The pipe flow in a re-circulating fountain corresponds to the figure. TV pipes a

Question

The pipe flow in a re-circulating fountain corresponds to the figure. TV pipes arc all circular, with the diameter of the wide pipes being D S cm, the diameter of the narrow pipe being d 2 cm. and in between there is a 60 degree contraction. The fluid is water (rho = 1000 kg/m3, v - 1.00 times 10-6m2/s). In the Urge diameter pipes, kf = 0.002D and the small diameter pipes arc smooth. The equivalent length of the square bend is L, - 0.8 m. The outlet velocity is U =12 m/s, the outlet pressure is atmospheric, alpha = 1.0 in all pipes, and g - 9.81 m/s2. Losses at the pump inlet arc incorporated into the pump efficiency. Including minor losses for the bend, contraction, and sudden outlet (expansion with D2>>D1). determine the total head loss through the pipe system. Compute the pump head required for this system. If the pump efficiency is 85%. what is the shaft work for this pump? If the conditions given correspond to the peak efficiency of the pump, and the specific speed of the pump is N1 - 1.00. determine ohm for the pump in terms of rpm (60 rpm - 2pi rad/s).

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