PPlease show steps for full credit! Water flows steadily through the galvanized

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PPlease show steps for full credit!

Water flows steadily through the galvanized iron pipe system at a rate of 0. 40 ft3/s. The pipe diameter is 2. 5 inches and the length of pipe between sections A and B is 75 ft. The loss coefficient of each filter is 7. 0 and all other minor losses are negligible. The height difference between points A and B is 11 ft. If the pump is 60% efficient at that flow rate, how much power must be supplied to the pump (lb middot ft/s) from a motor (not shown in figure) so that the pressure after the five filters (PB) is 20 psi (gage)?

Explanation / Answer

V = Q / A

V = 0.4 / ((3.14 / 4) *(2.5 / 12)^2)

V = 11.734 ft/s

Kinematic viscosity of water, neu = 1*10^-5 ft^2 /s

Reynolds number Re = V*d / neu

Re = 11.734*(2.5 / 12) / (1*10^-5)

Re = 2.44*10^5

For galvanized iron, roughness e = 0.0005 ft

Relative roughness e/d = 0.0005 / (2.5 / 12) = 0.0024

From Moody diagram, for Re = 2.44*10^5 and e/d = 0.0024, we get friction factor f = 0.024

Head loss = fL/d *V^2 / (2g) + Loss in 5 filters

= fL/d *V^2 / (2g) + 5*K*V^2 / (2g)

= [0.024*75 / (2.5 / 12) + 5*7] * 11.734^2 / (2*32.2)

= 93.3 ft

rho*Head loss = (Pa - Pb) + Pump pressure rise

62.4*93.3 = (62.4*11 - 20*12^2) + Pump pressure rise

Pump pressure rise = 8015.5 lb/ft^2

Pump power = 8015.5*flowrate = 8015.5*0.4 = 3206.2 lb-ft/s

Actual pump power = 3206.2 / 0.6 = 5343.6 lb-ft/s

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