/*! 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 8 24.8 You are interested in measu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 8

24.8 You are interested in measuring the fluid velocity in a narrow rectangular open channel carrying petroleum waste between locations in an oil refinery. You know that, because of bottom friction, the velocity varies with depth in the channel. If your technician has time to perform only two velocity measurements, at what depths would you take them to obtain the best estimate of the average velocity? State your recommendation in terms of the percent of total depth \(d\) measured from the fluid surface. For example, measuring at the top would be \(0 \% d\), whereas at the very bottom would be \(100 \%\).

Short Answer

Expert verified
The two velocity measurements should be taken at depths equal to \(33.33\%\) and \(66.67\%\) of the total depth \(d\) in the narrow rectangular open channel. These depths provide the best estimate of the average velocity for fully developed open channel flow.

Step by step solution

01

1. Identify the total depth of the channel

The total depth of the channel is given by \(d\). The fluid surface is at \(0\%d\), and the bottom of the channel is at \(100\%d\).
02

2. Determine the best theoretical depths for measuring velocity

Since the technician has time for only two measurements, we would ideally like to measure velocities at two depths that will best represent the average velocity across the depth of the channel. A well-known theorem in fluid mechanics, Torricelli's law, shows that the mean velocity in an open channel flow can be approximated by the average of the velocities at one-third and two-thirds of the channel depth for fully developed open channel flow. Therefore, it would be best to take velocity measurements at \(d_1 = \frac{1}{3}d\) and \(d_2 = \frac{2}{3}d\).
03

3. Convert depths to percentages

To state our recommendation in terms of percent of total depth, we need to convert \(d_1\) and \(d_2\) to percentages of the total depth \(d\). To do this, we divide each depth by the total depth and multiply by 100: - \(d_1\) percentage: \(\frac{1}{3}d \cdot \frac{100}{d} = 33.33\%d\) - \(d_2\) percentage: \(\frac{2}{3}d \cdot \frac{100}{d} = 66.67\%d\)
04

4. State the recommendation

We recommend taking the two velocity measurements at depths equal to \(33.33\%\) and \(66.67\%\) of the total depth \(d\). This will provide the best estimate of the average velocity in the narrow rectangular open channel carrying petroleum waste.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Torricelli's Law
Torricelli's law is a fundamental principle in fluid mechanics that applies to fluids in motion, specific to scenarios where the fluid exits through an orifice or a hole in a container. Essentially, this law relates the speed of the efflux of the fluid to the height of the fluid in the container. The law states that the velocity (\( v \) of the fluid flowing from the hole at a depth (\( h \) below the surface of the liquid is given by the equation: \[ v = \sqrt{2gh} \] where \(g\) is the acceleration due to gravity.

For the exercise in question, Torricelli's law can be creatively applied to approximate the average velocity of the flow in an open channel. This law supports the assumption that the flow velocity at different depths can be averaged to estimate the velocity of the entire fluid column, lending theoretical support for taking measurements at specific fractions of the total depth.
Open Channel Flow
Open channel flow refers to the fluid flow that has a free surface exposed to the atmosphere, as opposed to being entirely enclosed by boundaries, such as in a pipe. Examples of open channel flow include rivers, canals, and in this case, a channel carrying petroleum waste in an oil refinery. In fluid mechanics, understanding the behavior of open channel flow is paramount, especially when determining flow velocities, depths, and rates.

The principle used in the textbook solution stems from the fact that velocity in an open channel flow is not uniform across the depth. Typically, due to friction with the channel bottom and sides, maximum flow velocity occurs just below the surface, and minimum velocity is at the bottom. Averaging the velocities at specific strategic depths, such as at one-third and two-thirds, provides a practical means to estimate the average velocity for the entire depth of the channel.
Fluid Mechanics
Fluid mechanics is the study of fluids (liquids and gases) and the forces acting on them. It is a branch of physics that is incredibly relevant to various fields, including engineering, meteorology, oceanography, and even medicine. Two critical areas within fluid mechanics are the study of fluid statics (fluids at rest) and fluid dynamics (fluid movement).

