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Chapter 12: Problem 49

At a certain point in a horizontal pipeline, the water's speed is \(2.50 \mathrm{~m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4} \mathrm{~Pa}\). 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 in the pipeline is \(1.875 * 10^4 Pa\).

Step by step solution

01

Use the continuity equation

The first step in this exercise is to find the speed of the water at the second point using the continuity equation \(A_1v_1=A_2v_2\), where \(A_1\) and \(A_2\) are the cross-sectional areas at the first and second points respectively and \(v_1\) and \(v_2\) are the speeds at these points. Here, \(A_2\) is twice \(A_1\), and \(v_1\) is given as \(2.50 ms^{-1}\). Solving for \(v_2\), we have \(v_2=A_1v_1/A_2\)\(=2.50 ms^{-1}/2 = 1.25 ms^{-1}\)
02

Use Bernoulli's equation

After finding the speed at the second point, we will use Bernoulli's equation to find the gauge pressure at the second point. Bernoulli's equation is \(P_1 + 0.5蟻v_1^2 = P_2 + 0.5蟻v_2^2\). Here \(P_1\) is the pressure at the first point, \(蟻\) is the density of the fluid (here we cancel it in the calculation as it's same at both points), \(v_1\) is the velocity at the first point and similarly \(P_2\) and \(v_2\) are the pressure and velocity at the second point.
03

Calculate the pressure at the second point

Inserting the known values into the equation from step 2: \(P_1 + 0.5 v_1^2 = P_2 + 0.5 v_2^2\). We are required to find \(P_2\), so we rearrange the equation as follows: \(P_2 = P_1 + 0.5 v_1^2 - 0.5 v_2^2\). The question states that \(P_1\) = \(1.8 * 10^4 Pa\), \(v_1\) = \(2.5 ms^{-1}\), and we found \(v_2\) = \(1.25 ms^{-1}\) in step 1. Substituting these values gives: \(P_2=1.8*10^4 Pa + 0.5 * (2.5 ms^{-1})^2 - 0.5 * (1.25 ms^{-1})^2\). Solving this equation will yield the pressure at the second point.

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Key Concepts

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

Continuity Equation in Fluid Dynamics
The concept of the continuity equation is fundamental in fluid dynamics. It helps us understand how fluid flows through a pipe. Essentially, it states that the mass flow rate of a fluid must remain constant from one cross-section of the pipe to another.

To mathematically express this, we use:
  • Continuity equation: \(A_1v_1 = A_2v_2\)
Here:
  • \(A_1\) and \(A_2\) are the cross-sectional areas.
  • \(v_1\) and \(v_2\) are the fluid velocities at these points.
This equation tells us that if the area increases, the velocity must decrease, and vice versa, to keep the flow rate constant. For example, when the cross-sectional area doubles, as in the original exercise, the speed of the water is halved.
Understanding Fluid Dynamics
Fluid dynamics is the study of how fluids behave when they are in motion. It covers a broad range of concepts, from the simple flow of water through a garden hose to complex air flow patterns around an aircraft. The principles of fluid dynamics allow us to predict and control the behavior of fluids in various environments.

Some key aspects include:
  • Flow Rate: Measures how much fluid is passing a point in a given time.
  • Turbulence: Refers to chaotic changes in pressure and flow velocity.
  • Viscosity: Describes the internal friction of the fluid, affecting its flow.
In the context of the given problem, fluid dynamics provides the framework to understand how changes in cross-sectional area affect flow velocity and pressure, as described by both the continuity and Bernoulli's equations.
Gauge Pressure Explained
Gauge pressure is a way of expressing pressure in a system. It's the pressure in a given system relative to atmospheric pressure.

To break it down:
  • Gauge Pressure: Measures pressure that is not total pressure, ignoring atmospheric pressure.
  • Formula: \(P_{gauge} = P_{absolute} - P_{atmospheric}\)
Gauge pressure is especially useful in cases where relative pressure differences are more informative than absolute pressure values, as seen in the original exercise. It provides a straightforward means of determining how much pressure is being added by the system itself, without the added complexity of atmospheric factors. Thus, in fluid systems like pipelines, knowing the gauge pressure helps in assessing and maintaining the system's integrity and efficiency.

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Most popular questions from this chapter

Estimate the fraction of your body's total volume that is above the surface of the water when you float in seawater (density \(1030 \mathrm{~kg} / \mathrm{m}^{3}\) ). (a) Use this estimate and your weight to calculate the total volume of your body. (b) What is your average density? How does your average density compare to the density of seawater?

A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

On another planet that you are exploring, a large tank is open to the atmosphere and contains ethanol. A horizontal pipe of cross sectional area \(9.0 \times 10^{-4} \mathrm{~m}^{2}\) has one end inserted into the tank just above the bottom of the tank. The other end of the pipe is open to the atmosphere. The viscosity of the ethanol can be neglected. You measure the volume flow rate of the ethanol from the tank as a function of the depth \(h\) of the ethanol in the tank. If you graph the volume flow rate squared as a function of \(h,\) your data lie close to a straight line that has slope \(1.94 \times 10^{-5} \mathrm{~m}^{5} / \mathrm{s}^{2} .\) What is the value of \(g,\) the acceleration of a free-falling object at the surface of the planet?

Water runs into a fountain, filling all the pipes, at a steady rate of \(0.750 \mathrm{~m}^{3} / \mathrm{s}\). (a) How fast will it shoot out of a hole \(4.50 \mathrm{~cm}\) in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

A barrel contains a \(0.120 \mathrm{~m}\) layer of oil floating on water that is \(0.250 \mathrm{~m}\) deep. The density of the oil is \(600 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?
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