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At one point in a pipeline, the water's speed is \(3.00 \mathrm{~m} / \mathrm{s}\) and the gauge pressure is \(4.00 \times 10^{4} \mathrm{~Pa}\). Find the gauge pressure at a second point in the line \(11.0 \mathrm{~m}\) lower than the first if the pipe diameter at the second point is twice that at the first.

Short Answer

Expert verified
The gauge pressure at the second point is \(-6415.25 \ \mathrm{Pa}.\)

Step by step solution

01

Identify Given Values

We are provided with the following values at the first point in the pipeline: speed of water, \( v_1 = 3.00 \ \mathrm{m/s} \), and gauge pressure, \( p_1 = 4.00 \times 10^4 \ \mathrm{Pa} \). It is also stated that the second point in the pipeline is \( 11.0 \ \mathrm{m} \) lower than the first point, and the diameter at the second point is twice that of the first.
02

Relation of Areas from Diameters

Since the diameter at the second point is twice that at the first, the area \( A_2 \) is four times \( A_1 \). This is because the area is proportional to the square of the diameter: \[ A_2 = \pi \left(2 \times \frac{d_1}{2}\right)^2 = 4 \times \pi \left(\frac{d_1}{2}\right)^2 = 4A_1. \]
03

Apply Continuity Equation to find Speed at Second Point

By the Continuity Equation (\( A_1 v_1 = A_2 v_2 \)), substitute the given relation for \( A_2 \) to solve for the speed at the second point:\[ v_2 = \frac{A_1 v_1}{A_2} = \frac{A_1 v_1}{4A_1} = \frac{v_1}{4}. \]Using \( v_1 = 3.00 \ \mathrm{m/s} \), we find \( v_2 = \frac{3.00}{4} = 0.75 \ \mathrm{m/s}.\)
04

Apply Bernoulli's Equation

Use Bernoulli’s equation for the two points:\[ p_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = p_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2. \] Assuming water density \( \rho = 1000 \ \mathrm{kg/m^3} \), with \( h_1 - h_2 = 11.0 \ \mathrm{m} \), and given that \( h_2 = h_1 - 11.0 \):\[ p_1 + \frac{1}{2}(1000)(3.00)^2 = p_2 + \frac{1}{2}(1000)(0.75)^2 + (1000)(9.81)(11.0). \]
05

Solve for Gauge Pressure at Second Point

Substitute numbers into the Bernoulli equation and solve for \( p_2 \):\[ 4.00 \times 10^4 + 4500 = p_2 + 281.25 + 107910. \]Combine and rearrange terms:\[ 44500 = p_2 + 281.25 + 107910. \]\[ p_2 = 44500 - 281.25 - 107910. \]\[ p_2 = -6415.25 \ \mathrm{Pa}. \] The negative result implies relative to atmospheric pressure, the pressure is lower at the second point, due to increased gravitational potential energy.

