Page - 55 - in Clean Water Using Solar and Wind - Outside the Power Grid

Water supply 55 The required hydraulic power is: P Whydr = ⋅ ⋅ ⋅ = 1 3600 10 1000 9 81 27. It is necessary to take some head loss in the piping system into consideration. We can assume this to be 10%. The required electric power, given an 80% motor/pump efficiency and a 10% head loss, then becomes P Wel = ⋅ = ≈ 27 0 8 0 9 37 5 38 . . . The power is proportional to the head (depth) so a 20 m lift would need twice as much hydraulic power (54 W) and 76 W of electric power. Doubling the flow rate also means doubling the required power. Thus, to lift 10 m3 per hour from a depth of 100 m will require a hydraulic power of P W kWhydr = ⋅ ⋅ ⋅ = ≈ 10 3600 100 1000 9 81 2725 2 7. . and around 2.7/(0.9 ⋅ 0.8) ≈ 3.8 kW of electric power. To get the energy (in joules) the Q is replaced by the total volume V (m3) in the calculations. Example 4.2: Energy to Pump Groundwater Let us calculate the energy to lift 1 m3 of water from a depth of 10 m. This will require: Ehydr = 1 ⋅ 10 ⋅ 1000 ⋅ 9.81 = 9.8 ⋅ 104 J where J is the energy measured in joules. With 10% pipe loss and 80% motor/pump efficiency taken into consideration the required electric energy input is around 13.6 ⋅ 104 J. Translating J (=Ws or watt seconds) to kWh as a more commonly used unit for electric energy: 1 kWh = 1000 (W) ⋅ 3600 (seconds) = 3.6 ⋅ 106 Ws = 3.6 ⋅ 106 J = 3.6 MJ so, the electric energy of 13.6 ⋅ 104 J corresponds to 3.8 ⋅ 10−2 kWh. Another way to calculate the energy is to simply multiply the power (38 W) by time: E Ws kWhel = ⋅ = =38 3600 38 1000 3600 3600 0 038. Downloaded from https://iwaponline.com/ebooks/book-pdf/520710/wio9781780409443.pdf by IWA Publishing user