF/mC = Flux: ( )12124 10 100 12.4 nCQ CV −= = × × =


At point P1:

Average 12.5 V 12.5 V/cm1.0 cm

E = =

1.25 kV/m = potential gradient= Flux density 2 oD E Eε ε= = ( )12 32 8.85 10 1.25 10−= × × × × 222.125 nC/m= (b) Insulation resistance: Each unit square has resistance ρ= . Use a typical 1310ρ = Ωm units then total R for one metre is:

124 1.43 10 m28

R ρ ×= = × Ω

At point P1: 1.25 kV/mE = as before

2 A/mj Eσ=

( )3 10 213

1 1.25 10 1.25 10 A/m10

−= × × = ×

2125 pA/m= (c) Inductance: If total I through centre conductor = 100A No. of flux tubes = 4 No. of equi-potentials = 28


Inductance of each square oµ µ= = (H/m)

Total inductance 7

74 4 4 10 1.8 10 H/m28 28µ π −

−× ×= = = ×

Flux density: Total flux LIψ = ( )71.8 10 100−= × × 51.8 10−= × webers

Hence: 64.5 104ψψ −∆ = = × webers

At P1:

Flux density ( )6

44.5 10 2

2.25 10 tesla0.01 1

B−

−×

= = ××

= 2.25 gauss

Potential gradient 4

7

2.25 104 10o

BHµ π

−

−

×= =

×

= 179 ampere turns/metre

[ or ( )100 2 28

1790.01

×= = ]

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