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ECE 504:
Advanced Power Systems Protection Spring 2009
C 1 C =6.6315 × 10 −^4
ω ⋅XCt
XCt :=3 Im Zc⋅ ( ) XCt =− 4
L Xt L =0. ω
Xt :=2 Im Zsp⋅ ( )+ 2 Im Zlp⋅ ( )+ Im Zs0( )+Im Zl0( ) Xt =22.
R :=2 Re Zsp⋅ ( )+ 2 Re Zlp⋅ ( )+ Re Zs0( )+Re Zl0( ) R =4.
simulate an AG fault
Zc 4 Zc =−1.3333i 3
:= ⋅e i⋅^ −^90 ⋅DR
Zs0 :=1 e⋅ i 80⋅^ ⋅DR Zs0 =0.1736 +0.9848i
Zsp :=1 e⋅ i 85⋅^ ⋅DR Zsp =0.0872 +0.9962i
Zl0 :=12 e⋅ i 75⋅^ ⋅DR Zl0 =3.1058 +11.5911i
Zlp :=4 e⋅ i 80⋅^ ⋅DR Zlp =0.6946 +3.9392i
DR π ω := 2 ⋅π ⋅ 60 ω =376. 180
v(t)
Zs Zc
vbus
Zl
Series capacitor solution:
file name: SeriesCap_2.mcd
ECE 504:
Advanced Power Systems Protection Spring 2009
β 2 b (^) β 2 =153. a 2 4
β 1 :=ω := −
α 1 := 0 α^2 α^2 =40.
a 2
[( 1 )^2 12 ] [( 2 )^222 ]
() ( cos sin )
s s
Is n s s
b 1 b =2.5326 × 10 4 L C⋅
a R a =81. L
n V n =1.5914 × 10 3 L
( ) ( cos sin )
s^2 2 s^2 a s b
I s n s s
Use the ' trick of Papoulis ' method to solve by Partial Fractions
( )(^1 )
( cos sin )
( )(^1 )
() ( cos sin )
() cos sin ()^1
2 2 2
2 2
2 2
L L C
s s s R
s
L
V s
Is
s C
s R s L
Is V s
s C
I s R s L
s
Vs V s
= ⋅ ⋅ + = ⋅ + dt ⋅ L + C ⋅∫ it ⋅ dt
v ( t ) V sin(ω t θ) i ( t ) R di ( t )^1 ()
t :=0 0.001, ..0.
V := 67 ⋅ 2 θ := 0
ECE 504:
Advanced Power Systems Protection Spring 2009
(^200) 0.017 0.033 0.05 0.067 0.083 0.1 0.12 0.13 0.15 0.
0
20
Fault Current
Time (seconds)
f1 t( ) f2 t( )
t
(^300) 0.017 0.033 0.05 0.067 0.083 0.1 0.12 0.13 0.15 0.
20
10
0
10
20
Current
i t( )
t
ECE 504:
Advanced Power Systems Protection Spring 2009
vbus t( ) :=f1 t( ) +f2 t( )
f2 t( ) M β 2
:= ⋅ e −^ α^2 ⋅t⋅sin (^ β2 t⋅ +A2)
A2 :=arg M( ) A2 =0.
M2 := M M2 =960.
M
V s⋅ L
Rl + s Ll⋅^1 s C⋅
⋅ ⋅(ω^ ⋅ cos (θ^ ⋅DR)+s sin⋅ (θ^ ⋅DR))
s^2 +ω^2
s :=−α 2 +i ⋅β 2
f1 t( ) M β 1
:= ⋅ e −^ α^1 ⋅t⋅sin (^ β1 t⋅ +A1)
A1 :=arg M( ) A1 =−0.
M1 := M M1 =3.0172 × 104
M
V s⋅ L
Rl + s Ll⋅^1 s C⋅
⋅ ⋅(ω^ ⋅ cos (θ^ ⋅DR)+s sin⋅ (θ^ ⋅DR))
s^2 + a s⋅ +b
s :=i ⋅ω
( )(^1 )
1 (^1 )( cos sin )
() () ()^1 ()
2 2 2
L L C
s s s R
s
s C
Rl s Ll
L
V s
sC
Vbuss Is Rl s Ll
it dt
C
Ll
dt
vbust it Rl dit
= ⋅ + ⋅ + ∫ ⋅
ω
ω θ θ
Ll Xl Ll =0. ω
Xl :=2 Im Zlp⋅ ( )+Im Zl0( ) Xl =19.
Rl :=2 Re Zlp⋅ ( )+Re Zl0( ) Rl =4.
ECE 504:
Advanced Power Systems Protection Spring 2009
I dt 1 32
t ←x dt⋅ C (^) x ←i t( )
for x ∈0 1, .. 320
C
K := 32
k :=0 1, .. 31
N := 6
a :=0 1, ..N 32⋅
(^400 50 100 150 )
20
0
20
I (^) a
a
V dt 1 32
t ←x dt⋅ C (^) x ←vbus t( )
for x ∈0 1, .. 320
C
(^1000 50 100 150 )
0
100
V (^) a
a
ECE 504:
Advanced Power Systems Protection Spring 2009
Vph M ←submatrix V x 2( , ⋅ , x 2⋅ + 31 , 0 , 0 )
val (^) x^2 K ⋅ 2 k
M (^) k e
− i⋅ 2 ⋅ πk K
⋅
for x ∈0 1, ..( N) 16⋅
val
Z0 :=Zl0 +Zc Z1 :=Zlp +Zc
Iph M ←submatrix I x 2( , ⋅ , x 2⋅ + 31 , 0 , 0 )
val (^) x^2 K ⋅ 2 k
M (^) k e
− i⋅ 2 ⋅ πk K
⋅
val (^) x val (^) x
(Z0 −Z1) val ⋅ (^) x 3 Z1⋅
for x ∈0 1, ..(N ) 16⋅
val
spiral impedance:
z
val (^) x
Vph (^) x Iphx
for x ∈0 1, ..(N ) 16⋅
val
zone 1 mho: m := 0 1, .. 360 reach :=2.
θm := m DR⋅ ang := 80
mhom reach 2
:= ⋅(^ e i⋅^ θ^ m+e i ang⋅^ ⋅DR)
steady-state impedance:
sz
Zlp Zlp +Zc
