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Lecture 9

Transformers, Per Unit

Professor Tom Overbye

Department of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

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1

Announcements

Be reading Chapter 3

HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class.

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2

Transformer Equivalent Circuit

Using the previous relationships, we can derive anequivalent 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

r e

e

a r r r r

x a x x x x

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3

Simplified Equivalent Circuit

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4

Calculation of Model Parameters

The 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 withsecondary open; measure the primary current and losses

(the test may also be done applying the voltage to the

secondary, calculating the values, then referring the

values back to the primary side).

short circuit test: with secondary shorted, apply voltage

to primary to get rated current to flow; measure voltage

and losses.

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5

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, contd

e

2

sc e2 2

e

2

e

100 30500 , R 60

200 500

P 500 kW R 2 ,

Hence X 60 2 60

200 410

200R 10,000 10,000

20

sc e

e sc

c

e m m

MVA kVI A jX

kV A

R I

kVR MkW

kVjX jX X

A

From the short circuit test

From the open circuit test

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Residential Distribution Transformers

Single phase transformers are commonly used inresidential distribution systems. Most distribution

systems are 4 wire, with a multi-grounded, common

neutral.

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Per Unit Calculations

A 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 transformers

This 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, 1f1. Pick a 1fVA 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 on 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)

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Per Unit Solution Procedure

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

already in per unit)

2. Solve

3. Convert back to actual as necessary

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Per Unit Example

Solve for the current, load voltage and load powerin 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.

Original Circuit

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Per Unit Example, contd

2

2

2

8 0.64100

8064

10016

2.56100

LeftB

MiddleB

RightB

kVZMVA

kVZ

MVAkV

ZMVA

Same circuit, with

values expressed

in per unit.

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Per Unit Example, contd

L

2*

1.0 0 0.22 30.8 p.u. (not amps)3.91 2.327

V 1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V I

Z

S

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Per Unit Example, contd

To convert back to actual values just multiply theper unit values by their per unit base

L

Actual

ActualL

ActualG

MiddleB

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 MVAI 1250 Amps

80 kV

I 0.22 30.8 Amps 275 30.8

V

S

S

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Three Phase Per Unit

1. Pick a 3fVA base for the entire system,

2. Pick a voltage base for each different voltagelevel, VB. Voltages are line to line.

3. Calculate the impedance base

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

3BS f

2 2 2, , ,

3 1 1( 3 )

3

B LL B LN B LNB

B B B

V V VZS S Sf f f

Exactly the same impedance bases as with single phase!

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Three Phase Per Unit, cont'd

4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3I I

3 3 3

B B B

B LL B LN B LN

S S S

V V V

f f ff f

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 3fpower 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 the1fexample voltages). Also assume the generator is

Y-connected so its line to line voltage is 13.8 kV.

Convert to per unit

as before. Note the

system is exactly the

same!

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3fPer Unit Example, cont'd

L

2*

1.0 00.22 30.8 p.u. (not amps)

3.91 2.327

V 1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V I

Z

S

Again, analysis is exactly the same!

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3fPer Unit Example, cont'd

L

Actual

Actual

LActualG

MiddleB

ActualMiddle

0.859 30.8 27.6 kV 23.8 30.8 kV

0.189 0 300 MVA 56.7 0 MVA0.22 30.8 300 MVA 66.0 30.8 MVA

300 MVAI 125 (same cur0 Amps

3138 kVI 0.22 30.

rent!)

8

V

SS

Amps 275 30.8

Differences appear when we convert back to actual values

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3fPer Unit Example 2Assume a 3fload of 100+j50 MVA with VLL of 69 kV

is connected to a source through the below network:

What is the supply current and complex power?

Answer: I=467 amps, S = 103.3 + j76.0 MVA

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Per Unit Change of MVA Base

Parameters for equipment are often given using

power rating of equipment as the MVA base

To analyze a system all per unit data must be on a

common power base

2 2

Hence Z /

Z

base base

OriginalBase NewBasepu actual pu

OriginalBase NewBase

pu puOriginalBase NewBaseBaseBase

NewBaseOriginalBase NewBaseBasepu puOriginalBase

Base

Z Z Z

V V

ZSS

SZ

S

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Per Unit Change of Base Example

A 54 MVA transformer has a leakage reactance or

3.69%. What is the reactance on a 100 MVA base?

