Power station 11

 Question = A power station has to meet the following demand : • group A - 200 kW between 8 a.m. and 6 p.m.  • group B - 100 kW between 6 a.m. and 10 a.m. • group C - 50 kW between 6 a.m. and 10 a.m. • group D - 100 kW between 10 a.m. and 6 pm and then between 6 pm and 6 am . 

Plot the daily load curve and determine (1) diversity factor (2) unit generated/day (3) load factor ? 

Solution =

It is clear that Total load on power station is 100 kW for 0-6 hours , 150 kW for 6 - 8 hours , 350 kw for 8 - 10 hours , 300 kW for 10 - 18 hours and 100 kW for 18 - 24 hours . Plotting the load on power station versus time , we get the daiy load curve as shown in fig..  it is clear from the curve that maximum demand on the station is 350 kw and occurs from 8 am to 10 am .

(1) diversity factor = sum of individual maximum demand of group ÷ max .. demand on station , = ( 200 +100 + 50 + 100 ) kw ÷ 350 kW , = 450 ÷ 350 = 1.286 answer ✓ 

(2) according to load curve , units generated/day = kw × Hour , = ( 100 ×6 ) + (150 × 2 ) + ( 350 × 2 ) + ( 300 × 8 ) + ( 100 × 6 ) , = 4600 kWh, answer ✓

(3) load factor = average loade ÷ max.. demand. , = And , average load = unit generated/day ÷ 24 hours , = 4600 kWh ÷ 24 hours , = 191.7 kw ,

» So , load factor = 191.7 kw ÷ 350 kW. = 0.548 , = × 100 for % = 0.548 × 100 = 54.8 % , answer ✓ 

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