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Water flows steadily from an open tank through a pipe in the lower portion of th

ID: 1627575 • Letter: W

Question

Water flows steadily from an open tank through a pipe in the lower portion of the side of the tank. The top of the fluid level in the cylindrical tank is point 1 and is 10.0 m above ground level. The pipe extends out horizontally a short distance from the side of the tank and then turns straight down and turns horizontally outward again at 2.00 m above ground level. In that last horizontal section there is at first a pipe with a cross sectional area of 4.80 X 10^-2 m^2 at point 2. The horizontal pipe necks down to a cross sectional area of 1.60 X 10^-2 m^2 at the discharge point 3. The cross sectional area of the tank is very large compared with the cross sectional areas of the pipe. Assuming that Bernoulli's equation applies, the discharge rate, delta V / delta t, that flows out of the pipe is computed to be 0.2 m^3 / s. What is the gauge pressure, pgauge, at point 2?

Explanation / Answer

a2 = area at point 2= 4.80*10-2 m2

a3 = area at point 3 = 1.60*10-2m2

discharge rate ( water flows out of the pipe) = del v / del t = 0.2m3 /s

v2 = velocity at point 2

v3 = velocity at point 3

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from continuity equation:

volume of liquid entering per second = volume of liquid leaving per second

a2*v2 = a3*v3 --------------------------1

volume of liquid leaving per second,a3*v3 = 0.2

v3 = 0.2/ 1.60*102 = 12.5 m/s

from eq 1: a2*v2 = 0.2

v2 = 0.2/ 4.80*10-2 = 4.17 m/s

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applying Bernoulli's equation at point 2&3

P2 + pgh2 +p*(1/2)*v22 = P3 + pgh3 +p*(1/2)*v32

P2 = pressure at point 2 = ?

P3 = pressure at point 3 = atmospheric pressure = 1.013*105Pa

P2 = P3 + pg(h3-h2) + p*(1/2)*( v32 - v22)

h3-h2 = 0 (pt 2 and 3 are at same height)

p= density of water = 1000 kg/m3

P2 = 1.013*105 + 0 + 1000*(1/2)*( 12.52 - 4.172)

P2 = 170.73*103 Pa

gauge pressure = 170.73*103 - 1.013*105 = 6.94*104 Pa