EE 369POWER SYSTEM ANALYSIS

Lecture 8Transformers, Per UnitTom Overbye and Ross Baldick

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AnnouncementsFor lectures 8 to 10 read Chapter 3Homework 6 is 5.9, 5.14, 5.24, 5.26, 5.27, 5.28, 5.29,

5.32, 5.33, 5.37, 5.38, 5.43; due October 10. Turn in to grader when you meet for class tour of UT power plant that day. Mid-term will cover through homework 6.

Mid-term review in class on October 15.Mid-term in class on October 17; you can bring in a

sheet of paper with notes, calculator, and writing materials. Mid-term includes hw 6.

Homework 7 is 5.19, 5.23, 3.1, 3.3, 3.4, 3.7, 3.8, 3.10, 3.12, 3.13, 3.14, 3.16; due October 24.

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Transformers Overview• Power systems are characterized by many different

voltage levels, ranging from 765 kV down to 220/110 volts.

• Transformers are used to transfer power between different voltage levels.

• The ability to inexpensively change voltage levels is a key advantage of ac systems over dc systems.

• In this section we’ll development models for the transformer and discuss various ways of connecting three phase transformers.

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Ideal TransformerFirst we review the voltage/current relationships for an

ideal transformer– no real power losses– magnetic core has infinite permeability– no leakage flux

We’ll define the “primary” side of the transformer as the side that usually receives power from a line etc, and the secondary as the side that usually delivers power to a load etc:– primary is usually the side with the higher voltage, but may

be the low voltage side on a generator step-up transformer.

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Ideal Transformer Relationships

1

1 1 2 2

1 21 1 2 2

1 2 1 1

1 2 2 2

Assume we have flux in magnetic material.

Then flux linking coil 1 having turns is:

, and similarly

,

= turns ratio

m

m m

m m

m

N

N N

d dd dv N v N

dt dt dt dtd v v V N

adt N N V N

Note that I2 and I2’ are in opposite directions

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Current Relationships

'1 1 2 2

'1 1 2 2

'1 1 2 2

To get the current relationships use ampere's law

with path around core having total length :

mmf

Assuming uniform flux density in the core

having area ,

L

d N i N i

H L N i N i

B LN i N i

A

H L

'1 1 2 2

then B A

LN i N i

A

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Current/Voltage Relationships'

1 1 2 2

'1 2 1 22 2'

1 2 12

1

2

1 2

1 2

If is infinite then 0 . Hence

1or , where

1Then: and:

0

10

N i N i

i N i Ni i

N i N ai

II a

aV V

I Ia

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Impedance Transformation Example

•Example: Calculate the primary voltage and current for an impedance load Z on the secondary

21

21

2 2 / and substituting:

0

10

a VVVIZa

I V Z

21 2 1

21

1

1

primary referred value of

secondary load impedance

VV aV I

a ZV

a ZI

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Real Transformers

• Real transformers– have losses– have leakage flux– have finite permeability of magnetic core

• Real power losses– resistance in windings (I2 R)– core losses due to eddy currents and hysteresis

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Transformer Core lossesEddy currents arise because of changing flux in core.Eddy currents are reduced by laminating the core

Hysteresis losses are proportional to area of BH curveand the frequency

These losses are reduced by using material with a “thin” BH curve

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Effect of Leakage Flux

1 1 1 1

2 2 2 2

1 1 1 2 2

Not all flux is within the transformer core

, where is the coil 1 leakage flux,

, where is the coil 2 leakage flux,

Assuming a linear magnetic medium we get

l m l

l m l

l l l l

N

N

L i L i

2

22

'

11 1 1 1 1

1

''

2 2 2 2 2

, including winding

resistance ,

, including resistance .

ml

ml

ddiv r i L N

dt dtr

di dv r i L N r

dt dt

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Effect of Finite Core Permeability

m

1 1 2 2 m

m 21 2

1 1

1 m

Finite core permeability means a non-zero mmf

is required to maintain in the core

,

where is the reluctance.

This effect is usually modeled as a magnetizing current

N i N i R

R

R Ni i

N N

Ni i

2 m2 m

1 1

where ,

modeled by resistance and inductance.

Ri i

N N

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Transformer Equivalent CircuitUsing the previous relationships, we can derive an equivalent circuit model for the real transformer

' 2 '2 2 1 2

' 2 '2 2 1 2

This model is further simplified by referring all

impedances to the primary side (and approximating

by swapping the referred elements and the shunts):

e

e

r a r r r r

x a x x x x

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Simplified Equivalent Circuit

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Calculation of Model ParametersThe parameters of the model are determined

based upon: – nameplate data: gives the rated voltages and power– open circuit test: rated voltage is applied to primary

with secondary open; measure the primary current and losses (the test may also be done applying the rated voltage to the secondary, calculating the values, then referring the values back to the primary side).

– short circuit test: with secondary shorted, apply (lower than rated) voltage to primary to get rated primary current to flow; measure voltage and losses.15

Transformer Example

•Example: A single phase, 100 MVA, 200/80 kV transformer has the following test data:

–open circuit: 20 amps, with 10 kW losses–short circuit: 30 kV, with 500 kW losses

•Determine the model parameters.

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Transformer Example, cont’d

2 2 2

2 2

2 2 2rated

100MVA 30 kV500 A, 60

200kV 500 A

500 kW / 500,000 /(500) 2 ,

Hence 60 2 60

( ) (200) (kV)4M

10 kW

sc e e

sc e sc e sc sc

e

coc

e

I R jX

P R I R P I

X

VR

P

R

From the short circuit test

From the open circuit test

rated 200 kV10,000 10,000

20 Ae m moc

VjX jX X

I

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Residential Distribution Transformers

Single phase transformers are commonly used in residential distribution systems. Most distributionsystems are 4 wire, with a multi-grounded, common neutral.

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Per Unit CalculationsA key problem in analyzing power systems is the large

number of transformers. – It would be very difficult to continually have to refer

impedances to the different sides of the transformersThis problem is avoided by a normalization of all

variables.This normalization is known as per unit analysis.

actual quantityquantity in per unit

base value of quantity

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Per Unit Conversion Procedure, 11. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB

5. Convert actual values to per unit

Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts) 20

Per Unit Solution Procedure

1. Convert to per unit (p.u.) (many problems are already in per unit)

2. Solve3. Convert back to actual as necessary

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Per Unit ExampleSolve for the current, load voltage and load power in the circuit shown below using per unit analysis with an SB of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV, respectively.

Original Circuit22

Per Unit Example, cont’d2 2

2 2

2 2

8 (kV)0.64

100MVA

80 (kV)64

100MVA

16 (kV)2.56

100MVA

LeftB

MiddleB

RightB

Z

Z

Z

Same circuit, withvalues expressedin per unit.

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Per Unit Example, cont’d

2*

1.0 00.22 30.8 p.u. (not amps)

3.91 2.327

1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

L

LL L L

G

Ij

V

VS V I

ZS

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Per Unit Example, cont’dTo convert back to actual values just multiply the per unit values by their per unit base

Actual

Actual

Actual

Middle

ActualMiddle

0.859 30.8 16 kV 13.7 30.8 kV

0.189 0 100 MVA 18.9 0 MVA

0.22 30.8 100 MVA 22.0 30.8 MVA

100 MVA1250 Amps

80 kV

0.22 30.8 275 30.8

L

L

G

B

V

S

S

I

I

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Three Phase Per Unit

1. Pick a 3 VA base for the entire system, 2. Pick a voltage base for each different voltage

level, VB,LL. Voltages are line to line. 3. Calculate the impedance base

Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases

3BS

2 2 2, , ,3 1 1

( 3 )

3B LL B LN B LN

BB B B

V V VZ

S S S

Exactly the same impedance bases as with single phase using the corresponding single phase VA base and voltage base!26

Three Phase Per Unit, cont'd

4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3

3 3 3B B B

B LL B LN B LN

S S SI I

V V V

Exactly the same current bases as with single phase!

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Three Phase Per Unit Example•Solve for the current, load voltage and load power in the previous circuit, assuming:

–a 3 power base of 300 MVA, –and line to line voltage bases of 13.8 kV, 138 kV and 27.6 kV (square root of 3 larger than the 1 example voltages)–the generator is Y-connected so its line to line voltage is 13.8 kV.

Convert to per unitas before. Note the system is exactly the same!

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3 Per Unit Example, cont'd

2*

1.0 00.22 30.8 p.u. (not amps)

3.91 2.327

1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

L

LL L L

G

Ij

V

VS V I

ZS

Again, analysis is exactly the same!29

3 Per Unit Example, cont'd

L

Actual

ActualL

ActualG

MiddleB

ActualMiddle

0.859 30.8 27.6 kV 23.8 30.8 kV

0.189 0 300 MVA 56.7 0 MVA

0.22 30.8 300 MVA 66.0 30.8 MVA

300 MVA125 (same cu0 Amps

3138 kV

0.22 30

rrent

8

!)

.

V

S

S

I

I

Amps 275 30.8

Differences appear when we convert back to actual values

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