1

Instructor: Kai Sun

Fall 2013

ECE 421/521 Electric Energy Systems

Power Systems Analysis I 3 – Generators, Transformers and

the Per-Unit System

2

Outline

• Synchronous Generators

• Power Transformers

• The Per-Unit System

3

Synchronous Generators

Stator

Salient-pole rotor Cylindrical/round rotor

Field current

Armature winding

Field winding

4

Types of Rotors • Salient pole rotors

– Concentrated windings on poles and non-uniform air gap

– Short axial length and large diameter – Hydraulic turbines operated at low speeds

(large number of poles) – Have damper windings to help damp out speed

oscillations • Round rotors

– 70% of large synchronous generators (150~1500MVA)

– Distributed winding and uniform air gap – Large axial length and small diameter to limit

the centrifugal forces – Steam and gas turbines, operated at high

speeds, typically 3600 or 1800rpm (2 or 4-pole) – Eddy in the solid steal rotor gives damping

effects

Round rotor generator under construction

16 poles salient-pole rotor (12 MW)

(Source: http://emadrlc.blogspot.com)

5

Generator Model

cosa N tλ φ ω=• Flux linkage with coil a (leading the axis of aa’ by ωt)

max

max

sin sin

cos( )2

aa

de N t E tdt

E t

λ ω φ ω ω

πω

= − = =

= −

max 2E N fNω φ π φ= =

| | 4.44E fNφ= (rms value)

• Induced voltage:

2 60P nf = (n: synchronous speed in rpm; P: the number of poles)

d

q Axis of coil a

(reference)

N

S

(occurring when ωt=π/2)

• Assume: ia is lagging ea by ψ (ia reaches the maximum when mn aligns with aa’)

n

m

ψ

max max max2 4sin( ) sin( ) sin( )3 3a b ci I t i I t i I tω ψ ω ψ π ω ψ π= − = − − = − −

ea

6

max

max

max

sin( ) sin( )2 2sin( ) sin( )3 34 4sin( ) sin( )3 3

a a m

b b m

c c m

F Ki KI t F t

F Ki KI t F t

F Ki KI t F t

ω ψ ω ψ

ω ψ π ω ψ π

ω ψ π ω ψ π

= = − = −

= = − − = − −

= = − − = − −

d

q Axis of coil a

(reference)

N

S

n

m

ψ

ψ ea

Fr

Fs

1 sin( ) cos( )2 2sin( )cos( )3 34 4sin( )cos( )3 3

m

m

m

F F t t

F t t

F t t

ω ψ ω ψ

ω ψ π ω ψ π

ω ψ π ω ψ π

= − −

+ − − − −

+ − − − −

• Total mmf Fs due to three phases of the stator

Component along mn:

1

2 4[sin 2( ) sin 2( ) sin 2( )] 0

2 3 3mF

F t t tπ π

ω ψ ω ψ ω ψ= − + − − + − − =

Component orthogonal to mn

22 2 4 4

sin( ) sin( ) sin( ) sin( ) sin( ) sin( )3 3 3 3m m mF t t F t t F t tF π π π π

ω ϕ ω ψ ω ψ ω ψ ω ψ ω ψ= − − + − − − − + − − − −

( )2sin 1 cos 2 / 2α α= −22 4 3

[3 cos 2( ) cos 2( ) cos 2( )]2 3 3 2 mmF

F t t t Fπ πω ψ ω ψ ω ψ= − − + − − + − − =

32s mF F= Fs, the resultant total armature/stator mmf has constant amplitude, and is

orthogonal to mn and revolving synchronously with Fr

• Magnetic motive forces (mmf’s) of three phases:

7

Phasor Diagram • No-load conditions:

– No phase currents and Fs=0 – In each phase, field mmf Fr produces no-load E

(excitation voltage) • Loading conditions:

– Armature carries three-phase currents – Fs≠0 is orthogonal to mn and induces emf Ear

