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Chapter 11: Problem 71

A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0 \mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400 \mathrm{m} .\) What volume flow rate will keep the pressures in the two horizontal pipes the same?

Short Answer

Expert verified
The volume flow rate is approximately \( 5.72 \times 10^{-3} \ m^3/s. \)

Step by step solution

01

Understanding the Problem

We need to find the volume flow rate that keeps the pressures the same in two horizontal pipes, connected by a vertical section. We will use the principle of conservation of mechanical energy, particularly Bernoulli鈥檚 equation.
02

Using Bernoulli's Equation

Bernoulli鈥檚 equation 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, \( \rho \) is fluid density, \( v \) is velocity, \( g \) is gravity, and \( h \) is height. Since \( P_1 = P_2 \), \( h_1 = 0 \), and \( h_2 = 10 \) m, modify the equation to focus on velocities.
03

Applying Continuity Equation

The continuity equation states \[ A_1v_1 = A_2v_2 \]where \( A \) is the cross-sectional area, \( v \) is velocity. For circular pipes, \( A = \pi r^2 \). Thus: \[ \pi (0.0200)^2 v_1 = \pi (0.0400)^2 v_2. \] Solve for \( v_2 \) in terms of \( v_1 \): \[ v_2 = \frac{1}{4}v_1. \]
04

Solving for Velocities

Substitute \( v_2 = \frac{1}{4}v_1 \) back into the simplified Bernoulli鈥檚 equation: \[ \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho \left( \frac{1}{4}v_1 \right)^2 + \rho g (10.0). \] Simplifying gives: \[ \frac{15}{32}v_1^2 = 10.0 \times 9.81. \] Solve for \( v_1 \), separate and simplify: \[ v_1 = \sqrt{\frac{10 \times 9.81 \times 32}{15}}. \]
05

Calculating Volume Flow Rate

Calculate \( v_1 \) and use it to find the volume flow rate \( Q \) using \( Q = A_1 v_1 \): \[ v_1 = \sqrt{\frac{312.32}{15}} \approx 4.57 \ m/s. \] Then, \[ Q = \pi (0.0200)^2 \times 4.57 = 5.72 \times 10^{-3} \ m^3/s. \]
06

Conclusion

Thus, the volume flow rate required to keep pressures equal in both horizontal pipes is approximately \( 5.72 \times 10^{-3} \ m^3/s. \)

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Key Concepts

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

Bernoulli's Equation
Understanding Bernoulli's Equation is essential in fluid dynamics, especially when analyzing how fluids behave in different pipe systems. Bernoulli's Equation articulates the conservation of mechanical energy in a fluid flow. It essentially states that the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline.

The equation is expressed as: \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]where:
  • \( P \) = pressure energy
  • \( \rho \) = fluid density
  • \( v \) = fluid velocity
  • \( g \) = acceleration due to gravity
  • \( h \) = height above a reference point
In the context of two horizontal pipes with a vertical connection, Bernoulli's Equation helps determine the relationship between velocities and pressures at different points within the system. Since pressure in both pipes is the same in this setup, the equation can be simplified, focusing solely on kinetic and potential energy changes to solve for the required velocities.
Continuity Equation
The Continuity Equation is a principle in fluid dynamics that captures the idea of mass conservation within a flow. It asserts that the mass flow rate of a fluid must remain constant from one cross-section of a pipe to another, as long as there are no leaks or additions.

Mathematically, the Continuity Equation is expressed as: \[ A_1v_1 = A_2v_2 \]where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid.
  • In our problem, the first pipe has a radius of \( 0.0200 \, \text{m} \), and the second pipe has a radius of \( 0.0400 \, \text{m} \).
  • From the equation, the velocity \( v_2 \) is found to be a quarter of \( v_1 \), due to the squared relationship of the radii.
This equation ensures that when fluid moves through varying sections, its speed has to adjust inversely to the change in cross-sectional area to maintain a constant flow rate. For larger cross-sections, the velocity decreases; for smaller ones, it increases.
Volume Flow Rate
Volume Flow Rate is a key concept that encapsulates how much fluid passes through a point in a pipe per unit time. It's essential for calculating efficiencies in systems like plumbing, fuel transport, and fluid transportation networks.

Volume Flow Rate, denoted as \( Q \), is given by:\[ Q = Av \]where:
  • \( A \) = cross-sectional area of the pipe
  • \( v \) = velocity of the fluid
  • The units are typically in \( \text{m}^3/\text{s} \).
To maintain equal pressures in the given pipes, we calculated a specific volume flow rate that balances the described setup. Solving using both Bernoulli鈥檚 Equation and the Continuity Equation, we determined that the required volume flow rate is \( 5.72 \times 10^{-3} \, \text{m}^3/\text{s} \). This result ensures that pressure equilibrium is maintained, confirming an optimal and steady state within the system.

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Most popular questions from this chapter

At a given instant, the blood pressure in the heart is \(1.6 \times 10^{4} \mathrm{Pa}\). If an artery in the brain is \(0.45 \mathrm{m}\) above the heart, what is the pressure in the artery? Ignore any pressure changes due to blood flow.

Three fire hoses are connected to a fire hydrant. Each hose has a radius of \(0.020 \mathrm{m}\). Water enters the hydrant through an underground pipe of radius \(0.080 \mathrm{m} .\) In this pipe the water has a speed of \(3.0 \mathrm{m} / \mathrm{s} .\) (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

Crossing a River. You and your team come to a slow flowing river that you need to cross. The nearest bridge is 20 miles to the north, too dangerous and too far to trek with your group. You explore the area down river and discover an abandoned shed with a stash of 55 -gallon drums (all empty and weighing 35.0 lb each) and a stack of 10 -foot planks: You will build a raft. The six members of your team have a combined weight of 925 pounds (assuming everyone was truthful). You also have a four-wheeler (all-terrain vehicle), which weighs 450 lb, and other gear, which adds another 315 lb. Your simple raft design is as follows: a platform of planks with barrels strapped to the bottom. You estimate that the planks weigh about \(45.0 \mathrm{lb}\) each, and you will need 20 of them to make a platform that can accommodate everything for the one-way trip. You measure the dimensions of the cylindrical drums and find they have diameter \(D=22.5\) inches and height \(h=33.5\) inches (note: not exactly \(^{455}\) gallons"). (a) What is the minimum number of barrels that you will need so that the raft will float when fully loaded? (b) What is the minimum number of barrels you will need if you want the platform to be at least 6 inches above the water when fully loaded?

A hand-pumped water gun is held level at a height of \(0.75 \mathrm{m}\) above the ground and fired. The water stream from the gun hits the ground a horizontal distance of \(7.3 \mathrm{m}\) from the muzzle. Find the gauge pressure of the water gun's reservoir at the instant when the gun is fired. Assume that the speed of the water in the reservoir is zero and that the water flow is steady. Ignore both air resistance and the height difference between the reservoir and the muzzle.

In the human body, blood vessels can dilate, or increase their radii, in response to various stimuli, so that the volume flow rate of the blood increases. Assume that the pressure at either end of a blood vessel, the length of the vessel, and the viscosity of the blood remain the same, and determine the factor \(R_{\text {dilated }} / R_{\text {normal }}\) by which the radius of a vessel must change in order to double the volume flow rate of the blood through the vessel.
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