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Power System Stability: Out-of-Step Protection - Session 33e Phasor Diagrams, Study notes of Electrical and Electronics Engineering

University of Idaho (U of I)Electrical and Electronics Engineering

Phasor diagrams and calculations for the analysis of power system stability during out-of-step conditions. The diagrams illustrate the relationships between eg, v, vs, and i, as well as the corresponding voltages and impedances. A handout for session 33e of ece 504: power system stability course.

Typology: Study notes

Pre 2010

Uploaded on 08/19/2009

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ECE 504

Power System Stability Out-of-Step

Protection Session 33e

I. Define the Situation j1−:=

Xg0.25:= XL0.10:= Xs0.653:=

jXs

+ VD -

jXgjXL

δ0 deg⋅:=

Egδ

()

1.2ejδ⋅

:= Vs1ej0⋅deg⋅

II. Draw Phasor Diagrams that Include Eg, V, Vs, and I

XTXgXL

+Xs

+:=

EgVDVs

+=jX

⋅I⋅Vs

+=VV

sjX

LXs

()

⋅I⋅+=

Iδ

()

Egδ

()

−

⋅

Vδ

()

VsjX

LXs

()

⋅Iδ

()

⋅+:=

Zδ

()

Vδ

()

Iδ

()

C:\JLAW\CLASSES\S07 ECE 504

\HANDOUTS\Out-of-Step\s33e.mcd Page 1 of 9 March 9, 2007

Discover Study notes of Electrical and Electronics Engineering University of Idaho (U of I)

Partial preview of the text

Download Power System Stability: Out-of-Step Protection - Session 33e Phasor Diagrams and more Study notes Electrical and Electronics Engineering in PDF only on Docsity!

Power System Stability Protection

I. Define the Situation j := − 1

Xg := 0.25 XL := 0.10 Xs :=0.

E g V S

jX s

I

+ V D -

jX g jX L

V

δ :=0 deg⋅

Eg ( )δ^ 1.2e

j ⋅δ

:= Vs 1e

j 0⋅ ⋅deg

II. Draw Phasor Diagrams that Include E g , V, V s, and I

XT :=Xg + XL+Xs

Eg = VD +Vs = j X⋅ T⋅ I+Vs V = Vs +j ⋅( X L +Xs)⋅I

I ( )δ^

Eg ( )δ^ −Vs

j X⋅ T

V ( )δ^ :=Vs +j ⋅ ( XL +Xs)⋅I ( )δ

Z ( )δ^

V( ) δ

I ( )δ

C:\JLAW\CLASSES\S07 ECE 504

Page 1 of 9 March 9, 2007

Power System Stability Protection

δ :=20 deg⋅

I ( )δ^ = 0.409 −0.127i I ( )δ^ = 0.429 arg I(^ ( )δ)^ =−17.274 deg

V ( )δ^ = 1.096 +0.308i V( ) δ^ = 1.138 arg V(^ ( )δ)^ =15.705 deg

Z ( )δ^ = 2.23 +1.45i Z ( )δ^ = 2.656 arg Z(^ ( )δ)^ =32.979 deg

Eg ( δ)

V ( δ)

I (δ )

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

Z ( )δ^ =2.

arg Z(^ ( )δ)^ =32.979 deg

Za 0 :=Z ( )δ

Z ( δ)

arg Z( (δ ))

C:\JLAW\CLASSES\S07 ECE 504

Page 2 of 9 March 9, 2007

Power System Stability Protection

δ :=60 deg⋅

I ( )δ^ = 1.036 +0.399i I ( )δ^ = 1.11 arg I(^ ( )δ)^ =21.052 deg

V ( )δ^ = 0.7 +0.78i V( ) δ^ = 1.048 arg V(^ ( )δ)^ =48.114 deg

Z ( )δ^ = 0.84 +0.43i Z ( )δ^ = 0.944 arg Z(^ ( )δ)^ =27.062 deg

Za 2 :=Z ( )δ

Eg ( δ)

V ( δ)

I (δ )

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

Z ( δ)

arg Z( (δ ))

C:\JLAW\CLASSES\S07 ECE 504

Page 4 of 9 March 9, 2007

Power System Stability Protection

δ :=90 deg⋅

I ( )δ^ = 1.196 +0.997i I ( )δ^ = 1.557 arg I(^ ( )δ)^ =39.806 deg

V ( )δ^ = 0.249 +0.901i V( ) δ^ = 0.935 arg V(^ ( )δ)^ =74.535 deg

Z ( )δ^ = 0.49 +0.34i Z ( )δ^ = 0.6 arg Z(^ ( )δ)^ =34.729 deg

Za 3 :=Z ( )δ

Eg ( δ)

V ( δ)

I (δ )

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

Z ( δ)

arg Z( (δ ))

C:\JLAW\CLASSES\S07 ECE 504

Page 5 of 9 March 9, 2007

Power System Stability Protection

δ :=150 deg⋅

I ( )δ^ = 0.598 +2.033i I ( )δ^ = 2.119 arg I(^ ( )δ)^ =73.605 deg

V ( )δ^ = −0.531 +0.45i V( ) δ^ = 0.696 arg V(^ ( )δ)^ =139.689 deg

Z ( )δ^ = 0.13 +0.3i Z ( )δ^ = 0.329 arg Z(^ ( )δ)^ =66.085 deg

Za 5 :=Z ( )δ

Eg ( δ)

