Generating stations 20

 Question = calculate the continuous power that will be available from hydroelectric plant having an available head of 300 meter ,  catchment area of 150 km² ,  annual rainfall 1.25 meter ,  yield factor 50% . assume penstock , turbine and generator efficiency to be 96% , 86% and 97% respectively . if the load factor is 40% what should be the rating of the generators installed ? 

Solution =»  catchment area = 150 km² , or. 150 × 10³ × 10³ m² , or , 150 × 10⁶ m² , 

» head = 300 Meter , » annual rainfall = 1.25 meter , » Load factor = 40% , or , 0.4  , » yield factor ( K ) = 50% , or , 0.5 , » efficiency ( η ) overall = 96 % × 86 % ×97 % = 0.800832 % , 

» volume of water which can be utilised per annum =  catchment area × annual rainfall ×  yield factor , = 150 × 10⁶ × 1.25 × 0.5 , = 93.75 × 10⁶ m³ , 

» weight of water available per annum is W = 93.75 ×10⁶ × 1000 × 9.81 , = 919.6875 × 10⁹ N , 

» electrical energy available per annum = W × H ×  η overall , = 919.6875 ×10⁹ N × 300 m × 0.800832 , Watt - sec.. 

» for kilo = ÷ 1000 , and , hour = ÷ 3600 , 

» for kWh = ( 919.6875 ×10⁹ N × 300 m × 0.800832 ) ÷ 1000 × 3600 kWh , 

» 61.376265 × 10⁶ kWh , 

» hours in 1 year = 24 hours × 365 days , = 8760 hours , 

» average Power = 61.376265 × 10⁶ kWh ÷ 8760 = 7006.42 kw /hour , 

» max.. demand = average demand / load factor , = 7006.42/0.4 , = 17516 kw , 

» therefore , the maximum capacity of the generator should be 17516 kw , 

Or , 

» let , Cos Ø = 0.8 ( take ) , kva = kw /Cos Ø , = 17516/0.8 , = 21895 kva , or, 22000 KVA , answer ✓

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