A fountain with an opening of radius 0.018 m shoots a stream of water vertically

Question

A fountain with an opening of radius 0.018 m shoots a stream of water vertically from ground level at 6.1 m/s. The density of water is 1000 kg/m3.

(a) Calculate the volume rate of flow of water.

(b) The fountain is fed by a pipe that at one point has a radius of 0.031 m and is 2.2 m below the fountains opening. Calculate the absolute pressure in the pipe at this point.

(c) The fountain owner wants to launch the water 4.6 m into the air with the same volume flow rate. A nozzle can be attached to change the size of the opening. Calculate the radius needed on this new nozzle.

Explanation / Answer

a) Volume Flow Rate =

F = v*A

= (6.1)*(Pi*(0.018)^2)

= 0.006209 m^3/sec

b) (v_pipe)*(A_pipe) = (v_pipe)*(Pi*(0.031)^2) = F = 0.006209

(v_pipe) = 2.00 m/sec

(P_pipe) =
= (101325) + (1/2)**v^2 - (1/2)**(v_pipe)^2 + *g*h - *g*(0)

= (101325) + (1/2)*(1000)*(6.1)^2 - (1/2)*(1000)*(2.00)^2 + (1000)*(9.8)*(2.2)

= 96370 Pa

c) ) (1/2)**v^2 = *g*h

v = Sqrt[ 2*g*h ]

v = Sqrt[ 2*(9.8)*(4.6) ]

v = 9.495261976m/sec

Volume Flow Rate = F = 0.006209 =

= v*A

= (9.495261976)*(Pi*(R )^2)

R = 0.96647meters

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