02 ee m ee trafos a

Electric Equipment QP 2013-2014

POWER TRANSFORMERS

ELECTRIC EQUIPMENT

79

POWER TRANSFORMERS

Reversible electrical machine that converts electrical energy in the form U1, I1, f into another in the form U2, I2, f.

= dd = dd

Φ

μr = ∞

i1 i2

u1u2N1 N2

f f


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POWER TRANSFORMERS

As the magnetic flux is sinusoidal, it can be expressed as a phasor

And then, voltages are

Voltages are in phase. The ratio of transformation is defined as

= cos ⟶ = =

= = =

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POWER TRANSFORMERS

If you connect a load to the secondary will satisfy

The ideal transformer does not dissipate energy (neither active nor reactive) then

= −

∗ = − ∗ = ∗∗ = − = − 1i2

u2N2

f

Z2

Φ

μr = ∞

i1

u1 N1

f


82

POWER TRANSFORMERS

According to the signs convention currents are 180º out of phase and their modules satisfy

The impedance Z1 that is viewed from the primary side

= = 1 = = − 1 = = =

i2

u2N2

f

Z2

Φ

μr = ∞

i1

u1 N1

f

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POWER TRANSFORMERS

Nearly ideal transformer: μ r ≠ ∞

ℱ = + ℛ = = ℱℛ= dd = ℛ dℱd = ℛ dd + dd= dd = ℛ dℱd = ℛ dd + dd

= 4π10 Hm

Φi1 i2

u1u2N1 N2

f fB, l, s, μr


84

POWER TRANSFORMERS

Nearly ideal transformer: μ r ≠ ∞

We know that

We name mutual induction coefficient the following value

= ℛ = ℛ= ℛ

Φi1 i2

u1u2N1 N2

f fB, l, s, μr

85

POWER TRANSFORMERS

Nearly ideal transformer:

As we assume that all the flux into primary side is concatenated into the secondary side (perfect coupling) we have

and, therefore,

In matrix form:

= dd + dd = dd + dd= dddd

= =


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POWER TRANSFORMERS

Nearly ideal transformer:

In sinusoidal steady state results in

Magnetic coupling!

=

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POWER TRANSFORMERS

Nearly ideal trafo under no-loadUnder no-load secondary current is null (I2 =0). Hence

The current consumed by the transformer under no-load is called magnetizing current

= = = =

= =

= = = =


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POWER TRANSFORMERS

Nearly ideal trafo under no-loadWe define

Magnetizing currents satisfy

Magnetizing flux is

= = = 1 = 1

= ℛ = ℛ = ℛ = ℛ

= = == = =

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POWER TRANSFORMERS

Nearly ideal trafo under no-loadMagnetizing flux is exactly the same whatever side is the one connected to the power supply:

= ℛℛ = = 1


90

POWER TRANSFORMERS

Nearly ideal trafo under load

Voltages, in sinusoidal steady state, are

= − = + = +

− = + → = − +

i2

u2N2

f

Z

Φi1

u1 N1

f B, l, s, μr

91

POWER TRANSFORMERS

Nearly ideal trafo under load

Primary current is the sum of a fix part (magnetizing) and a part that depends on the load (current consumed by the ideal transformer with the same load).

= + = 1 + = + = +

= + − + = + + = − ++ = +


92

POWER TRANSFORMERS

Nearly ideal trafo under loadWhen the load of the transformer is large, we can consider that the transformer is ideal, since

Under no-load, in the contrary, transformer behaves like a large value inductance as

>>

=

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POWER TRANSFORMERS

Nearly ideal trafo under loadDoes not matter whether nearly ideal transformer is under load or under no-load, primary and secondary voltages satisfy rt: = + = + = −

= − = − → = 1 + = 1 + −

= + − = = =


94

POWER TRANSFORMERS

Nearly ideal trafo under loadAt what flux does nearly ideal transformer work?

Remembering

we have

Flux proportional to the voltage and independent of the load!

= ℱℛ = + ℛ = 1ℛ + + − = 1ℛ + − = ℛ + −

= ℛ + − = ℛ = ℛ = ℛ ℛ = → = = = −

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POWER TRANSFORMERS

Nearly ideal trafo under loadCurrents do not satisfy rt:

= 1 + 1 = 1 + 1 = − 1 + 1 → = − 1 + 1 = − 1 −

= = 1 = + = + = −


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POWER TRANSFORMERS

Nearly ideal trafo under loadBut primary effective current and secondary current YES!

