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At a certain point in a horizontal pipeline, the water's speed is 2.50 m/s and the gauge pressure is 1.80 \(\times\) 10\(^4\) 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 x 10^4 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 is twice of the first point.
02

Apply the Continuity Equation

According to the continuity equation, the product of the cross-sectional area and velocity is constant. Let \( A_1 \) and \( v_1 \) be area and velocity at the first point, and \( A_2 \) and \( v_2 \) at the second point. Since \( A_2 = 2A_1 \), by continuity: \[A_1 \cdot v_1 = A_2 \cdot v_2 \]\[A_1 \cdot 2.50 = 2A_1 \cdot v_2 \]Cancel \( A_1 \) from both sides and solve for \( v_2 \):\[2.50 = 2v_2 \Rightarrow v_2 = \frac{2.50}{2} = 1.25 \text{ m/s}\]
03

Apply Bernoulli's Equation

Bernoulli's equation relates pressures and velocities at two points in a fluid:\[P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\] where \( \rho \) is the fluid density, \( v_1 = 2.50 \text{ m/s} \), and \( v_2 = 1.25 \text{ m/s} \). Solve for \( P_2 \) (the gauge pressure at the second point):\[1.80 \times 10^4 + \frac{1}{2} \cdot \rho \cdot (2.50)^2 = P_2 + \frac{1}{2} \cdot \rho \cdot (1.25)^2\]Simplify and solve for \( P_2 \):\[P_2 = 1.80 \times 10^4 + \frac{1}{2} \cdot \rho \cdot (2.50)^2 - \frac{1}{2} \cdot \rho \cdot (1.25)^2 \]
04

Simplify the Equation

Calculate the terms:\[\frac{1}{2} \cdot \rho \cdot (2.50)^2 - \frac{1}{2} \cdot \rho \cdot (1.25)^2 = \frac{1}{2} \cdot \rho \cdot (6.25 - 1.5625)\]\[= \frac{1}{2} \cdot \rho \cdot 4.6875\]Substitute this back into the expression for \( P_2 \):\[P_2 = 1.80 \times 10^4 + \rho \cdot 2.34375\]With \( \rho = 1000 \text{ kg/m}^3 \) (density of water):\[P_2 = 1.80 \times 10^4 + 1000 \cdot 2.34375 = 1.80 \times 10^4 + 2343.75 = 2.034375 \times 10^4 \text{ Pa}\]
05

Round Off the Answer

The calculated gauge pressure at the second point is \( 2.034375 \times 10^4 \text{ Pa} \), which can be rounded to significant figures traditionally provided by initial conditions; hence:\[P_2 \approx 2.03 \times 10^4 \text{ Pa}\]

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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 stating that the mass flow rate of a fluid must remain constant from one cross-section of a pipe to another. This is expressed in the equation as:\[ A_1 \cdot v_1 = A_2 \cdot v_2 \]where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid. In the context of our exercise, this equation helps us understand how fluid velocity changes as the cross-sectional area of the pipe changes. If the area at the second point is twice that of the first, the fluid's velocity must decrease to half to maintain the same flow rate. By rearranging the continuity equation, we can determine the new velocity, \( v_2 \), which is found to be 1.25 m/s when the initial velocity \( v_1 \) is 2.50 m/s. Utilizing this equation is essential for solving problems where you need to ensure that the flow of fluid stays constant despite changes in pipe dimensions.
fluid dynamics
Fluid dynamics is the study of fluids (liquids and gases) in motion. It revolves around principles like the Continuity Equation and Bernoulli's Equation, both key to understanding how fluids behave in different scenarios. These principles allow us to predict and calculate the behavior of fluids in various applications, from natural phenomena to engineering systems.
In our horizontal pipeline example, fluid dynamics is utilized to model how water flows through the pipe. These calculations give insights into important parameters such as pressure and velocity at different points. It involves applying laws of physics to ensure efficient and accurate predictions of fluid behavior. Fluid dynamics also enables us to solve complex problems in design and analysis of pipelines, ensuring desired performance in systems like water supply networks. Understanding these concepts provides a solid foundation for applying physical laws to practical situations.
gauge pressure
Gauge pressure refers to the pressure of a fluid in a system above the atmospheric pressure. It is different from absolute pressure, as it does not consider atmospheric pressure in its measurement. Instead, it measures the pressure relative to atmospheric pressure.In the context of our pipeline problem, gauge pressure allows us to calculate changes in pressure due to the fluid's velocity as it moves through sections of the pipeline with different cross-sectional areas. By using Bernoulli's Equation, we can relate the gauge pressure at two points in the pipeline. Bernoulli's Equation is:\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]Here, \( P_1 \) and \( P_2 \) are the gauge pressures at points 1 and 2, while \( \rho \) represents fluid density. Through solving this equation, we find the gauge pressure at the second point. This is important for ensuring safe and efficient operations in systems like pipelines or any fluid-handling equipment.
pipe flow
Pipe flow is the study of fluid movement within a closed conduit, typically a pipe. It is crucial in numerous applications ranging from household plumbing to extensive industrial and municipal water systems.
Pipe flow analysis helps determine how fluids travel through pipes, focusing on parameters like pressure, velocity, and flow rate. Bernoulli's and the Continuity Equations crucially aid in such analyses by providing a mathematical framework for predicting fluid behavior as conditions change along the pipe's length.
In the exercise, understanding how pipe diameter affects flow dynamics is essential. When the cross-sectional area doubles, the velocity decreases to maintain constant flow, leading to changes in pressure as described by Bernoulli's Equation. This emphasizes the interconnectedness of velocity, pressure, and area in pipe flows and underscores the importance of ensuring proper pipe design and operation for efficient fluid transport.

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Most popular questions from this chapter

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 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 m\(^3\)/s? (b) At a second point in the pipe the water speed is 3.80 m/s. What is the radius of the pipe at this point?

A rock has mass 1.80 kg. When the rock is suspended from the lower end of a string and totally immersed in water, the tension in the string is 12.8 N. What is the smallest density of a liquid in which the rock will float?

A barrel contains a 0.120-m layer of oil floating on water that is 0.250 m deep. The density of the oil is 600 kg/m\(^3\). (a) What is the gauge pressure at the oil\(-\)water interface? (b) What is the gauge pressure at the bottom of the barrel?

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-mhigh, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-mdeep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg/m\(^3\). If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (\(Hint:\) Consider the outward force on a circular ring of the tank wall of width \({dy}\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

Ballooning on Mars. It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 kg/m\(^3\) (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 5.00 g. We inflate them with a very light gas whose mass we can ignore. (a) What should be the radius and mass of these balloons so they just hover above the surface of Mars? (b) If we released one of the balloons from part (a) on earth, where the atmospheric density is 1.20 kg/m\(^3\), what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down? (c) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

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