The demand for electric power dips over night while people s

The demand for electric power dips over night while people sleep, and power companies will often sell overnight power at substantially lower prices to encourage consumers to use available power generation capacity rather than build multiple expensive power plants used only during peak hours. Accordingly, power companies often purchase power produced during the day by private parties at a good price. Lets consider a power company that sells electric power for $0.06/kWh at night and pays $0.13/kWh for power during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir 50 m above the lake level, pumping water from the lake to the reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump-motor operates as a turbine-generator during reverse flow. Preliminary analysis shows that a water flow rate of 2 m3/s can be used in either direction, and the irreversible head loss of the piping system is 4 m. The combined pump-motor and turbine-generator efficiencies are expected to be 75 percent each. Assuming the system operates for 10 h each in the pump and turbine modes during a typical day, determine the potential revenue this pump-turbine system can generate per year.

Solution

Net head

H = 50 - 4 = 46 m

m = p V = 1000 * 2=2000 Kg /s

Mech power delivered by turbine

= mgh =2000 X 9.81 X 46
  
= 902520 W/s

Actual electricity generated is = Efficiency X Power developed by turbine

= 0.75 X 902520 = 676890 W /s

FOR 10 hours(at day hours) =10 X 676890 X 3600 = 10 X 2.436 X 10⁹ Wh
= 10 X 2.436 X 10⁶ KWh

The same power will be requied for pumping the water from lake to the reservior

(at night hours) = 10 X 2.436 X 10⁹ Wh
= 10 X 2.436 X 10⁶ KWh

Electricity cost in day hours = 0.13 $/KWh

Electricity cost in night hours =0.06 $ KWh

so Potential revenue this system generate for the year

=(Work b the system - work done on system by pump ) X 365 DAYS

= (0.13 - 0.06) X 10 X 2.436 X 10⁶ KWh X 365

= 622398000 $