In downtown DC after a torrential rain storm you hear that the storm drain cover

Question

In downtown DC after a torrential rain storm you hear that the storm drain covers are blowing off and water is spouting out of them. If the water flow in the storm drain pipes, with a cross-sectional area of 9.42 m^2. are moving with a speed of 4m/s. What will the speed of the water be exiting the drain cover? How high h will the water spout if the storm drain covers have an area of 3.14 m^2? Assume that the storm drain pipes and the covers are at 0m height. Assume the applied force is gravity only. Based on this explain how the water spouts so high in the movies when a fire hydrant is broken off.

Explanation / Answer

From continuity equation A1V1 = A2V2 ====> V2 = A1V1/A2 = 9.42*4/3.14 = 12 m/s

FROM Bernouli's equation

P1+ 1/2 rho V1^2 = P2 + 1/2 rho V2^2

P1 -P2 = rhog(h1-h2)

rho gh2 = 1/2 rho(V2^2 -V1^2)

h2 = 1/2 rho (V2^2 -V1^2)/ rho*g

       = 1/2 (V2^2 -V1^2)/ g

     = 1/2 (144 -16)/9.8

    h2 = 6.530 m

          

             

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