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Water flows at \(1 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(1 \mathrm{~m}\) and depth \(1 \mathrm{~m}\). What is the maximum contraction of the channel that will not choke the flow?

Short Answer

Expert verified
The maximum contraction width is approximately 0.32 meters.

Step by step solution

01

Understand the Problem

The problem requires us to find the maximum contraction width of a rectangular channel where water flows at a steady rate of 1 cubic meter per second, with initial dimensions of 1 meter wide and 1 meter deep, without causing choking (restriction of flow).
02

Apply Continuity Equation

The continuity equation states that the flow rate must remain constant. Initially, the flow rate is calculated as the product of cross-sectional area and velocity: \( Q = A \times v = b \times h \times v \), where \( b = 1 \) meter, \( h = 1 \) meter, and \( v \) is the velocity. Given \( Q = 1 \mathrm{~m}^{3}/ ext{s} \), we find the initial velocity \( v \) by: \( 1 = 1 \times 1 \times v \). Thus, \( v = 1 \mathrm{~m/s} \).
03

Consider the Effect of Contraction

In a contracted section, the width changes but the flow rate remains the same. The new cross-sectional area is \( A' = b' \times h \), where \( b' \) is the new width. Using the continuity equation: \( Q = A' \times v' = b' \times 1 \times v' \). For maximum contraction \( b' \), \( v' \) must reach critical velocity.
04

Identify Critical Conditions

The critical velocity \( v_c \) for flow occurs when \( Fr = 1 \) (Froude number equals 1). The Froude number is given by: \( Fr = \frac{v}{\sqrt{g \times h}} \). Solving for critical velocity, set \( Fr = 1 \): \( 1 = \frac{v_c}{\sqrt{g \times 1}} \). Thus, \( v_c = \sqrt{g} \approx 3.13 \mathrm{~m/s} \) (assuming \( g = 9.81 \mathrm{~m/s}^{2} \)).
05

Calculate Maximum Contraction

Using \( v' = v_c = 3.13 \mathrm{~m/s} \) and the flow rate \( Q = 1 \mathrm{~m}^{3}/ ext{s} \), apply the continuity equation: \( 1 = b' \times 1 \times 3.13 \). Solving for \( b' \), \( b' = \frac{1}{3.13} \approx 0.32 \mathrm{~m} \).

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Key Concepts

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

Continuity Equation
In fluid mechanics, the continuity equation is foundational. It helps us understand how fluid flows in different sections of a channel must balance.
The equation for steady flow is expressed as: \[ Q = A \times v = b \times h \times v \] where: - \( Q \) is the flow rate - \( A \) is the cross-sectional area - \( v \) is the velocity of flow - \( b \) and \( h \) are the channel width and depth
In our problem, the initial flow equation becomes \( 1 = 1 \times 1 \times v \), which simplifies to \( v = 1 \ \mathrm{m/s} \). This is our starting velocity. Understanding the continuity equation allows us to track changes in flow characteristics when other parameters, like channel width, are altered.
Flow Rate Calculation
Flow Rate is another essential idea in fluid mechanics, indicating volume of fluid passing a point in a unit of time. It's measured in cubic meters per second (m³/s).
In the given exercise, the flow rate \( Q \) is a constant 1 \( \mathrm{m}^3/\mathrm{s} \).
Since fluid keeps flowing at this rate, irrespective of channel modifications, we can use this value to calculate new channel dimensions or velocity values. This constant flow rate helps maintain a balance in fluid behavior, considering it's never disrupted.
Consequentially, whether you have a wide channel or a narrow one, the flow rate inspires change in another variable, ensuring the product of area and velocity remains the same. Thus, it provides a consistent baseline for further fluid dynamics calculations.
Critical Velocity
Critical Velocity corresponds to the velocity at which flow becomes critical or reaches a critical state.
It's the minimum velocity needed to sustain a particular state of flow without transitioning into another flow regime.
In open channel flows, critical velocity is interconnected with the Froude number. Our exercise specifies that for maximum contraction, the velocity should be critical to avoid "choking" the flow.
The critical velocity \( v_c \) is calculated in terms of the gravitational factor \( g \) and given by: \[ v_c = \sqrt{g \times h} \] Using \( g = 9.81 \ \mathrm{m/s^2} \), we compute that \( v_c \approx 3.13 \ \mathrm{m/s} \), which becomes the target velocity in the contracted section.
This concept is vital in determining how much a channel can be constricted, ensuring the flow remains unchoked.
Froude Number
The Froude number is a dimensionless figure in fluid mechanics. Named after William Froude, it describes the relation between flow inertia and gravitational force, expressed as: \[ Fr = \frac{v}{\sqrt{g \times h}} \] An important detail of the problem is that at the critical state, the Froude number \( Fr \) equals 1.
This tells us that the flow has reached its critical velocity, where the specific energy is at a minimum.
  • When \( Fr < 1 \), flow is subcritical (gravity dominates)
  • When \( Fr = 1 \), flow is critical
  • When \( Fr > 1 \), flow is supercritical (inertia dominates)
Evaluating the Froude number helps in deciding the flow state, ensuring channels are designed to maintain desired flow conditions effectively.
Channel Flow Contraction
Channel Flow Contraction occurs when a channel's dimensions shrink, impacting the flow. In this problem, we evaluate how much the width of the channel can decrease, impacting no flow disruption.
Recognizing that the channel begins at 1 meter width, using continuity and critical velocity, the contraction limit exists at 0.32 meters.
This contraction represents the balance between reducing flow area while keeping velocity below a level that would turn the flow supercritical, potentially choking it. By applying fluid mechanics principles, such as understanding continuity and Froude numbers, you can decide how far channels can be contracted before adverse effects take place.
Practically knowing this limit ensures water flows smoothly and efficiently, regardless of architectural constraints.

