Developed by
Project RPA
Last updated: 08.03.2025
e-mail: proRpaLab@gmail.com
GRID MODEL 20/0.4kV
Circuit diagram
SC LV-1
No
АВС
АВС0
AB
BC
CA
AB0
BC0
CA0
A0
B0
C0
OPC LV-1
Нет
АВС
AB
BC
CA
A
B
C
neutral (N)
R
SC
, R
OPC
SC MV-1
No
АВС
АВС0
AB
BC
CA
AB0
BC0
CA0
A0
B0
C0
OPC MV-1
No
АВС
AB
BC
CA
A
B
C
R
SC
, R
OPC
SC LV-2
No
АВС
АВС0
AB
BC
CA
AB0
BC0
CA0
A0
B0
C0
OPC LV-2
No
АВС
AB
BC
CA
A
B
C
neutral (N)
R
SC
, R
OPC
SC MV-2
No
АВС
АВС0
AB
BC
CA
AB0
BC0
CA0
A0
B0
C0
OPC MV-2
No
АВС
AB
BC
CA
A
B
C
R
SC
, R
OPC
Currents [1]
I
A
I
B
I
C
I
1
I
2
I
0
I
N
I
AB+
I
BC+
I
CA+
I
AB
I
BC
I
CA
Voltage [1]
U
A
U
B
U
C
U
1
U
2
U
0
3U
0
U
AB+
U
BC+
U
CA+
U
AB
U
BC
U
CA
Phasor notation
remove numbers
Currents [2]
I
A
I
B
I
C
I
1
I
2
I
0
I
N
I
AB+
I
BC+
I
CA+
I
AB
I
BC
I
CA
Voltages [2]
U
A
U
B
U
C
U
1
U
2
U
0
3U
0
U
AB+
U
BC+
U
CA+
U
AB
U
BC
U
CA
Phasor notation
remove numbers
Currents [3]
I
A
I
B
I
C
I
1
I
2
I
0
I
N
I
AB+
I
BC+
I
CA+
I
AB
I
BC
I
CA
Voltages [3]
U
A
U
B
U
C
U
1
U
2
U
0
3U
0
U
AB+
U
BC+
U
CA+
U
AB
U
BC
U
CA
Phasor notation
remove numbers
Currents [4]
I
A
I
B
I
C
I
1
I
2
I
0
I
N
I
AB+
I
BC+
I
CA+
I
AB
I
BC
I
CA
Voltages [4]
U
A
U
B
U
C
U
1
U
2
U
0
3U
0
U
AB+
U
BC+
U
CA+
U
AB
U
BC
U
CA
Phasor notation
remove numbers
0.4 кV (А)
Load on the 0.4 kV busbar, Ubase
S=
+ j
kVA
U
base
=
kV
0.4 kV power line
through Z
0
through Z
S
Z
1(2)
=
+ j
mΩ/m
Z
S
=
mΩ/m
Z
0
=
+ j
mΩ/m
l=
m
Load on the 0.4 kV busbar
S
A
=
+ j
kVA
S
B
=
+ j
kVA
S
C
=
+ j
kVA
20 kV, T, Ubase, coupling (A)
Power system
U=
kV
ψ
1
=
°
U
base
=
kV
through Z
through I
k
through S
k
X
1(2)
=
Ω
I
k3
=
kA
S
k3
=
MVA
Z
N
=
+ j
Ω
C
∑
=
µF
Load on the 20 kV busbar
through Z
through S (Z
0
=inf)
Z
1(2)
=
+ j
Ω
P=
MW
Q=
Mvar
Power line 20 kV
I know Z
0
Z
1(2)
=
+ j
Ω/km
L=
km
Z
0
=
+ j
Ω/km
C=
µF/km
Transformer 20 kV
V group
Dyn11
Yzn11
Yyn0
Dyn5
S=
kVA
P
k
=
kW
uk=
%
V
ratio
=
X
0
/X
1
=
as seen from the 0.4kV side
Connection between the power systems on the 20 kV side
I know X
0
X
1(2)
=
Ω
X
0
=
Ω
0.4 kV (B)
Load on the 0.4 kV busbar
S=
+ j
kVA
0.4 kV power line
through Z
0
through Z
S
Z
1(2)
=
+ j
mΩ/m
Z
S
=
Ω/km
Z
0
=
+ j
mΩ/m
l=
m
Load connected to the 0.4 kV power line
S
A
=
+ j
kVA
S
B
=
+ j
kVA
S
C
=
+ j
kVA
20 kV, Т (B)
Power system
U=
kV
ψ
2
=
°
through Z
through I
k
through S
k
X
1(2)
=
Ω
I
k3
=
kA
S
k3
=
MVA
Z
N
=
+ j
Ω
C
∑
=
µF
Load on the 20 kV busbar
through Z
through S (Z
0
=inf)
Z
1(2)
=
+ j
Ω
P=
MW
Q=
Mvar
20 kV power line
I know Z
0
Z
1(2)
=
+ j
Ω/km
l=
km
Z
0
=
+ j
Ω/km
C=
µF/km
Transformer 20 kV
V group
Dyn11
Yzn11
Yyn0
Dyn5
S=
kVA
P
k
=
kW
uk=
%
V
ratio
=
X
0
/X
1
=
as seen from the 0.4kV side
