1

Instructor: Kai Sun

Fall 2014

ECE 421/599 Electric Energy Systems

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

• Magneto-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

31sin3

sP XE Vφδ −

=

|E|∠δ |V|∠0

14

Example 3.2 |E|∠δ |V|∠0

θ3

a b

θ3

|V|=17.32kV P=40MW

E with δ=90o

15

E∠δ V∠0

16

Comparison

E δ Ia θ 23.6kV 10.6o 807.5A -54.43o (lagging) 18.7kV 21.8o 769.8A 0 14.0kV 29.7o 962.3A 36.87o (leading)

6.9KV (minimum) 90o 2073A 68.2o (leading)

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

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

Fix Pmax,3φ=40MW and |V|=30kV

17

•V Curve of a synchronous machine: – |Ia|-|If|: armature current |Ia| as the function of 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 polar axis (d axis) is less than that in the inter-polar axis (q axis)

– No such a constant Xs being independent of the rotor position

– 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

( ) ( ) ( )

ar a a a

a a ar a a s a

E V jX I R jX IV R I j X X I V R jX I

= + + += + + + = + +

d

q Axis of coil a

(reference)

N

S

n

m

ψ

E

Note: When the direction of ea (axis of coil) is parallel with q axis

of the rotor, ea reaches the maximum, so in the phasor

diagram, q axis is the axis of E

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:

Consider a simple case ignoring Ra and Xl

q axis

d axis

20

Power Transformer •Ideal transformers

– Winding resistance is negligible (no power loss due to winding)

– No leakage flux – Permeability of the core is infinite

such that net mmf to establish flux in core is zero.

(mmf balance) – No core loss (no power loss due to core)

•Real transformers: – Windings have resistance – Windings do not link the same flux – Permeability of the core is finite – Core losses (hysteresis losses and

eddy current losses due to time varying flux)

F=I⋅N=Φ⋅R=Φ/P

(Source: wikipedia.org)

21

Ideal Transformer • Primary winding (assuming sinusoidal flux)

• Secondary winding – 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 max4.44 90E fN= Φ ∠ Note: e1 is the load voltage of the primary side and e2 is the source voltage of the secondary side

1 1 2 2I N I N= 1 2 1

2 1 2

E I NE I N

= =

2 2 max4.44 90E fN= Φ ∠ ( )2 2 2max cos 90de N E tdtφ ω= = + °

22

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

I2=I’2=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 1 2 1

2 2 2

E I NE I N

= =′

23

2 2 2 2E V Z I= +

11 2

2

1 12 2 2

2 2

21

22

2

2

12

2

2 2

N VN

NE ENN NV Z I

N ZN

Z

N N

IV

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

Ideal Transformer 1 2 1

2 2 2

E I NE I N

= =′

Note: V1≈E1, so Z1 may be moved to the right side of the shunt branch and combined with Z’2

24

Approximate Equivalent Circuits • Utilize 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 circuit referred to the secondary side

( )1 2 2 2 2e eV V R jX I′ = + +V2

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:

V1

26

Determination of Equivalent Circuit Parameters

•Open-circuit (no-load) test – Neglect (R1+jX1)I0 (since |R1+jX1|<<|Rc1//jXm1|) – 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 • Efficiency: 95% ~ 99% • Given |V2,rated|, |S|=3|V2,rated||I2,rated| (full-load rated VA) and PF • Actual load is |I2|=n|I2,rated| where n is the fraction of the full load power

Pcu=3Re2|I2,rated|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

P nP n P n P

η = =+ +

2

0 0d dd I dnη η

= ⇔ = 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 • A bank of three single-phase transformers connected in Y or ∆ arrangements • 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)

– Y-Y and △-△: HV/LV ratio is same for line and phase voltages; Y-Y is rarely used due to the 3rd harmonics problem.

