Transformers Change voltage, frequency remains

Principle of Function

dtdN

dtdue i

Φ⋅=

Ψ==−

Maxwell-Faraday law of induction

Single-Phase Transformer Magnetic circuit

u1 ui10

i10

Φ1h0

ui20 u20

N1

N2

Input (primary) winding

Output (secondary) winding

u1 ui1

ui2 u2 i1

i2

L1σ

L2σ

11

11111

1111 iudtdiLiR

dtd

NdtdiLiRu ++⋅=

Φ++⋅= σ

µσ

22

22222

2222 iudtdiLiR

dtd

NdtdiLiRu +−⋅−=

Φ+−⋅−= σ

µσ

hh 21 Φ+Φ=Φµ

Fundamental Transformer Equation

X2σ

X1σ h

Real Transformer with Sinusoidal Supply

111111ˆˆˆˆ

iUIXjIRU +⋅⋅+⋅= σ 222222ˆˆˆˆ

iUIXjIRU +⋅⋅−⋅−= σ

1U 1ˆ

iU

2ˆ

iU 2U

2I

1I

Electromotive Force Ui in Winding Φh = Φhm . sin ωt

ui = N . dΦ /dt = N .Φhm . ω. cos ωt

ω = 2πf

2mUU = for sinusoidal waveform

12

1max11 ⋅⋅⋅Φ⋅= ωµNUi Ui1 = 4,44· f1·N1· Φμmax

Watch out!

Ui … effective value

Φμmax … maximum value

44,42

2=

πUi2 = 4,44· f1·N2· Φμmax

2

1

2

1

i

i

UU

NNp ==

Equivalent Circuit of Transformer

Φ1h

Φ1σ

No-Load Operation

µσ

IXU

XXRUI

hh

=≅++

=1

102

1121

1010

)(

10110110 ˆˆˆ iUjXIRU ++⋅= σ

1020 ˆˆ iUU =′

)%105(10 −≈i

Current Waveform in No-Load Operation

Iron Losses ΔPFe ΔPFe = f ( Bhm , f ) Losses in equivalent circuit are represented

by appropriately placed resistor. Equivalent resistance RFe is connected with main reactance X1h in parallel.

010 FePP ∆≈

FeIII 1110 ˆˆˆ += µ

201 ˆˆˆ III ′+=

Complete Equivalent Circuit of Transformer

Complete Equivalent Circuit of Transformer

hFe

hFejXRjXRZ

1

10ˆ

+⋅

=

σ111ˆ jXRZ +=

σ222ˆ XjRZ ′+′=′

00111 ˆˆˆˆˆ IZIZU ⋅+⋅=

00222 ˆˆˆˆˆ IZIZU ⋅+′⋅′−=′

Phasor Diagram of Transformer with Load

Short-circuit operation

1k

I1k = I1N

1k

)13,004,0(1 −≈ku

Short circuit voltage in per unit system:

kN

k

N

Nk

N

kk z

ZZ

UIZ

UUu 1

1

1

1

11

1

11 ==

⋅==

Stable short-circuit current by nominal voltage:

k

N

k

N

Nk

NN

k

Nk u

IzI

ZZZU

ZUI

1

1

1

1

11

11

1

11 ====

Nk II )258(1 −≈

Short-circuit operation

1k

I1k = I1N

1k R1k = R1 + R′2

X1k = X1σ + X′2σ

Z1k = R1k + jX1k

k

kkk ZR 111 cosϕ=

kkk ZX 111 sinϕ=Per unit no-load losses ΔpkN :

kNN

kN

N

N

N

k

N

kk p

SP

II

ZR

ZRr ∆=

∆=⋅== 2

1

21

1

1

1

11

dNNjNjkN PPPP ∆+∆+∆=∆ 21

Additional losses caused by leakage fluxes in construction and container.

(ΔPFek is neglected)

Total short-circuit losses by nominal current:

Short-Circuit Operation

kjjFe PPPPPP ∆+∆=∆+∆+∆=∆ 021

k

k

kp PPPPP

PPPP

PP

∆+∆+∆+∆

−=∆+∆+

==0

0

0

1η

kNN

kN

PiPSiPiP

∆⋅+∆+⋅∆+∆

−= 202

20

cos1

ϕη

Losses within load operation:

Efficiency:

Losses and Efficiency

For per unit load i :

02 PPPi kkN ∆=∆=∆⇒0=

didη

Maximum efficiency by load operation is estimated from condition:

kNPPi

∆∆

= 0

Maximum η occurs when losses in winding are equal to no-load losses.

For a common value of losses 04 PPkN ∆⋅≈∆5,0≈iwe get maximum efficiency for per unit load

Losses and Efficiency

PN

P

η

Typical waveform of transformer efficiency

Losses and Efficiency

Is estimated as a difference between input voltage in no-load operation and input voltage by particular load current I2 and cosφ2 .

( )

+=

+≅∆

Nkk

N

kk

IIxr

UIXIRu

1

12121

1

211211

sincos

sincos

ϕϕ

ϕϕ

(Valid approximately for z1k < 4 %)

Voltage Drop over Load

Three-Phase Transformer with Independent Magnetic Circuit

ia ib ic

Symmetrical Three-Phase System

ia + ib + ic = 0

Φa + Φb + Φc = 0

Three-Phase Transformer with Dependent Magnetic Circuits

Magnetic Circuit of Three-Phase Transformer

Star (wye) Connection

UUV UU ⋅= 3

Uuv

Triangle (delta) Connection

UUV II ⋅= 3

Zigzag (Interconnected Star) Connection

UUVU

UZ UUUUU ⋅≅⋅=

+= 866,0

2330cos

22

(indicates the windings configurations and the difference in phase angle between them)

