/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A golf course sprinkler system d... [FREE SOLUTION] | 91Ó°ÊÓ

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A golf course sprinkler system discharges water from a horizontal pipe at the rate of \(7200 \mathrm{~cm}^{3} / \mathrm{s}\). At one point in the pipe, where the radius is \(4.00 \mathrm{~cm},\) the water's absolute pressure is \(2.40 \times 10^{5} \mathrm{~Pa}\). At a second point in the pipe, the water passes through a constriction where the radius is \(2.00 \mathrm{~cm} .\) What is the water's absolute pressure as it flows through this constriction?

Short Answer

Expert verified
The water's absolute pressure at the constriction is approximately 223327 Pa.

Step by step solution

01

Calculate the cross-sectional areas

First, calculate the cross-sectional areas at both points in the pipe using the formula for the area of a circle: \[ A = \pi r^2 \]For the first point with radius \( r_1 = 4.00 \text{ cm} = 0.04 \text{ m} \):\[ A_1 = \pi (0.04)^2 = 0.005024 \text{ m}^2 \]For the second point with radius \( r_2 = 2.00 \text{ cm} = 0.02 \text{ m} \):\[ A_2 = \pi (0.02)^2 = 0.001256 \text{ m}^2 \]
02

Apply the Continuity Equation

The continuity equation states that the volumetric flow rate must remain constant throughout the pipe:\[ A_1 v_1 = A_2 v_2 = Q \]We are given \(Q = 7200 \text{ cm}^3/\text{s} = 0.0072 \text{ m}^3/\text{s}\). Substitute \(A_1\) to find \(v_1\):\[ v_1 = \frac{Q}{A_1} = \frac{0.0072}{0.005024} = 1.433 \text{ m/s} \]Now find \(v_2\):\[ v_2 = \frac{Q}{A_2} = \frac{0.0072}{0.001256} = 5.731 \text{ m/s} \]
03

Bernoulli's Equation for Pressure Determination

Bernoulli's equation states:\[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \]Assume the pipe is horizontal, so \(h_1 = h_2\) and simplify:\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]Substitute the known values. Given \(P_1 = 2.40 \times 10^5 \text{ Pa}\) and water density \(\rho = 1000 \text{ kg/m}^3\):\[ 2.40 \times 10^5 + \frac{1}{2} \times 1000 \times (1.433)^2 = P_2 + \frac{1}{2} \times 1000 \times (5.731)^2 \]
04

Solve for P2

Rearrange Bernoulli's equation to solve for \(P_2\):\[ P_2 = 2.40 \times 10^5 + \frac{1}{2} \times 1000 \times (1.433)^2 - \frac{1}{2} \times 1000 \times (5.731)^2 \]Calculate:\[ P_2 = 2.40 \times 10^5 + 1026.145 - 16433.395 \]\[ P_2 = 2.40593706 \times 10^5 - 16407.25 \]\[ P_2 = 223327.36 \text{ Pa} \]
05

Validate and Interpret Result

The calculated pressure at the constriction is lower than the pressure at the wider part, which aligns with Bernoulli's principle: as fluid velocity increases, pressure decreases. This validates our calculations.

