Power station 14

 Question = A power station has a daily load cycle as under: 260 MW for 6 hours , 200 MW for 8 hours , 160 MW for 4 hours , 100 MW for 6 hours . If the power station is equipped with 4 set of 75 MW each , calculate (1) daily load factor , (2) plant capacity factor , (3) daily requirement if the calorific value of oil used were 10,000 kcal/kg and the average heat reat of station were 2860 kcal/kWh .

Solution = (1) load factor = average load ÷ maximum demand , 

» so, maximum demand = 260 MW , 

» average load = unit generated per day ÷ 24 hours , 

= Unit generated per day = [ (260 × 6 ) + ( 200 × 8 ) + ( 160 × 4 ) + ( 100 × 6 ) ] MWH , 

»4400 MWH , or , 4400 × 10³ kWh , 

» average load = 4400 ÷ 24 ,=  183.333 MW , 

» load factor = 183.333 MW ÷ 260 MW , = 0.70512 , 

» for % = × 100 = 0.70512 × 100 = 70.5 % , answer ✓ 

(2) plant capacity factor = average demand per day ÷ station capacity , 

» average demand per day =Unit generated per day  ÷ 24 hour , = 4400 MWH ÷ 24 H  = 1,83,333 KW ,  

AND 

» Station capacity = 75 × 10³ kW × 4 = 300 × 10³ kW , 

» So ,P.C.F = ( 1,83,333 ) ÷ ( 300 × 10³ ) = 0.6111 , 

» for %  = × 100 , = 0.6111 × 100 = 61.1% , answer ✓ 

(3) for daily recruitment = heat required per day = 1 kWh = 2860 kcal ,  » so, 4400 × 10³ kWh = 2860 kcal × 4400 × 10³ kWh , 

» Fule required per day = 10,000 kcal = 1 kg , 

» 1 kcal = 1÷ 10,000 kg , 

» so , 2860 kcal × 4400 × 10³ kWh = ( 1 × 2860 kcal × 4400 × 10³ kWh ) ÷ ( 10,000 ) , = 1258.4 × 10³ kg , or , 1258.4 tons , answer ✓ 

• 1 ton = 1000 kgs =  ( 10³ ) ,