– The sum of Fr and Fs gives mmf Fsr in air gap

– The resultant air gap flux φsr induces emf Esr=E+Ear

– Define the reactance of the armature reaction Xar= -Ear/(jIa)

– Terminal voltage V, resistance Ra and leakage

reactance Xl

Xs=Xl+Xar is known as the synchronous reactance

d

q Axis of coil a

(reference)

N

S

n

m

ψ

ψ ea

Fr

Fs

Fsr

Fs

Fsr

Fr

sr ar aE E jX I= +

[ ( )] ( )a ar a a s aE V R j X X I V R jX I= + + + = + +

E Esr

Ear

(Here, we ignore the difference between d and q axes, so it is good for round-rotor generators)

8

Circuit Model (per Phase)

• When Fr is ahead Fsr by δr, the machine is operating as a generator

• When Fr falls behind of Fsr by δr, (Ia changes the direction) the machine acts as a motor

• Synchronous condenser (δr=0), used to supply or absorb VARs to regulate line voltages.

• Since Xl ≈0, δr ≈δ • Generated real power:

• δr or δ is known as the power angle

Fs

Fsr

Fr

E

Load

jXsIa

3 3 sins

E VP

Xφ δ≈

9

• Consider different power factors

Load

E ( )a s aE V R jX I= + +

0a

s

E VI

Zδ

γ∠ − ∠

=∠

10

Real and Reactive Power Outputs

• Consider a generator connected to an infinite bus, which is an equivalent bus of a large power network and has constant voltage V (magnitude, angle and frequency).

s aE V jX I= +E∠δ V∠0

3 3 sins

E VP

Xφ δ=

max(3 ) 3s

E VP

Xφ =

*3 3 aS VIϕ =

( )3 3 cos 3 ( )s s

V VQ E V E V

X Xφ δ= − −

3 3 sins

P E VT

Xφ δω ω

= =

max 3s

E VT

Xω=

11

*3 3 3 cos .a aP VI V I constφ θ = ℜ = =

cos | | .aI ab constθ = =

θ3

a b

θ3

1 1 1 1 .| | sin coss a constcd E X Iδ θ == =

3 3 ( )s

VQ E V

Xφ −

E∠δ V∠0

θ1>0, θ2=0, θ3<0

• Consider constant real power output:

• Reactive power output can be controlled by means of the rotor excitation (adjusting E) while maintaining a constant real power output

– If |E|>|V|, the generator delivers reactive power to the bus, and the generator is said to be overexcited.

– If |E|<|V|, the generator is absorbing reactive power from the bus – Generators are main source of reactive power, so they are normally operated in

the overexcited mode – δ ↑ |E|↓ Minimum excitation: when δ=δmax=90o, i.e. the stability limit

12

Examples 3.1 E∠δ V∠0

13

E∠δ V∠0

31sin3

sP XE Vφδ −

=

14

Example 3.2 E∠δ V∠0

15

E∠δ V∠0

16

Comparison E δ Ia θ 23.6kV 10.6o 807.5A -54.43o

18.7kV 21.8o 769.8A 0 14.0kV 29.7o 962.3A 36.87o

6.9KV (minimum) 90o 2073A 68.2o

• When excitation decreases, – armature current first decreases and then increases

(V-shape function) – power factor changes from lagging to leading

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• V Curve of a synchronous machine: – the curve of the armature current Ia as the function of the field

current If – Example 3.3: for the generator of Example 3.1, construct the V

curve for the rated power of 40MW with varying field excitation from 0.4 P.F. leading to 0.4 P.F. lagging. Assume the open-circuit characteristics in the operating region is given by |E|=2000|If| (V)

30 (kV)3

V =

3 40 (kA)3 cos 30 3 cosa

PI

Vφ

θ θ= =

40(MW)P =

30 9 | | (kV)3s a aE V jX I j I θ= + = + ∠

| | | | 1000 / 2000(A)fI E= ×

θ=cos-1(0.4)~ -cos-1(0.4)

18

Salient-Pole Synchronous Generators

• The model developed in previous slides assumes uniform magnetic reluctance in the air gap (round rotors).

