Generating stations 17

 Question = A run of river hydroelectric plant with pondage has the following data ;  installed capacity =  10 MW , water head , H =  20 meter , efficiency overhaul = 80% , load factor = 40% , 

(1) determine the river discharge in m³/sec.  required for the plant , (2)  if on a particular day ,  the river flow is 20 m³/sec. , What load factor can the plant supply ?

Solution = consider the duration to be of one week unit generated / week = max demand × L.F. × hours in a week , 

» (10 × 10³) × (0.4) × (24×7) kWh = 67.2 ×10⁴ kWh. (1) 

» let Q m³/sec. be the rever discharge required = 

» weight of water available/sec. , W = Q × 9.81 × 1000 , = 9810 Q Newtown (N) , 

» average Power produced = W × H ×  η overall ,  = 9810 Q × 20 × 0.8 W , = 156960 Q watt , or , 156.96 Q Kw 

» ( 24 hours × 7 days = 168 hours )

»  unit generated/week = 156.86 Q × 168 kWh , = 26369 Q kWh (2) 

» equations (1) and (2) , We get , 

» 26,369 Q = 67.2 × 10⁴ , = Q = 67.2 × 10⁴ /26369 , = 25.48 m³/sec , answer ✓

(2) formula , 

» load factor = unit generated on that day kilowatt ÷ 10⁴ × hour in on day , 

» unit generated on that day = power developed × 24 , 

» power developed = average Power produced × river discharge , = 156.96 ×20 , = 3139.2 kw , 

» unit generated on that day = 3139.2 kw × 24 = 75341 kWh , 

»( 10 MW = 10000 kw , or , 10⁴ kw )

» So , load factor =( 75341/10⁴ × 24 ) × 100 , = 31.4 % , answer ✓