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Question

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A pump is designed to pump water at a flow rate of 3 gallons per second through a 1 inch diameter hose at zero elevation. The water flows into a very long hose (of constant cross-section) that can withstand large pressures. You walk the end of the hose uphill and notice that the flow rate decreases. You continue walking uphill until the flow rate drops to zero. To what elevation have you walked, and what is the pressure at the pump? (Hint: the pump is supplying the same energy per volume to the water, but now that energy is going into maintaining a pressure difference at the pump, rather than to accelerating the water.) You use this pump to pump water up 1000 ft. What is the flow rate through the end of the hose at this elevation?

Explanation / Answer

Solution:

Volume flow rate of water = 3 gallons/s =0.0114 m^3/s

(1 gallon = 0.00379 m^3/s)

Radius of the hose pipe at =1/2 in= 0.0127 m

Volume Flow rate = area x velocity = Area x velocity v =0.0114

A=area= pi r^2 =pi (0.0127)^2

=> velocity v = 0.0114/A = 22.4 m/s

Since the flow rate at the elevation drops to 0,

pressure differnce between initial and final points = 1/2 x density of water x velocity ^2

=> pressure = 1/2 x 1000 x (22.4)^2

                  = 251753 Pascals = 2.49 atm

Pressure= density x gravity x height

using this , we find the height as h = 251753 /(9.8)(1000) = 25.7 meters

(Flow rate becomes zero for h>25.7 m)

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