• For a salient-pole rotor, the reluctance in the rotor direct axis (d axis) is less than that in the quadrature axis (q axis) – Represent the machine by two circuits

respectively for d and q axes – Define d-axis and q-axis reactances

(Xd>Xq) to replace a single synchronous reactance Xs

– Resolve Ia into d-axis and q-axis components: Id and Iq

[ ( )] ( )a ar a a s aE V R j X X I V R jX I= + + + = + +

d

q Axis of coil a

(reference)

N

S

n

m

ψ

ea

19

cos d dE V X Iδ= +

3 3 cosaP V Iφ θ=

cos cos sina q dI ab de I Iθ δ δ= + = +

sin q qV X Iδ =

sinq

q

VI

Xδ

=

cosd

d

E VI

Xδ−

=

( ) 23 3 cos sin 3 sin 3 sin 2

2d q

q dd d q

X XE VP V I I V

X X Xφ δ δ δ δ−

= + = +

3 3 sind

E VP

Xφ δ≈Approximately, the second item can be ignored:

20

Power Transformer

• Ideal transformers – Winding resistance is

negligible – No leakage flux – Permeability of the core is

infinite (net mmf to establish flux in core is zero.)

– No core loss

• Real transformers: – Windings have resistance – Windings do not link the

same flux – Permeability of the core is

not infinite – Core losses (hysteresis

losses and eddy current losses due to time varying flux)

21

Ideal Transformer • Assuming sinusoidal flux

• For an ideal transformer – No flux leakage:

– Zero reluctance in core (exact mmf balance between the primary and

secondary)

max cos tφ ω= Φ

( )1 1 1 max

1max

sin

cos 90

de N N tdt

E t

φ ω ω

ω

= = − Φ

= + °

1max 1 max2E fNπ= Φ

1 1 max4.44E fN= Φ

Note: Flux is lagging the induced voltage by 90o

2 2 max4.44E fN= Φ

2 1 2 2I N I N′ =

1 2 1

2 2 2

E I NE I N

= =′

i’2

22

Real Transformers • Modeling the current under no-load conditions (due to finite core permeability and core losses)

I2=0 I1=I0 ≠0 I0=Im +Ic

Im is the magnetizing current to set up the core flux

– In phase with flux and lagging E1 by 90o, modeled by E1/(jXm1) Ic supplies the eddy-current and hysteresis losses in the core

– A power component, so it is in phase with E1, modeled by E1/(Rc1)

• Modeling flux leakages – Primary and secondary flux leakage reactances: X1 and X2

• Modeling winding resistances

– Primary and secondary winding resistances: R1 and R2

Ideal Transformer

23

2 2 2 2E V Z I= +

11 2

2

1 12 2 2

2 22

1 12 2 2

2 2

2 2 2

NE ENN NV Z IN N

N NV Z IN NV Z I

=

= +

′= +

′ ′ ′= +

2 2 22 2

1 12 2

2 2

Z R jX

N NR j XN N

′ ′ ′= +

= +

Exact equivalent circuit referred to the primary side

Exact Equivalent Circuit

24

• For no-load conditions: V1≈E1

R1 and X1 can be combined with R’2 and X’2 to obtain equivalent Re1 and Xe1

Approximate equivalent circuit referred to the primary side

( )1 2 1 1 2e eV V R jX I′ ′= + +

2

11 1 2

2e

NR R RN

= +

2

11 1 2

2e

NX X XN

= +

*

223LSI

V ∗′ =

′

Approximate Equivalent Circuits

Approximate equivalent circuit referred to the secondary side

( )1 2 2 2 2e eV V R jX I′ = + +

25

Simplified Circuits Referred to One Side

• Power transformers are generally designed with very high permeability core and very small core loss.

