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Instructor: Kai Sun

Fall 2013

ECE 421/521 Electric Energy Systems

Power Systems Analysis I 2 – Basic Principles

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Outline

• Power in a 1-phase AC circuit

• Complex power

• Balanced 3-phase circuit

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Single Phase AC System

( ) cos( )m vv t V tω θ= +

( ) cos( )m ii t I tω θ= +

Z + - Load

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Phasor Representation

• V=|V|∠θv I=|I|∠θi – RMS phasors of v(t) and i(t)

• A phasor is a complex number that carries the amplitude and phase

angle information of a sinusoidal function

• Phasors represent sinusoidal signals of a common frequency (ω) as vectors in the complex plane w.r.t. a chosen reference signal.

• Phasor representation is a mapping from the time domain to complex number domain.

( ) cos( ) 2 | | cos( )

( ) cos( ) 2 | | cos( )m v v

m i i

v t V t V t

i t I t I t

ω θ ω θ

ω θ ω θ

= + = +

= + = +

Reference

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[ ]

( ) ( ) ( ) cos( ) cos( )1 cos( ) cos(2 )21 cos( ) cos[2( ) ( )]21 cos cos[2( ) ]21 cos cos 2( ) cos sin 2( ) s2

m m v i

m m v i v i

m m v i v v i

m m v

m m v v

p t v t i t V I t t

V I t

V I t

V I t

V I t t

ω θ ω θ

θ θ ω θ θ

θ θ ω θ θ θ

θ ω θ θ

θ ω θ θ ω θ

= = + +

= − + + +

= − + + − −

= + + −

= + + + +[ ]inθ

| | / 2, | | / 2m mV V I I= =

• Instantaneous power delivered to the load:

v iθ θ θ= −

( ) ( )

( ) cos [1 cos 2( )] sin sin 2( )R X

v v

p t p t

p t V I t V I tθ ω θ θ ω θ= ⋅ + + + ⋅ +

[ ]

Using trigonometric identity1

cos cos cos( - ) cos( )2

A B A B A B= + +

Impedance angle >0 for inductive load and <0 for capacitive load

Root-mean-square (RMS) values

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Calculate the RMS current phasor: ( ) 100cosv t tω= 1.25 60 ΩZ = ∠ °

1 100 0 1 80 60 1.25 602 2

I ∠ °= = ∠ − °

∠ °

0 90 180 270 360-100

-50

0

50

100v(t)=Vm cos ωt, i(t)=Im cos(ωt -60)

ωt, degree0 90 180 270 360

-2000

0

2000

4000

6000p(t)=v(t) i(t)

ωt, degree

0 90 180 270 3600

1000

2000

3000

4000pR(t) Eq. 2.6

ωt, degree 0 90 180 270 360-4000

-2000

0

2000

4000pX(t) Eq. 2.8

ωt, degree

Example 2.1

( ) 80cos( 60 )oi t tω= −

Assume

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• Observations on pR(t): – Oscillating at frequency 2*ω (twice of the source frequency) – Changing between 0 and 2|V|⋅|I|cosθ (always positive) – Average value is |V|⋅|I|cosθ – It is the energy flow into the circuit

• Observations on pX(t):

– Oscillating at frequency 2*ω (twice of the source frequency) – Changing between -|V|⋅|I|sinθ and |V|⋅|I|sinθ (either positive or negative) – Average value is zero – It is the energy borrowed & returned by the circuit. It does no useful

work in the load but takes some line capacity

( ) ( )

( ) cos [1 cos 2( )] sin sin 2( )R X

v v

p t p t

p t V I t V I tθ ω θ θ ω θ= ⋅ + + + ⋅ +

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Real and Reactive Powers

• Real power or active power (average power) • Unit ~ watt or W • Power factor (PF): cosθ= cos(θv-θi)

– Lagging or leading when θ =θv-θi >0 or <0

def| | | | cosP V I θ= ⋅

def| | | | sinQ V I θ= ⋅

( ) ( )

( ) cos [1 cos 2( )] sin sin 2( )R X

v v

p t p t

p t V I t V I tθ ω θ θ ω θ= ⋅ + + + ⋅ +

• Reactive power • Unit ~ var (volt-ampere reactive). Some people use Var, VAR or VAr • Q>0 or <0 when θ=θv-θi >0 or <0

