Generating stations 18

 Question = the weekly discharge of a typical hydroelectric plant is as under , 

• days/discharge ( m³/sec ) = sun/500 , mon/520 , Tues/850 , wed/800 , thur/875 , fri/900 , sat/546 , 

The plant has an effective head of 15 m and overall efficiency of 85% . if the plant operates on 40% load factor ,  estimate (1)  the average daily discharge , (2) pondage required and (3)  installed capacity of proposed plant . 

Solution = is given , 

» Head (H) = 15 m ,  η overall = 85 % , load factor = 40 % , 

» show the plot of weekly discharge is taken along y- axis and day along x - axis .

» (1) average daily discharge = ( 500 + 520 + 850 + 800 + 875 + 900 + 546 ) ÷ 7 days , = 4991 ÷ 7 = 713 m³/sec.. , answer ✓ 

(2) it is clear from graph that on three days (viz ,  Sunday , Monday and Saturday ) the discharge is less than the average discharge . 

»  volume of water required on these 3 day = ( 500 + 520 + 546 ) × 24 × 3600 m³ , = 1566 × 24 × 3600 m³/sec . , 

» ( 1 hour = 3600 sec.. , 24 hours = 24 × 3600 sec.. ) 

» volume of water required on these 3 days = 3  × 713 × 24 × 3600 m³ , = 2139 × 24 × 3600 m³ , 

» pondage required = (2139 - 1566 ) × 24 × 3600 m³ , = 495 × 10⁵ m³ , answer ✓ 

(3) weight of water available/sec. = W = 713 × 1000 ×9.81 N , 

» average Power produced = W × H × η overall ,  = ( 713 × 1000 ×9.81 N ) × (15) × (0.85 ) watt , 

» ( mass of 1 m³ of water is 1000 kgs ) 

» 89180 × 10³ Watt or , 89180 kW , 

» installed capacity of the plant = output power ÷ load factor , 

» 89180 kW ÷ 0.4 , = 223 × 10³ kw , or , 223 MW , answer ✓