• If ignoring the shunt branch:

26

Determination of Equivalent Circuit Parameters

• Open-circuit (no-load) test – neglect (R1+jX1)I0 – Measure input voltage V1, current I0, power P0 (core/iron loss)

21

10

cVRP

= 1

1c

c

VIR

= 2 20m cI I I= −

11m

m

VXI

=

P0

27

• Short-circuit test – Apply a low voltage VSC to create rated current ISC

– Neglect the shunt branch due to the low core flux

1sc

esc

VZI

=( )1 2

sce

sc

PRI

= 2 21 1 1e e eX Z R= −

ISC

VSC

PSC

28

Transformer Performance • Efficiencies vary from 95% to 99%

|S|=3|V2||I2| full-load rated VA

n: fraction of the full load power |S|×PF

Pcu=3Re2|I2|2 full-load copper loss (current dependent)

Pc: core/iron loss at rated voltage (mainly voltage dependent, almost constant)

• Maximum efficiency (constant PF) occurs when

var

output powerinput power

out

out const

PP P P

η = =+ +

2

0dd Iη

= c

cu

PnP

=Learn Example 3.4

( ) ( )2

| | | || | | | /cu c cu c

n S PF S PFn S PF n P P S PF n P P n

× × ×= =

× × + × + × + × +

29

Voltage Regulation • % change in terminal voltage from no-load to full load (rated)

– Generator

– Transformer

Utilizing the equivalent circuit referred to the primary/secondary side

VR= 100nl rated

rated

V VV−

×

VR= 100rated

rated

E VV−

×

2 2

2

VR= 100nl rated

rated

V VV−

×

1 2

2

VR= 100V V

V′−×

′

1 2

2

VR= 100V V

V′ −

×

( )1 2 1 1 2e eV V R jX I′ ′= + +

( )1 2 2 2 2e eV V R jX I′ = + +

30

Three-Phase Transformer Connections • Four possible combinations: Y-Y, -, Y- and -Y

– Y-connection: lower insulation costs, with neutral for grounding, 3rd harmonics problem (3rd harmonic voltages/currents are all in phase, i.e. van3=vbn3=vcn3=Vmcos3ωt)

– -connection: more insulation costs, no neutral, providing a path for 3rd harmonics (all triple harmonics are trapped in the loop), able to operate with only two phases (V-connection)

– In Y-Y or - connection, HV/LV ratio is same for line and phase voltages – Y- and -Y are commonly used as voltage step-down and step-up

transformers, respectively

31

• Y- and Y- connections will result in a phase shift of 30o between the primary and secondary line-to-line voltages

• According to the American Standards Association (ASA), the windings are arranged such that the HV side line voltage leads the LV side line voltage by 30o – E.g. Y- Connection:

Y- and -Y Connections

| || |

AnYP Y

P ab

VV N aV V N∆ ∆

= = =

| | 3| |

YL AB

YP An

V VV V

= =

| |L P abV V V∆ ∆= =

| | 3| |

YL AB

L ab

V V aV V∆

= =

32

Per-Phase Model for Y- or -Y Connection • Neglect the shunt branch • Replace the connection by an equivalent Y connection

(ZY=Z∆/3) • Work with only one phase (equivalent impedances are

line-to-neutral values)

33

Autotransformers • The primary and secondary coils are electrically connected

– A conventional two-winding transformer can be changed into an autotransformers by connecting its two windings in series

– Secondary is either manually or automatically variable.

20/25 MVA (69/40kV) McGraw-Edison Substation Auto-Transformer (Y-Y) (Source:

http://www.tucsontransformer.com)

(Source: EPRI Power System Dynamic Tutorial)

34

Autotransformer Model

1 2 1

2 1 2

V I N aV I N

= = =

1 12 1 2 2

2 2

(1 )H L L LN NV V V V V V V a VN N

= + = + = + = +

IL

11 2 1 1

2

(1 )L HNI I I I I a IN

= + = + = +

( ) 21 2 1 1 1

1

2 2

(1 )

1 (1 )

auto

w w conducted

NS V V I V IN

S S Sa

= + = +

= + = +

Power rating advantage Equivalent Circuit (if the equivalent impedance is

referred to the N1-turn side)

35

Advantages and Disadvantages of Autotransformers

• Advantages – Smaller and more efficient

• Lower leakage reactance • Lower loss • Lower exciting current • Higher kVA rating

– Variable output voltage when sliding contact is used.

