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Chapter 14: Problem 42

At one point in a pipeline the water's speed is 3.00 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(5.00 \times 10^{4} \mathrm{Pa}\) . 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 approximately 162,129 Pa.

Step by step solution

01

Understand Bernoulli's Equation

Bernoulli's equation relates the speed, pressure, and height of a fluid at two points along a streamline. It is given by: \[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \], where \( P \) is pressure, \( \rho \) the fluid density, \( v \) the fluid velocity, and \( h \) the height above a reference level.
02

Apply Equation to Given Points

At point 1, we have speed \( v_1 = 3.00 \; \text{m/s} \), gauge pressure \( P_1 = 5.00 \times 10^4 \; \text{Pa} \), and height \( h_1 = 11.0 \; \text{m} \). At point 2, the height is \( h_2 = 0 \; \text{m}\) since it is 11.0 m lower, and the pipe's diameter being twice implies \( v_2 = \frac{v_1}{4} = \frac{3.00}{4} = 0.75 \; \text{m/s} \), due to conservation of mass in the continuity equation \( A_1v_1 = A_2v_2 \) and \( A_2 = 4A_1 \).
03

Simplify Bernoulli's Equation for Known Values

Since the only unknown is \( P_2 \), rearrange Bernoulli's equation: \[ P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 + \rho g h_1 \]. Substitute \( g = 9.81 \; \text{m/s}^2 \) and \( \rho = 1000 \; \text{kg/m}^3 \) for water.
04

Substitute Values into the Equation

Insert the known values: \[ P_2 = 5.00 \times 10^4 + \frac{1}{2} \times 1000 \times (3.00)^2 - \frac{1}{2} \times 1000 \times (0.75)^2 + 1000 \times 9.81 \times 11.0 \]. Calculate each term to find \( P_2 \).
05

Calculate the Pressure at Point 2

First, calculate each term: \( \frac{1}{2} \times 1000 \times 9.00 = 4500 \), \( \frac{1}{2} \times 1000 \times 0.5625 = 281.25 \), and \( 1000 \times 9.81 \times 11.0 = 107910 \). Substitute these into the equation to get: \[ P_2 = 50000 + 4500 - 281.25 + 107910 = 162128.75 \; \text{Pa} \].

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is the study of how fluids (liquids and gases) move and behave. In this exercise, we're looking at water flowing through different sections of a pipeline. Understanding fluid dynamics helps us predict and calculate variables like speed and pressure at different points. This area of science is crucial for many applications, from designing pipelines to predicting weather patterns.
This problem is solved using principles of Bernoulli's equation and the continuity equation, which are fundamental to understanding fluid dynamics. As we break down the equation, notice how the speed and height of the fluid impact pressure changes. By understanding these relationships, it becomes easier to visualize how fluids move under different conditions and constraints.
Pressure Calculation
Pressure in fluids reflects the force exerted by the fluid per unit area. In pipelines or any enclosed fluid system, pressure calculations are essential. Knowing the pressure at one point allows us to find the pressure at another if we know the fluid speed, density, pipe diameter, and height difference.
Bernoulli's equation connects pressure with fluid speed and height, demonstrating the energy conservation within the fluid. This exercise involves starting with a known gauge pressure and calculating an unknown by manipulating Bernoulli's equation. Gauge pressure is the pressure relative to atmospheric pressure, often used for practical measurements because it excludes constant atmospheric pressure. By substituting known values into Bernoulli's equation, we can simplify and solve for the unknown pressure at another point in the pipeline.
Continuity Equation
The continuity equation is a cornerstone of fluid dynamics. It states that for an incompressible fluid (like water) flowing through a pipeline, the flow rate must remain constant. This is expressed as: \(A_1v_1 = A_2v_2\), where \(A\) represents the cross-sectional area and \(v\) the fluid velocity.
In this exercise, the pipe's diameter changes from one point to another, affecting the fluid velocity. If the diameter at the second point is twice the first, the cross-sectional area is four times larger (\(A_2 = 4A_1\)). Since the flow rate must be constant, the velocity at the second point \(v_2\) becomes a quarter of the velocity at the first point \(v_1\). This interconnection highlights why the continuity equation is essential for calculating how changes in the pipeline diameter affect fluid speed and, subsequently, pressure.

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Most popular questions from this chapter

Assume that crude oil from a supertanker has density 750 \(\mathrm{kg} / \mathrm{m}^{3}\) . The tanker runs aground on a sandbar. To refloat the tanker, its oil cargo is pumped out into steel barrels, each of which has a mass of 15.0 \(\mathrm{kg}\) when empty and holds 0.120 \(\mathrm{m}^{3}\) of oil. You can ignore the volume occupied by the steel from which the barrel is made. (a) If a salvage worker accidentally drops a filled, sealed barrel overboard, will it float or sink in the seawater? (b) If the barrel floats, what fraction of its volume will be above the water surface? If it sinks, what minimum tension would have to be exerted by a rope to haul the barrel up from the ocean floor? (c) Repeat parts (a) and (b) if the density of the oil is 910 \(\mathrm{kg} / \mathrm{m}^{3}\) and the mass of each empty barrel is 32.0 \(\mathrm{kg}\) .

A barge is in a rectangular lock on a freshwater river. The lock is 60.0 \(\mathrm{m}\) long and 20.0 \(\mathrm{m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{N}\) load of scrap metal is put onto the barge. The metal has density \(9000 \mathrm{kg} / \mathrm{m}^{3} .\) (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

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) What is the average density of the sun? (b) What is the average density of a neutron star hat has the same mass as the sun but a radius of only 20.0 \(\mathrm{km}\) ?

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?
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