Generating stations 15

 Question =A hydroe-electrical power station has a reservoir of area 2.4 km² and capacity 5 × 10⁶ m³ . The effective head of water is 100 meter .  the pen stocks , turbine and generation efficiency are respectively 95% ,  90% and 85% . 

(1) calculate the total electrical energy that can be generated from the power station .

 (2) if a load of 15000 kilowatt has been supplied for 3 hour , find the fall in reservoir level . 

Solution = is given , 

• Reservoir area = 2.4 km² ,or , 2.4 × 10⁶ m²  • water head , H = 100 meter , •  volume of reservoir = 5 × 10⁶ m³ ,  • efficiency ( η ) penstock = 95% ,  •  η turbine = 90% , •  η generation = 85% , 

»( 1) electrical energy that can be generated =  W× H × η  over all , 

» w , or weight of water available = volume of reservoir × weight of 1m³ water = 5 ×10⁶ m³ × 1000 kg , = 5 × 10⁶ m³ × 1000 kg , or 5 × 10⁹ kg , or , 5 × 10⁹ × 9.81 N , 

» overall η = 0.95 × 0.9 × 0.85 , = 0.726 , 

» energy generated = (  5 × 10⁹ × 9.81 N × 100 m × 0.726 ) ÷ ( 1000 × 3600 ) , kWh , = 9,89,175 kWh , answer ✓ 

» (2) let , 3 hours in fall in reservoir level x meter , 

So , 1 hours in fall in reservoir level x/3 , 

And , average discharge/sec. =( Area of reservoir × x ) ÷ ( 3 × 3600 ) , 

» 2.4 km² or , 2.4 × 10⁶ m² 

» ( 2.4 × 10⁶ ×X ) ÷ ( 3 × 3600 ) , = 222.2 x m³ , 

 » weight of water available/sec , w = 222.2 x × 1000 × 9.81 = 21.8 x × 10⁵ N , 

» average Power produced = W ×H × η overall ,  = 21.8x × 10⁵ × 100 × 0.726 watt , = 15.84x ×10⁷ watts , or , 15.84x ×10⁴ kw , 

» but , kw produced = 15000 kw , ( given )

» So , 15.84x ×10⁴ kw = 15000 kw , 

» x = (15000 ) ÷ ( 15.84 × 10⁴ ) , = 0.0947 m or , × 100 cm = 9.47 cm ,  therefore, the level of reservoir will fall by 9.47 cm. Answer ✓ 

(2) 2nd method , 

» level of reservoir = volume of reservoir ÷ area of reservoir , = 5 × 10⁶m³ ÷ 2.4 ×10⁶ m³, = 2.083 m , 

» kWh generated in 3 hours = 15000 × 3 = 45000 kWh , 

» if kWh generated are 9,89,175 kWh , fall in reservoir level = 2.083 m 

» kWh generated are 1 kWh , fall in reservoir level =  2.083 m ÷ 9,89,175 kWh , 

» kWh generated are 45000 kWh , fall in reservoir level = ( 2.083 m × 45000 kWh ) ÷ ( 9,89,175 kwh ) , = 0.0947 m , or , × 100 cm = 9.47 cm. Answer ✓