Question

Problem #1: A 4.5 kW pump operating at steady state draws in liquid water at 1 bar, 15 °C and delivers the water at 5 bar to a location whose elevation is 6 m above that of the inlet. There is no significant change in velocity between the inlet and exit, and the local acceleration of gravity is 9.8 m/s? Would it be possible to pump 5.5 m in 10 min or less? Explain. (25 pts)

Answer 1

Given Dataran Power of pump (P) = 4.5 kW = 4500 W 2, + P = 1 bar = 105 Pa. - P₂ = 5 bar = 5x105 Pa. R2-2. = 6m. V = V₂=v. Sol:- Hydraulic head (H) is, Here Vi = V₂ =V. . .81 H = 46.7747 m

.: Power of = pump (P) = 89 QH a 4500 = 10°x9,81% Qx 46.7747 Q = 4500 19103 x 9.81X 46. 7747 - Q = 9.8069 X 10-3 m3/s. in 10 min. - Volume of water pumped is (v) = Qxd = (9.8069% 10%)x (10 x 60) , V = 5.8841 m3 g here by given pump 5.8841 ms of water is pumped in 10 min cohich is higher than 5.5m3 ₃. 5.5m3 of water can be pumped in less than to min.
it is possible to pump 5.5 m​​​​​​3 of water in less than 10 min.