The main water line into a tall building has a pressure of600kPaat 5mbelow ground level. Apump brings the pressure up so the water can be delivered at 200kPaat the top floor 150maboveground level. The volumetric flow rate is 0.01m3/s. Consider the liquid water as an incompressiblesubstance. Assume steady state operation, negligible changes in kinetic energy, and the water temperature remains constant. Determine the required pumpingpower [kW, hp].

Required pump power = - 11.205 kW

Explanation:

Pi = 600 kPa

Zi ( initial height ) = - 5m

P2 = 200 kPa

Ze ( top floor above ground level ) = 150m

Flowrate = 0.01 m^3/s = 10 kg/s

Temperature = 10°C  ( assumed value )  remains constant

calculate the value of pumping power required ( i.e. work done )

Applying the energy equation

Hi + 1/2(vi)^2 + gZi = He + 1/2(Ve)^2 + gZe + W

given negligible internal/kinetic  energy  difference in "h's = the pv terms

W = ( Pi Vi - Pe V2 ) + g( Zi - Z2 )

   = ( (600 * 0.001)  - (200 * 0.001) ) + (9.81/1000) ( - 5 - 150 )

   = - 1.1205 Kg/kJ

required pump power = flowrate * -1.1205

                                     = 10 * ( - 1.1205 ) = - 11.205 kW

b. check your rearview mirror

explanation: