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Chapter 13: Problem 43

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At area, \(1,\) the cross-sectional area of the pipe is 0.070 \(\mathrm{m}^{2}\) and the magnitude of the fluid velocity is 3.50 \(\mathrm{m} / \mathrm{s} .\) What is the fluid speed at points in the pipe where the cross-sectional area is \((a) 0.105 \mathrm{m}^{2},\) (b) 0.047 \(\mathrm{m}^{2}\) ?

Short Answer

Expert verified
For (a), fluid speed is 2.333 m/s; for (b), it is 5.213 m/s.

Step by step solution

01

Understanding Continuity Equation

In fluid dynamics, the Continuity Equation ensures that the mass flow rate remains constant across any cross-section of a pipe. This can be expressed as \( A_1v_1 = A_2v_2 \), where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid. This equation will help us find the velocity at different points in the pipe.
02

Applying Continuity Equation to Part (a)

For point (a) where the area \( A_2 = 0.105 \ \mathrm{m}^2 \), we apply the continuity equation \( A_1v_1 = A_2v_2 \). With \( A_1 = 0.070 \ \mathrm{m}^2 \) and \( v_1 = 3.50 \ \mathrm{m/s} \), calculate \( v_2 \):\[0.070 \times 3.50 = 0.105 \times v_2\]Divide both sides by 0.105 to solve for \( v_2 \).\[v_2 = \frac{0.070 \times 3.50}{0.105}\]
03

Calculating Fluid Velocity for Part (a)

Complete the calculation for \( v_2 \):\[v_2 = \frac{0.245}{0.105} = 2.333 \ \mathrm{m/s}\]Thus, the fluid velocity at the point with an area of \( 0.105 \ \mathrm{m}^2 \) is \( 2.333 \ \mathrm{m/s} \).
04

Applying Continuity Equation to Part (b)

For point (b) where the area \( A_2 = 0.047 \ \mathrm{m}^2 \), use the same continuity equation:\[0.070 \times 3.50 = 0.047 \times v_2\]Divide both sides by 0.047 to find \( v_2 \).\[v_2 = \frac{0.070 \times 3.50}{0.047}\]
05

Calculating Fluid Velocity for Part (b)

Complete the calculation for \( v_2 \):\[v_2 = \frac{0.245}{0.047} = 5.213 \ \mathrm{m/s}\]Thus, the fluid velocity at the point with an area of \( 0.047 \ \mathrm{m}^2 \) is \( 5.213 \ \mathrm{m/s} \).

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Key Concepts

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

Continuity Equation
Fluid dynamics is a fascinating area of physics that explores the behavior of fluids, whether liquids or gases, in motion. One of the core concepts in fluid dynamics is the Continuity Equation. This equation is a statement of conservation of mass, ensuring that the mass of fluid entering a pipe equals the mass leaving it, assuming no mass is added or removed. This principle is especially crucial when dealing with pipes of varying diameters. It's mathematically expressed as:\[ A_1v_1 = A_2v_2 \]In this equation:- \( A \) represents the cross-sectional area- \( v \) stands for fluid velocityThe equation tells us that the product of the cross-sectional area and the velocity at any two points along the pipe must remain constant. Hence, if the area changes, the velocity must adjust accordingly to ensure a constant mass flow rate.
Mass Flow Rate
Mass flow rate is a measure of the mass of fluid passing through a given point in the system per unit time. It combines the concepts of fluid density, velocity, and cross-sectional area. In mathematical terms, it's given by the expression:\[ \dot{m} = \rho Av \]Where:- \( \dot{m} \) is the mass flow rate- \( \rho \) is the fluid's density- \( A \) is the cross-sectional area- \( v \) is the fluid velocityIn many practical applications, the density \( \rho \) remains constant, especially for incompressible fluids like water. This means that any changes to the cross-sectional area or velocity directly affect the flow rate. A larger cross-section decreases velocity, provided the flow rate remains constant, and vice versa.
Cross-Sectional Area
The cross-sectional area of a pipe is the surface area of the cut surface of the pipe, perpendicular to the direction of flow. It plays a crucial role in determining the fluid velocity inside the pipe. If you picture slicing a pipe in half, the flat face that meets your eye is the cross-sectional area. - Larger cross-sectional areas result in slower fluid speeds. - Smaller cross-sectional areas speed up the flow. This relationship is integral to the Continuity Equation, as we've discussed. Changes in the cross-sectional area necessitate adjustments in velocity to maintain a stable mass flow rate, making cross-sectional area a pivotal factor in fluid flow calculations.
Fluid Velocity
Fluid velocity is the speed at which a fluid moves in a particular direction through a pipe or channel. It's a dynamic quantity adjusted frequently based on the cross-sectional area's changes.Velocity is measured in units of distance per time, such as meters per second (\(m/s\)). - As the pipe narrows, velocity increases.- When the pipe widens, velocity decreases.This inverse relationship between velocity and cross-sectional area ensures a constant flow rate, as dictated by the Continuity Equation. Fluid velocity is essential for processes that depend on the rate of fluid transfer, impacting everything from simple water plumbing to industrial fluid systems.

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Most popular questions from this chapter

A hollow plastic sphere is held below the surface of a fresh- water lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 \(\mathrm{m}^{3}\) and the tension in the cord is 900 \(\mathrm{N} .\) (a) Calculate the buoyant force exerted by the water on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. 13.42\()\) , because as the depth increases, so does the pressure difference, which the difference, if any.tends to collapse the diver's lungs. since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external-internal pressure difference when the diver's lungs are at a depth of 6.1 \(\mathrm{m}\) (about 20 \(\mathrm{ft}\) ? Assume that the diver is in fresh- water. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

Water is flowing in a cylindrical pipe of varying circular cross-sectional area, and at all points the water completely fills the pipe. (a) At one point in the pipe, the radius is 0.150 \(\mathrm{m} .\) What is the speed of the water at this point if the volume flow rate in the pipe is 1.20 \(\mathrm{m}^{3} / \mathrm{s} ?\) (b) At a second point in the pipe, the water speed is 3.80 \(\mathrm{m} / \mathrm{s}\) . What is the radius of the pipe at this point?

Blood pressure. Systemic blood pressure is defined as the ratio of two pressures, both expressed in millimeters of mercury. Normal blood pressure is about \(\frac{120 \mathrm{mm}}{80 \mathrm{mm}},\) which is usually just stated as \(\frac{120}{80}\) . (See also Problem \(24 . )\) What would normal systemic blood pressure be if, instead of millimeters of mercury, we expressed pressure in each of the following units, but continued to use the same ratio format? (a) atmospheres, (b) torr, (c) Pa, (d) \(\mathrm{N} / \mathrm{m}^{2},\) (e) psi.

Compressible fluids. Throughout this chapter, we have dealt only with incompressible fluids. But under very high pressure, fluids do, in fact, compress. (a) Show that the continuity condition for compressible fluids is \(\rho_{1} A_{1} v_{1}=\rho_{2} A_{2} v_{2}\) where \(\rho\) is the density of the fluid. (b) Show that your result reduces to the familiar result for incompressible fluids.
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