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Chapter 12: Problem 51

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 absolute pressure as the water flows through the constriction in the pipe can be calculated with the steps above by using Bernoulli's equation. Remember to convert all units to SI before substituting them into formulas.

Step by step solution

01

Find the Initial Velocity

The initial velocity \(v1\) can be calculated from the volumetric flow rate equation which is \( Q = A1*v1 \) where \( Q = 7200 \, cm^{3} / s \), \( A1 = \pi (4 \, cm)^2 \), and \( v1\) is the initial velocity. So, it can be written as \( v1 = Q / A1 \). Remember to convert the radii and flow rate from cm to m and cm^3/s to m^3/s respectively before substitution in formula.
02

Calculate the Final Velocity

In this case, the flow rate \(Q\) remains constant before and after the constriction. Hence the final velocity \( v2 \) can be calculated using \( Q = A2*v2 \) where \( Q = 7200 \, cm^{3} / s \), \( A2 = \pi (2 \, cm)^2 \), and \( v2 \) is the final velocity. So, it can be written as \( v2 = Q / A2 \).
03

Find the Absolute Pressure

Now we apply Bernoulli's equation which is \( P1 + 0.5*蟻*v1^{2} = P2 + 0.5*蟻*v2^{2} \), where \( P1 = 2.40 \times 10^{5} \, Pa \), \( v1 \) is the initial velocity, \( 蟻 = 1000 \, kg/m^{3} \) is the density of water, and \( v2 \) is the final velocity. \( P2 \) is the pressure at the pipe's constriction. Rearrange the equation and solve for \( P2 \).

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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 a fundamental principle in fluid mechanics that explains the relationship between pressure, velocity, and elevation in a flowing fluid. It is essentially a statement of the conservation of energy for flowing fluids. The equation is represented as:\[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]where:
  • \( P \) is the pressure within the fluid.
  • \( \rho \) is the fluid density.
  • \( v \) is the flow velocity.
  • \( g \) is the gravitational acceleration.
  • \( h \) is the height above a reference point.
In scenarios where we analyze a horizontal pipe, the height \( h \) does not change, so the equation simplifies. For the given problem, using Bernoulli鈥檚 equation allows us to calculate changes in pressure due to differences in velocity at two points along the pipe by neglecting any elevation changes.
Continuity Equation
The Continuity Equation is crucial in fluid dynamics as it expresses the principle of conservation of mass in a flowing system. The equation asserts that the mass flow rate must remain constant from one cross-section of a pipe to another, assuming the fluid is incompressible and there is a steady flow.It is expressed as:\[ A_1 v_1 = A_2 v_2 \]where:
  • \( A_1 \) and \( A_2 \) are the cross-sectional areas at two different points.
  • \( v_1 \) and \( v_2 \) are the fluid velocities at those points.
For an incompressible fluid like water, this equation helps in determining the velocities at different points in a pipe with variable radii. By keeping the flow rate consistent, the equation allows us to solve for one unknown if the others are known. This is how we derive the velocity after the constriction point in the scenario.
Flow Rate
Flow Rate is the volume of fluid that passes through a given surface per unit of time, an important factor in evaluating fluid dynamics. For incompressible fluids, the flow rate \( Q \) is described by:\[ Q = A \times v \]where:
  • \( Q \) is the volumetric flow rate.
  • \( A \) is the cross-sectional area.
  • \( v \) is the fluid velocity.
In the given problem, the flow rate is constant and provided as 7200 cm鲁/s.To maintain this rate in different sections of the pipe, fluid velocity must adjust when the pipe's diameter changes, as per the Continuity Equation. Understanding flow rate helps predict how fast fluids move through constricted areas of a pipe, impacting pressures calculated with Bernoulli鈥檚 Equation.
Pressure Calculations
Pressure Calculations in fluid flow involve determining the pressure changes when movement involves differences in speed and cross-sectional area.Using Bernoulli鈥檚 principle, we find:\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]Given initial pressure \( P_1 \) and velocities \( v_1 \) and \( v_2 \), you can solve for the unknown pressure \( P_2 \) at another point.Steps include:
  • Find initial velocity \( v_1 \) with the flow rate and area \( A_1 \), using \( v_1 = \frac{Q}{A_1} \).
  • Calculate final velocity \( v_2 \) through the constriction with area \( A_2 \), using \( v_2 = \frac{Q}{A_2} \).
  • Substitute these into the simplified Bernoulli鈥檚 equation to find \( P_2 \).
This approach lets you understand how fluid velocity calculations directly influence pressure changes, offering insight into fluid behavior in confined systems.

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Most popular questions from this chapter

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 \(7200 \mathrm{~kg} / \mathrm{m}^{3}\). (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?

Estimate the fraction of your body's total volume that is above the surface of the water when you float in seawater (density \(1030 \mathrm{~kg} / \mathrm{m}^{3}\) ). (a) Use this estimate and your weight to calculate the total volume of your body. (b) What is your average density? How does your average density compare to the density of seawater?

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density \(\left.1050 \mathrm{~kg} / \mathrm{m}^{3}\right)\) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is \(5980 \mathrm{~Pa}\), what is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 12.6 ) of the fluid.

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

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is \(0.070 \mathrm{~m}^{2},\) and the magnitude of the fluid velocity is \(3.50 \mathrm{~m} / \mathrm{s}\). (a) What is the fluid speed at points in the pipe where the cross-sectional area is (a) \(0.105 \mathrm{~m}^{2}\) and (b) \(0.047 \mathrm{~m}^{2}\) ? (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.
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