Ahorizontal 2-m-diameter conduit is half filled with a liquid (sg=1.6 ) and is capped at both ends with plane vertical surfaces. the air pressure in the conduit above the liquid surface is 200kpa. determine the resultant force of the fluid acting on one of the ends caps, and locate this force relative to the bottom of the conduit.

Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit

Explanation:

The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.

The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.

Therefore,

Ft = F1 + F2

According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.

To get F1,

F1 = p x A

= p x (πr²)

Substituting values,

F1 = 200 x π x 1²

F1 = 628.32 kN

This resultant force acts at the center of the plate.

To get F2,

F2 = Π x hc x A

F2 = Π x (4r/3π) x (πr²/2)

Π - weight density of oil,

A - area on which oil pressure is acting,

hc - the distance between the axis of the conduit and the centroid of the semicircular area

Π = Specific gravity x 9.81 x 1000

Therefore

F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)

F2 = 10.464 kN

Ft = F1 + F2

Ft = 628.32 + 10.464

Ft = 638.784 kN

The resultant force on the surface is 639 kN

Taking moments of the forces F1 and F2 about the centre,

Mo = Ft x y

Ft x y = (F1 x r) + F2(1 - 4r/3π)

Making y the subject,

y = (628.32 + 10.464(1 - 4/3π)/ 638.784

y = 0.993m

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