/*! 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 18 As described in Prob. \(24.17,\)... [FREE SOLUTION] | 91Ó°ÊÓ

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As described in Prob. \(24.17,\) the cross-sectional area of a channel can be computed as $$A_{c}=\int_{0}^{B} H(y) d y$$ where \(B=\) the total channel width \((\mathrm{m}), H=\) the depth \((\mathrm{m}),\) and \(y=\) distance from the bank (m). In a similar fashion, the average flow \(Q\left(\mathrm{m}^{3} / \mathrm{s}\right)\) can be computed as $$Q=\int_{0}^{B} U(y) H(y) d y$$ where \(U=\) water velocity \((\mathrm{m} / \mathrm{s})\). Use these relationships and a numerical method to determine \(A_{c}\) and \(Q\) for the following data: $$\begin{array}{l|cccccc} y, \mathrm{m} & 0 & 2 & 4 & 5 & 6 & 9 \\ \hline H_{,} \mathrm{m} & 0.5 & 1.3 & 1.25 & 1.7 & 1 & 0.25 \\ \hline U_{,} \mathrm{m} / \mathrm{s} & 0.03 & 0.06 & 0.05 & 0.12 & 0.11 & 0.02 \end{array}$$

Short Answer

Expert verified
The cross-sectional area \(A_c\) of the channel is approximately \(8.25 \, \mathrm{m}^2\), and the average flow \(Q\) is about \(1.58 \, \mathrm{m}^3/\mathrm{s}\).

Step by step solution

01

Review the relationships

The given relationships for \(A_c\) and \(Q\) are: \(A_c = \int_{0}^{B} H(y) dy\) \(Q = \int_{0}^{B} U(y) H(y) dy\) Step 2: Approximate the integrals using the trapezoidal rule
02

Trapezoidal rule

The trapezoidal rule can be used to approximate an integral as follows: \(\int_{a}^{b} f(x) dx \approx \frac{h}{2}\left[f(a) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(b)\right]\) Step 3: Calculate the cross-sectional area
03

Applying the trapezoidal rule to \(A_c\)

Using the trapezoidal rule for \(A_c\), we have: \(A_c \approx \frac{1}{2}\left[H(0) + 2H(2) + 2H(4) + 2H(5) + 2H(6) + H(9)\right]\) Substitute the given values for \(H\) at each distance \(y\): \(A_c \approx \frac{1}{2}\left[0.5 + 2(1.3) + 2(1.25) + 2(1.7) + 2(1) + 0.25\right]\) \(A_c \approx 8.25 \, \mathrm{m}^2\) Step 4: Calculate the average flow
04

Applying the trapezoidal rule to \(Q\)

Using the trapezoidal rule for \(Q\), we have: \(Q \approx \frac{1}{2}\left[U(0)H(0) + 2U(2)H(2) + 2U(4)H(4) + 2U(5)H(5) + 2U(6)H(6) + U(9)H(9)\right]\) Substitute the given values for \(U\) and \(H\) at each distance \(y\): \(Q \approx \frac{1}{2}\left[0.03(0.5) + 2(0.06)(1.3) + 2(0.05)(1.25) + 2(0.12)(1.7) + 2(0.11)(1) + 0.02(0.25)\right]\) \(Q \approx 1.58 \, \mathrm{m}^3/\mathrm{s}\) In summary, the cross-sectional area \(A_c\) of the channel is approximately \(8.25 \, \mathrm{m}^2\), and the average flow \(Q\) is about \(1.58 \, \mathrm{m}^3/\mathrm{s}\).

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Key Concepts

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

Trapezoidal Rule
Numerical integration often refers to finding the approximate value of an integral. The Trapezoidal Rule is a simple method for doing this. It approximates the area under a curve using trapezoids, making it easy to apply without complex calculations.
The general formula for applying the Trapezoidal Rule is given by:
  • \( \int_{a}^{b} f(x)\, dx \approx \frac{h}{2}\left[f(a) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(b)\right] \)
Here, \(h\) is the distance between consecutive points, and \(f(x)\) is the function being integrated. When using this rule, you calculate the function's value at each point and multiply it by \(h/2\) to get the area of each trapezoid.
This rule is especially useful when you have a set of discrete data points. It helps to approximate the integral even when it can't be solved analytically.
Cross-sectional Area Calculation
Calculating the cross-sectional area, particularly in hydraulic engineering, is crucial for understanding how much space is available for a fluid to flow through a channel. This is denoted as \(A_c\) and can be found using the formula:
  • \( A_c = \int_{0}^{B} H(y) \, dy \)
In this expression, \(H(y)\) represents the depth of the channel at a distance \(y\) from the bank. By using the Trapezoidal Rule, the integral can be approximated, giving us an estimate of how much cross-sectional area the channel has.
This calculation is vital in designing and assessing waterways, as it impacts the flow capacity and can indicate potential flooding risks. Engineers use this information to modify channels and ensure safe and efficient fluid movement.
Hydraulic Engineering
Hydraulic engineering deals with the flow and conveyance of fluids, primarily water. This field utilizes different principles and mathematical strategies to manage water resources and design projects like dams, levees, and channels.
In hydraulic engineering, accurate calculations related to water flow, velocity, and channel dimensions are key to creating effective water management systems. By understanding channel cross-sections and flow rates, engineers can predict how water will behave in various conditions.
Techniques from numerical integration, like the Trapezoidal Rule, play a crucial role. They help in predicting and planning for real-world scenarios where water needs to move efficiently and safely through engineered structures.
Average Flow Calculation
The average flow, denoted as \(Q\), is a measure of how much fluid is passing through a section of the channel over time. It's crucial for ensuring that channels can handle the anticipated flow without overflow or erosion.
The formula for calculating average flow is:
  • \( Q = \int_{0}^{B} U(y) H(y) \, dy \)
Here, \(U(y)\) is the velocity of the water at distance \(y\) from the bank, while \(H(y)\) represents the channel's depth at that point.
Again, the Trapezoidal Rule is often employed to estimate \(Q\) when exact solutions are not feasible due to complex or discrete data points. This allows engineers to make necessary adjustments to the channel's design, preventing damages and ensuring effective water management.

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Most popular questions from this chapter

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