Generating stations 21

 Question = a hydro-electric plant has a reservoir of area 2 km² and of capacity 5 million cubic meters . The net head of water at the turbine is 50 meters .  If the efficiency of turbine and generator are 85% and 95% respectively , calculate the total energy in kwh that can be generated from this station . If a load of 15000 kilowatt has been supplied for 4 hour , find the fall in reservoir.?

Solution = (1) weight of water available , W = volume of reservoir × weight of 1 m³ of water , = 5 ×10⁶ × 1000 = 5 ×10⁹ × 9.81 N , 

» η overall= 0.85 × 0.95 = 0.8075 , 

» electrical energy that can be generated = W × H × η overall , watt sec... ,=   5 ×10⁹ × 9.81 N × 50 × 0.8075 /1000 ×3600 kWh , = 550109.375 kWh or , 5.5 × 10⁵ kWh , answer ✓ 

» (2) let , x meter be the fall in reservoir level in 4 hours , 

» 2 km²  or , 2 × 10⁶ m² , 

» average discharge/second = ( area of reservoir × x ) ÷ ( 4 ×3600 ) , =( 2 × 10⁶ × x ) ÷ ( 4 × 3600 ) , = 138.888 x m³ , 

» weight of water available per second , W = 138.888 × 1000 ×9.81 N , = 1362.499 × 10³ N , 

» average Power produced = W × H × η overall , watt sec... , = 1362.499 × 10³ ×50 m × 0.8075 , = 55010.8971 x × 10³  Watt sec .. , or , 55010.8971 x kW , 

» but , kW produced = 15000 kw , 

» 55010.8971 x kW = 15000 kW , 

» x = 15000/ 55010.8971 , = 0.2726 m , or , × 100 cm , = 27.26 cm , 

» therefore , the level of reservoir will fall by 27.26 cm , answer ✓