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Chapter 14: Problem 36

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.

Short Answer

Expert verified
(a) 2.33 m/s; (b) 5.21 m/s; (c) 882 m鲁.

Step by step solution

01

Understand the Problem

We are dealing with a situation where water is flowing through a pipe with varying cross-sectional areas. The given variables include the cross-sectional area and the velocity of the fluid at different points. Our goal is to find the fluid velocity at different cross-sectional areas and the total volume of water discharged in an hour.
02

Apply the Continuity Equation

For incompressible fluid flow, the principle of conservation of mass is used. The continuity equation states that the product of the cross-sectional area and the fluid velocity at one section must equal the product at any other section: \[ A_1 v_1 = A_2 v_2 \]Given:- At point 1: \( A_1 = 0.070 \, \text{m}^2 \), \( v_1 = 3.50 \, \text{m/s} \)- At point 2 (Unknown speed for \( A_2 = 0.105 \, \text{m}^2 \)):Solve for \( v_2 \):\[ v_2 = \frac{A_1 \cdot v_1}{A_2} = \frac{0.070 \, \text{m}^2 \times 3.50 \, \text{m/s}}{0.105 \, \text{m}^2} \]
03

Calculate Fluid Velocity for 0.105 m虏

Calculate \( v_2 \) using the values:\[ v_2 = \frac{0.070 \, \text{m}^2 \times 3.50 \, \text{m/s}}{0.105 \, \text{m}^2} = 2.33 \, \text{m/s} \]This is the fluid speed at a cross-sectional area of 0.105 \( \text{m}^2 \).
04

Calculate Fluid Velocity for 0.047 m虏

Now apply the same principle to a different section with \( A_2 = 0.047 \, \text{m}^2 \):\[ v_2 = \frac{0.070 \, \text{m}^2 \times 3.50 \, \text{m/s}}{0.047 \, \text{m}^2} \]Calculate:\[ v_2 \approx 5.21 \, \text{m/s} \]
05

Calculate Total Discharge Volume

To find the total volume discharged in one hour, use the formula:\[ \text{Volume} = A_1 \times v_1 \times t \]where \( t = 1 \, \text{hour} = 3600 \, \text{seconds} \).Calculate: \[ \text{Volume} = 0.070 \, \text{m}^2 \times 3.50 \, \text{m/s} \times 3600 \, \text{s} \]\[ \text{Volume} = 882 \, \text{m}^3 \]

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Key Concepts

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

Incompressible Fluid Flow
Incompressible fluid flow is a fundamental concept in fluid dynamics. It relates to fluids that do not change in density when in motion. Water, as used in our exercise, is a common example of an incompressible fluid. This simplification means that the volume of water entering any section of the pipe is the same as the volume exiting that section, assuming no leakage.

This assumption is crucial when analyzing fluid flow in systems like pipes because it allows us to apply the continuity equation. Whether in small diameter pipes or larger ones, as long as water remains incompressible, this understanding of flow dynamics simplifies calculations and keeps the math manageable.

When dealing with varying cross-sectional areas in pipes, the incompressibility implies that any change in velocity must compensate for a change in area. This interaction between velocity and cross-sectional area is the backbone of how we calculate changes in water flow within different parts of the pipe.
Conservation of Mass
The principle of conservation of mass is a core concept that ties directly into our exercise about fluid flow. The law states that mass in a closed system must remain constant over time. Translated into fluid dynamics, this means that as long as the fluid is incompressible, the mass flowing into an area equals the mass flowing out.

This law is essential for deriving the continuity equation:
  • In a pipe with different sections, the product of cross-sectional area and velocity at one section equals the product at another.
  • Mathematically expressed as: \( A_1 v_1 = A_2 v_2 \)
This equation forms the backbone of our calculations. It helps determine the velocity of fluid in different sections of a pipe as shown by applying the given values in the exercise. Conserving mass ensures that when the area of a pipe segment increases or decreases, fluid velocity adjusts to keep the mass flow rate steady.
Pipe Flow Calculations
Pipe flow calculations involve using the continuity equation to solve for unknown variables like fluid velocity and volume of flow, given certain conditions. Considering a pipe with variable cross-sections, these calculations help to determine how quickly the fluid is moving at different parts of the pipe.

To solve such problems, you can follow these steps:
  • Identify the given conditions. In our exercise, we have initial conditions at point 1: area and velocity.
  • Use the continuity equation \( A_1 v_1 = A_2 v_2 \) to find velocities at other points when area changes.
  • Calculate the velocity for different areas using proper substitution.
  • Finally, find the volume flow rate to determine how much fluid is discharged over a period of time. This is found by multiplying the initial area and velocity by time.
In our solved exercise, these calculations revealed velocities at \(0.105 \, ext{m}^2\) and \(0.047 \, ext{m}^2\) sections, as well as the total volume discharged from the pipe. Pipe flow calculations thus serve as a comprehensive method of understanding fluid behavior in varied conditions.

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Most popular questions from this chapter

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 \(\mathrm{Pa}\) . (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

The densities of air, helium, and hydrogen (at \(p=1.0\) atm and \(T=20^{\circ} \mathrm{C} )\) are \(1.20 \mathrm{kg} / \mathrm{m}^{3}, 0.166 \mathrm{kg} / \mathrm{m}^{3},\) and 0.0899 \(\mathrm{kg} / \mathrm{m}^{3}\) , respectively. (a) What is the volume in cubic meters displaced by a hydrogen-filled airship that has a total "Iift" of 120 \(\mathrm{kN}\) ? (The "lift" is the amount by which the buoyant force exceeds the weight of the gas that fills the airship. \((b)\) What would be the "lift" if helium were used instead of hydrogen? In view of your answer, why is helium used in modern airships like advertising blimps?

A swimming pool is 5.0 \(\mathrm{m}\) long, 4.0 \(\mathrm{m}\) wide, and 3.0 \(\mathrm{m}\) deep. Compute the force exerted by the water against (a) the bottom; and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h\) , and integrate this over the end of the pool.) Do not include the force due to air pressure.

You cast some metal of density \(\rho_{\mathrm{m}}\) in a mold, but you are worried that there might be cavities within the casting. You measure the weight of the casting to be \(w\) , and the buoyant force when it is completely surrounded by water to be \(B\) . (a) Show that \(V_{0}=\) \(B /\left(\rho_{\text { water }} g\right)-w /\left(\rho_{\text { m }} g\right)\) is the total volume of any enclosed cavities. (b) If your metal is copper, the casting's weight is 156 \(\mathrm{N}\) , and the buoyant force is 20 \(\mathrm{N}\) , what is the total volume of any enclosed cavities in your casting? What fraction is this of the total volume of the casting?

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?
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