− = − 1 + 1 − − = − 1 + 1 − − = − 1 + 1 + − = − − + = − 1

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POWER TRANSFORMERS

Nearly ideal trafo under loadPower balance

= − ∗ = − ∗∗ = ∗ = +

= = + =

= ∗ = ∗ + ∗ = ∗ + ∗ = ∗− + ∗ = + ∗ = + ∗ = +


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POWER TRANSFORMERS

Nearly ideal trafo under loadEquivalent circuit (referred to primary):

The values Z’, I’2 and U’2 are the secondary ones referred to primary

= = =

= = + = − = − 1 = −

99

POWER TRANSFORMERS

N-i trafo with R in windingsWindings resistance can be considered concentrated and outside the transformer

They dissipate power, copper losses:

= + = + = + = + = +

i1 i2

u1 u2N1 N2

f f

R1 R2

v1 v2

Φ

B, l, s, μr


100

POWER TRANSFORMERS

N-i trafo with R in windings

Due to windings resistance, we have

However,

As R1 and R2 are (and must be) small (voltage drop and losses)

≠ ≠ − 1≈ ≈ − 1

i1 i2

u1 u2N1 N2

f f

R1 R2

v1 v2

Φ

B, l, s, μr

101

POWER TRANSFORMERS

Trafo values referred to primary

=

= = = = = = = ≈ = = −


102

POWER TRANSFORMERS

Trafo values referred to secondary

=

= = = = − = =

= ≈ = ≠ ≈

103

POWER TRANSFORMERS

N-i T. with leakage in windingsPart of the flux is closed with air path and it is not concatenated by both windings!

= dd + dd= dd + dd= + ℛ = ℛ = ℛTypically ℛ ≫ ℛ ℛ ≫ ℛ

ΦC

B, l, s, μr

i1 i2

u1u2N1 N2

ff

Φ1 Φ2


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POWER TRANSFORMERS

N-i T. with leakage in windings

We define

= ℛ + +ℛ= ℛ + + ℛ= ℛ ; = ℛ ; = ℛ ; = ; = ℛ ; = ℛ

ΦC

B, l, s, μr

i1 i2

u1u2N1 N2

ff

Φ1 Φ2

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POWER TRANSFORMERS


Substituting in the previous equations results in:

Magnetic coupling is not ideal. Global inductances are defined as: = + ; = + = + + +

= + dd + dd ; = dd + + dd

= − − − <

ΦC

B, l, s, μr

i1 i2

u1u2N1 N2

ff

Φ1 Φ2


106

POWER TRANSFORMERS


In sinusoidal steady state:

Coupling factor is defined as:

From now on we will use the following notation:

= ≤ 1

= dd + dd ; = dd + dd =

= = = == =

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POWER TRANSFORMERS

N-i T. with leakage in windingsEquivalent circuit referred to primary

≈ ≈ −


108

POWER TRANSFORMERS

N-i trafo with losses in ironIron losses are of two types:

- Hysteresis

where Kh is a constant that depends on material

- Eddy currents

where Kf is a constant that depends on material

= à,

= à

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POWER TRANSFORMERS

N-i trafo with losses in ironGlobal iron losses are

So we can associate a fictitious R for iron losses

≈ − à ≈ à2π à = à ≈ à2 = à2π , + à2π

= 12π , à, , + 12π à≈ à = ≈


110

POWER TRANSFORMERS

N-i trafo with losses in ironEquivalent circuit referred to primary

=

= =

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POWER TRANSFORMERS

N-i trafo with losses in ironEquivalent circuit referred to secondary

= =

=


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POWER TRANSFORMERS

Equivalent circuit of real trafo

Referred to primary

= = = = = ≈

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POWER TRANSFORMERS


Referred to secondary

= = = = = ≈


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POWER TRANSFORMERS


Cases in which simplify equivalent circuit of real transformer is correct and the accuracy is not greatly compromised

Calculation purposes

Simplifications Conditions

No-Load PFe, Img R1 ≈ 0 Xd1 ≈ 0 I 2 = 0

Under Load ΔU, PCu RFe ≈ ∞ Xmg ≈∞SHORT-CIRCUIT

ICC

RFe ≈∞; Xmg ≈∞;typically R1, R2 << Xd1, Xd2

V 2 = 0

115

POWER TRANSFORMERS

Transformer under no-loadResults from no-load test:

= → == = → = − → =

≈ dueto , ≪ ,

U1

A1

V1

W1

Tran

sfor

mer

V2


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POWER TRANSFORMERS

Transformer under loadFrom a certain level of secondary transferred power, values of I’mg and I’Fe are very small in comparison to I1 and, therefore, simplification of equivalent circuit can be accepted.

