Solved Example for Power System Stability II | ECE 504, Study notes of Electrical and Electronics Engineering

Material Type: Notes; Class: ST:Control Systems and Critical Infrastructures; Subject: Electrical & Computer Engr; University: University of Idaho; Term: Spring 2005;

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ECE 504 ST:
Power System Stability II
Example 12.3 University of Idaho
Session 15d
Ignore amortisseur windings. Lmdp
1
1
Lmdu
1
LLfd
+
:=
Given: P 0.9:= Q 0.3:= Vas 1.0 ej36deg
:=
Lmdp 0.14=
SPjQ+:= S 0.9 0.3i+=
Xmdp Lmdp
:=
φarg S():= φ18.4349 deg=
Ias
S
Vas
:= Ias 0.9487=arg Ias
()
17.5651 deg=
αarg Vas
()
:= α36 deg=
Ψat0 Vas Ias RajX
L
+
()
+:=
ΨI0 Asat e
Bsat Ψat0 ΨTI
()
:= ΨI0 0.1884=
Ksd
Ψat0
Ψat0 ΨI0
+
:= Ksd 0.8491=
Ksq Ksd
:= Ksq 0.8491=
kVA kW:= MVA 106W:= rev 1:= j1:= m10 3:=
ORIGIN 1:=
Example 12.3 All values are in pu unless noted otherwise.
Xdu 1.81:= Xqu 1.76:= Xdp 0.3:= XL0.16:=
Ra0.003:= Tdop 8.0 sec:= H 3.5:= KD0.0:=
Asat 0.031:= Bsat 6.93:= ΨTI 0.8:= Ldp Xdp
:=
Lmdu 1.65:= Lmqu 1.60:= LLXL
:=
Rfd 0.0006:= LLfd 0.153:=
C:\JLAW\CLASSES\S05PSSII
\HANDOUTS\s15d.mcd
Page 1 of 6 2/21/2005
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Power System Stability II Session 15d

Ignore amortisseur windings.

L

mdp

L

mdu

L

Lfd

Given: (^) P := 0.9 Q := 0.3 V

as

1.0 e

j 36⋅ ⋅deg

:= ⋅

L

mdp

S := P +j Q⋅ S =0.9 +0.3i

X

mdp

L

mdp

φ := arg S( ) φ =18.4349 deg

I

as

S

V

as

:= I

as

= 0.9487 arg I as

( )

=17.5651 deg

α arg V as

( )

:= α =36 deg

at

V

as

I

as

R

a

j X L

( )

I

A

sat

e

B sat

Ψ at

Ψ TI

− ( )

I

K

sd

at

at

I

:= K

sd

K

sq

K

sd

:= K

sq

kVA := kW MVA 10

6

:= W rev := 1 j := − 1 mΩ 10

− 3

:= ⋅Ω

ORIGIN := 1

Example 12.3 All values are in pu unless noted otherwise.

X

du

:= 1.81 X

qu

:= 1.76 X

dp

:= 0.3 X

L

R

a

:= 0.003 T

dop

:= 8.0 sec⋅ H := 3.5 K D

A

sat

:= 0.031 B

sat

TI

:= 0.8 L

dp

X

dp

L

mdu

:= 1.65 L

mqu

:= 1.60 L

L

X

L

R

fd

:= 0.0006 L

Lfd

C:\JLAW\CLASSES\S05PSSII Page 1 of 6 2/21/

Power System Stability II Session 15d

arg i

( fdphasor0)

=−10.8745 deg

i fdphasor

i =1. fdphasor

E

as

j X md

arg E

( as0)

E =79.1255 deg as

E =2.

as

E

qas

j X d

X

q

⋅ I

ad

arg I

( aq0)

I =79.1255 deg aq

I =0.

aq

I

as

cos δ i

⋅ e

j θ r

π

2

arg I

( ad0)

I =−10.8745 deg ad

I =0.

ad

I

as

sin δ i

⋅ e

j θ r

θ r

θ =−10.8745 deg r

arg E

( qas)

π

δ i

δ =43.13 deg i

arg E

( qas)

:= −α

L

md

K

sd

L

mdu

:= ⋅ L

md

L

mq

K

sq

L

mqu

L

mq

L

d

L

md

L

L

:= + L

d

L

q

L

mq

L

L

:= + L

q

X

md

L

md

X

q

L

q

:= X

d

L

d

E

B

V

as

j 0.65⋅ I as

:= − ⋅ E

B

= 0.9951 arg E

( B)

=− 0 deg

E

qas

V

as

R

a

j X q

I

as

E

qas

= 1.998 arg E

( qas)

=79.1255 deg

C:\JLAW\CLASSES\S05PSSII Page 2 of 6 2/21/

Power System Stability II Session 15d

D R

T

2

X Tq

X

Td

X

Td

X =0.

Td

X

E

X

mdpi

+ X

L

X

mdpi

L

mdpi

X :=

Tq

X =1.

Tq

X

E

X

qi

L

mdpi

X

di

L

di

X :=

qi

L

qi

X

mdi

L

mdi

L

mdpi

L

mdi

L

Lfd

L

qi

L =0.

qi

L

mqi

L

L

L

di

L =0.

di

L

mdi

L

L

L

mqi

md

L

md

I

fd

I

d

( )

md

mq

L

mq

− I

q

mq

δ 0

δ i

:= +α

δ 0

=79.13 deg

Small Signal

K

sdinc

1 B

sat

A

sat

⋅ e

B sat

Ψ at

Ψ TI

− ( )

K

sdinc

K

sqinc

K

sdinc

R

E

R

T

R

a

R

E

:= + R

T

− 3

= ×

X

E

L

mdi

K

sdinc

L

mdu

:= ⋅ L

mdi

L

mqi

K

sqinc

L

mqu

C:\JLAW\CLASSES\S05PSSII Page 4 of 6 2/21/

Power System Stability II Session 15d

a

a

3 3,

− 2 ⋅ 60 ⋅ πR fd

L

Lfd

L

mdpi

L

Lfd

− m 2

L

mdpi

a

3 2,

− 2 ⋅ 60 ⋅ πR fd

L

Lfd

m 1

⋅ L

mdpi

a := ⋅

3 1,

a

2 3,

a := 0

2 2,

a := 0

2 1,

:=2 60⋅ ⋅π

a

1 3,

K

2 H⋅

a :=

1 2,

K

2 H⋅

a :=

1 1,

K

D

2 H⋅

K

K =0.

n 2

md

L

mqi

I

d

⋅ m 2

mq

L

mdpi

I

q

L

mdpi

L

Lfd

I

q

K

K =0.

n 1

md

L

mqi

I

d

⋅ m 1

mq

L

mdpi

I

q

n 2

n =0. 2

R

T

D

L

mdi

L

mdi

L

Lfd

m 2

m =0. 2

X

Tq

D

L

mdi

L

mdi

L

Lfd

n 1

n =0. 1

E

B

R

T

sin δ

⋅ X

Td

cos δ

D

m 1

m =1. 1

E

B

X

Tq

sin δ

⋅ R

T

cos δ

D

C:\JLAW\CLASSES\S05PSSII Page 5 of 6 2/21/