Y

y

Vector group (hour angle)

Vector group (hour angle)

Y

d

Vector group (hour angle)

Hodinový úhel Vector group (hour angle)

Parallel operation of transformers

Balancing current I2 = 0

Requirement:

0=∆⇒ U

Requirements for parallel no-load operation:

1. Same nominal voltages

3. Same vector groups (hour angles) 4. Same voltage drop caused by no-load current

2. Same turns ratio

Parallel Operation of Transformers

Requirements for parallel operation of loaded transformers:

1. Same nominal voltages

2. Same turns ratio

3. Same vector groups (hour angles)

4. Same voltage drop caused by no-load current

5. Same voltage drop when loaded

2211 IZIZ kk =

Parallel Operation of Transformers

Parallel Operation of 3 Transformers

332211 IZIZIZ kkk ==

332211 iziziz kkk ==

5. Same voltage drop when loaded

321 kkk zzz ==

321 kkk uuu ==

Autotransformer

Part of the winding common for primary and secondary

→ lower weight, costs, losses, I0 . Input – output galvanically connected.

For 12

1 >=UUp

Current I1 flows only through Aa part of the winding Lower current flows through common part of the winding

21ˆˆˆ III +=

For normal transformer:

0ˆˆˆ2211 ≅=+ µFININ

0ˆˆ 2

1 ≅+pII

Current in the common part of the autotransformer winding:

( ) ( )pIpII 111 21 −=−=

Autotransformer

For 12

1 >=UUp

Power to secondary side is transmitted in form of:

22111 IUIUS ⋅≈⋅=

Typical power: (magnetic transmission)

( ) ( )pSIUIUIUUS 111121112112 −=−=−=

Electric power: (galvanic transmission)

pSSSSel 11121 ⋅=−=

Autotransformer

Current Instrument Transformer

Load: ammeter ~ short-circuit

Range: 5 A

Current instrument transformer must not be disconnected on output side!

Load: voltmeter ~ no-load

Range: 100 V

Voltage instrument transformer must not be short-circuited on output side!

Voltage Instrument Transformer

Connection of Transformer in No-Load Operation to Grid

dtdNiR

dtdiRu 11

111111ϕ

⋅+⋅=Ψ

+⋅=

Supposing: ( )αω

ϕ+⋅=

⋅= tUu

LNi sin, max11

11

1111

then: ( )αωϕϕ +⋅=+⋅= tN

Udt

dLR

Nu sin

1

max11111

11

1

1

1

Solution:

( )t

LR

zb

tLR

eet 11

1

11

1

coscosmax1111

−−

±

+−−= ϕααωϕϕ

Steady magnetic flux DC component of magnetic flux

Remanent flux

0;2

)1 == zbϕπα

tt ωϕπωϕϕ sin2

cos max11max1111 =

+−=

Connection of Transformer in No-Load Operation to Grid

0;0)2 == zbϕα)cos( 11

1

max1111

tLR

et−

+−= ωϕϕ

α

Connection of Transformer in No-Load Operation to Grid

0;0)2 == zbϕα

)cos( 11

1

max1111

tLR

et−

+−= ωϕϕ

Connection of Transformer in No-Load Operation to Grid

0;0)2 == zbϕα)cos( 11

1

max1111

tLR

et−

+−= ωϕϕ

Connection of Transformer in No-Load Operation to Grid

φ11mez ≈ 2,5 φ11max

Highest current by connection.

Connection of Transformer in No-Load Operation to Grid

Short Circuit of Transformer in Steady-State No-Load Operation

Fundamental voltage equation:

( )αω +=+= tUdtdiLiRu kk sinmax1

111

Solution for i1 = 0 at t = 0

( ) ( )

−−−+⋅=

− tLR

kkkk

k

etIi ϕαϕαω sinsin21

Angle α is determined by instant of short-circuit.

Extreme cases: 1)α = 0 2)α = φk + π/2

Extreme Cases of Short-Circuit

kϕα =tIi k ωsin2 11 ⋅=Solution to this is the steady

short-circuit current

φk

2πϕα += k tIi k ωcos.2 11 ⋅=′

Extreme Cases of Short-Circuit

2πϕα += k

tLR

kk

k

eIi−

=′′ 11 2

tIi k ωsin2 11 ⋅=′ tIi k ωcos2 11 ⋅⋅=′

tLR

kk

k

eIi−

−=′′ 11 2

Extreme Cases of Short-Circuit

2πϕα += k tIi k ωcos.2 11 ⋅=′

tLR

kk

k

eIi−

−=′′ ..2 11

+−=

− πk

kXR

kmez e

ZUI 12 1

1

111 iii ′′+′=

Extreme Cases of Short-Circuit

Limitation of Short-Circuit Current

11lim 2 1

k

k

RX

k

UI eZ

π− = − +

( ) ( )1lim 1 11,7 1,8 2 . 30 40k NI I tj I= ÷ ⋅ ÷ ⋅

Big transformers:

Small transformers:

( ) ( )1lim 1 11, 2 1,3 2 . 15 20k NI I tj I= ÷ ⋅ ÷ ⋅

Mechanical Stress of Transformers Forces ~ I1·I2 or more precisely I2

During short-circuit → SHORT-CIRCUIT FORCES

RADIAL

AXIAL BETWEEN THE WINDINGS

Mechanical Stress of Transformers

INSIDE THE WINDING

Mechanical Stress of Transformers

630 kVA

6 / 0,4 kV

1500 kVA

22 / 1,5 kV

400 kVA

22 / 0,4 kV

1000 kVA

22 / 0,4 kV

250 – 630 kVA

22 kV

50 – 1000 kVA

35 kV

25 MVA

110 ± 8 x 2 % / 23 kV

250 MVA 400 / 128 kV