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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 is 20343.75 Pa.

Step by step solution

01

Understand the Problem

We need to find the gauge pressure at a second point in a horizontal pipeline where the cross-sectional area doubles. Initially, the water speed is 2.50 m/s and the gauge pressure is 18000 Pa.
02

Use Continuity Equation

The continuity equation for an incompressible fluid states that \( A_1 v_1 = A_2 v_2 \), where \( A \) is the cross-sectional area and \( v \) is the flow speed. Given \( A_2 = 2A_1 \), the equation becomes \( v_1 = 2v_2 \). With \( v_1 = 2.5 \, \text{m/s} \), solve for \( v_2 \): \( v_2 = \frac{2.5}{2} = 1.25 \, \text{m/s} \).
03

Apply Bernoulli's Equation

Bernoulli's equation for a fluid flow is given by:\[P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh\]Since the pipeline is horizontal, \( h \) does not change, so we simplify to:\[P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2\]
04

Calculate Gauge Pressure at Second Point

Rearrange Bernoulli's equation to find \( P_2 \):\[P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2\]Assuming water density \( \rho = 1000 \, \text{kg/m}^3 \), substitute the known values:\[P_2 = 18000 + \frac{1}{2} (1000)(2.5^2) - \frac{1}{2} (1000)(1.25^2)\]\[P_2 = 18000 + 3125 - 781.25 = 20343.75 \, \text{Pa}\]

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Key Concepts

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

Continuity Equation
The Continuity Equation is a fundamental principle in fluid dynamics that describes the conservation of mass in fluid flow. It applies to incompressible fluid flows and can be expressed as:
  • \( A_1 v_1 = A_2 v_2 \)
where:
  • \( A \) is the cross-sectional area of the flow.
  • \( v \) is the flow speed.
Essentially, this equation tells us that the product of the cross-sectional area and the velocity of fluid at one point in a pipeline must equal that at another point farther along the pipeline.
This results from the principle that what flows into a point must flow out.
In the problem, we learned that the area at the second point is twice the first area. Hence, the velocity at the second point decreases by half when the area doubles.
This reduction leads to solving the equation \( v_2 = \frac{v_1}{2} \), resulting in a slower flow speed at the larger cross-sectional area.
Bernoulli's Equation
Bernoulli's Equation is a core concept in fluid dynamics that relates the pressure, velocity, and height energetics in fluid flow.
The equation is given by:
  • \( P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh \)
In this problem, the flow is horizontal, so the height term, \( \rho gh \), can be ignored. Bernoulli's principle essentially tells us that the sum of the pressure energy, kinetic energy, and potential energy remains constant across different points in a streamline.

Since the fluid flow area increases as we move from the first point to the second, but elevation remains unchanged, the increase in cross-sectional area results in a decrease in flow speed as per the continuity equation.
When using Bernoulli's equation specifically for this problem, it helps us find how the kinetic energy difference resulting from the change in velocity affects the gauge pressure by rearranging the formula to solve for \( P_2 \).
This gives us a direct measure of how pressure changes with varying flow speeds.
Gauge Pressure
Gauge Pressure is a measure of fluid pressure relative to the atmospheric pressure. It is important in systems where the atmospheric pressure is not part of the system's pressure consideration, like closed pipelines.
Gauge pressure is defined as:
  • \( P_{gauge} = P - P_{atm} \)
In this exercise, the initial gauge pressure provided was 18000 Pa. In fluid dynamics problems, especially those involving closed systems such as the pipeline in this problem, gauge pressure allows us to focus on variations due to the fluid's kinetic and potential energy changes without atmospheric influences.
As we calculated through Bernoulli's equation, the new gauge pressure at the second point is influenced primarily by the difference in velocity, given no height change. Resultantly, fluid slows down due to increased area, increasing pressure as the kinetic energy converts back into pressure energy, showcasing the conservation of mechanical energy principles.

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Most popular questions from this chapter

The earth does not have a uniform density; it is most dense at its center and least dense at its surface. An approximation of its density is \(\rho(r)=A-B r,\) where \(A=12,700 \mathrm{kg} / \mathrm{m}^{3}\) and \(B=\) \(1.50 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{4}\) . Use \(R=6.37 \times 10^{6} \mathrm{m}\) for the radius of the earth approximated as a sphere. (a) Geological evidence indicates that the densities are \(13,100 \mathrm{kg} / \mathrm{m}^{3}\) and \(2,400 \mathrm{kg} / \mathrm{m}^{3}\) at the earth's center and surface, respectively. What values does the linear approximation model give for the densities at these two locations? (b) Imagine dividing the earth into concentric, spherical shells. Each shell has radius \(r\) , thickness \(d r\) , volume \(d V=4 \pi r^{2} d r,\) and mass \(d m=\rho(r) d V .\) By integrating from \(r=0\) to \(r=R,\) show that the mass of the earth in this model is \(M=\frac{4}{3} \pi R^{3}\left(A-\frac{3}{4} B R\right)\) (c) Show that the given values of \(A\) and \(B\) give the correct mass of the earth to within 0.4\(\%\) (d) We saw in Section 12.6 that a uniform spherical shell gives no contribution to \(g\) inside it. Show that \(g(r)=\frac{4}{3} \pi G r\left(A-\frac{3}{4} B r\right)\) inside the earth in this model. (e) Verify that the expression of part (d) gives \(g=0\) at the center of the earth and \(g=9.85 \mathrm{m} / \mathrm{s}^{2}\) at the surface. (f) Show that in this model \(g\) does not decrease uniformly with depth but rather has a maximum of \(4 \pi G A^{2} / 9 B=10.01 \mathrm{m} / \mathrm{s}^{2}\) at \(r=2 A / 3 B=5640 \mathrm{km} .\)

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 \(\mathrm{m}^{3} / \mathrm{s} ?(\mathrm{b})\) 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?

A shower head has 20 circular openings, each with radius 1.0 \(\mathrm{mm}\) . The shower head is connected to a pipe with radius 0.80 \(\mathrm{cm} .\) If the speed of water in the pipe is \(3.0 \mathrm{m} / \mathrm{s},\) what is its speed as it exits the shower-head openings?

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is 15.0 \(\mathrm{cm}\) (Fig. 14.37 . (a) What is the gauge pressure at the water-mercury interface? (b) Calculate the vertical distance \(h\) from the top of the mercury in the right-hand arm of the tube to the top of the water in the left-hand arm.

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.

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