P

Q

| | | |V I⋅• Apparent power • Unit ~ volt ampere or VA (kVA or MVA)

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Characteristics of instantaneous power p(t)

For a pure resistor load (thermal load), • θ=θv-θi =0, PF=1 (unity PF) • P=|V||I|, so all electric energy becomes thermal energy For a pure inductive load, • θ=θv-θi =90, PF=0 • P=0, so electric energy = constant (no transformation to other forms) • p(t) oscillates between the source and the magnetic field associated with the

inductive load For a pure capacitive load, • θ=θv-θi =-90, PF=0 • P=0, so electric energy = constant (no transformation to other forms) • p(t) oscillates between the source and the electric field associated with the

capacitive load

( ) ( )

( ) cos [1 cos 2( )] sin sin 2( )

[1 cos 2( )] sin 2( )R X

v v

p t p t

v v

p t V I t V I t

P t Q t

θ ω θ θ ω θ

ω θ ω θ

= ⋅ + + + ⋅ +

= + + + +

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( ) 100cosv t tω=1.25 60 ΩZ = ∠ °

Example 2.1 ( ) 80cos( 60 )oi t tω= −

( )( )( ) ( )

( ) cos [1 cos 2 ] sin sin 2 4000cos 60 [1 cos 2 ] 4000sin 60 sin 2XRR X

o o

p tp tp t p t

p t V I t V I t t tθ ω θ ω ω ω= ⋅ + + ⋅ = + +

100 800 , 602 2

V I= ∠ ° = ∠− °

0 90 180 270 360-100

-50

0

50

100v(t)=Vm cos ωt, i(t)=Im cos(ωt -60)

ωt, degree0 90 180 270 360

-2000

0

2000

4000

6000p(t)=v(t) i(t)

ωt, degree

0 90 180 270 3600

1000

2000

3000

4000pR(t) Eq. 2.6

ωt, degree 0 90 180 270 360-4000

-2000

0

2000

4000pX(t) Eq. 2.8

ωt, degree

P

Q

Borrowing

Returning

RMS phasors:

P Q

Q

P

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What does “absorbing” or “generating” reactive power mean? • “Reactive power” with a circuit element does not

lead to real energy consumption • Starting from the time of v(t)=Vm,

– An inductive element (lagging PF) first absorbs and then returns the same amount of electric energy. Such a pattern is considered “absorbing” reactive power, and it helps reduce oscillations of power

– A capacitive element (leading PF) first generates and then absorbs the same amount of electric energy. Such a pattern is considered “generating” reactive power, and it helps maintain voltage

• For a real-world power system with long-distance transmission, low voltage at receiving ends is usually a big reliability concern. Shunt/series capacitors can be added as “sources” of reactive power or var

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Complex Power

• |S| is the apparent power • θ=tan-1(Q/P)

– When θ>0 (θi <θv, i.e. lagging PF), Q>0 i.e. absorbing Q (inductive load)

– When θ<0 (θi >θv, i.e. leading PF), Q<0 i.e. generating Q (capacitive load)

Reference

def= S P jQ+

cos sinV I j V I V Iθ θ θ= + = ∠

* v iV I VIθ θ= ∠ − =

2 2 | |S P Qθ θ= ∠ = + ∠

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PF=cosθ=P/|S| (lagging/leading is told by +/- sign of Q) P=|S|⋅ cosθ=|S|⋅ PF

2

2

| | sin | | 1 , if lagging=

| | sin | | 1 , if leading

S S PFQ

S S PF

θ

θ

⋅ = ⋅ −− ⋅ = − ⋅ −

R

X |Z| If the load impedance is Z, i.e .V=ZI, then

S = VI*= ZII*= Z|I|2= R|I|2+jX|I|2 = P + jQ

P=R|I|2 Q=X|I|2

• The load impedance angle θ is also called power angle (ϕ in some literature)

• Apparent power |S|∝|I|2 indicates heating and is used as a rating unit of power equipment

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• Some useful equations:

• If Z is purely resistive

• If Z is purely reactive

2*2* *

* *

VVVS VI V YZ Z

= = = =2

*

VZ

S=

2VP

R=

2 2 2

or 1/

V V VQ

X L Cω ω= =

−

*

2SYV

=

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• Due to the conservation of energy, the total complex power delivered to the loads in parallel is the sum of complex powers delivered to each (by KCL).

