- 131 Views
- Uploaded on
- Presentation posted in: General

Lesson 8 Symmetrical Components

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Lesson 8Symmetrical Components

Notes on Power System Analysis

- 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

- 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

Notes on Power System Analysis

where

a = 1/120°= (-1 + j 3)/2

a2 = 1/240°= 1/-120°

a3 = 1/360°= 1/0 °

Notes on Power System Analysis

Vp = A VsVs= A-1Vp

Notes on Power System Analysis

- 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

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

a

Zy

b

Zy

c

Zn

g

- 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

or in more compact notation Vp= ZpIp

Notes on Power System Analysis

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

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

V1 = ZyI1 = Z1 I1

V2 = ZyI2 = Z2 I2

Notes on Power System Analysis

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

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

- 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

- General case unsymmetrical impedances

Notes on Power System Analysis

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

- Special case symmetrical impedances

Notes on Power System Analysis

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

Sp = VagIa* + VbgIb* + VcgIc*

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

Sp = VpTIp*

= (AVs)T(AIs)*

= VsTATA* Is*

Notes on Power System Analysis

Sp = VpTIp* = VsT ATA* Is*

Sp = 3 VsT Is*

Notes on Power System Analysis

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

- 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

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

N

V1

Positive

I1

Z

N

V2

I2

Negative

Ungrounded Y load

Z

G

V0

Zero

I0

Z

N

Notes on Power System Analysis

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

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

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

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

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

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

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

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

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

- 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

- 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