lec2 power system

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POWER SYSTEMS ILecture 2

06-88-590-68

Electrical and Computer Engineering

University of Windsor

Dr. Ali Tahmasebi



1

Per Phase Analysis

l Per phase analysis allows analysis of balanced 3f systems with the same effort as for a single phase

system

l Balanced 3f Theorem: For a balanced 3f systemwith

– All loads and sources Y connected

– No mutual Inductance between phases



2

Per Phase Analysis, cont’d

l Then

– All neutrals are at the same potential

– All phases are COMPLETELY decoupled

– All system values are the same sequence as sources. Thesequence order we’ve been using (phase b lags phase a

and phase c lags phase a) is known as “positive”

sequence; later in the course we’ll discuss negative and

zero sequence systems.



3

Per Phase Analysis Procedure

To do per phase analysis

1. Convert all D load/sources to equivalent Y’s

2. Solve phase “a” independent of the other phases

3. Total system power S = 3 Va Ia*

4. If desired, phase “b” and “c” values can be

determined by inspection (i.e., ±120° degree phase

shifts)5. If necessary, go back to original circuit to determine

line-line values or internal Dvalues.



4

Per Phase Example

Assume a 3f , Y-connected generator with Van = 1Ð 0°volts supplies a D-connected load with ZD = -jWthrough a transmission line with impedance of j0.1W

per phase. The load is also connected to aD-connected generator with Va”b” = 1Ð 0°through asecond transmission line which also has an impedanceof j0.1W per phase.

Find

1. The load voltage Va’b’

2. The total power supplied by each generator, SY and SD



5

Per Phase Example, cont’d

First convert the delta load and source to equivalent

Y values and draw just the "a" phase circuit



6


' ' 'a a a

To solve the circuit, write the KCL equation at a'1

(V 1 0)( 10 ) V (3 ) (V j3

j j- Ð - + + - Ð - 30°)(- 10 )=0



7


' ' 'a a a

'a

' 'a b

' 'c ab

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

3

10(10 60 ) V (10 3 10 )3

V 0.9 volts V 0.9 volts

V 0.9 volts V 1.56

j j

j j j j

- Ð - + + - Ð - 30°)(- 10 ) =0

+ Ð °= - +

= Ð - 10.9° = Ð - 130.9°

= Ð 109.1° = Ð 19. volts1°



8


*'*

ygen

*

" '"

S 3 5.1 3.5 VA0.1

3 5.1 4.7 VA0.1

a aa a a

a agen a

V V V I V j

j

V V S V j j

D

æ ö-= = = +ç ÷

è ø

æ ö-= =- -ç ÷è ø



9

Power in Balanced 3 Circuits

In this balanced three-

phase circuit, T can be a

voltage source, a motor

or any impedance load.

For phase ‘a’: = 2 cos +

= 2 cos +



10

Power in Balanced 3 Circuits, cont’d

d - b = q is the phase shift between phase voltage and

phase current.

= () .()

= cos − + cos 2 + +

∅ = () + () + () = 3 cos − =

∅ (NOT a function of time)

In terms of line-to-line voltages:

∅ = 3 cos − = ∅



11


• For a generator, delivered power is calculated from

these 2 equations, for a load, these equations

calculate absorbed power.

• Advantage: for example, power delivered to a 3Fac motor is always constant, which results in

constant speed and torque. In contrast, a single

phase ac motor may have torque pulsations.



12


Complex power:

For phase ‘a’:

= Ð d , ̅ = Ð b ® ̅ = ̅∗ = Ð (d-b)

̅∅ = ̅ + ̅ + ̅ = 3 ̅= 3 ̅∗ = 3Ð (d−

b)

= 3 cos + sin = 3 cos + sin

̅∅

= ∅

+ ∅

Total apparent power:

̅∅ = ∅ = 3= 3 [VA]

NOTE: ̅ is a complex number but not a phasor!



13

Power System Components

l All power systems have three major components:

Generation, Load and Transmission/Distribution.

l Generation: Creates electric power.

l Load: Consumes electric power.

l Transmission/Distribution: Transmits electric power

from generation to load.

– Lines/transformers operating at voltages above 100 kV

are usually called the transmission system. The

transmission system is usually networked.

– Lines/transformers operating at voltages below 100 kV

are usually called the distribution system (radial).



