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 6.5 m/s. The density of water is 1020 kg/m3.

(a) Calculate the volume rate of flow of water. Answer in m3/s

(b) The fountain is fed by a pipe that at one point has a radius of 0.035 m and is 2.7 m below the fountains opening. Calculate the absolute pressure in the pipe at this point. Answer in Pa

(c) The fountain owner wants to launch the water 3.3 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. Answer in m

Explanation / Answer

a) Volume flow rate = A v

= (pi x 0.018^2) (6.5 ) = 6.616 x 10^-3 m^3 /s

b) volume flow rate will be same.

6.616 x 10^-3 = p[i x 0.035^2 x v

v = 1.72 m/s

Applying bernoulli;s equation,

P + rho * g* h + rho*v^2 /2 = constant

Patm + (1020 x 9.8 x 2.7) + (1020 x 6.5^2 /2) = P + 0 + (1020 x 1.72^2 /2 )

P = 101325 + 26989.2 + 21547.5 - 1508.78

P = 148352.92 Pa

c) Applying energy conservation,

mgh = mv^2 / 2

2 x 9.8 x 3.3 = v^2

v = 8.04 m/s

volume flow rate = A v

6.616 x 10^-3 = (pi r^2 ) ( 8.04 )

r = 0.0162 m

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