Generating stations 24

 Question = A hydroelectric plant has a catchment area of 120 km² . the available run of is 50% with annual rainfall of 100 cm . A head of 250 m is available on the average . efficiency of the power plant is 70% find (1)  average power produce (2)  capacity of the power plant . ?  Assume a load factor of 0.6 .

Solution = is given , 

• catchment area = 120 km² , or , 120 × 10³ × 10³  m² , or , 120 × 10⁶ m² , • yield factor or , run off = 50% , or, 0.5 , • annual rainfall = 100 cm , or , ÷ 100 = 1 meter , • head = 250 m , • efficiency ( η) = 70 % or, 0.7 , • load factor = 0.6 , 

( 1)  average Power produced = kWh ÷ 8760 , 

»( yield factor or , run off  , is same )

» volume of water which can be utilised per annum =  catchment area × annual rainfall ×  yield factor , = 120 × 10⁶ m² × 1 m × 0.5 , = 60 × 10⁶ m³ , 

» weight of water available per annum is = W =  60 × 10⁶ × 9.81 × 1000 , = 588.6 ×10⁹ N , 

» electrical energy available per annum = W × H ×  efficency overall ,  Watt - second  = 588.6 × 10⁹ × 250 × 0.7 = 103005 × 10⁹ Watt - sec , 

» in kwh  = ( 103005 ×10⁹ ) ÷ ( 1000 × 3600 ) ,  =2861.25 ×10⁴ kWh , 

» 24 hour × 365 days = 8760 hours , 

» average Power = ( 2861.25 ×10⁴ kWh ) ÷ 8760 H , = 3266 kW , answer ✓ 

(2) max . demand = average demand ÷ Load factor , = 3266 kw ÷ 0.6 , = 5443 kW ,  therefore , the maximum capacity of the generator should be 5443 kW ,

 » If Cos Ø = 0.8 , KVA = kW ÷ Cos Ø , = 5443 ÷ 0.8 =6803 kva , or , 7000 kva , (generator installed ) answer ✓