-
What is the utilization factor of a power station which supplies the following loads ?
Load A: Motor load of 200 kW between 10 AM to 7 PM
Load B: Lighting load of 100 k W bet ween 7 PM to 11 PM
Load C: Pumping load of 110 kW between 3 PM to 10 AM
-
- 1.60
- 1.32
- 1.00
- 2.56
Correct Option: B
Sum of individual maximum demands
= (200 + 100 + 110) kW = 410 kW.
| Time | Load |
|---|---|
| 10 AM to 3 PM | 200 kW |
| 3 PM to 7 PM | (200 + 110) kW = 310 kW |
| 7 PM to 11 PM | (100 + 110) kW = 210 kW |
| 11 PM to 10 AM | 110 kW |
Maximum demand of the whole system is 310 kW. Utiligation factor
| = | = | = 1.32 | ||
| Maximum demand of whole system | 132 |
