Electrical Power Systems

Load factor,
L =
Maximum load = 3 MW
\ L = 1575 .
3
Average load
Maximum load
Structure of Power Systems and ew Other Aspects 9
= 0.525
(b) Maximum demand = 3 MW. Therefore, 4 generating units of rating 1.0 MW each may be
selected. During the period of maximum demand 3 units will operate and 1 unit will remain as
stand by.
(c) Plant capacity = 4 × 1.0 = 4.0 MW
Reserve capacity = 4 – 3 = 1 MW
rom eqn. (1.3),
Plant actor =
Actual energy produced = 37.80 MWhr
Maximum plant rating = 4 MW
Time duration T = 24 hours
\ Plant actor =
Actual energy produced
Maximum plant rating ´ T
37. 80
4´ 24
= 0.39375.
(d) Operating schedule will be as follows:
One generating unit of 1 MW:— 24 hours
Second generating unit of 1 MW:— 6 AM – 9 PM (15 hours)
Third generating unit of 1 MW:— 9 AM – 12 Noon
2 PM – 6 PM
(7 hours)
Example 1.2: A generating station has a maximum demand of 80 MW and a connected load of
150 MW. If MWhr generated in a year are 400 × 103 , calculate (a) load factor (b) demand factor.
Solution:
Maximum demand = 80 MW
Connected load = 150 MW
Units generated in one year = 400 ×10 3 MWhr
Total number of hours in a year T = 8760
\ Average load =
Load factor, L =
\ L =
400 ´ 10
8760
3
Average load
Maximum load
45. 662
80
= 45.662 MW
= 0.57