Exercise 1.
A feeder has three loads : street lighting, residential and commercial. Le load profile of each
load is presented in the table below
residential
Commercial
Time
street lighting kw kw
kw
12:00 AM
100
250
300
1:00 AM
100
250
300
2:00 AM
100
250
300
3:00 AM
100
250
300
4:00 AM
100
250
300
5:00 AM
100
250
300
6:00 AM
100
250
300
7:00 AM
0
350
300
8:00 AM
0
450
400
9:00 AM
0
550
600
10:00 AM
0
550
1100
11:00 AM
0
550
1100
12:00 PM
0
600
1100
1:00 PM
0
600
1100
2:00 PM
0
600
1300
3:00 PM
0
600
1300
4:00 PM
0
600
1300
5:00 PM
0
650
1300
6:00 PM
0
750
900
7:00 PM
0
900
500
8:00 PM
100
1100
500
9:00 PM
100
1100
500
10:00 PM
100
900
300
11:00 PM
100
700
300
12:00 AM
100
350
300
1.) Draw the load curve of the three loads and the system one.
2.) Calculate the daily average load of the system
3.) Give the maximum demand of each load (Di) as well as the feeder one (Dg)
4.) Calculate the Load factor (Fld)
1
5.) Assume that the total connected load are 150kW, 2500kW and 2500kW for the street
lighting, residential and commercial respectively. Calculate the Demand factor (DF) of each load
and for all the system.
6.) The peak power is a 5pm. calculate the contributions (Ci) factors for each load
7.) Calculate the Diversity Factor (FD) of the system
8.) Deduce the Coincidence Factor (Fc)
9.) Calculate the load diversity of the system
Exercise 2
Repeat the same questions for the load below
Time
street lighting residential
commercial
12:00 AM
120
200
350
1:00 AM
120
200
350
2:00 AM
120
200
350
3:00 AM
120
200
350
4:00 AM
120
200
350
5:00 AM
120
200
350
6:00 AM
120
200
350
7:00 AM
0
300
350
8:00 AM
0
500
450
9:00 AM
0
550
1200
10:00 AM
0
550
1200
11:00 AM
0
550
1200
12:00 PM
0
600
1200
1:00 PM
0
600
1200
2:00 PM
0
600
1350
3:00 PM
0
600
1350
4:00 PM
0
600
1350
5:00 PM
0
650
1350
6:00 PM
0
750
1000
7:00 PM
0
900
600
8:00 PM
120
1000
600
9:00 PM
120
1000
600
10:00 PM
120
900
350
11:00 PM
120
700
350
12:00 AM
120
350
350
2
Exercise 3:
The feeder in the exercise 1 has a peak loss of 72kW at peak load and annual loss factor of 0.14
1.) Calculate the average power loss of the feeder
2.) The total annual energy loss of the feeder
3
Solutions :
Exercise 1
The total can be calculated by adding all the loads
Time
12:00 AM
1:00 AM
2:00 AM
3:00 AM
4:00 AM
5:00 AM
6:00 AM
7:00 AM
8:00 AM
9:00 AM
10:00 AM
11:00 AM
12:00 PM
1:00 PM
2:00 PM
3:00 PM
4:00 PM
5:00 PM
6:00 PM
7:00 PM
8:00 PM
9:00 PM
10:00 PM
11:00 PM
12:00 AM
street
lighting residencial commercial Total
100
250
300
650
100
250
300
650
100
250
300
650
100
250
300
650
100
250
300
650
100
250
300
650
100
250
300
650
0
350
300
650
0
450
400
850
0
550
600
1150
0
550
1100
1650
0
550
1100
1650
0
600
1100
1700
0
600
1100
1700
0
600
1300
1900
0
600
1300
1900
0
600
1300
1900
0
650
1300
1950
0
750
900
1650
0
900
500
1400
100
1100
500
1700
100
1100
500
1700
100
900
300
1300
100
700
300
1100
100
350
300
750
4
1.) Draw the load curve of the three loads and the system one.
5
As it can be seen the pick is at 5pm
2.) Calculate the daily average load of the system
a) firs method :
From the curve the average load can be given as
ave. Load=
(650*7+850+1150+1650*2+1700*2+1900*3+1950+1400+1700*2+1300+1100+750)/24
ave. Load=30500/24
ave. Load=1270.83kw
b) second method
from the table we do not count the first value ( because if we do the total hours will be 25) than
we average the rest on 24 that gives the same value
(650+650+650+650+650+650+650+850+1150+1650+1650+1700+1700+1900+1900+1900+1950+
1650+1400+1700+1700+1300+1100+750)/24
ave. Load=1270.83kw
3.) Give the maximum demand of each load (Di) as well as the feeder one (Dg)
From the curve or the table the maximum demand of each load (Di) as well as the feeder one
(Dg) are given as:
𝐷𝑠𝑡𝑟𝑒𝑒𝑡 = 100𝑘𝑤
𝐷𝑟𝑒𝑠 = 1100𝑘𝑤
𝐷𝑐𝑜𝑚 = 1300𝑘𝑤
𝐷𝑠𝑦𝑠 = 1950𝑘𝑤
4.) Calculate the Load factor (Fld)
From equation 6
6
𝐹𝐿𝐷 =
1270
= 𝟎. 𝟔𝟓
1950
5.) Assume that the total connected load are 150kW, 2500kW and 2500kW for the street
lighting, residential and commercial respectively. Calculate the Demand factor (DF) of each
load and for all the system.
The demand factor is given from equation 1
𝐷𝐹𝑠𝑡𝑟𝑒𝑒𝑡 =
100
= 𝟎. 𝟔𝟕
150
𝐷𝐹𝑟𝑒𝑠 =
1100
= 𝟎. 𝟒𝟒
2500
𝐷𝐹𝑐𝑜𝑚 =
1300
= 𝟎. 𝟓𝟐
2500
𝐷𝐹𝑠𝑦𝑠𝑡𝑒𝑚 =
1950
= 𝟎. 𝟑𝟖
2500 + 2500 + 150
6.) The peak power is a 5pm. calculate the contributions (Ci) factors for each load
The contributions (Ci) factors for each load is given by equation 18.a
𝐶𝑠𝑡𝑟𝑒𝑒𝑡 =
𝐶𝑟𝑒𝑠 =
0
=𝟎
100
650
= 𝟎. 𝟓𝟗
1100
𝐶𝑐𝑜𝑚 =
1300
=𝟏
1300
7.) Calculate the Diversity Factor (FD) of the system
7
a) first way : The Diversity Factor (FD) of the system is given by equation 11
𝐹𝐷 =
100 + 1100 + 1300
= 𝟏. 𝟐𝟖
1950
b) second way: The Diversity Factor (FD) of the system is given by equation 11 and equation 18.b
𝐹𝐷 =
100 + 1100 + 1300
= 𝟏. 𝟐𝟖
0 × 100 + 0.59 × 1100 + 1 × 1300
8.) Deduce the Coincidence Factor (Fc)
The Coincidence Factor (Fc) is given by equation 16
𝐹𝑐 =
1
= 𝟎. 𝟕𝟖
1.28
9.) Calculate the load diversity of the system
The load density of the system is given by equation 17
8
𝐿𝐷 = (100 + 1100 + 1300) − 1950 = 𝟓𝟓𝟎𝒌𝑾
All the results are given in this table
Street Residential commercial
Average load KW
Total connected load: Given (kW)
Individual maximum demand (Di) (kW)
Load factor (Fld)
Demand factor (DF)
Contribution factors Ci at (the pick 5pm)
Diversity Factor (FD)
Diversity Factor (FD) using Ci
Coincidence Factor (Fc)
Load diversity LD (kW)
150
100
2500
1100
2500
1300
0,67
0
0,44
0,59
0,52
1
Total
1270,83
5150
1950
0,65
0,38
1,28
1,28
0,78
550
Exercise 2
The results are given by the same equations of exercise 1
Street Residential commercial
Average load KW
Total connected load: Given (kW)
Individual maximum demand (Di) (kW)
Load factor (Fld)
Demand factor (DF)
Contribution factors Ci at (the pick 5pm)
Diversity Factor (FD)
Diversity Factor (FD) using Ci
Coincidence Factor (Fc)
Load diversity LD (kW)
150
120
2500
1000
2500
1350
0,80
0,00
0,40
0,65
0,54
1,00
Total
1348,75
5150
2000
0,67
0,39
1,24
1,24
0,81
470
Exercise 3:
9
The feeder in the exercise 1 has a peak loss of 72kW at peak load and annual loss factor of 0.14
1.) Calculate the average power loss of the feeder
The average power loss of the feeder is given by equation 25
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑠 = 𝐹𝐿𝑆 × 𝑝𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑠 𝑎𝑡 𝑝𝑒𝑎𝑘 𝑙𝑜𝑎𝑑
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑠 = 0.14 × 72 = 𝟏𝟎. 𝟖𝒌𝑾
2.) The total annual energy loss of the feeder
To pass from the power to the energy we multiply by the time
annual energy loss = 10.8 × 8760ℎ𝑟/𝑦𝑒𝑎𝑟 = 𝟖𝟖𝟑𝟎𝟎𝒌𝑾𝒉
10