EE2022 Electrical Energy Systems - MiXeDsIgNaL Electrical Energy Systems Panida Jirutitijaroen Department of Electrical and Computer Engineering Lecture 11: ... • Glover, Sarma,

EE2022 Electrical Energy Systems

Panida Jirutitijaroen Department of Electrical and Computer Engineering

Lecture 11: Transformer and Per Unit Analysis 17-02-2012

2/9/2012 1 EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen

Detailed Syllabus

2/9/2012 EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen 2

20/01/2012 Three-phase circuit analysis: Introduction to three-phase circuit. Balanced three-phase systems.

20/01/2012 Three-phase circuit analysis: Delta-Wye connection, Relationship between phase and line quantities

27/01/2012 Three-phase circuit analysis: Per-phase analysis, Three-phase power calculation

27/01/2012 Guest Lecture “Energy & Environment, Smart Grid & Challenges Ahead” by Prof. J Nanda (IIT Delhi,

IEEE Fellow)

30/01/2012 Generator modeling: Simple generator concept

03/02/2012 Generator modeling: Equivalent circuit of synchronous generators

03/02/2012 Generator modeling: Operating consideration of synchronous generators, i.e. Excitation voltage

control, real power control, and loading capability

06/02/2012 Generator modeling: Principle of asynchronous generators

10/02/2012 Transmission line modeling: Overhead VS Underground cable

10/02/2012 Transmission line modeling: Four basic parameters of transmission line

13/02/2012 Transmission line modeling: Long transmission line model, Medium-length transmission line model,

Short transmission line model

17/02/2012 Transmission line modeling: Operating consideration of transmission lines i.e. voltage regulation,

line loadability, efficiency

17/02/2012 Transformer and per unit analysis: Principle of transformer, Single-phase transformer

27/02/2012 Transformer and per unit analysis: Single-phase per unit analysis

02/03/2012 Transformer and per unit analysis: Three-phase transformer, Three-phase per unit analysis

02/03/2012 Review : if time permits.

IN THIS LECTURE

Learning outcomes

Outline

Reference


Learning Outcomes

• Formulate equivalent circuits of various components in electrical energy systems.

– Equivalent circuit of transformer

• Explain basic operations of different components in electrical energy systems.

– Short circuit/ open circuit test


Outline

• Fundamental concept of transformer

• Single Phase Transformer – Ideal Transformer

– Reflected load

– Maximum power transfer

– Practical Transformer

• Transformer operation – Short circuit test

– Open circuit test


References

• Glover, Sarma, and Overbye, “Power System Analysis and Design”.

– Chapter 3

2/9/2012 EE2022: Three-phase circuit by P. Jirutitijaroen 6

FUNDAMENTAL CONCEPT OF TRANSFORMER

Magnetic flux

Electromagnetic induction

Dot notation

Ampere’s Law

Faraday’s Law


Magnetic Flux

• DC source Constant magnetic flux

• AC source Varying magnetic flux

Source: http://www.lanl.gov/news/index.php/fuseaction/1663.

article/d/20085/id/13276

DC/AC What will happen if we have

another coil to link the varying magnetic flux?


Electromagnetic Induction

• Recall Faraday’s law:

• When we link Coil 2 to the magnetic flux generated by coil 1, if the flux is varying, there will be induced electromotive force (EMF) at Coil 2. The voltage, V2, will be generated by the magnetic force across wire.


Source http://yourelectrichome.blogspot.com/2011/07/introduction-to-coupled-circuit.html

Dot Notation • The direction of induced EMF depends on the direction of magnetic

flux i.e. location that the coil links magnetic flux. Dot notation is used to indicate the direction of current out of Coil 2 in the equivalent circuit.

● ● +

-

●

● +

-


Magnetic Core

• We can better link the magnetic flux by using magnetic core.

• Magnetic flux “Ф” is now confined in the core and links both windings.

EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen 2/9/2012 11

We are now interest to relate V1 and V2, and relate i1 and i2.

Note that ‘N’ refers to number of turns.

Faraday’s Law

• Recall that:

• Let

• Then

• We can write the above equation in a Phasor form.

• Since magnetic flux , – B = flux density (Weber/m²) , A = cross-sectional area (m²).

• We can write:


BA

t cos2

90cos2sin2 tNtNe

jNE

BAjNE

Faraday’s Law

• For ideal transformer, we assume that the flux linkage at coil 1 and coil 2 is the same i.e. there is no flux linkage loss.

• We can now find relationship between the voltage at two sides of the transformer as follows.


BAjNjNV 111

BAjNjNV 222

2

1

2

1

N

N

V

V

Ampere’s Law

• “Current passing through a conductor creates magnetic field around it ”

• B = μH

• B = Magnetic flux density (Weber/m² or Tesla)

• H = Magnetic field intensity (A/m)

• μ = Magnetic core permeability (H/m)


enclosedIHdl

Ampere’s Law Applied to Transformer

• “Magnetic flux along the path equals the net current enclosed by the path”


enclosedpath IHl

i₁N₁ -i₂N₂

Flu

x

Flu

x

2211 NiNiHlpath

2211 NiNiBlpath

Magnetic permeability

Magnetic Core Permeability

• Magnetic core permeability represent the ‘resistance’ that the magnetic core will allow the magnetomotive force to pass through.

• For ideal transformer, the ideal value of the permeability is infinity.

• We can now see the relationship of the current from both sides of the transformers.


2211 NiNiBlpath

2211 NiNi

SINGLE PHASE TRANSFORMER

Ideal transformer

Reflected load

Impedance matching

Practical transformer


Ideal Transformer

• Assumptions:

1. No resistance in both windings.

2. No leakage flux around the core.

3. No core resistive loss.

4. Core permeability is infinite.


2211 NiNi

2

1

2

1

N

N

V

V

Primary side Secondary side

An Ideal Transformer Model

• We represent an equivalent circuit of an ideal transformer as shown below.

• Define turn ratio as:

• From Faraday’s and Ampere’s Law:


22

2

11 aVV

N

NV

2

1

N

Na

11

2

12 aii

N

Ni

a:1

Complex Power

• Complex power at primary side,

• is the same as the complex power at secondary side.

• This means that ideal transformer has no real/reactive power losses.


2

*

22

*

22

*

111 SIVa

IaVIVS

Example 1

• From the circuit below, what is the current at the secondary and primary side?


10:1

Z = 100 Ω 100

V (rms)

V 1010

10 12

2

1 V

VV

V

A 1.0100

22

Vi

A 01.021

a

ii

Reflected Load

• We can reflect the load from one side of a transformer to the other side of a transformer.

• This trick allows us to combine the two separate primary/secondary circuits for easy(?) calculation.


a:1

Z₂

Interest to find reflected load “Z₁”

2

2

2

22

2

2

1

11 Za

i

Va

ai

aV

i

VZ

Z₁

Impedance Matching

• When a voltage source V with internal resistance Rs is connected to a load R, the amount of power at the load depends on the value of load resistance R.

• Maximum power transfer occurs when R = Rs. • In the case that we need to connect the voltage source to a load

that does not satisfy the above condition, we can use transformer to match impedance for maximum power transfer.

• To find an appropriate transformer, we let Rs = a²R and find a transformer turn ratio.


~ R

Rs a:1

R ~

Rs a²R

When Rs ≠ R

Ideal VS Practical Transformer

Ideal transformer

1. Zero resistance in the both windings.

2. No leakage flux around the core.

3. No core resistive loss.

4. Core permeability is infinite

Practical transformer

1. Winding losses (copper losses) represented as resistance in both windings.

2. Leakage flux around the core.

3. Core resistive losses (hysteresis loss + eddy current loss)

4. Magnetic core permeability is finite.


How can we represent this effect in the circuit?

Finite Magnetic Core Permeability

• For practical transformer,

• Recall that ,

• We call this ‘Magnetizing current’. • The ratio between the voltage across the coil (E₁) and magnetizing

current can be written as jω(..). Thus we use an inductor to represent the effect of finite magnetic core permeability in the equivalent circuit.


