Problem 3 A water tank with a cross-sectional area A1 = 0.75 m2 open to the atmo

Question

Problem 3 A water tank with a cross-sectional area A1 = 0.75 m2 open to the atmosphere has an outflow pipe at the bottom with a cross-sectional area A2 0.25 m2 as seen in the figure. Water (mass density: ,-1000 kg/m") fills the tank up to h 9.5 m above the outflow pipe. Al = 0.75m2 h = 9.5 m A2 0.25 m2 outflow p r, tan [10 pts] a) Initially the outflow pipe is closed with a stopper and no water flows out. What is the total pressure and what is the total force exerted by the water on the b) The stopper is then removed. Just after the outflow pipe is opened, what is the c) Just after the outflow pipe is opened, what is the velocity at which the water stopper? [10 pts] velocity of the water that flows out of the open pipe at the bottom? [5 pts] level at the top of the tank falls?

Explanation / Answer

(A) P = Patm + rho g h

P = (101325) + (1000 x 9.8 x 9.5)

P = 194425 Pa ........Ans

F = P A = (194425) (0.25)

F = 4.86 x 10^4 N .........Ans

(b) Applying bernoulli's equation,

P1 + rho g h1 + rho v^2 /2 = P2 + rho g h 2 + rho v^2 / 2

and A1 v1 = A2 v2  

v1 = (0.25/0.75)v2 = v2 / 3

Patm + (1000 x 9.8 x 9.5) + (1000)(v2/3)^2 /2 = Patm + 0 + 1000 v2^2 / 2

v2 = 14.5 m/s

(c) v1 = v2 / 3 = 4.82 m/s

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