– Y-△ : commonly used as voltage step-down transformers – △-Y : commonly used as voltage step-up transformers

31

Y-△ and △ -Y Connections • Y-△ and Y-△ connections result in a 30o phase shift 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-△ (HV-LV) Connection with the ratio of turns a= NH/NX>1

,

,

H P An H

X P ab X

V V N aV V N

= = =

,

,

3 30H L AB

H P An

V VV V

= = ∠

, ,X L X PV V=

,

,

3 30H L AB

X L ab

V V aV V

= = ∠

HV Side (indicated by “H”) in Y connection:

LV Side (indicated by “X”) in △ connection:

=VH,P

=VX,P =VX,L

=VH,L

(Complex ratio)

32

Per-Phase Model for Y-△ or △ -Y Connection

•Neglect the shunt branch •Replace the △ connection by an equivalent Y connection •Work with only one phase (equivalent impedances are line-to-

neutral values ZeY=Ze∆/3)

E.g. for Y-△ Connection, V1 is the phase voltage of the Y side and V2 is the line-to-line voltage of the △ side

V1

33

Autotransformers • A conventional two-winding transformer can be changed into an autotransformer

by connecting its two coils in series. • The connection may use a sliding contact to providing variable output voltage. • An autotransformer has kVA rating increased but loses insulation between primary

and secondary windings

20MVA (115/69kV) 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 1

2 1 2 22 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 HV side)

Transformed power (thru

EM induction)

Conducted power (V2I1)

• Apparent power:

35

Example 3.5

=1440x250x10-3

36

True or False? A two-winding transformer is first operated as a conventional two-winding transformer and then as an autotransformer to supply the same load. When it is operated as the autotransformer, it has

– a higher rating, – a higher efficiency and – a lower loss

than it is operated as the conventional two-winding transformer. •Answer: True. For the autotransformer, rating (not the actual

loading) will be increased, loss will decrease since currents in two windings will decrease to generate output current IL equal to output current I2 of the two-winding transformer. Then the efficiency will increase considering the loading is the same.

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

• 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:

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 in transmission systems and many distribution transformers have taps in one or more windings for changing turns ratios

• Off-load tap changing transformers – They are adjusted only 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

ULTC Model

• 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+

′= + ′ ′

P+jQ

I

42

Limitations of ULTC in Voltage Control • Normally, when the turns ratio is adjusted, Mvar flow across the transformer also changes • 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 (V2 ↑ and I2↑) and the weaker the primary side (high impedance), 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) (Source: EPRI Dynamic Tutorial)

43

Regulating Transformers – Voltage Magnitude Control

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

(Source: EPRI Dynamic Tutorial)

Choose 20kV, 345KV and 138kV as base voltages

48

• Four base quantities are required to completely define a per-unit system

• Usually, two independent bases are selected, and the other two are calculated – SB or MVAB Three-phase base volt-ampere

– VB or kVB Line-to-line base voltage

• 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 BB

BB B B

V kVVZI S MVA

= = =

pu pu puS V I ∗= pu pu puV Z I=

3 3 P PS V Iφ∗=

22

3 3

3 P L LPP

P

VS

VVZI S φ φ

−∗ ∗== =

( )

2 23

23

2

L LP B L L

B B

pup

puBu

B

VS

SZ S VVZ

ZV SS

Vφ

φ∗

∗− −∗= = = =

puB

SSS

= puB

VVV

= puB

III

= puB

ZZZ

=

49

ULTC Per-Unit Model

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

• 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+

′= + ′ ′

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 MVA base can be calculated using the above simplified equation.

( )2

oldold Bpu old old

B B

SZZ ZZ V

= =

( )2

newnew Bpu new new

B B

SZZ ZZ V

= =

2new oldnew old B Bpu pu old new

B B

S VZ ZS V

=

newpunew old

pu pu oldB

SZ Z

S=

(Simplified equation if VB is the same)

Eliminate Z

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 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

53

Example 3.7

VB1

VB1 VB2 VB3 VB4

VB5 VB6

ZB4

ZB2

ZB5

54

Example 3.8