Power station 23

 Question = A power station has to meet the following load demand: 


(1) Diversity factor (2) unit generated per day (3) Load factor . 

Solution = the given load cycle can be tabulated as under : 


From this  table it's clear that total load on power station is 20 kW for 0 to 10 hours , 50 kW for 10 to 4 pm hours , 70 kW for 4 pm to 6 pm hours , 50 kW for 6 pm to 10 pm hours and 20 kW for 10 pm to 12 pm , hours . 


Plotting the load on power station versus time , it is clear from the curve that maximum demand on the station is 70 kW and occurs from 4 pm to 6 pm . 

(1) Diversity factor = sum of individual maximum demand of groups ÷ M.D , 

» where , sum of individual maximum demand of groups = 50 + 30 + 20 = 100 kW ,

» maximum demand show in figure = 70 kW , 

» D.F. = 100 ÷ 70 = 1.42 answer ✓ 

(2) unit generated per day = kW × hours ,= [ ( 20 × 10 ) + ( 50 × 6 ) + ( 70 × 2 ) + ( 50 × 4 ) + ( 20 × 2 ) ,= 880 kWh , answer ✓ 

(3) » Load factor = average load ÷ maximum demand , 

» Average load = unit generated per day ÷ 24 h = 880 kWh ÷ 24 h = 36.666 kW , 

» So , L.F.= 36.666 kW ÷ 70 kW = 0.5238 , 

» for % ( × 100 ) = 0.5238 × 100 = 52.38 % , answer ✓ 

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