100

0.0369 0.0683 p.u.54eX

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Transformer Reactance

Transformer reactance is often specified as a

percentage, say 10%. This is a per unit value

(divide by 100) on the power base of the

transformer. Example: A 350 MVA, 230/20 kV transformer has

leakage reactance of 10%. What is p.u. value on

100 MVA base? What is value in ohms (230 kV)?

2

1000.10 0.0286 p.u.350

2300.0286 15.1

100

eX

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Three Phase Transformers

There are 4 different ways to connect 3ftransformersY-Y -

Usually 3ftransformers are constructed so all windings

share a common core

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3fTransformer Interconnections-Y Y-

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Y-Y Connection

Magnetic coupling with An/an, Bn/bn & Cn/cn

1, ,An AB A

an ab a

V V Ia a

V V I a

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Y-Y Connection: 3fDetailed Model

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Y-Y Connection: Per Phase Model

Per phase analysis of Y-Y connections is exactly the

same as analysis of a single phase transformer.

Y-Y connections are common in transmission systems.

Key advantages are the ability to ground each side

and there is no phase shift is introduced.

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- Connection

Magnetic coupling with AB/ab, BC/bb & CA/ca

1 1, ,AB AB A

ab ab a

V I Ia

V I a I a

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- Connection: 3fDetailed Model

To use the per phase equivalent we need to use

the delta-wye load transformation

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- Connection: Per Phase Model

Per phase analysis similar to Y-Y except impedances

are decreased by a factor of 3.

Key disadvantage is D-Dconnections can not be

grounded; not commonly used.

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-Y Connection

Magnetic coupling with AB/an, BC/bn & CA/cn

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-Y Connection V/I Relationships

, 3 30

30 30Hence 3 and 3

For current we get

1

13 30 303

130

3

AB AB an ab anan

AB Anab an

ABa AB

ab

A AB AB A

a A

V Va V V V V a

V VV V

a a

II a I

I a

I I I I

a I

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-Y Connection: Per Phase Model

Note: Connection introduces a 30 degree phase shift!

Common for transmission/distribution step-down sincethere is a neutral on the low voltage side.

Even if a = 1 there is a sqrt(3) step-up ratio

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Y- Connection: Per Phase Model

Exact opposite of the D-Y connection, now with a

phase shift of -30 degrees.

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Load Tap Changing Transformers

LTC transformers have tap ratios that can be varied

to regulate bus voltages

The typical range of variation is 10% from the

nominal values, usually in 33 discrete steps(0.0625% per step).

Because tap changing is a mechanical process, LTC

transformers usually have a 30 second deadband to

avoid repeated changes.

Unbalanced tap positions can cause "circulating

vars"

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LTCs and Circulating Vars

slack

1 1.00 pu

2 3

40.2 MW

40.0 MW

1.7 Mvar

-0.0 Mvar

1.000 tap 1.056 tap

24.1 MW

12.8 Mvar

24.0 MW-12.0 Mvar

A

MVA

1.05 pu0.98 pu

24 MW

12 Mvar

64 MW

14 Mvar

40 MW

0 Mvar

0.0 Mvar

80%A

MVA

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Phase Shifter Example 3.13

slack

Phase Shifting Transformer

345.00 kV 341.87 kV

0.0 deg

216.3 MW 216.3 MW

283.9 MW 283.9 MW

1.05000 tap

39.0 Mvar 6.2 Mvar

93.8 Mvar125.0 Mvar

500 MW

164 Mvar 500 MW100 Mvar

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ComED Control Center

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ComED Phase Shifter Display

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Autotransformers

Autotransformers are transformers in which the

primary and secondary windings are coupled

magnetically and electrically.

This results in lower cost, and smaller size andweight.

The key disadvantage is loss of electrical isolation

between the voltage levels. Hence auto-

transformers are not used when a is large. Forexample in stepping down 7160/240 V we do not

everwant 7160 on the low side!