In order to calculate power and efficiency, transversal branch is never negligible! = ∗ + +

= + = +

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POWER TRANSFORMERS

Transformer under short-circuitResults from short-circuit test:% = 100 = → = =

= = → = − → = = = +

= 0

= − = pu = pu

= pu (ifidealgrid)A1

V1

W1

Tran

sfor

mer

A2


118

POWER TRANSFORMERS

Transformer nameplate

The information displayed on the nameplate of the transformer are:

ratedfrequency, , ratedvoltagesandratedapparentpower, no−loadlossesandshort−circuitvoltage, ratedcurrents, no−loadcurrent primary andratedcopperlosses

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POWER TRANSFORMERS

Transformer efficiencyThe ratio of the output active power (trafo delivered power) to the input active power (trafo consumed power) is called the efficiency of the transformer:

Usually it is given in %

efficiency = DeliveredactivepowerConsumedactivepower= = + +

(%) = 100 = 100 + +


120

POWER TRANSFORMERS

Voltage drop in a transformer

Voltage drop in a transformer is defined as the difference between the ideal transformer expected voltage and the real voltage. Hence, the voltage drop in % is defined as:

Clearly, if the transformer is supplied with the primary rated voltage, the expected voltage in the secondary is the rated secondary voltage.

Δ % = 100 −

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POWER TRANSFORMERS

AutotransformersAdvantages: Less volume of iron and copper, and thus more lightweight and economical.

Disadvantage: Galvanic isolation is lost. If you break a turn in the common part, the entire input voltage appears at the output!

Transformer Autotransformer

I1

I2N1 N2U1U2

I1

I2

N1-N2

N2

U1

U2


122

POWER TRANSFORMERS

ExerciseThe figure represents the single line diagram of a single phase installation

TR: SR = 10 kVA UR = 440/120 V εSC = 5 % PCuR = 1,5 % I0 = 2 % P0 = 0,6 %

Line: R = 0,05 ΩLoad: PR = 6 kW fdpR = 0,8 (i) UR = 110 V

LoadLine

TR

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POWER TRANSFORMERS

Exercise

Knowing that the voltage in the load is the rated one, determine:1) Transformer currents

2) Percentage voltage drops in the line and in the transformer

3) Line, transformer and total efficiencies

4) Primary and secondary transformer currents in case of a short-circuit at the secondary

5) Primary and secondary transformer currents in case of a short-circuit at the end of the line

6) Current and active and reactive powers in case of load disconnection

LoadLine

TR


a

n

124

POWER TRANSFORMERS

ExerciseFrom single line diagram we can get the single phase circuit:

= pu = 0,05 12010000 = 0,072Ω= pu = 0,015 12010000 = 21,6mΩ

= − = 68,68mΩ= 2 = 0,1Ω = 110V

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POWER TRANSFORMERS

Exercise

= ∗∗ = 1 − sincos∗ = 6000 1 − 1 − 0,80,8110 = 54,55 − 40,91A= 68,18A = = 68,18 120440 = 18,6A

a

n


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POWER TRANSFORMERS

Exercise

= + = 0,1 54,55 − 40,91 + 110 = 115,45 − 4,1V → = 115,53V = + =115,45 − 4,1 + 0,0216 + 0,06868 54,55 − 40,91 = 119,44 − 1,23V= 119,45V = = 440120 119,45 = 437,98V∆ = 100 − = 100 115,53 − 110120 = 4,61%

∆ = 100 − = 100 119,45 − 115,53120 = 3,27%∆ = ∆ + ∆ = 7,88 %

a

n

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POWER TRANSFORMERS

Exercise

With respect to UR:∆ = 100 − = 100 115,53 − 110110 = 5,03%∆ = 100 − = 100 119,45 − 115,53110 = 3,56%

∆ = ∆ + ∆ = 8,59%

a

n


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POWER TRANSFORMERS

Exercise

= = 0,1 · 68,18 = 464,85W= pu = 0,006 · 10000 437,98440 = 59,45W= = 0,0216 · 68,18 = 100,41W% = 100 + = 100 60006000 + 464,85 = 92,81%

