Lesson 8 symmetrical components
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Lesson 8 Symmetrical Components. Symmetrical Components. Due to C. L. Fortescue (1918): a set of n unbalanced phasors in an n-phase system can be resolved into n balanced phasors by a linear transformation The n sets are called symmetrical components

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Lesson 8 Symmetrical Components

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Lesson 8 symmetrical components

Lesson 8Symmetrical Components

Notes on Power System Analysis


Symmetrical components

Symmetrical Components

  • Due to C. L. Fortescue (1918): a set of n unbalanced phasors in an n-phase system can be resolved into n balanced phasors by a linear transformation

    • The n sets are called symmetrical components

    • One of the n sets is a single-phase set and the others are n-phase balanced sets

    • Here n = 3 which gives the following case:

Notes on Power System Analysis


Symmetrical component definition

Symmetrical component definition

  • Three-phase voltages Va, Vb, and Vc (not necessarily balanced, with phase sequence a-b-c) can be resolved into three sets of sequence components:

    Zero sequence Va0=Vb0=Vc0

    Positive sequence Va1, Vb1, Vc1 balancedwith phase sequence a-b-c

    Negative sequence Va2, Vb2, Vc2 balanced with phase sequence c-b-a

Notes on Power System Analysis


Lesson 8 symmetrical components

Notes on Power System Analysis


Lesson 8 symmetrical components

where

a = 1/120°= (-1 + j 3)/2

a2 = 1/240°= 1/-120°

a3 = 1/360°= 1/0 °

Notes on Power System Analysis


Lesson 8 symmetrical components

Vp = A VsVs= A-1Vp

Notes on Power System Analysis


Lesson 8 symmetrical components

  • We used voltages for example, but the result applies to current or any other phasor quantity

Vp = A VsVs= A-1Vp

Ip = A Is Is= A-1Ip

Notes on Power System Analysis


Lesson 8 symmetrical components

Va = V0 + V1 + V2

Vb = V0 + a2V1 + aV2

Vc = V0 + aV1 + a2V2

V0 = (Va + Vb + Vc)/3

V1 = (Va + aVb + a2Vc)/3

V2 = (Va + a2Vb + aVc)/3

These are the phase a symmetrical (or sequence) components. The other phases follow since the sequences are balanced.

Notes on Power System Analysis


Sequence networks

a

Zy

b

Zy

c

Zn

g

Sequence networks

  • A balanced Y-connected load has

    three impedances Zy connected line to neutral

    and one impedance Zn connected neutral to ground

Notes on Power System Analysis


Sequence networks1

Sequence networks

or in more compact notation Vp= ZpIp

Notes on Power System Analysis


Lesson 8 symmetrical components

a

Zy

b

Zy

c

Zn

g

Zy

Vp = ZpIp

Vp = AVs = ZpIp= ZpAIs

AVs = ZpAIs

Vs= (A-1ZpA) Is

Vs = Zs Is where

Zs = A-1ZpA

n

Notes on Power System Analysis


Lesson 8 symmetrical components

V0 = (Zy + 3Zn) I0 = Z0 I0

V1 = ZyI1 = Z1 I1

V2 = ZyI2 = Z2 I2

Notes on Power System Analysis


Lesson 8 symmetrical components

I2

I0

I1

Zy

Zy

Zy

n

a

a

a

V2

V0

V1

3 Zn

n

g

n

Zero-

sequence

network

Positive-

sequence

network

Negative-

sequence

network

Sequence networks for Y-connected load impedances

Notes on Power System Analysis


Lesson 8 symmetrical components

I2

I0

I1

ZD/3

ZD/3

ZD/3

n

a

a

a

V2

V0

V1

n

g

n

Positive-

sequence

network

Negative-

sequence

network

Zero-

sequence

network

Sequence networks for D-connected load impedances.

Note that these are equivalent Y circuits.

Notes on Power System Analysis


Remarks

Remarks

  • Positive-sequence impedance is equal to negative-sequence impedance for symmetrical impedance loads and lines

  • Rotating machines can have different positive and negative sequence impedances

  • Zero-sequence impedance is usually different than the other two sequence impedances

  • Zero-sequence current can circulate in a delta but the line current (at the terminals of the delta) is zero in that sequence

Notes on Power System Analysis


Lesson 8 symmetrical components

  • General case unsymmetrical impedances

Notes on Power System Analysis


Lesson 8 symmetrical components

Z0 = (Zaa+Zbb+Zcc+2Zab+2Zbc+2Zca)/3

Z1 = Z2 = (Zaa+Zbb+Zcc–Zab–Zbc–Zca)/3

Z01 = Z20 = (Zaa+a2Zbb+aZcc–aZab–Zbc–a2Zca)/3

Z02 = Z10 = (Zaa+aZbb+a2Zcc–a2Zab–Zbc–aZca)/3

Z12 = (Zaa+a2Zbb+aZcc+2aZab+2Zbc+2a2Zca)/3

Z21 = (Zaa+aZbb+a2Zcc+2a2Zab+2Zbc+2aZca)/3

Notes on Power System Analysis


Lesson 8 symmetrical components

  • Special case symmetrical impedances

Notes on Power System Analysis


Lesson 8 symmetrical components

Z0 = Zaa+ 2Zab

Z1 = Z2 = Zaa– Zab

Z01=Z20=Z02=Z10=Z12=Z21= 0

Vp = ZpIpVs = ZsIs

  • This applies to impedance loads and to series impedances (the voltage is the drop across the series impedances)

