/*! 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 11 A horizontal oriented tube \(A B... [FREE SOLUTION] | 91Ó°ÊÓ

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A horizontal oriented tube \(A B\) of length \(l=2.5 \mathrm{~m}\) rotates with a constant angular velocity \(\omega=5 \mathrm{rad} / \mathrm{s}\) about a stationary vertical axis \(O O\) ' passing through the end \(A\). Initially the tube is filled with an ideal fluid. The end \(A\) of the tube is open, the closed end \(B\) has a very small orifice. Find the velocity of the fluid (in \(\mathrm{m} / \mathrm{s}\) ) relative to the tube when the liquid column length in tube reduce to \(h=1 \mathrm{~m}\).

Short Answer

Expert verified
The velocity of the fluid is approximately 11.46 m/s.

Step by step solution

01

Identify the given values

We are given the tube length \(l = 2.5\, \mathrm{m}\), angular velocity \(\omega = 5\, \mathrm{rad/s}\), and reduced liquid column length \(h = 1\, \mathrm{m}\). The tube rotates about a stationary vertical axis passing through point \(A\).
02

Apply the Hydrostatic Pressure principle

Since the tube is rotating about an axis, the centrifugal force creates a pressure difference along the tube. The fluid exerting a force towards the farthest point of rotation would result in pressure, where \(P_1\) at point \(A\) equals atmospheric pressure, and therefore at any point \(x\) within the liquid, the pressure is given by \(P = \omega^2 x^2 \, \frac{\rho}{2} + P_0\), where \(\rho\) is the density of the liquid and \(P_0\) is atmospheric pressure.
03

Establish Bernoulli's equation for flow

Apply Bernoulli's principle between point \(A\) and point \(B\). Since both ends are open to the atmosphere, their pressures equalize. Thus \(\frac{1}{2} \rho v^2 = \frac{1}{2} \rho \omega^2 h^2 + \frac{1}{2} \rho \omega^2 l^2\). Simplifying, we use \(v^2 = \omega^2 (l^2 - h^2)\) to find the fluid velocity \(v\).
04

Substitute known values and solve for velocity

Plug the known values into the derived equation: \(v = \sqrt{5^2 (2.5^2 - 1^2)}\). Calculate \(v = \sqrt{25 \times 5.25}=\sqrt{131.25}\).
05

Calculate the final velocity

The value obtained from the calculation is approximately \(11.46\, \mathrm{m/s}\).

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Key Concepts

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

Understanding Angular Velocity
Angular velocity is a key concept in fluid mechanics, especially when dealing with rotating systems. It refers to how fast an object rotates or spins around a particular axis. Unlike linear velocity, which tells us how far something moves in a certain time, angular velocity, denoted as \(\omega\), gives us the angle that an object covers per unit time, measured in radians per second (rad/s).
For our problem, the tube is rotating with an angular velocity of 5 rad/s. This means that every second, the tube covers an angle of 5 radians around the vertical axis. Understanding this helps us predict how forces like centrifugal force behave, which influences how the liquid moves inside the tube.
Bernoulli's Principle in Rotating Systems
Bernoulli's principle is a fundamental concept in fluid mechanics that explains the behavior of fluid flow. In a rotating system, Bernoulli's principle provides insights into the energy balance along a streamline.
We apply Bernoulli's equation in our case to understand the change in velocity of the fluid from point \(A\) to point \(B\) within the rotating tube. The principle states that in a steady flow, the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant. Since both ends of the tube at \(A\) and \(B\) are open, they are at atmospheric pressure. Hence, any increase in one form of energy results in a decrease in another.
For example, in our exercise, this means that the kinetic energy due to velocity increases as the fluid column shortens, while maintaining the balance through pressure and potential energy changes. This reflects in the equation used to find the velocity of the fluid as it exits through the orifice.
Centrifugal Force within the Fluid
Centrifugal force plays a crucial role when a container or tube is rotating around an axis. It's the outward force that acts on any mass when it moves in a circular path, and it acts perpendicular to the axis of rotation.
In the context of our tube, centrifugal force is what causes the liquid to experience a pressure difference along the length of the tube from point \(A\) towards point \(B\). As the tube rotates, each particle of the fluid experiences this force pushing it outwards, contributing to the pressure gradient. This is calculated using the formula, where pressure at a distance \(x\) from the axis is given by \(\frac{1}{2} \rho \omega^2 x^2 + P_0\). This helps us understand how centrifugal force affects hydrostatic pressure.
Hydrostatic Pressure and Its Effects
Hydrostatic pressure is the pressure exerted by a fluid at rest, due to the force of gravity. In the rotating tube experiment, however, we are not dealing with a stationary fluid. Instead, the rotation induces variations in hydrostatic pressure across the tube due to centrifugal force.
As a result of the centrifugal force, the fluid pressure along the tube increases from point \(A\) to point \(B\). At point \(A\), the pressure equals atmospheric pressure, and as we move towards \(B\), the pressure increases following the equation \(P = \omega^2 x^2 \frac{\rho}{2} + P_0\). Understanding this pressure gradient is key to predicting how the velocity of the fluid will change as the length of the fluid column reduces, as required in the original exercise. This helps account for the balance of forces and energy within the rotating system.

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Most popular questions from this chapter

A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side until water rises by \(25 \mathrm{~cm}\) on the other side. If the density of oil is \(0.8\), the oil level will stand higher than the water level by (1) \(6.25 \mathrm{~cm}\) (2) \(12.50 \mathrm{~cm}\) (3) \(31.75 \mathrm{~cm}\) (4) \(62.50 \mathrm{~cm}\)

An open vessel containing liquid is moving with constant acceleration \(a\) on a levelled horizontal surface. For this situation mark out the correct statement(s). (1) The maximum pressure is at the leftmost bottom corner. (2) Along a horizontal line within the liquid as we move from left to right the pressure decreases. (3) The pressure at all points on a line making an angle of \(\tan ^{-1}\left(\frac{a}{g}\right)\) with horizontal would be same. (4) Along a horizontal line within the liquid as we move from left to right, the pressure remains same.

A solid cone of height \(H\) and base radius \(H / 2\) floats in a liquid of density \(\rho\). It is hanging from the ceiling with the help of a string. The force by the fluid on the curved surface of the cone is \(\left(P_{0}=\right.\) atmospheric pressure \()\) (1) \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{3}\right)\) (2) \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{6}\right)\) (3) \(\frac{\pi H^{2}}{4}\left(\frac{P_{0}}{4}+\rho g H\right)\) (4) \(\frac{\pi H^{2}}{4}\left(P_{0}+\rho g H\right)\)

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weighs \(0.25 \mathrm{~kg}\) in air and is observed to displace \(0.015\) litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, \(19.3\) and \(8.9\). The approximate percentage of copper in the ornament is (1) \(5 \%\) (2) \(10 \%\) (3) \(15 \%\) (4) \(25 \%\)

Water from a tap emerges vertically downwards with an initial velocity \(v_{0}\). Assuming pressure is constant throughout the stream of water and the flow is steady, find the distance from the tap at which cross-sectional area of stream is half of the cross-sectional area of stream at the tap. (1) \(v_{0}^{2} / 2 g\) (2) \(3 v_{0}^{2} / 2 g\) (3) \(2 v_{0}^{2} / 2 g\) (4) \(5 v_{0}^{2} / 2 g\)

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