% = 100 ++ + + = 100 6464,856464,85 + 59,45 + 100,41 = 97,59%% = 100 pu pu = 100 · 0,9281 · 0,9759 = 90,57 %

a

n

129

POWER TRANSFORMERS

Exercise

= = 119,450,072 = 1659A = = 1659 120440 = 452,5A= + = 119,450,1216 + 0,06868 = 855,3A

= = 855,3 120440 = 233,3A

a

n


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POWER TRANSFORMERS

Exercise

= = 59,45W= − = pu − pu = 10000 0,02 − 0,006 = 190,8var= = 190,8 437,98440 = 189,1var

= = = pu = pu = 0,02 10000440 437,98440 = 0,452A

a

n

131

POWER TRANSFORMERS

Three-phase transformers

In order to transform a three-phase system in a different voltage one, three identical single-phase transformers can be used (TR1 = TR2 = TR3). It can be connected in wye or in delta on either side.

TR1

TR2

TR3

U1

U1

U1

U2

U2

U2


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POWER TRANSFORMERS

Three-phase transformersThe ratio of transformation of the three-phase bank does not necessarily coincide with the ratio of transformation of a single transformer. For instance:

= = 3= = 3

TR1

TR2

TR3

U1

U1

U1

U2

U2

U2

a

b

c

n

A

B

C

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POWER TRANSFORMERS


Three-phase banks are only used in very specific installations as they can cause problems in front of unbalanced load.

A three-phase transformer with columns (3 or 5, but windings only in the three central columns) has a primary winding and a secondary winding in each column.

Primary and secondary winding connections can be in wye, delta or zigzag. Depending on the connections we will obtain different properties and ratios of transformation.


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POWER TRANSFORMERS


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POWER TRANSFORMERS


Wye connection:

The neutral can be accessible (there may be phase to neutral voltages) and the line current and the branch current (the current of one winding) are equal.

Each winding has to withstand phase to neutral voltage.


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POWER TRANSFORMERS


Delta connection:

Each winding is made to withstand phase to phase voltage, but inside it only circulates the line current divided by root 3.

We need root 3 times more turns than in wye connection.

There is no neutral wire.

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POWER TRANSFORMERS


Zigzag connection:

Two windings per phase are needed. The line current will flow through these windings. We need more turns than in the case of wye connection (two divided by root 3 more turns).

The neutral is accessible and it is only used when the load is strongly asymmetric.


138

POWER TRANSFORMERS

Three-phase transformersPhase displacement:

The winding voltages of each column will be in phase or with 180º phase displacement (depending on where are the homologous points of the windings).

Due to three-phase system symmetric it can be shown that any phase displacement in wye, delta or zigzag windings is a multiple of π/6 regardless of the connection. Given that there are only 12 possible lags, this was assimilated to a clock that had the long handle to twelve (high voltage side, primary) and the short handle corresponding to the same secondary voltage.

139

POWER TRANSFORMERS

Three-phase transformersThe phase displacement assumes the transformer fed by the high voltage side with a direct sequence three-phase system. A change in the sequence implies a change in the phase displacement.

Uab

Ubc

Uca

12

UANUan

6

39

1

UAN

UBNUCN


140

POWER TRANSFORMERS

Three-phase transformersIEC vector group code:

IEC vector group code of a transformer provides a simple way of indicating how the internal connections of a transformer are arranged. Capital letter for high voltage side and lower case letters for low voltage side. The phase displacement is also indicated. If the neutral wire is present it has to be indicated with the letter ‘n’ (capital or lower case accordingly).

Example: Dyn11

High voltage side is delta connected, low voltage side is wye connected (and has neutral wire) and the phase displacement is 11·30º, that is, 330º.

141

POWER TRANSFORMERS

Three-phase transformersConnection of transformers in parallel:

In order to connect two transformers in parallel the following requirements have to be fulfilled:

- The same rt (and the same UN if possible)

- The same εCC (and the same cos ϕCC if possible)

- The same phase displacement

These conditions imply that with the same level of loading the output voltages are equal in module. Are equal in phase if the conditions in brackets are also fulfilled. It is recommended that the power ratio of the transformers is less than 2.


142

POWER TRANSFORMERS

Exercise

The figure represents the single line diagram of a three-phase installation.

