Planning of distribution system
By. Dr. Youcef Berrouche
Primary feeder
380kV
380kV
33kv or 13kv
110/200V
Primary feeder
Radial Primary feeder
The tie
Loop Type
Voltage of the feeder
V=1pu
V=2pu
V=1pu
V=1pu
Voltage of the feeder
Example 1)
A feeder has a voltage of 13kV Line to neutral and a length of
1000 m. This feeder is feeding a 514 kvar of loads
a) If we increase the voltage to 33kV, What is the new length if we
keep the same load
b) If we increase the voltage to 33kV and the load to 700kVar
What is the new length
c) If we increase the voltage to 33k and we keep the same length.
What is the new load that we can add
Design of the feeder
Case 1 : radial feeder with uniform load
The main
dI/dx=k
Laterals
l
Design of the feeder
Case 1 : radial feeder with uniform load
Is : sending current
Ir: receiving current =0
Design of the feeder
Case 1 : radial feeder with uniform load
k=Is/l
I(x)=Is(1-x/l)
VD(x)=Is.z.x.(1-x/2l)
VD(l)=0.5.Is.z.l
PLosses =1/3.Is.Is.r.l
Design of the feeder
Example :
A 11kv feeder has a length`` l=1km``. It is feeding 40 3-phases loads
as presented in the schematic bellow. Assume that all the loads
have the same power of 125kvar. The power factor is assumed 0.9
The impedance of the line is given as :z=(0.31+0.06J)/km
Vl-n=13kv
Design of the feeder
Example :
1.Calculate the current in each lateral
2.Calculate the sending current of the main
3.Calculate the constant k
4.Calculate the voltage drop on the main.
5.Calculate the current of the main at 500m
Design of the feeder
Example 1 :
Solution
1. The current of each lateral is given as :
So : Ilateral = 125*4/(1.73*13)=22.2A
2. The current of the main is : Is=5* Ilateral =111A
3. k=Is/l==111/1000=0.111 A/m
4. z=squrt (R2+X2)=0.316
Therfore : VD=0.316*111/2=17.538v
5. I (500m)=111(1-500/1000)=55.5A
Design of the feeder
Case 1 : radial feeder with non uniform load
l
Design of the feeder
Case 1 : radial feeder with non uniform load
Design of the feeder
Case 1 : radial feeder with non uniform load
Design of the feeder
Case 1 : radial feeder with non uniform load
Example 2 :
A feeder has a main length of 1.5 km and a 33kv. It is loading
6000kvar. It has z=(0.12+0.044J)/km. Calculate the voltage drop of
its main if is has non uniform distribution
Design of the feeder
Case 1 : radial feeder with non uniform load
Example 2 :
Is=6000/33=181A
Z=sqrt(0.12*0.12+0.044*0.044)=0.126 ohm
VD=2/3*0.126*1.5/1*181=22.8 V
AWG Tables
AWG Tables
Current
Resistanc Load
DiameterDiameter Square
e
Ratings (
amps)1)
AWG
(ohm/10
(mm)
(in)
(mm2)
A
00m)
24
22
20
18
16
14
13
12
10
8
6
4
3
2
1
0 (1/0)
00 (2/0)
0.51
0.64
0.81
1.02
1.29
1.63
1.80
2.05
2.59
3.25
4.115
5.189
4,752
6.543
7.348
8.252
9.266
0.02
0.025
0.032
0.04
0.051
0.064
0.072
0.081
0.10
0.13
0.17
0.20
0,22
0.26
0.29
0.33
0.37
0.20
0.33
0.50
0.82
1.3
2.0
2.6
3.3
5.26
8.30
13.30
21.15
26.65
33.62
42.41
53.49
67.43
87.5
51.7
34.1
21.9
13.0
8.54
6.76
5.4
3.4
2.2
1.5
0.8
0,6
0.5
0.4
0.31
0.25
3.5
5.0
6.0
9.5
20
24
30
34
52
75
95
120
154
170
180
200
225
Design of the feeders : main and lateral
Getting Information : loads picks, Main length,
lateral length....
Design of the lateral : the
section conductor TAB 1
Design of the Main: the
section conductor TAB 1
No
Is VD ok
Yes
Design of the feeders : main and lateral
Example :
A 11kv feeder has a length`` l=1km``. It is feeding 40 3-phases loads
as presented in the schematic bellow. Assume that all the loads
have the same power of 125kvar. The power factor is assumed 0.9
The impedance of the line is given as :z=(0.31+0.06J)/km