1

Lec 2: Power System Analysis 1

Dr. Malabika Basu

Transmission Lines

Equivalent Circuit• An equivalent circuit refers to the simplest form of a circuit that retains

all of the electrical characteristics of the original (and more complex) circuit. In its most common form, an equivalent circuit is made up of linear, passive elements. However, more complex equivalent circuits are used that approximate the nonlinear behavior of the original circuit as well. (source wikipedia)

Transformers Overview• Power systems are characterized by many different

voltage levels, ranging from 400 kV down to 230 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 develop models for the transformer and discuss various ways of connecting three phase transformers.

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.

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 Nd d d dv N v Ndt dt dt dt

d v v v N adt N N v N

φλ φ λ φ

λ φ λ φ

φ

= =

= = = =

= = → = =

2

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

lengthlength

Assuming uniform flux density in the corelengtharea

d N i N i

H N i N iB N i N i

N i N i

µ

φµ

Γ= = +

× = +×

= +

× = +×

∫ H Li

Current/Voltage Relationships'

1 1 2 2

1 2 1 2'

1 2 12

1 2

1 2

If is infinite then 0 . Hence1or

Then010

N i N ii N i N

N i N ai

av vi i

a

µ = +

= − = =

=

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

2121

010

a vvvi Za

=

21 2 1

21

1

1 vv av ia Z

v a Zi

= =

=

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

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

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

NN

L i L i

ddir i L Ndt dtdi dr i L Ndt dt

λ λ φλ λ φ

λ λ

φ

φ

= += +

= + +

= + +

3

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

NThis value is usually modeled as a magnetizing current

where im

i N i

Ni iN N

Ni i iN N

φφ

φ

φ

− = ℜ

ℜ= +

ℜ= + =

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 allimpedances to the primary side

r e

e

a r r r r

x a x x x x

= = +

= = +

Simplified Equivalent Circuit Calculation of Model Parameters• The parameters of the model are determined based upon

– Name plate 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.

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.

Transformer Example, cont’d

e

2sc e

2 2e

2

e

100 30500 , R 60200 500

P 500 kW R 2 ,

Hence X 60 2 60

200 410

200R 10,000 10,00020

sc e

e sc

c

e m m

MVA kVI A jXkV A

R I

kVR MkW

kVjX jX XA

= = + = = Ω

= = → = Ω

= − = Ω

= = Ω

+ + = = Ω = Ω

From the short circuit test

From the open circuit test

4

Residential Distribution TransformersSingle phase transformers are commonly used in residential distribution systems. Most distributionsystems are 4 wire, with a multi-grounded, commonneutral.

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

=

Per Unit Conversion Procedure, 1φ1. Pick a 1φ 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 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)

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

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.

Original Circuit

Per Unit Example, cont’d2

2

2

8 0.64100

80 6410016 2.56

100

LeftB

MiddleB

RightB

kVZMVA

kVZMVA

kVZMVA

= = Ω

= = Ω

= = Ω

Same circuit, withvalues expressedin per unit.

5

Per Unit Example, cont’d

L

2*

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

V 1.0 0 0.22 30.8p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V IZ

S

∠ °= = ∠ − °+

= ∠ ° − ∠ − °× 2.327∠90°= 0.859∠ − 30.8°

= = =

= ∠ °× ∠ ° = 0.22∠ °

Per Unit Example, cont’d

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

LActual

ActualLActualG

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 MVA100 MVAI 1250 Amps

80 kVI 0.22 30.8 Amps 275 30.8

V

S

S

= ∠ − °× = ∠ − °

= ∠ °× = ∠ °

= ∠ °× = ∠ °

= =

= ∠ − °×1250 = ∠ − ° Α

1. Pick a 3φ 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

Three Phase Per UnitProcedure is very similar to 1φ except we use a 3φVA base, and use line to line voltage bases

3BS φ

2 2 2, , ,3 1 1

( 3 )3

B LL B LN B LNB

B B B

V V VZ

S S Sφ φ φ= = =

Exactly the same impedance bases as with single phase!

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 I3 3 3

B B B

B LL B LN B LN

S S SV V V

φ φ φφ φ= = = =

Exactly the same current bases as with single phase!

