λ (J 1 ) 1 (J 1 ) 2 (I C ) 1 = 270A U 1 U 2 1 = 6300 370 U 1 U 2 2 = 6300 370 I SARE , (I 1 ) SARE F SARE Δ λ (I C ) 2 = 270A λ Ψ 1 =0 o Ψ 2 =30 o
λ
(J1)1 (J1)2
(IC)1 = 270A
€
U1
U2
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
=6300370
€
U1
U2
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
=6300370
ISARE, (I1)SARE FSARE
Δ λ
(IC)2 = 270A
λ
Ψ1=0o Ψ2=30o
€
U1
U2
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
=6300370
⇒n1n2
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
=U1
U2
=6300370
3=17 3
U1
U2
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
=6300370
⇒n1n2
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
=
U13
U23
=6300
3370
3=17
λ
Δ
λ
λ
€
⇒n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
= 3 ⋅ n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
30 60 90 120 150 180 210 240 270 300 330 360
(VL)1
(Is1)1
(V13)1 (V23)1 (V21)1 (V31)1 (V32)1 (V12)1
(Ip1)1
IC
-IC
€
n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC( )1 = x
€
−n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC( )1 = −x
(V32)1
(J1)1
(IC)1 = 270A
Δ
λ
Ψ1=0o
30 60 90 120 150 180 210 240 270 300 330 360
(VL)1
(Is3)1
(V13)1 (V23)1 (V21)1 (V31)1 (V32)1 (V12)1
(Ip3)1
IC
-IC
€
−n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC = −x
(V32)1
€
n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC = x
(J1)1
(IC)1 = 270A
Δ
λ
Ψ1=0o
(J1)1
(IC)1 = 270A
Δ
λ
Ψ1=0o (Ip3)3
(Ip1)3
(J1)3 = (Ip1)3 - (Ip3)3
€
x€
2x
€
−x
€
−2x
30 60 90 120 150 180 210 240 270 300 330 360
€
n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC = x
€
−n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC = −x
€
−n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC = −x
€
n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC = x
30 60 90 120 150 180 210 240 270 300 330 360
(VL)2
(Is1)2
(V13)2 (V23)2 (V21)2 (V31)2 (V12)2
IC
-IC
€
n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
⋅ IC( )2 = 3 ⋅ n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC( )1 = 3 ⋅ x
(V32)2
(J1)2
(IC)2 = 270A
λ
Ψ2=30o
λ
(J1)2 = (Ip1)2
€
−n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
⋅ IC( )2 = − 3 ⋅ n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC( )1 = − 3 ⋅ x
ISARE
30 60 90 120 150 180 210 240 270 300 330 360
(J1)2
(J1)1
€
x =n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
⋅ IC( )1 =1
17 3⋅ 270 = 9.17A
1+ 3( ) ⋅ x = 1+ 3( ) ⋅ 9.17 = 25A
2 + 3( ) ⋅ x = 2 + 3( ) ⋅ 9.17 = 34.22A
ISARE =12π
⋅ 2 ⋅ π3⋅ 9.172 + 2 ⋅ π
3⋅ 252 + 2 ⋅ π
3⋅ 34.222
⎡
⎣ ⎢ ⎤
⎦ ⎥
=9.172 + 252 + 34.222
3
ISARE = 25A
€
x€
2x
€
−x
€
−2x
€
3 ⋅ x
€
− 3 ⋅ x
€
−x
€
x
€
− 1+ 3( ) ⋅ x
€
2 + 3( ) ⋅ x
€
1+ 3( ) ⋅ x
€
− 2 + 3( ) ⋅ x
€
VLC( )1 = VLC 0( )1 − ΔVX( )1 = 500 − 60 = 440V
VLC 0( )1 =VLC 0 PD3 =3 3VO PD3
π=
3 3 370 23
⎛
⎝ ⎜
⎞
⎠ ⎟
π= 500v
VO PD3 = 2 ⋅ U23
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 370 2
3
λ
Δ
λ
€
VLC( )2 = VLC '( )2 = VLC 0 '( )2 − ΔVX( )2 = 433− 60 = 373V
VLC 0 '( )2 = VLC 0( )2 ⋅ cosψ2 = 500 ⋅ cos30º= 433V
VLC 0( )2 =VLC 0 PD3=3 3VO PD3
π=
3 ⋅ 3 ⋅ 370 23
⎛
⎝ ⎜
⎞
⎠ ⎟
π= 500v
VO PD3 = 2 ⋅ U23
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 370 2
3
λ
€
FSARE =PLC( )1 + PLC( )23 ⋅U1 ⋅ ISARE
=VLC( )1 ⋅ (IC )1[ ] + VLC( )2 ⋅ (IC )2[ ]
3 ⋅U1 ⋅ ISARE=440 ⋅ 270[ ] + 373 ⋅ 270[ ]
3 ⋅ 6300 ⋅ 25= 0.80
€
cos ϕ1( )1 =1− ΔVX
VLC 0
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
cos ϕ1( )1 =1− 0.12 = 0.88
ϕ1( )1 = arccos 0.88( ) = 28.35º
cos ϕ1( )2 = cosψ2 −ΔVX
VLC 0
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
cos ϕ1( )2 = cos30º− 60500
= 0.74
ϕ1( )2 = arccos 0.74( ) = 41.75º
VR
(I1)1 (I1)2
€
ϕ1( )1 ≠ ϕ1( )2
€
(I1)SARE = (I1)SARE ,X2 + (I1)SARE ,Y
2 = 20.112 +14.112 = 24.56A
(I1)SARE ,X = (I1)1,X + (I1)2,X =10.88 + 9.23 = 20.11A
(I1)1,X = (I1)1 ⋅ cos ϕ1( )1 =12.37 ⋅ cos28.35 =10.88A
(I1)2,X = (I1)2 ⋅ cos ϕ1( )2 =12.37 ⋅ cos41.75 = 9.23A
(I1)SARE ,Y = (I1)1,Y + (I1)2,Y = 5.87 + 8.24 =14.11A
(I1)1,Y = (I1)1 ⋅ sin ϕ1( )1 =12.37 ⋅ sin28.35 = 5.87A
(I1)2,Y = (I1)2 ⋅ sin ϕ1( )2 =12.37 ⋅ sin41.75 = 8.24A
(I1)1 (I1)2
€
I1( )1 =VLC 0( )1 ⋅ (IC )1
3 ⋅U1
=500 ⋅ 2703 ⋅ 6300
=12.37A
I1( )2 =VLC 0( )2 ⋅ (IC )2
3 ⋅U1
=500 ⋅ 2703 ⋅ 6300
=12.37A
(I1)2,X (I1)1,X
(I1)1,Y (I1)2,Y