2

1

2

1

1 iN

N

N

Bli

path

BAjNE 11

2211 NiNiBlpath

2

1

211 i

N

N

j

Ei

A Practical Transformer Model


Copper losses

Leakage reactance

Iron losses (core)

Magnetizing susceptance

Reflected copper losses and leakage reactance of secondary winding.

Note that in Chapter 3 [Glover, Sarma, and Overbye, “Power System Analysis and Design”], the core losses are represented as ‘shunt admittance’, Y = G –jB where G and B is positive. The imaginary part is negative to represent inductive property.

A Simplified Model

• Z₁ and Z₂ are series impedances representing the resistive loss and flux linkage loss in the two windings.

• Y is a shut admittance representing iron core loss and magnetizing susceptance.

• Typically Y is very small i.e. resistance is very large. This means that the currents flowing through Z₁ and a²Z₂ are almost the same. We can simply combine Z₁ and a²Z₂ to “Zeq”, the equivalent series impedance.


Z₁

Y

a:1 a²Z₂

simplified

a:1 Zeq = Z₁+a²Z₂

Y

Transformer Parameter Tests

Short circuit test • To find equivalent series

impedance. • Short circuit the secondary

side. • Apply rated current at the

primary side. • Measure real power and

voltage at the primary side.

Open circuit test

• To find equivalent shunt admittance.

• Open circuit the primary side.

• Apply rated voltage at the secondary side.

• Measure real power and current at the secondary side.


a:1 Zeq

Y ~

a:1 Zeq

Y ~ P₂,I₂ P₁, V₁

Example 2: Short Circuit Test

• Consider a single-phase 20kVA, 480/120 V 60 Hz transformer. During short circuit test, rated current is applied to the primary side. The voltage of 35 V and real power of 300 W are measured. Find equivalent series impedance of this transformer.


A 667.41480

1020 3

,1

,1

rated

rated

ratedV

SI

1728.0667.41

30022

,1

1

rated

eq

I

PR

84.0667.41

35

,1

1

rated

eqI

VZ

0.8221728.084.0 22

eqX

480:120 Zeq = Req + jXeq

Y ~ P₁ = 300 W V₁ = 35 V

Example 3: Open Circuit Test

• Consider the same transformer as Example 2. During open circuit test: rated voltage applied to secondary side, then I₂ = 12 A and P₂ = 200 W. Find equivalent shunt admittance Y of this transformer.


V 480 V, 120 ,1,2 ratedrated VV

SV

PG

rated

eq 000868.0480

20022

,1

1

SV

aI

V

IY

ratedrated

00625.0667.41

412

,1

2

,1

1

SB 0.00619000868.000625.0 22

480:120 Zeq

Y = G-jB ~ P₂ = 200 W I₂ = 12 A

Saturation

• In practical transformer model, we assume constant core permeability and linear relationship between B and H follows.

• In fact, the B-H curve for ferromagnetic materials used for transformer core is nonlinear and has multiple values.

• As H increases, the core become saturated i.e. the magnetic flux density B increase at a much lower rate.

• This effect is NOT included in the equivalent circuit. 2/9/2012 EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen 31

B-H curve is approximated by a dashed line.

Summary • A turn ratio (a:1) is the ratio between the number of

turns on the primary side of transformer and that on the secondary side.

• A load on the secondary side of transformer can be reflected to the primary side of the transformer.

• We can use transformer for impedance matching by choosing the turn ratio that makes reflected load equal to internal resistance of voltage source.

• A practical transformer contains series impedance and shunt admittance. – Series impedance represents winding losses and flux

leakage losses – Shunt admittance represents iron core losses and

magnetizing susceptance.

• Short circuit test is used to find seires impedance by short circuit the secondary side.

• Open circuit test is used to find shunt admittance by open circuit the primary side.


a:1 Zeq = Z₁+a²Z₂

Y

Next Lecture

• Single-phase per unit analysis

– We use this analysis to eliminate transformer model in the circuit.


Documents

EE2022 Electrical Energy Systems - MiXeDsIgNaL Electrical Energy Systems Panida Jirutitijaroen Department of Electrical and Computer Engineering Lecture 11: ... • Glover, Sarma,