Generating stations 14

 Question = calculate the average Power in kilowatt that can be generated in a hydroelectric project from the following data catchment area =  5 × 10⁹ m² , mean head = H = 30 m , annual rain fall , F = 1.25 m , yield factor. K = 80% , overall efficiency = 70% , if the load factor is 40% so what is the rating of generator installed ? 

Solution = average Power = electrical energy available per annum = W × H ×  η overall , watt sec . 

» w , or weight of water available per annum is = volume of water which can be utilise per annum × 9.81 × 1000 , 

» volume of water which can be utilised per annum = catchment area × annual rainfall × yield factor , = 5 ×10⁹ m² × 1.25 m × 0.8 = 5× 10⁹ m³ , 

» So , W = 5 × 10⁹ m³ × 9.81 ×1000 = 49.05 ×10¹² N , 

» electrical energy available per annum = 49.05 × 10¹² N × 30 m ×0.7 watt sec.. , 

(For kWh = ÷ 1000 and , for hour = ÷3600 , )

» ( 49.05 × 10¹² N × 30 m ×0.7 watt sec.. ) ÷ ( 1000 × 3600 ) , kWh , 

» 2.86 ×10⁸ kWh , and , average Power = 2.86 × 10⁸/8760 ,  = 32648 kw/hour, answer ✓ 

इसके अलावा  , for generator value, 

» maximum demand = average demand ÷ load factor , = 32648 ÷ 0.4 , = 81620 kw , therefore , the maximum capacity of the generator should be 81620 kw. ,

» kva = kw/Cos Ø , = 81620/0.8 = 102025 kva , (generator value )  , answer ✓