ADAMA SCIENCE AND TECHNOLOGY UNIVERSITY
Assignment#2
Distribution System Engineering
Prepared by: Tahir Kedir
PGR/24532/014
Submitted To: Dr.Ing Getechew B
Submission Date: 05/06/2022
Assignment 2
Distribution System Engineering
I. Two transformers each serving four customers are shown in the Figure.
The following table gives the time interval and kVA demand of the four customer
demands during the peak load period of the year. Assume a power factor of 0.9
lagging.
Time #1 #2 #3 #4 #5 #6 #7 #8
6.00 20 10 30 20 37 39 28 0
6.30 25 5 35 15 39 42 29 6
7.00 30 15 10 10 24 45 30 8
7.30 35 17 15 25 30 35 20 12
8.00 10 20 19 22 22 30 18 9
8.30 5 25 12 12 26 41 16 8
9.00 6 30 16 18 27 42 19 7
10.00 8 34 17 22 28 45 20 9
For each transformer determine the following:
(a) the maximum kVA demand during the period
(b) Load factor
(c) Diversity factor
(d) Select corresponding transformer rating from standard
(e) Energy (kWh) during the 4-hour period
1
Solution
Table1: Transformer#1
Total
Customer Customer Customer Customer Demand
Time #1 #2 #3 #4 (KVA)
6:00 20 10 30 20 80
6:30 25 5 35 15 80
7:00 30 15 10 10 65
7:30 35 17 15 25 92
8:00 10 20 19 22 71
8:30 5 25 12 12 54
9:00 6 30 16 18 70
10:00 8 34 17 22 81
Max.Demand per Customer (KVA) 35 34 35 25 92
a) The maximum kVA demand during the period. From table1 maximum demand for
customer#1, customer#2, customer#3 and customer#4 is 35KVA, 34KVA, 35KVA and
25KVA respectively. Therefore, for transformer #1, the 30min KVA is maximum at time
7:30 and the maximum KVA demand is 92KVA. A load demand characteristic is shown
by the following graph.
100
90
80
Load Demand(KVA)
70
60 Customer#1
50 Customer#2
40 Customer#3
30 Customer#4
20 Total Demand(KVA)
10
0
6:00 6:30 7:00 7:30 8:00 8:30 9:00 10:00
Time Interval
b) Load factor: Load factor is the ratio of the average load demand to the maximum load
demand. Average load demand can be determined as:
∑
2
c) Diversity factor
d) Select corresponding transformer rating from standard
Transformer rating: is sum individual maximum demand of KVA per Diversity factor.
(35+34+35+25) KVA/1.4=92.1429KVA
Therefore, suitable transformer rating could be
e) Energy (kWh) during the 4-hour period
Total energy consumed by Customer#1, #2, #3, #4 is the sum of individual KVA time’s power
factor times the given period.
( )
Table 2: Transformer#2
Total
Customer Customer Customer Customer Demand
Time #5 #6 #7 #8 (KVA)
6:00 37 39 28 0 104
6:30 39 42 29 6 116
7:00 24 45 30 8 107
7:30 30 35 20 12 97
8:00 22 30 18 9 79
8:30 26 41 16 8 91
9:00 27 42 19 7 95
10:00 28 45 20 9 102
Max.Demand per Customer(KVA) 39 45 30 12 116
a) The maximum kVA demand during the period. From table2 maximum demand for
customer#5, customer#6, customer#7 and customer#8 is 39KVA, 45KVA, 30KVA
and 12KVA respectively. Therefore, for transformer #2, the 30min KVA is maximum
at time 6:30 and the maximum KVA demand is 116KVA. A load demand
characteristic is shown by the following graph.
3
140
120
Load Demand(KVA)
100
Customer#5
80
Customer#6
60
Customer#7
40 Customer#8
Total Demand(KVA)
20
0
6:00 6:30 7:00 7:30 8:00 8:30 9:00 10:00
Time Interval
b) Load factor: Load factor is the ratio of the average load demand to the maximum load
demand. Average load demand can be determined as:
∑
c) Diversity factor
d) Select corresponding transformer rating from standard
Transformer rating is the sum of individual maximum KVA per Diversity factor.
(39+45+30+12) KVA/1.086=116KVA
Therefore, suitable transformer rating could be
e) Energy (kWh) during the 4-hour period
Energy consumed is equal to the area under the load curve
Total energy consumed for Customer#5, #6, #7, #8 is sum of individual KVA times power factor
( )