In addressing problems such as the one proposed, fluid dynamics principles are employed to understand and predict how fluids behave when they move. By applying theoretical principles such as Torricelli's law, and considering the specific characteristics of the flow, such as being an open channel flow, fluid mechanics allows us to derive useful solutions and recommendations for practical scenarios like measuring the average velocity in an industrial application.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Compute work as described in Sec. \(24.4,\) but use the following equations for \(F(x)\) and \(\theta(x)\) $$\begin{array}{l} F(x)=1.6 x-0.045 x^{2} \\ \theta(x)=-0.00055 x^{3}+0.0123 x^{2}+0.13 x \end{array}$$ The force is in newtons and the angle is in radians. Perform the integration from \(x=0\) to \(30 \mathrm{m}\)

Soft tissue follows an exponential deformation behavior in uniaxial tension while it is in the physiologic or normal range of elongation. This can be expressed as $$\sigma=\frac{E_{o}}{a}\left(e^{a \varepsilon}-1\right)$$ where \(\sigma=\) stress, \(\varepsilon=\operatorname{strain},\) and \(E_{o}\) and \(a\) are material constants that are determined experimentally. To evaluate the two material constants, the above equation is differentiated with respect to \(\varepsilon\) which is a fundamental relationship for soft tissue $$\frac{d \sigma}{d \varepsilon}=E_{o}+a \sigma$$ To evaluate \(E_{o}\) and \(a\), stress-strain data is used to plot \(d \sigma / d \varepsilon\) versus \(\sigma\) and the slope and intercept of this plot are the two material constants, respectively. The table contains stress-strain data for heart chordae tendineae (small tendons use to hold heart valves closed during contraction of the heart muscle). This is data from loading the tissue; different curves are produced on unloading. $$\begin{array}{l|cccccccccccc} \sigma \times 10^{3} \mathrm{N} / \mathrm{m}^{2} & 87.8 & 96.6 & 176 & 263 & 350 & 569 & 833 & 1227 & 1623 & 2105 & 2677 & 3378 & 4257 \\ \hline \varepsilon \times 10^{-3} \mathrm{m} / \mathrm{m} & 153 & 198 & 270 & 320 & 355 & 410 & 460 & 512 & 562 & 614 & 664 & 716 & 766 \end{array}$$ (a) Calculate the derivative \(d \sigma / d \varepsilon\) using finite differences that are second-order accurate. Plot the data and eliminate the data points near the zero points that appear not to follow the straight-line relationship. The error in this data comes from the inability of the instrumentation to read the small values in this region. Perform a regression analysis of the remaining data points to determine the values of \(E_{o}\) and \(a\). Plot the stress versus strain data points along with the analytic curve expressed by the first equation. This will indicate how well the analytic curve matches the data. (b) Often the previous analysis does not work well because the value of \(E_{o}\) is difficult to evaluate. To solve this problem, \(E_{o}\) is not used. A data point is selected \((\bar{\sigma}, \bar{\varepsilon})\) that is in the middle of the range used for the regression analysis. These values are substituted into the first equation, and a value for \(E_{o} / a\) is determined and substituted into the first equation: $$\sigma=\left(\frac{\bar{\sigma}}{e^{a \bar{\varepsilon}}-1}\right)\left(e^{a \varepsilon}-1\right)$$ Using this approach, experimental data that is well defined will produce a good match of the data points and the analytic curve. Use this new relationship and again plot the stress versus the strain data points and the new analytic curve.

Use numerical integration to compute how much mass leaves a reactor based on the following measurements. $$\begin{array}{l|cccccccc} t, \min & 0 & 10 & 20 & 30 & 35 & 40 & 45 & 50 \\ \hline Q, \mathrm{m}^{3} / \mathrm{min} & 4 & 4.8 & 5.2 & 5.0 & 4.6 & 4.3 & 4.3 & 5.0 \\ \hline \mathrm{c}, \mathrm{mg} / \mathrm{m}^{3} & 10 & 35 & 55 & 52 & 40 & 37 & 32 & 34 \end{array}$$

Suppose that the current through a resistor is described by the function $$i(t)=(60-t)^{2}+(60-t) \sin (\sqrt{t})$$ and the resistance is a function of the current, $$R=10 i+2 i^{2 / 3}$$

Integration provides a means to compute how much mass enters or leaves a reactor over a specified time period, as in $$M=\int_{t_{1}}^{t_{2}} Q c d t$$ where \(t_{1}\) and \(t_{2}=\) the initial and final times, respectively. This formula makes intuitive sense if you recall the analogy between integration and summation. Thus, the integral represents the summation of the product of flow times concentration to give the total mass entering or leaving from \(t_{1}\) to \(t_{2}\). If the flow rate is constant, \(Q\) can be moved outside the integral: $$M=Q \int_{t_{1}}^{t_{2}} c d t$$ Use numerical integration to evaluate this equation for the data listed below. Note that \(Q=4 \mathrm{m}^{3} / \mathrm{min}\). $$\begin{array}{l|cccccccc} t, \min & 0 & 10 & 20 & 30 & 35 & 40 & 45 & 50 \\ \hline c, \mathrm{mg} / \mathrm{m}^{3} & 10 & 35 & 55 & 52 & 40 & 37 & 32 & 34 \end{array}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.