• In other words, a balance must be maintained all the time for both real power and reactive power

* * * *1 2 3 1 2 3 1 2 3

1 1 2 2 3 3

[ ]S VI V I I I VI VI VI S S SP jQ P jQ P jQ P jQ

∗= = + + = + + = + ++ = + + + + +

Complex Power Balance

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Example 2.2 1

2 3

1200 0 V, 60 0 , 6 12 , 30 30

V Z jZ j Z j= ∠ ° = + Ω= + Ω = − Ω

Find the power absorbed by each load and the total complex power

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22

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1200 0 1200 20 0 A60 0 601200 0 200 200(1 2) 40 80 A6 12 1 2 5

1200 0 40 40(1 ) 20 20 A30 30 1 2

VI jZ jV jI jZ j jV jI jZ j j

∠ °= = = = +

+∠ ° −

= = = = = −+ +∠ ° +

= = = = = +− −

1 2 3 96,000 W+ 72,000 varS S S S j= + + =

*1 1

*2 2

*3 3

1200 0 (20 0) 24,000 W+ 0 var

1200 0 (40 80) 48,000 W+ 96,000 var

1200 0 (20 20) 24,000 W 24,000 var

S VI j jS VI j jS VI j j

= = ∠ ° − =

= = ∠ ° + =

= = ∠ ° − = −

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• Other approaches

1 2 3

*

(20 0) (40 80) (20 20)80 60 100 36.87 A

(1200 0 )(100 36.87 ) 120,000 36.87 VA 96,000 W 72,000 var

I I I I j j jj

S VIj

= + + = + + − + += − = ∠ − °

= = ∠ ° ∠ ° = ∠ °= +

2 2

1 *1

2 2

2 *22 2

3 *3

(1200) 24,000 W 0 var60

(1200) 48,000 W 96,000 var6 12

(1200) 24,000 W 24,000 var30 30

VS j

Z

VS j

Z j

VS j

Z j

= = = +

= = = +−

= = = −+

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Theorem of Conservation of Complex Power

For a network supplied by independent sources all at the same frequency (all voltages and currents are assumed to be sinusoids), the sum of the complex power supplied by the independent sources equals the sum of the complex power received by all the other branches of the network

• For a single source with elements in series or parallel, it is proved by

Kirchhoff’s voltage or current law (KVL/KCL)

• For a general case, it is proved by Tellegen’s theorem. • Application of the theorem: a part of the network can be replaced by

an equivalent independent source

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• Some examples on Bergen and Vittal’s book

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Homework #2

• Read through Saadat’s Chapter 2.1~2.4 • ECE521: 2.1~2.6 • ECE421: 2.1, 2.3, 2.4, 2.5 • Due date: 9/11 (Wednesday) submitted in class or by

email

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Power Factor Correction

P=|V||I|⋅ cosθ=|S|⋅ PF • When PF<1, current I needs to increase by 1/PF to deliver

the same P as that with PF=1.

• Thus, the major loads of the system are expected to have close to unity PFs. That is not a problem for residential and small commercial customers.

• However, industrial loads are inductive with low lagging PFs, so capacitors may be installed to improve PFs

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Example 2.3 (a) Find P, Q, I and PF at the source (b) Find capacitance C of the shunt capacitor

to improve the overall PF to 0.8 lagging

P+jQ

C

1

2

*1 1

*2 2

200 0 2 0 A100

200 0 4 8 A10 20

200 0 (2 0) 400 W+ 0 var

200 0 (4 8) 800 W 1600 var

I

I jj

S VI j jS VI j j

∠ °= = ∠ °

∠ °= = −

+

= = ∠ ° − =

= = ∠ ° + = +

Solution: (a)

*

*

1200 1600 2000 53.13 VA2000 53.13 10 53.13 A

200 0

S P jQ jSIV

= + = + = ∠ °

∠ − °= = = ∠ − °

∠ °

cos(53.13 ) 0.6 laggingPF = =

(b)

1

2 2

*

6

cos (0.8) 36.87tan 1200 tan(36.87 ) 900 var

1600 900 700 var

(200) 57.14 700

10 46.42 μF2 (60)(57.14)

c

cc

Q PQ

VZ j

S j

C

θθ

π

−′ = = °′ ′= = ° == − =

= = = − Ω

= =

New power angle

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*

*

1200 900 1500 36.871500 36.87 7.5 36.87

200 0

S jSIV

′ = + = ∠ °

′ ∠ − °′ = = = ∠ − °∠ °

New apparent power and current

10→7.5

2000→1500

High-voltage capacitor bank (150kV - 75MVAR)

(Source: wikipedia.org) Learn Example 2.4 (high-voltage)

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Complex Power Flow

1 1 1V V δ= ∠ 2 2 2V V δ= ∠

1 1 2 212

1 21 2

V VI

Z

V VZ Z

δ δγ

δ γ δ γ

∠ − ∠=

∠

= ∠ − − ∠ −

1 2*12 1 12 1 1 1 2[ ]

V VS V I V

Z Zδ γ δ γ δ= = ∠ ∠ − − ∠ −

21 1 2

1 2

V V VZ Z

γ γ δ δ= ∠ − ∠ + −

21 1 2

12 1 2cos cos( )V V V

PZ Z

γ γ δ δ= − + −

21 1 2

12 1 2sin sin( )V V V

QZ Z

γ γ δ δ= − + −

P12+jQ12

1 212 1 2sin( )

V VP

Xδ δ= −

112 1 2 1 2[ cos( )]

VQ V V

Xδ δ= − −

If R=0, i.e. |Z|=X and γ=90o

P21+jQ21

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• Since R=0, there are no transmission line losses, and P12=-P21

• In a normal power systems, |Vi|≈100% and |δ1-δ2| is small (e.g. 5~30o). • P12∝sin(δ1-δ2)≈δ1-δ2

– Real power flow is governed mainly by the angle difference of the terminal voltages.

– For example, if V1 leads V2, i.e. δ1-δ2>0, the real power flows from node 1 to node 2.

– Theoretical maximum power transfer (i.e. static transmission capacity):

• Q12 ∝|V1|-|V2|cos(δ1-δ2) ≈|V1|-|V2| – Reactive power flow is determined by the magnitude difference of

terminal voltages

1 212 1 2sin( )

V VP

Xδ δ= −

1 2 1 212 max sin 90

V V V VP

X X= =

112 1 2 1 2[ cos( )]

VQ V V

Xδ δ= − −

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Example 2.5

1 2120 5 V 100 0 V, 1 7 ΩV V Z j= ∠ − ° = ∠ ° = +，

12

21

120 5 100 0 3.135 110.02 A1 7

100 0 120 5 3.135 69.98 A1 7

Ij

Ij

∠ − ° − ∠ °= = ∠ − °

+∠ ° − ∠ − °

= = ∠ °+

I21

*12 1 12

*21 1 21

376.2 105.02 97.5 W 363.3 var

313.5 69.98 107.3 W 294.5 var

S V I jS V I j

= = ∠ ° = − +

= = ∠ − ° = −

S12=P12+jQ12 S21=P21+jQ21

12 21 9.8 W 68.8 var =L L LS S S j P jQ= + = + +

2 212 (1)(3.135) 9.8 WLP R I= = =

2 212 (7)(3.135) 68.8 varLQ X I= = =

-40 -30 -20 -10 0 10 20 30-1000

-800

-600

-400

-200

0

200

400

600

800

1000

P12

P21

PL

Source #1 Voltage Phase Angle, δ1

P, W

atts

Determine the real and reactive powers supplied or received by each source and the power loss in the line

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Balanced Three-Phase Circuits • A balanced source: three sinusoidal voltages are

generated having the same amplitude but displaced in phase by 120o

– Instantaneous power delivered to the external loads is constant

Positive phase sequence Negative phase sequence

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An p AE E α= ∠

120Bn p AE E α= ∠ − °

240Cn p AE E α= ∠ − °

An An G aV E Z I= −

Balanced Y-connected Loads

an An L aV V Z I= −

0an pV V= ∠ ° 120bn pV V= ∠ − ° 240cn pV V= ∠ − °

• Let Van be the reference:

