Generating stations 19

 Question = a hydroelectric station has an average available head of 100 m and reservoir capacity of 50 million cubic metres .  calculate the total energy in kwh that can be generated , assuming hydraulic efficiency  of 85% and electric efficiency of 90% .

Solution = weight of water available is W = volume of water × density , = 50 × 10⁶  ×1000 kg , 

» overall efficiency = 0.85 × 0.9 = 0.765 , 

» work = ( 500 × 10⁶ ) × ( 1000 ) kg × ( 9.81 N ) , 

» So , electrical energy available = W × H ×  η overall , watt - sec.. , 

» 50 × 10⁶ × 1000 × 9.81 × 100 × 0.765 / 3600 ×1000 , kWh , 

» 37523250 × 10⁶/3600 × 1000 , kWh , 

» 10.423 × 10⁶ kWh , answer ✓ 

(1 kWh = 1000 Watt × 3600 seconds , = 36 ×10⁵ joules , ) 

( For , hour = ÷ 3600 , and , kilo = ÷ 1000 )