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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, asserting that the mass flow rate of a fluid must remain constant from one cross-section of a pipe to another. This concept is particularly useful in understanding how fluids behave in varying pipe sizes.
For the exercise involving the sprinkler system, we apply the equation: \[ A_1 v_1 = A_2 v_2 = Q \] Where:
  • \(A_1\) and \(A_2\) are the cross-sectional areas at different points in the pipe
  • \(v_1\) and \(v_2\) are the velocities of the water at those points
  • \(Q\) is the flow rate, which is constant
In the example, it illustrates the principle that when the pipe diameter decreases, the velocity of water must increase to keep the flow rate unchanged. This automatic adjustment in velocity ensures that the volume of water passing through the constricted area remains consistent with that flowing through the wider section.
Bernoulli's Equation
Bernoulli's Equation provides a way to understand the energy distribution within a flowing fluid and is crucial when calculating pressure differences within a pipe. It can be written as: \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \] For horizontal flow, the gravitational term \(\rho gh\) cancels out, simplifying to: \[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \] Where:
  • \(P_1\) and \(P_2\) are the pressures at the two points
  • \(v_1\) and \(v_2\) are the fluid velocities
  • \(\rho\) is the fluid's density
In the example, using the velocities computed from the Continuity Equation helps us solve for the unknown pressure \(P_2\) at the constriction. This principle shows that as the fluid speeds up in the narrowed part of the pipe, the pressure decreases, aligning with the conservation of energy.
Flow Rate
Flow Rate refers to the volume of fluid which passes a point in the system per unit time. It's an essential parameter in fluid dynamics and is often expressed in units like cubic meters per second (m\(^3\)/s).Given in the exercise as \(7200 \text{ cm}^3/\text{s}\), it can also be expressed in terms of \(\text{m}^3/\text{s}\) as \(0.0072 \text{ m}^3/\text{s}\). Understanding flow rate is key to managing fluid systems, especially when dealing with different pipe sizes, as it affects how quickly and efficiently a fluid can be transported or dispersed.The constant flow rate helps assure us that even as the pipe width changes, the amount of water flowing through the system remains unchanged. This underlying principle is part of what ensures systems like sprinkler networks work reliably and predictably under various configurations.
Pressure Calculation
Pressure Calculation in the context of fluid flow involves calculating the force exerted by the fluid per unit area. It integrates concepts like velocity and flow rate to provide insights into the operational dynamics of fluid systems. Using data from the exercise, we determined pressures using Bernoulli's Equation, emphasizing that with a higher velocity at a narrower pipe section, the resultant pressure drops. This intricacy stems from the balancing of kinetic and potential energy within the system. The practical significance is underscored in applications like sprinkler systems where predicting pressure is vital for ensuring water reaches desired destinations effectively. Correct pressure calculations ensure all parts of the system receive adequate water supply without causing damage from too high pressure.

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Most popular questions from this chapter

Water discharges from a horizontal cylindrical pipe at the rate of \(465 \mathrm{~cm}^{3} / \mathrm{s}\). At a point in the pipe where the radius is \(2.05 \mathrm{~cm},\) the absolute pressure is \(1.60 \times 10^{5} \mathrm{~Pa}\). What is the pipe's radius at a constriction if the pressure there is reduced to \(1.20 \times 10^{5} \mathrm{~Pa}\) ?

A hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of \(0.650 \mathrm{~m}^{3}\) and the tension in the cord is \(900 \mathrm{~N}\). (a) Calculate the buoyant force exerted by the water on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

In seawater, a life preserver with a volume of \(0.0400 \mathrm{~m}^{3}\) will support a \(75.0 \mathrm{~kg}\) person (average density \(980 \mathrm{~kg} / \mathrm{m}^{3}\) ), with \(20 \%\) of the person's volume above water when the life preserver is fully sub-merged. What is the density of the material of the life preserver?

You are designing a manned submersible to withstand the pressure of seawater at the bottom of the Mariana Trench, which is one of the deepest parts of the earth's oceans. The bottom of the trench is at a depth of \(10,900 \mathrm{~m}\). (a) What is the gauge pressure at this depth? (You can ignore the small changes in the density of the water with depth.) (b) What is the inward force on a circular glass window \(5.0 \mathrm{~cm}\) in diameter due to the water outside? (c) If the internal pressure is 1 atm, what outward force does this produce on the window? (You may ignore the small variation in pressure over the surface of the window.)

Elephants under pressure. An elephant can swim or walk with its chest several meters underwater while the animal breathes through its trunk, which remains above the water surface and acts like a snorkel. The elephant's tissues are at an increased pressure due to the surrounding water, but the lungs are at atmospheric pressure because they are connected to the air through the trunk. Figure 13.47 shows the gauge pressures in an elephant's lungs and abdomen when the elephant's chest is submerged to a particular depth in a lake. In this situation, the elephant's diaphragm, which separates the lungs from the abdomen, must maintain the difference in pressure between the lungs and the abdomen. The diaphragm of an elephant is typically \(3.0 \mathrm{~cm}\) thick and \(120 \mathrm{~cm}\) in diameter. If the elephant were to snorkel in saltwater, which is more dense than freshwater, would the maximum depth at which it could snorkel be different from the depth in freshwater? A. Yes; that depth would increase because there is less pressure at a given depth in saltwater than in freshwater. B. Yes; that depth would decrease because there is greater pressure at a given depth in saltwater than in freshwater. C. No, because pressure differences within the submerged elephant depend only on the density of air, not on the density of the water. D. No, because the buoyant force on the elephant would be the same in both cases.

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