Question

"Floating ball water fountain" was developed in Finland consists of massive 9072 kg, 1.8 diameter stone sphere is supported by pressure force on the curved surface within a pedestal as shown in Figur Due to limited gap, the flow is essentially that a laminar flow between parallel plates. When the pump is turned on and the water pressure within the pedestal reaches 55 kPa, the sphere rises off the pedestal, creating a 1.3 X 10-2 gap through 1.3 x 10 cm which the water flows 1.8 m 10 cm 1.2 m p - 55 kPa (a) Estimate the pump flowrate, Q. (in m3/s) required to accomplish this (b) Show the effect if the pump flowrate were reduced half of Q. Pump

Answer 1

(a) Calculate the flow rate of water per unit width. 2h Ap 9 3 ul (1.3x10+ (55x10) 2 3(1.002)(0.1) =1.005x10-mº/s per unit width Here, viscosity of water is taken at 20°C. Calculate the pump flow rate. Q=9(1.27) =(1.005 x 10-7)(1.27) = 3.79x10-7 m/s Therefore, the pump flow rate is 3.79x10-7 m/s. (6) If the flow rate is reduced to half of the pump flow rate, then the gap between pedestal and ball should be increased in order to maintain the 55 kPa pressure.

Calculate the new gap between pedestal and ball. 3 9 h - h 0.59 h 9 1.3x10+ 2. h = 5.16x10m Therefore, the gap between pedestal and ball should be increased to 2(5.16x10-9 m)=1.032x10-4m.