(1 0 1 120 ) 3 30ab an bn p pV V V V V= − = ∠ ° − ∠ − ° = ∠ °

(1 120 1 240 ) 3 90bc bn cn p pV V V V V= − = ∠ − ° − ∠ − ° = ∠ − °

(1 240 1 0 ) 3 150ca cn an p pV V V V V= − = ∠ − ° − ∠ ° = ∠ °

Phase voltages (line-to-neutral voltages):

Line voltages (line-to-line voltages):

| | 3L pV V=

Vbc

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• On the neutral line (return line) In=Ia+Ib+Ic=0

• The neutral line may not actually exist (one line per phase)

• The balanced 3-phase power system problems can be solved on a “per-phase” basis, e.g. for Phase A only. The other two phases carry identical currents except for phase shifts.

| | 0| |

an ana p

p p

V VI IZ Z

θθ

∠= = = ∠ −

∠

120bnb p

p

VI IZ

θ= = ∠ − ° −

240cnc p

p

VI IZ

θ= = ∠ − ° −

L pI I=

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Balanced -connected Loads

L pV V=

0ab pI I= ∠ °

120bc pI I= ∠ − °

240ca pI I= ∠ − °

• Let Iab be the reference:

(1 0 1 240 ) 3 30a ab ca p pI I I I I= − = ∠ ° − ∠ − ° = ∠ − °

(1 120 1 0 ) 3 150b bc ab p pI I I I I= − = ∠ − ° − ∠ ° = ∠ − °

(1 240 1 120 ) 3 90c ca bc p pI I I I I= − = ∠ − ° − ∠ − ° = ∠ °

| | 3L pI I=

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-Y Transformation

• Idea: compare Ia=f∆(Van, Z∆) and Ia=fY(Van, ZY)

– Find Ia=f∆(Van, Z∆):

Z∆ Z∆

Z∆

n

ab ac ab aca

V V V VIZ Z Z∆ ∆ ∆

+= + =

3 30 3 30ab ac an anV V V V+ = ∠ °+ ∠− °

3 ( 3 / 2 / 2 3 / 2 / 2) 3an anV j j V= + + − =

3 ana

VIZ∆

=

– Find Ia=fY(Van, Z∆):

ana

Y

VIZ

=

3YZZ ∆=

Vca

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Balanced Three-Phase Power

2 cos( )an p vv V tω θ= +

2 cos( 120 )bn p vv V tω θ= + − °

2 cos( 240 )cn p vv V tω θ= + − °

2 cos( )an p ii I tω θ= +

2 cos( 120 )bn p ii I tω θ= + − °

2 cos( 240 )cn p ii I tω θ= + − °

For balanced loads

3 an a bn b cn cp v i v i v iφ = + +

Total instantaneous power:

2 cos( )cos( )p p v iV I t tω θ ω θ= + +

2 cos( 120 )cos( 120 )p p v iV I t tω θ ω θ+ + − ° + − °

2 cos( 240 )cos( 240 )p p v iV I t tω θ ω θ+ + − ° + − °

3 [cos( ) cos(2 )]p p v i v ip V I tφ θ θ ω θ θ= − + + +

[cos( ) cos(2 240 )]p p v i v iV I tθ θ ω θ θ+ − + + + − °

[cos( ) cos(2 480 )]p p v i v iV I tθ θ ω θ θ+ − + + + − °

3 3 3 cosp pp P V Iφ φ θ= =

Zero Total instantaneous power is constant:

Where is reactive power?

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• Extend the concept of complex power to 3-phase systems

3 3 sinp pQ V Iφ θ=

*3 3 3 3 p pS P jQ V Iφ φ φ= + =

• Expressed in terms of line voltage VL and line current IL

– Y-connection: – ∆-connection:

/ 3p LV V= p LI I=

p LV V= / 3p LI I=

3 3 cosL LP V Iφ θ=

3 3 cosL LQ V Iφ θ=

θ is the angle between the phase voltage and the phase current for the same phase (e.g. Phase A)

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Example 2.7 (a) The current, real power and reactive

powers drawn from the supply (b) The line voltage at the combined loads (c) The current per phase in each load (d) The total real and reactive powers in

each load and the line

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V2=110-j20

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Homework #3

• Read through Saadat’s Chapter 2.5~2.12 • ECE521: 2.7~2.10, 2.14~2.16 • ECE421: 2.7, 2.8, 2.10, 2.15, 2.16 • Due date: 9/20 (Friday) submitted in class or by email