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Chapter 13: Problem 55

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 approximately \(2.03 \times 10^4 \, \mathrm{Pa}\).

Step by step solution

01

Identify Known Values

From the problem statement, note that the water's speed at the first point is \( v_1 = 2.50 \, \mathrm{m/s} \) and the gauge pressure is \( P_1 = 1.80 \times 10^{4} \, \mathrm{Pa} \). The cross-sectional area at the second point is twice that of the first point, which means \( A_2 = 2A_1 \).
02

Use Continuity Equation

The continuity equation for incompressible, steady flow of a liquid like water is given by: \( A_1v_1 = A_2v_2 \). Since \( A_2 = 2A_1 \), substituting we get: \( A_1v_1 = 2A_1v_2 \). Simplifying, we find the speed of water at the second point: \( v_2 = \frac{v_1}{2} = \frac{2.50 \, \mathrm{m/s}}{2} = 1.25 \, \mathrm{m/s} \).
03

Apply Bernoulli's Equation

Bernoulli's Equation for the flow of an incompressible fluid is given by: \( P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \). Here, \( \rho \) is the density of the water. We are asked to find \( P_2 \), the gauge pressure at the second point. Rearrange this to derive: \( P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2 \).
04

Calculate Gauge Pressure at Second Point

Assuming the density of water \( \rho = 1000 \, \mathrm{kg/m^3} \), substitute known values into the equation: \( P_2 = 1.80 \times 10^4 \, \mathrm{Pa} + \frac{1}{2}(1000 \, \mathrm{kg/m^3})(2.50 \, \mathrm{m/s})^2 - \frac{1}{2}(1000 \, \mathrm{kg/m^3})(1.25 \, \mathrm{m/s})^2 \). Calculate to find: \( P_2 = 1.80 \times 10^4 \, \mathrm{Pa} + 3125 \, \mathrm{Pa} - 781.25 \, \mathrm{Pa} = 2.03 \times 10^4 \, \mathrm{Pa} \).

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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 the behavior of fluids in motion. This area of physics helps us understand how liquids and gases flow in various situations. In our exercise, water is flowing through a pipeline. The movement of fluids is determined by forces like pressure, gravity, and the shape of the path they take. These forces affect the speed and pressure of fluid at different points in the flow.

Considerations in fluid dynamics involve:
  • Speed: How fast the fluid is moving.
  • Pressure: The force exerted by the fluid per unit area.
  • Path: The shape and nature of the container or space through which the fluid is flowing.
In analyzing fluid dynamics, we use equations such as Bernoulli鈥檚 equation and the continuity equation. These mathematical tools help solve problems where fluid properties change along their path.
Continuity Equation
The continuity equation is a fundamental principle that describes the conservation of mass in fluid dynamics. For incompressible fluids, like water in our exercise, the equation ensures that the mass flow rate is constant along the flow path.

The equation itself is: \[ A_1v_1 = A_2v_2 \]
Here, \( A_1 \) and \( A_2 \) are the cross-sectional areas of the pipeline at different points, and \( v_1 \) and \( v_2 \) are the fluid speeds at these points. This means that if the cross-sectional area of the pipe narrows, the speed of the fluid must increase, and vice versa.

In the given exercise, the cross-sectional area at the second point is twice as large as at the first point. Therefore, the speed is halved at the second point, maintaining the product \( A imes v \) constant. This powerful equation simplifies many fluid flow problems by allowing us to link changes in area with changes in velocity.
Gauge Pressure
Gauge pressure is the pressure relative to atmospheric pressure. It is the pressure that acts within the flow system minus the atmospheric pressure. In practical terms, it鈥檚 what you measure with most types of pressure gauges, which don鈥檛 account for atmospheric pressure.

In our exercise, gauge pressure is used to describe the pressure within the pipeline at two different points. Knowing how gauge pressure changes along the flow path is crucial, especially when dealing with equipment and pipelines that are designed to operate at specific pressure ranges.

When solving pressure problems in fluid dynamics, we frequently use Bernoulli鈥檚 equation, which considers pressure, speed, and height. Though in our problem, since the pipe is horizontal, height differences cancel out. This leaves speed and pressure as the main factors linked by the equation.
Incompressible Flow
Incompressible flow is an idealization used in fluid dynamics under which a fluid's density is constant. The water flow in the exercise behaves as if it is incompressible. This assumption is generally valid for liquids like water, contributing to the simplification of analysis and calculations.

When applying Bernoulli鈥檚 equation to incompressible flows, we see:
  • The density (\( \rho \)) is uniform and constant.
  • The changes in kinetic energy and pressure within the flow are easily relatable.
By considering the flow as incompressible, complex changes in density and related calculations are avoided, making it possible to directly apply equations like the continuity equation and Bernoulli's equation efficiently.

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Most popular questions from this chapter

A \(975-\mathrm{kg}\) car has its tires each inflated to "32.0 pounds." (a) What are the absolute and gauge pressures in these tires in \(1 \mathrm{~b} /\) in. \(^{2}, \mathrm{~Pa},\) and atm? (b) If the tires were perfectly round, could the tire pressure exert any force on the pavement? (Assume that the tire walls are flexible so that the pressure exerted by the tire on the pavement equals the air pressure inside the tire.) (c) If you examine a car's tires, it is obvious that there is some flattening at the bottom. What is the total contact area for all four tires of the flattened part of the tires at the pavement?

A small circular hole \(6.00 \mathrm{~mm}\) in diameter is cut in the side of a large water tank. The top of the tank is open to the air. The water is escaping from the hole at a speed of \(10 \mathrm{~m} / \mathrm{s}\). How far below the water surface is the hole?

Blood pressure. Systemic blood pressure is expressed as the ratio of the systolic pressure (when the heart first ejects blood into the arteries) to the diastolic pressure (when the heart is relaxed). Normal systemic blood pressure is \(\frac{120}{80}\). (a) What are the maximum and minimum forces (in newtons) that the blood exerts against each square centimeter of the heart for a person with normal blood pressure? (b) Blood pressure is normally measured on the upper arm at the same height as the heart. Due to therapy for an injury, a patient's upper arm is extended \(30.0 \mathrm{~cm}\) above his heart. In that position, what should be his systemic blood-pressure reading, expressed in the standard way, if he has normal blood pressure? The density of blood is \(1060 \mathrm{~kg} / \mathrm{m}^{3}\).

Exploring Europa's oceans. Europa, a satellite of Jupiter, appears to have an ocean beneath its icy surface. Proposals have been made to send a robotic submarine to Europa to see if there might be life there. There is no atmosphere on Europa, and we shall assume that the surface ice is thin enough that we can ignore its weight and that the oceans are freshwater having the same density as on the earth. The mass and diameter of Europa have been measured to be \(4.78 \times 10^{22} \mathrm{~kg}\) and \(3130 \mathrm{~km},\) respectively. (a) If the submarine in tends to submerge to a depth of \(100 \mathrm{~m}\), what pressure must it be designed to withstand? (b) If you wanted to test this submarine before sending it to Europa, how deep would it have to go in our oceans to experience the same pressure as the pressure at a depth of \(100 \mathrm{~m}\) on Europa?

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 \(15.0 \mathrm{~cm}\) (Figure 13.42 ). (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.
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