• Disadvantage – No isolation between primary and secondary.

36

Example 3.5

=1440x250x10-3

37

Three-Winding Transformers • Primary (P), secondary (S) and tertiary

(T) windings • Typical applications:

– The same source supplying two independent loads at different voltages

– Interconnection of two transmission systems of different voltages

– In a substation, the tertiary winding is used to provide voltage for auxiliary power purposes or to supply a local distribution system

– Reactive power compensation by switched reactors or capacitors connected to the tertiary winding

38

Three-Winding Transformer Model • Estimate impedances by short-circuit tests:

Zps = measured P impedance, with S short-circuited and T open Zpt = measured P impedance, with T short-circuited and S open Z’st = measured S impedance, with T short-circuited and P open

• Referring Z’st to the P side:

• Zp, Zs and Zt are impedances of three windings referred to the P side:

2p

st sts

NZ Z

N

′=

ps p s

pt p t

st s t

Z Z ZZ Z ZZ Z Z

= +

= +

= +

( )( )( )

/ 2

/ 2

/ 2

p ps pt st

s ps st pt

t pt st ps

Z Z Z Z

Z Z Z Z

Z Z Z Z

= + −

= + −

= + −

39

Voltage Control of Transformers

• Voltage magnitude → Reactive power flow • Voltage phase angle → Real power flow

• Two commonly used methods – Tap changing Transformers – Regulating transformers

40

Tap Changing Transformers

• Practically all power transformers and many distribution transformers have taps in one or more windings for changing the turns ratio

• Off-load tap changing transformers – Can only be adjusted when the transformer current flow has been

completely interrupted – Typically, 4 taps with 2.5% of rated voltage each tap to allow an operating

range of -5% ~ +5% of rated voltage

• Under-load tap changing (ULTC) transformers – More flexible when the transformer is in-service – May have built-in voltage sensing circuitry for automatic tap changing to

keep voltage constant – Typically, off-load tap changers at the HV side for -5% ~ +5% – ULTC at the LV side with 32 taps of 0.625% each for an operating range

of -10% ~ +10%

41

• How to adjust tS and tR to maintain V’1 and V’2 at desired levels?

Since phase shift δ is small

( )R SV V R jX I= + +

cos

S S

R

V V

V ab de

δ≈

= + +

cos sin

cos sin

R

R R RR R

RR

V I R I XR XV I V I V

V VRP XQ

VV

φ φ

θ θ

θ θ

= + +

= + +

+= +

1 22

S RR

RP XQt V t V

t Vφ φ+

′ ′= +′

21 2

1s R

R

RP XQt t V

V t Vφ φ+

′= + ′ ′

ULTC Model P+jQ

I

42

• Normally, when the turns ratio is adjusted, the Mvar flow across the transformer is also adjusted

• However since a transformer itself absorbs Mvar to build its internal magnetic field, when its secondary voltage is raised via a tap change, its Mvar usage increases and its primary voltage often drops. The greater the tap change and the weaker the primary side, the greater the primary voltage drop.

• If the primary side is weak, the tap change may not necessarily increase the secondary voltage. Therefore, spare Mvar must be available for a tap change to be successful.

