/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 A horizontal 2 -m-diameter condu... [FREE SOLUTION] | 91Ó°ÊÓ

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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Gravity
Specific Gravity (SG) is a vital concept in fluid mechanics. It denotes the ratio of the density of a fluid compared to the density of water. This measurement is dimensionless and helps in understanding how fluid substances relate to one another under varying gravitational fields. In this problem, the specific gravity of the liquid is given as 1.6. This tells us that the liquid is 1.6 times denser than water.

Here's how it works in practice:
  • Density of water is approximately 1000 kilograms per cubic meter (\(kg/m^3\)).
  • The density of the liquid in the conduit can be found by multiplying the specific gravity by the density of water. In this case, it is \(1600 \, kg/m^3\).
Using specific gravity helps engineers quickly assess how forces like pressure move or shift in fluids. It's essential when calculating pressures and forces in submerged or partially submerged bodies, as we've seen in the problem's solution.
Resultant Force
In fluid mechanics, resultant force is the total force exerted by a fluid on a surface, taking both magnitude and direction into account. Calculating this force is crucial in determining how structures will react under fluid pressure, particularly in the design of vessels, dams, or pipes.

To find the resultant force on the end cap of the conduit in our problem, we:
  • Determine the pressure at the bottom end of the conduit, which includes both the air pressure above the fluid and the hydrostatic pressure of the fluid itself.
  • The pressure at the bottom is calculated as \(215.696 \, kPa\).
  • The area of the end cap, being circular, has an area of \(\pi \, m^2\).
  • The resultant force is then found by multiplying pressure by area: \(677.221 \, kN\).
This force tells us how much pressure is applied and can help in assessing how structures will withstand such conditions. Understanding resultant forces enables engineers to design safe and efficient structures in varied fluid environments.
Centre of Pressure
The centre of pressure is a point on an immersed surface where the total pressure force is considered to act. It's a critical concept for anyone designing systems where fluids exert pressure on submerged surfaces, like the walls of containers or dam gates.

For a semicircular surface like the end cap of the conduit, the vertical position of the centre of pressure can be found using the relation \(\frac{4R}{3\pi}\), where \(R\) is the radius.
  • In this situation, \(R = 1 \, m\), giving a centre of pressure that is \(0.425 \, m\) above the conduit base.
Understanding the centre of pressure is crucial for engineering integrity and stability, as it defines how the fluid force can produce rotational effects on submerged bodies. Accurately locating this point helps in aligning design features effectively to counteract or utilize forces exerted by fluids.

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Most popular questions from this chapter

A studant needs to measure the air pressure inside a compressed air tenk but does not have ready access to a pressure gage. Using materials already in the lab, she builds a U-tube manometer using two clear 3 -ft-long plastic tubes, flexible hoses, and a tape measure. The only readily avai able liquids are water from a tap and a bottle of corn syrup. She selects the corn syrup because it has a larger density \((S G=1.4) .\) What is the maximum air pressure, in psia, that can be measured?

A child riding in a car holds a string attached to a floating, helium-filled balloon. As the car decelerates to a stop, the balloon tilts backwards. As the car makes a right-hand turn, the balloon tilts to the right. On the other hand, the child tends to be forced forward as the car decelerates and to the left as the car makes a right-hand turn. Explain these observed effects on the balloon and child.

On the suction side of a pump, a Bourdon pressure gage reads \(40 \mathrm{kPa}\) vacuum. What is the corresponding absolute pressure if the ocal atmospheric pressure is \(100 \mathrm{kPa}(\mathrm{abs}) ?\)

A 30 -ft-high downspout of a house is clogged at the bottom. Find the pressure at the bottom if the downspout is filled with \(60^{\circ} \mathrm{F}\) rainwater.

A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by \(W\). When in salt-free riverwater. it floats 0.25 ft deeper than when in seawater \(\left(\gamma=64 \mathrm{lb} / \mathrm{ft}^{3}\right) .\) Find the weight \(W\).

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