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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

A large, \(40.0 \mathrm{~kg}\) cubical block of wood with uniform density is floating in a freshwater lake with \(20.0 \%\) of its volume above the surface of the water. You want to load bricks onto the floating block and then push it horizontally through the water to an island where you are building an outdoor grill. (a) What is the volume of the block? (b) What is the maximum mass of bricks that you can place on the block without causing it to sink below the water surface?

A swimming pool is \(5.0 \mathrm{~m}\) long, \(4.0 \mathrm{~m}\) wide, and \(3.0 \mathrm{~m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 2200.355 -L cans per minute. At point 2 in the pipe, the gauge pressure is \(152 \mathrm{kPa}\) and the cross-sectional area is \(8.00 \mathrm{~cm}^{2}\). At point \(1,1.35 \mathrm{~m}\) above point \(2,\) the cross- sectional area is \(2.00 \mathrm{~cm}^{2}\). Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and \(2 ;\) (d) gauge pressure at point 1 .

Viscous blood is flowing through an artery partially clogged by cholesterol. A surgeon wants to remove enough of the cholesterol to double the flow rate of blood through this artery. If the original diameter of the artery is \(D,\) what should be the new diameter (in terms of \(D\) ) to accomplish this for the same pressure gradient?

How does the force the diaphragm experiences due to the difference in pressure between the lungs and abdomen depend on the abdomen's distance below the water surface? The force (a) increases linearly with distance; (b) increases as distance squared; (c) increases as distance cubed; (d) increases exponentially with distance.
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