Lecture 8Transformers, Per Unit

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

2

Announcements

For lectures 8 to 10 please be reading Chapter 3 Homework #3 is due now Homework #4 4.34, 4.35, 5.14, 5.26; due 9/25

3

Transformers Overview

Power systems are characterized by many different voltage levels, ranging from 765 kV down to 240/120 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.

4

Ideal Transformer

First 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 takes power, and the secondary as the side that usually delivers power.– primary is usually the side with the higher voltage, but

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

5

Ideal Transformer Relationships

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

= turns ratio

m

m m

m m

m

N N

d d d dv N v N

dt dt dt dtd v v v N

adt N N v N

6

Current Relationships

'1 1 2 2

'1 1 2 2

'1 1 2 2

'1 1 2 2

To get the current relationships use ampere's law

mmf

length

length

Assuming uniform flux density in the core

lengtharea

d N i N i

H N i N i

BN i N i

N i N i

H L

7

Current/Voltage Relationships

'1 1 2 2

1 2 1 2'

1 2 12

1 2

1 2

If is infinite then 0 . Hence

1or

Then

0

10

N i N i

i N i NN i N ai

av v

i ia

8

Impedance Transformation Example

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

21

21

0

10

a vvviZa

21 2 1

21

1

1 vv av i

a Zv

a Zi

9

Real Transformers

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

1. Real power losses– resistance in windings (i2 R)– core losses due to eddy currents and hysteresis

10

Transformer Core losses

Eddy 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

11

Effect of Leakage Flux

2

22

1 1 1

2 2 2

'1 1 1 2 2

11 1 1 1 1

''

2 2 2 2

Not all flux is within the transformer core

Assuming a linear magnetic medium we get

v

v

l m

l m

l l l l

ml

ml

N

N

L i L i

ddir i L N

dt dt

di dr i L N

dt dt

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Effect of Finite Core Permeability

m

1 1 2 2 m

m 21 2

1 1

2 m1 2 m

1 1

Finite core permeability means a non-zero mmf

is required to maintain in the core

N

This value is usually modeled as a magnetizing current

where im

i N i

Ni i

N N

Ni i i

N N

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Transformer Equivalent Circuit

Using 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

r e

e

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 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 with

secondary 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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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

e

2sc e

2 2e

2

e

100 30500 , R 60

200 500

P 500 kW R 2 ,

Hence X 60 2 60

2004

10

200R 10,000 10,000

20

sc e

e sc

c

e m m

MVA kVI A jX

kV A

R I

kVR M

kW

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 in residential distribution systems. Most distributionsystems 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, 1

1. 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 unitNote, 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 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.

Original Circuit

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Per Unit Example, cont’d

2

2

2

80.64

100

8064

100

162.56

100

LeftB

MiddleB

RightB

kVZ

MVA

kVZ

MVA

kVZ

MVA

Same circuit, withvalues expressedin per unit.

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Per 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

ZS

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Per Unit Example, cont’d

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

LActual

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 3 VA base for the entire system,

2. Pick a voltage base for each different voltage level, VB. 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!

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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 3B B B

B LL B LN B LN

S S S

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 of300 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). Also assume the generator is Y-connected so its line to line voltage is 13.8 kV.

Convert to per unitas before. Note thesystem is exactly thesame!

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3 Per 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

ZS

Again, analysis is exactly the same!

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3 Per Unit Example, cont'd

LActual

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 MVAI 125 (same cur0 Amps

3138 kV

I 0.22 30.

rent!)

8

V

S

S

Amps 275 30.8

Differences appear when we convert back to actual values

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3 Per Unit Example 2

Assume a 3 load 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