Notes on Power System Analysis


Power in sequence networks

Power in sequence networks

Sp = VagIa* + VbgIb* + VcgIc*

Sp = [VagVbgVcg] [Ia*Ib*Ic*]T

Sp = VpTIp*

= (AVs)T(AIs)*

= VsTATA* Is*

Notes on Power System Analysis


Power in sequence networks1

Power in sequence networks

Sp = VpTIp* = VsT ATA* Is*

Sp = 3 VsT Is*

Notes on Power System Analysis


Lesson 8 symmetrical components

Sp= 3 (V0 I0* + V1 I1* +V2 I2*) = 3 Ss

In words, the sum of the power calculated in the three sequence networks must be multiplied by 3 to obtain the total power.

This is an artifact of the constants in the transformation. Some authors divide A by 3 to produce a power-invariant transformation. Most of the industry uses the form that we do.

Notes on Power System Analysis


Sequence networks for power apparatus

Sequence networks for power apparatus

  • Slides that follow show sequence networks for generators, loads, and transformers

  • Pay attention to zero-sequence networks, as all three phase currents are equal in magnitude and phase angle

Notes on Power System Analysis


Lesson 8 symmetrical components

N

E

V1

Positive

I1

Z1

N

V2

I2

Negative

Zn

Z2

G

Y generator

V0

3Zn

Zero

I0

Z0

N

Notes on Power System Analysis


Lesson 8 symmetrical components

N

V1

Positive

I1

Z

N

V2

I2

Negative

Ungrounded Y load

Z

G

V0

Zero

I0

Z

N

Notes on Power System Analysis


Lesson 8 symmetrical components

Zero-sequence networks for loads

G

Y-connected load grounded through Zn

V0

3Zn

I0

Z

N

G

D-connected load ungrounded

V0

Z

Notes on Power System Analysis


Y y transformer

Y-Y transformer

H1

X1

A

a

Zeq+3(ZN+Zn)

A

a

B

I0

Va0

VA0

b

g

c

C

Zero-sequence

network (per unit)

N

n

ZN

Zn

Notes on Power System Analysis


Y y transformer1

Y-Y transformer

Zeq

H1

X1

A

a

A

a

I1

Va1

VA1

B

b

n

c

C

Positive-sequence

network (per unit)

Negative sequence

is same network

N

n

ZN

Zn

Notes on Power System Analysis


D y transformer

D-Y transformer

H1

X1

A

a

Zeq+3Zn

A

a

B

VA0

I0

Va0

b

g

c

C

Zero-sequence

network (per unit)

n

Zn

Notes on Power System Analysis


D y transformer1

D-Y transformer

H1

X1

A

a

Zeq

A

a

B

I1

Va1

VA1

b

n

c

C

Positive-sequence

network (per unit)

Delta side leads wye

side by 30 degrees

n

Zn

Notes on Power System Analysis


D y transformer2

D-Y transformer

H1

X1

A

a

Zeq

A

a

B

I2

Va2

VA2

b

n

c

C

Negative-sequence

network (per unit)

Delta side lags wye

side by 30 degrees

n

Zn

Notes on Power System Analysis


Lesson 8 symmetrical components

Three-winding (three-phase)

transformers Y-Y-D

H and X in grounded Y and T in delta

Zero sequence

Positive and negative

Neutral

Ground

ZT

X

ZX

ZH

ZH

H

H

X

ZX

ZT

T

T

Notes on Power System Analysis


Lesson 8 symmetrical components

Three-winding transformer data:

WindingsZBase MVA

H-X5.39%150

H-T6.44%56.6

X-T4.00%56.6

Convert all Z's to the system base of 100 MVA:

Zhx = 5.39% (100/150) = 3.59%

ZhT= 6.44% (100/56.6) = 11.38%

ZxT= 4.00% (100/56.6) = 7.07%

Notes on Power System Analysis


Lesson 8 symmetrical components

Calculate the equivalent circuit parameters:

Solving:

ZHX= ZH+ ZX

ZHT= ZH+ ZT

ZXT= ZX+ZT

Gives:

ZH= (ZHX+ ZHT- ZXT)/2 = 3.95%

ZX= (ZHX+ ZXT- ZHT)/2 = -0.359%

ZT= (ZHT+ ZXT- ZHX)/2 = 7.43%

Notes on Power System Analysis


Typical relative sizes of sequence impedance values

Typical relative sizes of sequence impedance values

  • Balanced three-phase lines:

    Z0 > Z1 = Z2

  • Balanced three-phase transformers (usually):

    Z1 = Z2 = Z0

  • Rotating machines: Z1 Z2 >Z0

Notes on Power System Analysis


Unbalanced short circuits

Unbalanced Short Circuits

  • Procedure:

    • Set up all three sequence networks

    • Interconnect networks at point of the fault to simulate a short circuit

    • Calculate the sequence I and V

    • Transform to ABC currents and voltages

Notes on Power System Analysis


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