Three Phase Per Unit Example•Solve for the current, load voltage and load power •in the previous circuit, assuming a 3φ 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 1φexample 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!

3φ Per Unit Example, cont'd

L

2*

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

V 1.0 0 0.22 30.8p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V IZ

S

∠ °= = ∠− °

+= ∠ °− ∠− °×2.327∠90°= 0.859∠−30.8°

= = =

= ∠ °× ∠ ° =0.22∠ °

Again, analysis is exactly the same!

6

3φ Per Unit Example, cont'd

LActual

ActualLActualG

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 MVA300 MVAI 125 (same cur0 Amps

3138 kV

I 0.22 30.

rent!)

8

V

S

S

= ∠ − °× = ∠ − °

= ∠ °× = ∠ °

= ∠ °× = ∠ °

= =

= ∠ − °× Amps 275 30.81250 = ∠ − ° Α

Differences appear when we convert back to actual values

3φ Per Unit Example 2•Assume a 3φ 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

‘a’ Operator)120sin120(cos1201

01200 jea j +==∠=

)240sin240(cos2401024002 jea j +==∠=

01 2 =++ aa

=

2

1

0

2

2

11

111

VVV

aaaa

VVV

c

b

a

=

c

b

a

VVV

aaaa

VVV

2

2

2

1

0

11

111

31

Vp = AVsVs = A-1Vp

Definition of Symmetrical Components

• Zero-sequence components: consisting of 3 phasors of equal magnitudes and with zero phase displacements

• Positive sequence components: consisting of 3 phasors with equal magnitudes, ±120 0 phase displacement, and +ve sequence

• Negative Sequence components: consisting of 3 phasors with equal magnitudes, ±120 0 phase displacement, and -ve sequence

Different sequence components

Vao Vbo Vco = Vo

a1

b1

c1a0 b0 c0

a2

b2

c2

Conversion Phase to Sequence

1

1 2

2

By taking the inverse we can convert from thephase values to the sequence values

1 1 11with 13

1Sequence sets can be used with voltages as wellas with currents

s

α α

α α

−

−

=

=

I A I

A

7

Example 8.1, 8.2 (Glover)

Symmetrical Component Example 2

1 2s

2

0Let

Then1 1 1 0 0

1 13

6.121

a

b

c

VVV

α α

α α

−

5∠9 ° = = 8∠150°

8∠ − 30°

5∠9 ° 1.67∠9 ° = = 8∠150° = 3.29∠ −135° 8∠ − 30° ∠68°

V

V A V

Symmetrical Component Example 30

2

2

10 0Let 10

Then1 1 1 10 01 10

1

s

s

I

I

I

α α

α α

+

−

∠ ° = = − ∠0° 5∠0°

∠ ° 5.0∠0° = = − ∠0° = 18.0∠46.1° 5∠0° 18.0∠ − 46.1°

I

I AI

Use of Symmetrical Components• Consider the following wye-connected load:

( )

( )

( )

n a b c

ag a y n n

ag Y n a n b n c

bg n a Y n b n c

cg n a n b Y n c

I I I IV I Z I Z

V Z Z I Z I Z I

V Z I Z Z I Z I

V Z I Z I Z Z I

= + += +

= + + +

= + + +

= + + +

ag y n n n a

bg n y n n b

ccg n n y n

V Z Z Z Z IV Z Z Z Z I

IV Z Z Z Z

+ = + +

Use of Symmetrical Components

1

1

3 0 0

0 0

0 0

ag y n n n a

bg n y n n b

ccg n n y n

s s

s s s s

y n

y

y

V Z Z Z Z IV Z Z Z Z I

IV Z Z Z Z

Z Z

Z

Z

−

−

+ = + +

= = =

= → =

+

=

V Z I V A V I A I

A V Z A I V A Z A I

A Z A

Networks are Now Decoupled0 0

0 0

3 0 0

0 0

0 0

Systems are decoupled

( 3 )

y n

y

y

y n y

y

V IZ Z

V Z I

ZV I

V Z Z I V Z I

V Z I

+ +

− −

+ +

− −

+

=

= + =

=