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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 foundational in fluid dynamics. It is particularly important when analyzing fluid flow in a pipeline. The equation expresses the conservation of energy within a fluid stream, describing how the speed, pressure, and height at two points of the same streamline relate to each other. It is expressed as:\[ p + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]where:
  • \( p \) is the pressure energy per unit volume
  • \( \rho \) is fluid density
  • \( v \) is fluid speed
  • \( g \) is gravitational acceleration
  • \( h \) is the height above a reference point
This equation is applied to understand how changes in elevation (such as in our pipeline problem) impact pressure and speed within the fluid. It also helps calculate the gauge pressure at various points in the system by establishing the relationship between these factors.
Continuity Equation
The Continuity Equation highlights a fundamental principle of fluid flow: the conservation of mass. For a fluid flowing through a pipeline, the mass flow rate must remain constant between any two points along the streamline. This is captured by the formula:\[ A_1 v_1 = A_2 v_2 \]Where \( A \) represents cross-sectional area and \( v \) is the fluid velocity. In the provided example, the second point has a pipe diameter twice that of the first point. Therefore, the cross-sectional area is four times larger at the second point (since area is proportional to the square of the diameter). This means the velocity must decrease to maintain the same flow rate, resulting in a reduced speed at the second point.
Gauge Pressure
Gauge pressure is used to measure pressure relative to atmospheric pressure, not absolute pressure. This makes it particularly useful in applications like pipeline flow where the difference in pressure is more significant than the overall pressure level. It is essentially the pressure within a system minus the atmospheric pressure. In calculations, it's often expressed in Pascals (Pa). In the problem context, you calculate the gauge pressure at different points along the pipeline by effectively using Bernoulli's Equation. Understanding and calculating gauge pressure is essential for analyzing how pressure variations influence flow characteristics in a fluid system.
Pipeline Flow
Pipeline flow refers to the movement of fluid within pipes, which can be analyzed using principles of fluid mechanics. Several factors influence how a fluid flows, including the pipe's diameter, length, and the fluid's velocity. In fluid dynamics, it's crucial to account for how changes in any of these factors can influence both pressure and speed. This comes into play in our example exercise, where the change in height and pipe diameter between two points affects the gauge pressure and velocity of water, showcasing practical applications of both Bernoulli's Equation and the Continuity Equation.
Water Speed
Water speed in a pipeline can vary significantly depending on several factors, such as changes in pipe diameter or the elevation difference between two points. The Continuity Equation helps determine how these factors cause velocity changes. In this scenario, the speed at the first point was given as 3.00 m/s, but since the diameter increased at the second point, the speed reduced to 0.75 m/s. Knowing the water speed is critical for solving pressure-related problems in fluid dynamics, using Bernoulli's Equation to find pressures at different sections of a pipeline.
Pressure Calculation
Pressure calculation is a critical component of fluid dynamics, especially when assessing fluid behavior in systems like pipelines. In the example exercise, you start by identifying both initial pressures and speeds. Bernoulli's Equation was then applied to find the gauge pressure at the second point by comparing energy across both points.To calculate effectively, you need to consider all forms of energy present: kinetic energy \( \left( \frac{1}{2} \rho v^2 \right) \), gravitational potential energy \( (\rho gh) \), and pressure energy \(p\). Accurate calculations are essential for ensuring efficient and safe pipeline operations in various engineering fields.

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Most popular questions from this chapter

There is a maximum depth at which a diver can breathe through a snorkel tube (Figure 13.41 ) because as the depth increases, so does the pressure difference, which 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 freshwater. (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.)

Glaucoma. Under normal circumstances, the vitreous humor, a jelly-like substance in the main part of the eye, exerts a pressure of up to \(24 \mathrm{~mm}\) of mercury that maintains the shape of the eye. If blockage of the drainage duct for aqueous humor causes this pressure to increase to about \(50 \mathrm{~mm}\) of mercury, the condition is called glaucoma. What is the increase in the total force (in newtons) on the walls of the eye if the pressure increases from \(24 \mathrm{~mm}\) to \(50 \mathrm{~mm}\) of mercury? We can quite accurately model the eye as a sphere \(2.5 \mathrm{~cm}\) in diameter.

BIO Standing on your head. (a) When you stand on your head, what is the difference in pressure of the blood in your brain compared with the pressure when you stand on your feet if you are \(1.85 \mathrm{~m}\) tall? The density of blood is \(1060 \mathrm{~kg} / \mathrm{m}^{3}\). (b) What effect does the increased pressure have on the blood vessels in your brain?

An open cylindrical tank of acid rests at the edge of a table \(1.4 \mathrm{~m}\) above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is \(75 \mathrm{~cm}\) deep?

Find the gauge pressure in pascals inside a soap bubble \(7.00 \mathrm{~cm}\) in diameter. The surface tension of this soap is \(0.0250 \mathrm{~N} / \mathrm{m}\).

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