14

Power Transformers

l Power systems are characterized by many different

voltage levels, ranging from 765 kV down to

240/120 volts.

l Transformers are used to convert voltage and transfer power between different voltage levels.

l The ability to inexpensively change voltage levels is

a key advantage of ac systems over dc systems.

l In this section we’ll development models for the

transformer and discuss various ways of connecting

three phase transformers.



15

Ideal Transformer

l First we review the voltage/current relationships for

an ideal transformer

– no winding resistance ® no real power losses

– magnetic core has infinite permeability

– no leakage flux

– no core loss

l We’ll define the “primary” side of the transformer

as the side that usually takes power, and thesecondary 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.



16

Ideal Transformer Equations

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 d v N v N dt dt dt dt

d v v v N a

dt N N v N

f

l f l f

l f l f

f

= =

= = = =

= = ® = =



17

Current Equations in Ideal Transformer

'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

B N i N i

N i N i

m

f

m

G= = +

´ = +

´= +

´ = +´

òH L



18

Current/Voltage Equations

'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 N

N i N ai

av v

i i

a

m = +

= - = =

é ùé ù é ùê ú=ê ú ê úê úë û ë û

û



19

Impedance Transformation Example

• Example: Calculate the primary voltage and current

for an impedance load on the secondary

21

21

01

0

a vv

vi Z a

é ùé ùé ù ê úê ú=ê ú ê úë û ê úûû

21 2 1

21

1

1 vv a v ia Z

va Z

i

= =

=



20

Real Transformers

• Real transformers

– have losses

– have leakage flux

– have finite permeability of magnetic core

• Add R1

and R2

to windings 1 and 2 to represent i2 R

losses of each winding

• Add X 1 to account for leakage flux of winding 1

(flux that links winding 1 but not winding 2)



21

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 curve

and the frequency

These losses are reduced

by using material with a

thin BH curve



22

Transformer Core Losses

• To represent both eddy current loss and Hysteresis

loss, we add a resistive shunt branch R C (or GC) that

carries the current iC = core loss current.



23

Effect of Finite Core Permeabili ty

m

1 1 2 2 m

m 21 2

1 1

2 m1 2 m1 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

N i i

N N

N i i i N N

f

f

f

f

- = Â

Â= +

Â= + =

Add X m

to represent this effect.



24

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

= = +

= = +



25

Simplified Equivalent Circuit



26

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 secondarywith primary open; measure the secondary current and

losses (Wattage).

– short circuit test: with secondary shorted, apply voltage

to primary to get rated current to flow at primary;

measure voltage and losses.



27

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.



28

Transformer Example, cont’d

e

2

sc e

2 2e

2

e

100 30500 , R 60

200 500

P 500 kW R 2 ,

Hence X 60 2 60

200410

200R 10,000 10,000

20

sc e

e sc

c

e m m

MVA kV I A jX

kV A

R I

kV R M kW

kV jX jX X

A

= = + = = W

= = ® = W= - = W

= = W

+ + = = W =

From the short circuit test

From the open circuit test



29

Residential Distribution Transformers

Single phase transformers are commonly used in

residential distribution systems. Most distribution

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

neutral.



30

Per Unit Calculations

l 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 transformersl This problem is avoided by a normalization of all

variables.

l This normalization is known as per unit analysis.

actual quantityquantity in per unit

base value of quantity=



31

Per Unit Conversion Procedure, 1

1. Pick a 1f 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)



32

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



33

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



34

Per Unit Example, cont’d

2

2

2

80.64

100

8064

100

162.56

100

Left B

Middle B

Right B

kV Z

MVA

kV Z

MVA

kV Z

VA

= = W

= = W

= = W

Same circuit, with

values expressed

in per unit.



35


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.

L L L L

G

I j

V S V I

Z

S

Ð °= = Ð - °+

= Ð °- Ð - °́ 2.327Ð 90°

= 0.859Ð - 30.8°

= = =

= Ð °́ Ð °=0.22Ð °



36


To convert back to actual values just multiply the

per 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 Amps80 kV

I 0.22 30.8 Amps 275 30.8

V

S

S

= Ð - °́ = Ð - °= Ð °́ = Ð °

= Ð °́ = Ð °

= =

= Ð - °́ 1250 = Ð - °A

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