TR: SR = 620 kVA UR = 10/0,44 kV εCC = 5 % PCuR = 1 % I0 = 1 % P0 = 0,6 % Dyn11

MV Line: R = 6 Ω X = 6 ΩLV Line: R = 0,01 Ω X = 0 ΩLoad: PR = 500 kW fdpR = 0,8 (i) UR = 400 V

LoadLV Line

TR

MV Line

143

POWER TRANSFORMERS

Exercise

Knowing that the voltage at the beginning of the installation is 10,5 kV, determine:1) Percentage voltage drops in lines and in the transformer

2) Lines, transformer and total efficiencies


4) Primary and secondary transformer currents in case of a short-circuit at the end of the LV line

LoadLV Line

TR

MV Line


144

POWER TRANSFORMERS

Exercise

a

n

= ∗ = 0,40,50,8 0,8 − 0,6 = 204,8 + 153,6mΩ= pu = 0,05 0,440,62 = 15,61mΩ= pu = 0,01 0,440,62 = 3,12mΩ= − = 15,3 mΩ

145

POWER TRANSFORMERS

Exercise

= = 6 0,4410 = 11,62mΩ = = 6 0,4410 = 11,62mΩ= = 10,5 44010 = 462V; = 3 = 4623 = 266,7V; = 266,7V

= + + + + = 266,711,62 + 11,62 + 3,12 + 15,3 + 10 + 204,8 + 153,6 = 717,9 − 564,6A= 913,3 A

a

n


146

POWER TRANSFORMERS

Exercise

= = 204,8 + 153,6 0,7179 − 0,5646 = 233,7 − 5,36V= 233,8V; = 3 = 404,96V = + = 233,7 − 5,36 + 0,01 717,9 − 564,6 = 240,9 − 11V= 241,2V; = 3 = 417,7V

= + = 240,9 − 11 + 3,12 + 15,3 0,7179 − 0,5646 = 251,8 − 1,8V= 251,8V; = 3 = 436,1V; = = 9,913kV

a

n

With respect to load rated voltage:

147

POWER TRANSFORMERS

Exercise

Δ % = 100 − = 100 417,7 − 404,96400 = 3,185%Δ % = 100 − = 100 436,1 − 417,7400 = 4,6%Δ % = 100 − = 100 462 − 436,1400 = 6,475%

a

n


With respect to transformer rated voltage:

148

POWER TRANSFORMERS

Δ % = 100 − = 100 417,7 − 404,96440 = 2,895%Δ % = 100 − = 100 436,1 − 417,7440 = 4,182%Δ % = 100 − = 100 462 − 436,1440 = 5,886%

Exercisea

n

149

POWER TRANSFORMERS

= 3ℜ = 3 · 0,2048 · 913,3 = 512,48kW= 3 = 3 · 0,01 · 913,3 = 25,02kW= 3 = 3 · 0,00312 · 913,3 = 7,81kW= pu = 0,006 · 620 9,91310 = 3,66kW= 3 = 3 · 0,01162 · 913,3 = 29,1kW

Exercisea

n


150

POWER TRANSFORMERS

% = 100 + + ++ + + + = 100 548,97548,97 + 29,1 = 94,97%% = 100 ++ + + = 100 537,5537,5 + 3,66 + 7,81 = 97,91%

% = 100 + = 100 512,48512,48 + 25,02 = 95,35%

% = 100 pu pu pu = 100 · 0,9497 · 0,9791 · 0,9535 = 88,66%

Exercisea

n

151

POWER TRANSFORMERS

Exercise

Current in the secondary:

Current in the primary:

= + + = 266,711,62 + 11,62 + 3,12 + 15,3 = 8,69kA= 8,69kA

= 8,69 0,4410 = 382,4A

a

n


Current in the secondary:

Current in the primary:

152

POWER TRANSFORMERS

Exercise

= + + + = 266,711,62 + 11,62 + 3,12 + 15,3 + 10 = 7,29kA= 7,29kA

= 7,29 0,4410 = 321A

a

n

153

POWER TRANSFORMERS

Exercise

TR1=TR2: SR = 4,5 MVA UR = 20/6 kV εSC = 5 % PCuR = 1 % I0 = 1,7 % P0 = 0,7 %

Line: R = 100 mΩ X = 300 mΩLoad: PR = 8000 kW fdpR = 0,96 (i) UR = 6 kV

LoadLine

TR1

TR2

The figure represents the single line diagram of a three-phaseinstallation.


154

POWER TRANSFORMERS

Exercise

Knowing that voltage at the load is 6 kV, determine:1) Percentage voltage drops in the line and in transformers

2) Line, transformers and total efficiencies

3) Primary and secondary transformer currents in case of a short-circuit at the load


LoadLine

TR1

TR2

02 ee m ee trafos a

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