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Chapter 2: Problem 107

A horizontal 2 -m-diameter conduit is half filled with a liquid \((S G=1.6)\) and is capped at both ends with plane vertical surfaces. The air pressure in the conduit above the liquid surface is \(200 \mathrm{kPa} .\) Determine the resultant force of the fluid acting on one of the end eaps, and locate this force relative to the bottom of the conduit.

Short Answer

Expert verified
The resultant force of the fluid acting on one of the end caps is 677.221kN and is located 0.425m above the bottom of the conduit.

Step by step solution

01

Determination of Pressure at the End Cap

First, the pressure at the surface of the fluid is equal to the air pressure, which is given as \(200kPa\). The pressure at the bottom of the conduit (at the center of the end cap) is the sum of air pressure and fluid pressure, which can be calculated using the formula: fluid pressure = density \(\times\) gravity \(\times\) height, where density = Specific Gravity \( \times \) density of water = \(1.6 \times 1000kg/m^3 = 1600kg/m^3\), and height = diameter / 2 = 2m / 2 = 1m. So, the fluid pressure = \(1600kg/m^3 \times 9.81m/s^2 \times 1m = 15.696kPa\). Thus, the pressure at the bottom of the conduit = air pressure + fluid pressure = \(200kPa + 15.696kPa = 215.696kPa\).
02

Calculation of Resultant Force

Next, the resultant force on the end cap is given by: resultant force = pressure \(\times\) area, and the diameter of the end cap is given as 2m, so the area \(A = \pi \times (d/2)^2 = \pi \times (2m/2)^2 = \pi m^2\). Thus, the resultant force = \(215.696kPa \times \pi m^2 = 677.221kN\).
03

Determination of Force Position

Finally, for a semicircular container, it has been established that the vertical distance of the centre of pressure (which is the point where the total sum of a pressure field acts) from the bottom of the conduit is \(4R/(3\pi)\), where R is the radius of the conduit. So, the vertical distance from the bottom to the centre of pressure is \(4 \times 1m/(3\pi) = 0.425m\).

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Most popular questions from this chapter

A 2 -ft-diameter hemispherical plexiglass "bubble" is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2 -ft- diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere.

Find the center of pressure of an elliptical area of minor axis \(2 a\) and major axis \(2 b\) where axis \(2 a\) is vertical and axis \(2 b\) is horizontal. The center of the ellipse is a vertical distance \(h\) below the surface of the water \((h>a)\). The fluid density is constant. Will the center of pressure of the ellipse change if the fluid is replaced by another constant-density fluid? Will the center of pressure of the ellipse change if the vertical axis is tilted back an angle \(\alpha\) from the vertical about its horizontal axis? Explain.

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