Power station 12

 Question = The daily demands of three consumer are given below . 

Plot the daily load curve and find (1) maximum demand of individual consumer (2) load factor of individual consumer (3) diversity factor (4) load factor of the station . ? ?? 

Solution =  load curve को जब हम horizontal addition करते हैं तो = 200 + 800 + 2400 + 800 + 400 ,watt पाते हैं । इसमें 2400 Watt maximum demand पर है । 

» (1) maximum demand of individual consumer , or consumer 1 = 800 Watt , answer ✓ 

» consumer 2 = 1000 watt , answer ✓ 

» consumer 3 = 1200 Watt , answer ✓ 

(2) Load factor = average load ÷ maximum demand , or , 

Load factor = ( energy consumed/day ) ÷ ( maximum demand × hours in a day ) , 

» So , load factor of consumer 1 =  ( 600 W× 6 H ) + ( 200 W ×2 H ) + ( 800 W × 6 H ) ÷ ( 800 W × 24 H ) , 

» 8800 WH ÷ 19200 WH = 0.45833 , 

» for % × 100 = 0.45833 × 100 = 45.833 % , answer ✓ 

• Load factor in consumer 2 = ( 200 W × 8 H ) + ( 1000 W × 2 H ) + ( 200 W × 2 H ) ÷ ( 1000 W × 24 H ) 

» 4000 WH ÷ 24000 WH , = 0.167 , 

» for % × 100 = 0.167 × 100 = 16.7 % , answer ✓

• load factor in consumer 3 = ( 200 W × 6 H ) + ( 1200 W × 2 H ) + ( 200 W × 2 H ) ÷ ( 1200 W × 24 H ) , 

» 4000 ÷ 28800 , = 0.138 , 

» for % × 100 = 0.138 × 100 = 13.8 % , answer ✓ 

(3) Diversity factor = sum of individual maximum demand ÷ max demand on station , 

» ( 800 + 1000 + 1200 ) Watt ÷ 2400 Watt , = 3000 ÷ 2400 , = 1.25 , answer ✓ 

(4) load factor = total energy consumed per day ÷ simultaneous max.. demand × 24 , 

» ( 8800 WH+ 4000 WH + 4000 WH) ÷ ( 2400 W× 24 H ) , = 16800 WH ÷ 57600 WH , = 0.291 , 

» for % × 100 = 29.1 % , answer ✓