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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 \(\mathrm{m} .\) What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.20 \(\mathrm{m}^{3} / \mathrm{s} ?(\mathrm{b})\) At a second point in the pipe the water speed is 3.80 \(\mathrm{m} / \mathrm{s}\) . What is the radius of the pipe at this point?

Short Answer

Expert verified
(a) 16.98 m/s; (b) 0.317 m.

Step by step solution

01

Understand the Problem

You are given a pipe with varying cross-sectional areas. Water flows through it at a constant rate. First, you need to find the speed of water where the radius is 0.150 m. Then, at another point where the water speed is 3.80 m/s, you need to find the radius.
02

Apply Continuity Equation for Part (a)

The continuity equation for incompressible fluid flow states that the volumetric flow rate remains constant. Thus, we use the formula \( Q = A \cdot v \), where \( Q \) is the volumetric flow rate, \( A \) is the cross-sectional area, and \( v \) is the velocity. Here, \( Q = 1.20 \, \text{m}^3/\text{s}\) and \( A = \pi \cdot (0.150)^2 \). Solve for \( v \).
03

Calculate Cross-sectional Area

Calculate the area at the first point using the formula for the area of a circle: \( A = \pi r^2 \). Substitute \( r = 0.150 \, \text{m} \) to find \( A \). \[ A = \pi \times (0.150)^2 = 0.0707 \, \text{m}^2 \]
04

Calculate Velocity at First Point

Using the continuity equation \( Q = A \cdot v \), and the known values of \( Q \) and \( A \), solve for \( v \):\[ v = \frac{Q}{A} = \frac{1.20}{0.0707} \approx 16.98 \, \text{m/s} \]
05

Apply Continuity Equation for Part (b)

Now, you have to find the radius when the velocity is given as 3.80 m/s. Use the same continuity equation: \( Q = A \cdot v \). Use the known values of \( Q \) and \( v \) to find \( A \).
06

Solve for Cross-sectional Area at Second Point

The volumetric flow rate is still 1.20 m^3/s, so using \( v = 3.80 \, \text{m/s} \), solve for \( A \):\[ A = \frac{Q}{v} = \frac{1.20}{3.80} \approx 0.316 \, \text{m}^2 \]
07

Calculate Radius at the Second Point

To find the radius at the second point, use the formula for area \( A = \pi r^2 \). Rearranging gives \( r = \sqrt{\frac{A}{\pi}} \). Substitute \( A = 0.316 \, \text{m}^2 \) to calculate \( r \):\[ r = \sqrt{\frac{0.316}{\pi}} \approx 0.317 \, \text{m} \]

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Key Concepts

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

Volumetric Flow Rate
When dealing with fluid dynamics, one important concept is the volumetric flow rate. It tells us how much fluid passes through a section of a pipe in a given time. This parameter is essential for understanding how fluids behave in various piping systems. The volumetric flow rate (\( Q \)) is defined by \( Q = A \cdot v \), where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid. Think of this as the "flow speed" of the fluid system.
To visualize, imagine filling a bucket with water. The volumetric flow rate helps you determine how quickly the bucket fills up. For example, if the flow rate is high, the bucket fills up fast. Conversely, if the flow rate is low, it takes longer. This straightforward concept is fundamental when analyzing how fluids travel through pipes. By knowing one aspect (e.g., speed of flow or shape of the pipe), we can compute the other parameters if the flow rate remains unchanged.
The volumetric flow rate being steady implies that despite changes in the shape or size of the pipe, the volume of fluid passing through per second remains constant. This forms the core of the continuity equation in incompressible fluid dynamics.
Cross-sectional Area Calculation
The cross-sectional area of a pipe is an essential part of determining how quickly fluid can flow through it. When calculating this, particularly for circular pipes, we use the formula \( A = \pi r^2 \). Here, \( r \) represents the radius of the pipe, and \( \pi \) (approximately 3.14159) is a mathematical constant. This formula arises from basic geometry.
Visualize the pipe as a stack of circular slices. Each slice having a radius \( r \), the area of these circular slices directly influences how much fluid the pipe can hold. For example, if the radius is 0.150 m, the area will be \( A = \pi \times (0.150)^2 = 0.0707 \, \text{m}^2 \). A larger radius results in a larger area, enabling more fluid to flow, assuming that other factors like velocity remain constant.
By using cross-sectional area calculations with the continuity equation, we can predict changes in velocity and flow rate as the pipe's radius changes. This concept is crucial for designing efficient piping systems in engineering and environmental sciences.
Incompressible Fluid Flow
Incompressible fluid flow is a principle that greatly simplifies fluid dynamics problems. It assumes that the density of the fluid remains constant throughout its course in a system. This means the fluid does not compress or expand, making calculations more straightforward. Water is a common example of a nearly incompressible fluid, as its density changes very little under usual conditions.
By applying the idea of incompressibility, fluid flow problems utilize the continuity equation \( Q = A \cdot v \), where \( Q \), the volumetric flow rate, remains constant. This allows us to interrelate changes in velocity and cross-sectional area with ease, as calculations assume the same amount of fluid material passes through different sections of a pipe each second.
The incompressible fluid flow assumption is indispensable when calculating the behavior of fluids in pipes of varying dimensions. It permits engineers to build models that reliably predict how changes in a system, like varying pipe diameters, will affect the velocity and flow rate. These predictions are vital for ensuring systems operate efficiently for purposes ranging from plumbing to irrigation and more complex industrial processes.

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Most popular questions from this chapter

At one point in a pipeline the water's speed is 3.00 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(5.00 \times 10^{4} \mathrm{Pa}\) . Find the gauge pressure at a second point in the line, 11.0 \(\mathrm{m}\) lower than the first, if the pipe diameter at the second point is twice that at the first.

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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.

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