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Most popular questions from this chapter

A trapezoidal irrigation channel is to be excavated to supply water to a farm. The design flow rate is \(1.8 \mathrm{~m}^{3} / \mathrm{s}\), the side slopes are 2: 1 (H:V), the longitudinal slope of the channel is \(0.1 \%\), Manning's \(n\) is 0.025 , and the geometry of the channel is to be such that the length of each channel side is equal to the bottom width. (a) Specify the dimensions of the channel required to accommodate the design flow under normal conditions. (b) If the channel lining can resist an average shear stress of up to \(4 \mathrm{~Pa}\), under what flow conditions is the channel lining stable?

A trapezoidal drainage channel of bottom width \(5 \mathrm{~m}\), side slopes 2: 1 (H:V), Manning's \(n\) of \(0.018,\) and longitudinal slope of \(0.1 \%\) terminates at a gate where the relationship between the flow through the gate \((Q)\), headwater elevation \((H W),\) tailwater elevation \((T W),\) and gate opening \((h)\) is given by $$ Q=13.3 h \sqrt{H W-T W} $$ where \(Q\) is in \(\mathrm{m}^{3} / \mathrm{s}\) and \(H W, T W,\) and \(h\) are in meters. If the elevation of the bottom of the channel at the gate location is \(0.00 \mathrm{~m},\) the tailwater elevation is \(1.00 \mathrm{~m},\) and the flow in the channel is \(20 \mathrm{~m}^{3} / \mathrm{s}\), estimate the minimum gate opening such that the water surface elevation \(100 \mathrm{~m}\) upstream of the gate does not exceed \(2.20 \mathrm{~m}\). (Note: The depth of flow \(100 \mathrm{~m}\) upstream of the gate is not \(2.20 \mathrm{~m} .)\)

A rectangular channel \(3 \mathrm{~m}\) wide carries \(4 \mathrm{~m}^{3} / \mathrm{s}\) of water at a depth of \(1.5 \mathrm{~m}\). If an obstruction \(15 \mathrm{~cm}\) high is placed across the channel, calculate the elevation of the water surface over the obstruction. What is the maximum height of the obstruction that will not cause a rise in the water surface upstream?

Water flows in a trapezoidal channel that has a bottom width of \(5 \mathrm{~m}\), side slopes of \(2: 1(\mathrm{H:V}),\) and a longitudinal slope of 0.0001 . The channel has an equivalent sand roughness of \(1 \mathrm{~mm}\). Calculate the uniform flow depth in the channel when the flow rate is \(18 \mathrm{~m}^{3} / \mathrm{s}\). Is the flow hydraulically rough, smooth, or in transition? Would the Manning equation be valid in this case? Explain.

Water flows at a depth of \(4.00 \mathrm{~m}\) in a trapezoidal concrete-lined channel with a bottom width of \(4 \mathrm{~m}\) and side slopes of \(3: 1(\mathrm{H}: \mathrm{V})\). The longitudinal slope of the channel is 0.0001 , and the water temperature is \(20^{\circ} \mathrm{C}\). Assess the validity of using the Manning equation, assuming that \(n=0.013\).

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