Water frictionless and incompressible flows steadily from a

Water (frictionless and incompressible) flows steadily from a large tank and exits through a vertical, constant diameter pipe as shown below. The air in the tank is pressurized to 50 kPa. Determine height, h, to which the water rises, the water velocity in the pipe, and the pressure in the horizontal part of the pipe.

Solution

1) The height to which the water rises depends upon the increase in pressure due to air and is given by,

p2 - p1 = rho x g (h - h1) where h1 = 4m while no air pressure.

50 x 10³ = 1000 x 9.81 ( h - 4)

50000 / 9810 + 4 = h

h = 10.19m

b) p1 - p2 = 0.5 x rho x (V₂² - V₁² )

V1 is zero and pressure diff is 50kPa

V2 = sqrt ( 50 x 10³ / 0.5 x 1000)

V2 = 5 m/s

c) using eqn in b) solve for p2

p2 = p1 - 0.5 x V2² where p1 = 50 x 10³ + 1000 x 9.81 x 2 = 69 kPa

So p2 = 69 - (0.5 x 25) = 69 - 12.5 = 56.5 kPa