Power station 22
Question = A generating station has the following daily load cycle.
Solution = Daily curve is drawn by taking the load y-axis and time along x-axis. For the given load cycle , the load curve is show in figure .(1) It is clear the load curve that maximum demand on the power station is 35 MW and occurs during the period 16 to 20 hours in maximum demand = 35 MW , answer ✓
(2) unit generated per day = [ ( 20 × 6 ) + ( 25 × 4 ) + ( 30 × 2 ) + ( 25 × 4 ) + ( 35 × 4 ) + ( 20 × 4 ) , MWH ,
» after calculate we found = 600 MWH or , 600 ×10³ kWh , answer ✓
(3) average load = unit generated per day ÷ 24 hours , = ( 600 ×10³ kWh ) ÷ ( 24 hours ) , = 25 × 10³ kW , answer ✓
(4) Load factor = average load ÷ maximum demand , = ( 25 × 10³ kW ) ÷ ( 35 × 10³ kW ) , = 0.714 or,
» for % ( × 100 ) = 0.714 × 100 = 71.4 % . answer ✓
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