V ( δ)

I (δ )

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

Z ( δ)

arg Z( (δ ))

C:\JLAW\CLASSES\S07 ECE 504

Page 7 of 9 March 9, 2007

Power System Stability Protection

δ :=180 deg⋅

I ( )δ^ = 2.193i I ( )δ^ = 2.193 arg I(^ ( )δ)^ =90 deg

V ( )δ^ = −0.652 V( ) δ^ = 0.652 arg V(^ ( )δ)^ =− 180 deg

Z ( )δ^ = 0.3i Z ( )δ^ = 0.297 arg Z(^ ( )δ)^ =90 deg

Za 6 :=Z ( )δ

Eg ( δ)

V ( δ)

I (δ )

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

Z ( δ)

arg Z( (δ ))

Power System Stability: Out-of-Step Protection - Session 33e Phasor Diagrams, Study notes of Electrical and Electronics Engineering

Related documents

Partial preview of the text

Download Power System Stability: Out-of-Step Protection - Session 33e Phasor Diagrams and more Study notes Electrical and Electronics Engineering in PDF only on Docsity!

I. Define the Situation j := − 1

Xg := 0.25 XL := 0.10 Xs :=0.

E g V S

jX s

I

+ V D -

jX g jX L

V

δ :=0 deg⋅

Eg ( )δ^ 1.2e

:= Vs 1e

II. Draw Phasor Diagrams that Include E g , V, V s, and I

XT :=Xg + XL+Xs

Eg = VD +Vs = j X⋅ T⋅ I+Vs V = Vs +j ⋅( X L +Xs)⋅I

I ( )δ^

Eg ( )δ^ −Vs

j X⋅ T

V ( )δ^ :=Vs +j ⋅ ( XL +Xs)⋅I ( )δ

Z ( )δ^

V( ) δ

I ( )δ

C:\JLAW\CLASSES\S07 ECE 504

Page 1 of 9 March 9, 2007

δ :=20 deg⋅

I ( )δ^ = 0.409 −0.127i I ( )δ^ = 0.429 arg I(^ ( )δ)^ =−17.274 deg

V ( )δ^ = 1.096 +0.308i V( ) δ^ = 1.138 arg V(^ ( )δ)^ =15.705 deg

Z ( )δ^ = 2.23 +1.45i Z ( )δ^ = 2.656 arg Z(^ ( )δ)^ =32.979 deg

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

Z ( )δ^ =2.

arg Z(^ ( )δ)^ =32.979 deg

Za 0 :=Z ( )δ

C:\JLAW\CLASSES\S07 ECE 504

Page 2 of 9 March 9, 2007

δ :=60 deg⋅

I ( )δ^ = 1.036 +0.399i I ( )δ^ = 1.11 arg I(^ ( )δ)^ =21.052 deg

V ( )δ^ = 0.7 +0.78i V( ) δ^ = 1.048 arg V(^ ( )δ)^ =48.114 deg

Z ( )δ^ = 0.84 +0.43i Z ( )δ^ = 0.944 arg Z(^ ( )δ)^ =27.062 deg

Za 2 :=Z ( )δ

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

C:\JLAW\CLASSES\S07 ECE 504

Page 4 of 9 March 9, 2007

δ :=90 deg⋅

I ( )δ^ = 1.196 +0.997i I ( )δ^ = 1.557 arg I(^ ( )δ)^ =39.806 deg

V ( )δ^ = 0.249 +0.901i V( ) δ^ = 0.935 arg V(^ ( )δ)^ =74.535 deg

Z ( )δ^ = 0.49 +0.34i Z ( )δ^ = 0.6 arg Z(^ ( )δ)^ =34.729 deg

Za 3 :=Z ( )δ

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

C:\JLAW\CLASSES\S07 ECE 504

Page 5 of 9 March 9, 2007

δ :=150 deg⋅

I ( )δ^ = 0.598 +2.033i I ( )δ^ = 2.119 arg I(^ ( )δ)^ =73.605 deg

V ( )δ^ = −0.531 +0.45i V( ) δ^ = 0.696 arg V(^ ( )δ)^ =139.689 deg

Z ( )δ^ = 0.13 +0.3i Z ( )δ^ = 0.329 arg Z(^ ( )δ)^ =66.085 deg

Za 5 :=Z ( )δ

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

C:\JLAW\CLASSES\S07 ECE 504

Page 7 of 9 March 9, 2007

δ :=180 deg⋅

I ( )δ^ = 2.193i I ( )δ^ = 2.193 arg I(^ ( )δ)^ =90 deg

V ( )δ^ = −0.652 V( ) δ^ = 0.652 arg V(^ ( )δ)^ =− 180 deg

Z ( )δ^ = 0.3i Z ( )δ^ = 0.297 arg Z(^ ( )δ)^ =90 deg

Za 6 :=Z ( )δ

arg V( s) , arg E( g ( δ)), arg V( ( δ)),arg I( ( δ))

C:\JLAW\CLASSES\S07 ECE 504

Page 8 of 9 March 9, 2007