(5/16x10%= +3.125% ~ 142.3kV)

(+10% ~ 151.8kV)

An example:

±10% / 33-position ULTC

(Neutral point: 0% ~ 138kV)

Limitations of ULTC in Voltage Control

(Source: EPRI Dynamic Tutorial)

43

Regulating Transformers – Voltage Magnitude Control

Series transformer

Exciting transformer

an an anV V V′ = + ∆ Using Van to induce a voltage (//) added to Van

44

Regulating Transformers – Phase Angle Control

• Phase shifting transformer (PST), phase angle regulator (PAR) or Quadrature booster

Exciting transformer

Series transformer

bcV∆an an bcV V V′ = + ∆

Using Vbc to induce a voltage (⊥) added to Van

400MVA 220/155kV phase-shifting transformer

(source: wikipedia.org)

45

Application of PST

(Source: EPRI Dynamic Tutorial)

240MW

290MW

46 (Source: EPRI Dynamic Tutorial)

47

The Per-Unit System • Quantity in per-unit = Actual quantity / Base value of quantity

• Why per-unit system?

– Neglecting different voltage levels of transformers, lines and generators

– Powers, voltages, currents and impedances are expressed as decimal fractions of respective base quantities

• A minimum of four base quantities are required to completely define

a per-unit system

puB

SSS

= puB

VVV

= puB

III

= puB

ZZZ

=

48

• Usually, two bases are selected: – Three-phase base volt-ampere SB or MVAB

– Line-to-line base voltage VB or kVB

• The other two are calculated accordingly

• The phase and line quantities expressed in per-unit are the same

• Relationship between the load power and per-unit impedance

3B

BB

SIV

=( ) ( )2 2

/ 3 B BBB

B B B

V kVVZI S MVA

= = =

pu pu puS V I ∗= pu pu puV Z I=

( )3 3 P PLS V Iφ∗=

22

(3 ) (3 )

3 P L LPP

P L L

VS

VVZI S φ φ

−∗ ∗== =

( ) ( ) ( ) ( )

2 222

2(3 ) 3 3

puL LP B L L B Bpu pu

B L B L L L puB

VS

VZ S V S SZ VZ V S S SVφ φ φ

− −∗ ∗ ∗ ∗= = = = =

49

• If all quantities are in per-unit, tS≈1pu and tR≈1pu • Assuming tStR=1

tR=1/tS

• Example 3.6: tS=1.08pu and tR=0.926pu

21 2

1s R

R

RP XQt t V

V t Vφ φ+

′= + ′ ′

ULTC Per-Unit Model

2

1

1 2

1S

VV

t RP XQV Vφ φ

′′

=+

−′ ′

50

Change of Base

• The nameplates of transformers and generators usually give impedance as x% based on its rated voltage and rated MVA. Also, it is usually operated under the same (or very close to) rated voltage level while the base MVA may be quite different. Then, the impedance under the new, system base can be calculated using the above simplified equation.

( )2

oldold Bpu old old

B B

Z SZ ZZ V

ΩΩ= =

( )2

newnew Bpu new new

B B

Z SZ ZZ V

ΩΩ= =

2new oldpunew old B

pu pu old newB B

S VZ ZS V

=

newpuold

pu oldB

SZ

S= (if VB is the same)

51

Advantages of the Per-Unit System

• Gives a clear idea of relative magnitudes of various quantities, e.g. voltage, current, power and impedance

• The per-unit impedance of equipment of the same general type based on their own ratings fall in a narrow range regardless of the rating of the equipment. Whereas their impedance in ohms vary greatly with the rating

• The per-unit values of impedance, voltage and current of a transformer are the same regardless of whether they are referred to the primary or the secondary side

• Idea for the computerized analysis and simulation of complex power system problems

• The circuit laws are valid in per-unit systems. For three-phase systems, factors of √3 and 3 are eliminated by properly selecting base quantities

52

Example 3.7

VB1

VB1 VB2 VB3 VB4

VB5 VB6

ZB4

ZB2

ZB5

53

Example 3.8

54

Homework #4

• Read through Saadat’s Chapter 3 • ECE521: 3.2, 3.4, 3.5, 3.6, 3.9, 3.10, 3.14, 3.15, 3.17 • ECE421: 3.2, 3.5, 3.6, 3.9, 3.10, 3.14, 3.15 • Due date: 10/4 (Friday) submitted in class or by email