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Chapter 12: Problem 45

At a certain point in a horizontal pipeline, the water's speed is 2.50 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4}\) Pa. Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.

Short Answer

Expert verified
The gauge pressure at the second point is approximately 20343.75 Pa.

Step by step solution

01

Understanding the Problem

We need to find the gauge pressure at a second point in a pipeline where the cross-sectional area is twice as large as the first point. We are given the water speed and gauge pressure at the first point.
02

Apply the Continuity Equation

The continuity equation for incompressible flow states that the product of cross-sectional area and velocity remains constant. Let the area and velocity at the first point be \(A_1\) and \(v_1\) and at the second point be \(A_2 = 2A_1\) and \(v_2\). Thus, \(A_1v_1 = A_2v_2\) implies \(v_2 = \frac{v_1}{2}\), since \(A_2\) is twice \(A_1\).
03

Calculation of Reduced Velocity

Given \(v_1 = 2.50 \ \mathrm{m/s}\), the equation \(v_2 = \frac{v_1}{2}\) gives \(v_2 = 1.25 \ \mathrm{m/s}\). The velocity at the second point is thus 1.25 m/s.
04

Apply Bernoulli's Equation

Bernoulli's equation relates pressure and velocity at two points in a horizontal pipe: \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \). Here \(P_1 = 1.80 \times 10^4 \ \mathrm{Pa}\), \(v_1 = 2.50\) m/s and \(v_2 = 1.25\) m/s. We need to solve for \(P_2\).
05

Solving Bernoulli's Equation for Pressure

Rearrange Bernoulli's equation to solve for \(P_2\): \(P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2)\). Assume the density of water \(\rho = 1000 \ \mathrm{kg/m^3}\).
06

Calculate the Pressure Difference

Calculate the pressure difference: \(\frac{1}{2}\rho (v_1^2 - v_2^2) = \frac{1}{2} (1000) ((2.50)^2 - (1.25)^2)\). Simplify this: \(\frac{1}{2} (1000) (6.25 - 1.5625) = 2343.75 \ \mathrm{Pa}\).
07

Final Calculation for Gauge Pressure at Second Point

Use the pressure difference to find \(P_2\): \(P_2 = 1.80 \times 10^4 \ \mathrm{Pa} + 2343.75 \) Pa. Thus, \(P_2 = 2.034375 \times 10^4 \ \mathrm{Pa}\).

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Key Concepts

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

Bernoulli's Equation
Bernoulli's Equation is a fundamental principle in fluid dynamics that describes the conservation of energy in a flowing fluid. It is expressed as: \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]
  • P: Pressure energy per unit volume of the fluid
  • \( \frac{1}{2} \rho v^2 \): Kinetic energy per unit volume, where \( v \) is the velocity of the fluid
  • \( \rho gh \): Potential energy per unit volume, with \( h \) as the height above a reference point
In our problem, the pipe is horizontal, negating gravitational effects, and simplifies to:\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] This equation helps determine how changes in velocity affect pressure at different points in a pipeline.
Continuity Equation
The Continuity Equation ensures mass conservation in fluid flow, particularly for incompressible fluids like water. It states:\[ A_1v_1 = A_2v_2 \] where:
  • \( A \): Cross-sectional area of the flow
  • \( v \): Velocity of the fluid
The equation asserts that the volume flow rate must remain constant across varying cross-sections:
  • For smaller areas, fluid must speed up
  • For wider areas, fluid must slow down
In the exercise, as the area doubles, the velocity reduces by half, exemplifying the inverse relationship between area and velocity.
Gauge Pressure
Gauge Pressure is the pressure in a system that is above the atmospheric pressure. It is important in fluid systems because it excludes atmospheric pressure, focusing solely on the pressure exerted by the fluid itself. In calculations:\[ \text{Gauge Pressure} = P_\text{system} - P_\text{atmosphere} \] This allows engineers to assess the actual working pressure within the pipeline without considering external atmospheric conditions. In the problem, knowing the initial gauge pressure and how it changes with velocity allows us to find the pressure at another pipeline point by using Bernoulli's equation.
Incompressible Flow
Incompressible Flow refers to a fluid flow where the fluid's density remains constant. For most practical scenarios involving liquids like water, incompressibility is a valid assumption. This simplification:
  • Allows use of the Continuity Equation without density changes
  • Ensures conservation of mass is easier to apply
In fluids with constant density, volume flow rate through varying cross-sectional areas is conserved, illustrating the direct application of the Continuity Equation in this scenario. Understanding incompressible flow is crucial for applying fundamental fluid dynamics principles effectively.

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Most popular questions from this chapter

A single ice cube with mass 9.70 g floats in a glass completely full of 420 \(\mathrm{cm}^{3}\) of water. You can ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 \(\mathrm{kg} / \mathrm{m}^{3} .\) What volume of salt water would the \(9.70-\mathrm{g}\) ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m? (Assume that the mains have a much larger diameter than the fire hose.)

Water is flowing in a pipe with a circular cross section but with varying 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 water is flowing into this pipe at a steady rate of 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?

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A .\) The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \omega. Gravitational forces are negligible. Consider a volume element of the fluid of area \(A\) and thickness \(d r^{\prime}\) a distance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime}\) . (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0},\) show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .\) (c) An object of volume \(V\) and density \(\rho_{\mathrm{ob}}\) has its center of mass at a distance \(R_{\mathrm{cmob}}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{ob}} R_{\mathrm{cmob}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}} .\) (e) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) . What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

(a) What is the average density of the sun? (b) What is the average density of a neutron star that has the same mass as the sun but a radius of only 20.0 km?
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