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An irrigation canal has a rectangular cross section. At one point where the canal is \(18.5 \mathrm{~m}\) wide and the water is \(3.75 \mathrm{~m}\) deep, the water flows at \(2.50 \mathrm{~cm} / \mathrm{s}\). At a second point downstream, but on the same level, the canal is \(16.5 \mathrm{~m}\) wide, but the water flows at \(11.0 \mathrm{~cm} / \mathrm{s} .\) How deep is the water at this point?

Short Answer

Expert verified
Depth at the second point is approximately 0.957 m.

Step by step solution

01

Understand the Problem

We have two points in a canal with different cross-section areas and flow velocities. We need to find the depth of the water at the second point given the depth at the first point and flow velocities at both points.
02

Apply the Continuity Equation

The conservation of mass principle for fluid flow in an incompressible fluid like water tells us that the flow rate must be constant throughout the canal. Therefore, \( A_1 v_1 = A_2 v_2 \), where \( A \) is the cross-sectional area and \( v \) is the flow velocity.
03

Calculate Initial Cross-Sectional Area

For the first point, the cross-sectional area (\( A_1 \)) is the width multiplied by the depth: \( A_1 = 18.5 \, \text{m} \times 3.75 \, \text{m} = 69.375 \, \text{m}^2 \).
04

Convert Velocity Units

Convert the flow velocities from cm/s to m/s. The initial flow velocity \( v_1 = 2.50 \, \text{cm/s} \) becomes \( 0.025 \, \text{m/s} \). Similarly, the latter flow velocity \( v_2 = 11.0 \, \text{cm/s} \) becomes \( 0.11 \, \text{m/s} \).
05

Calculate Initial Flow Rate

The initial flow rate \( Q_1 \) is \( A_1 \times v_1 = 69.375 \, \text{m}^2 \times 0.025 \, \text{m/s} = 1.734375 \, \text{m}^3/ ext{s} \).
06

Set Up Continuity Equation for Second Point

According to continuity, we set up \( Q_1 = A_2 \times v_2 \). Substitute \( A_2 = 16.5 \, \text{m} \times h_2 \), where \( h_2 \) is the unknown depth at the second point.
07

Solve for Depth

Now find \( h_2 \) using the equation \( 1.734375 = 16.5 \, \text{m} \times 0.11 \, \text{m/s} \times h_2 \). Thus, \( h_2 = \frac{1.734375}{16.5 \times 0.11} \approx 0.957 \text{ m} \).
08

Conclude the Solution

Therefore, the depth of the water at the second point is approximately \( 0.957 \, \text{m} \).

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is the study of the movement of liquids and gases. One crucial aspect of fluid dynamics is understanding how fluids behave as they travel through different environments. This involves analyzing factors such as pressure, velocity, and volume of fluid particles. The movement of water in an irrigation canal is a classic example. Here, we look at how the flow changes when the canal's dimensions or the incline changes. Fluid dynamics allows us to predict how the fluid will adjust to these changes, informing engineers and scientists to design better systems for water management. Understanding these principles also aids in resolving real-world problems, such as determining how water will flow and the levels it will reach under various conditions.
Cross-Sectional Area
The cross-sectional area of a canal is essential in determining how much fluid can flow through it at any given time. In the scenario with the canal, the cross-sectional area is calculated as the product of the width and the depth of the water. To find the cross-sectional area at any given point along the length of the canal, you multiply the water's depth by the canal's width at that point.
  • At the first point, the cross-sectional area is calculated as: \[A_1 = 18.5 \, \text{m} \times 3.75 \, \text{m} = 69.375 \, \text{m}^2\]
  • For the second point, you would use the formula: \[A_2 = \text{width} \times \text{depth}\]
Calculating the correct cross-sectional area is crucial as it directly impacts the flow rate and the velocity at which the water can move through the canal.
Flow Velocity
Flow velocity refers to how fast the fluid travels through the canal. This can significantly change depending on how wide or deep the canal is at different points along its path. In fluid dynamics, understanding flow velocity is vital to calculating how much fluid passes through a specific section of the canal over time. We use it to calculate the flow rate, which is simply the product of flow velocity and cross-sectional area.
  • In the given example, the initial flow velocity is converted from one unit to another. The conversion from cm/s to m/s is necessary to ensure consistent units when applying the continuity equation.
  • The initial flow velocity of the first section is calculated as \(0.025 \, \text{m/s}\), and at the second point, it’s \(0.11 \, \text{m/s}\).
Understanding flow velocity helps in solving the continuity equation to find unknowns, such as the depth of the fluid at different points.
Conservation of Mass
The conservation of mass principle, a fundamental concept in fluid dynamics, declares that mass cannot be created or destroyed within an isolated system. This is particularly important in calculating how fluids behave when moving from one section to another in systems like water canals.
This principle is mathematically expressed through the continuity equation: \[A_1v_1 = A_2v_2\]This equation ensures that the mass that enters a segment of the canal per unit time equals the mass exiting per unit time. Here,
  • \(A_1v_1\) is the flow at the first point with known dimensions and flow velocity.
  • \(A_2v_2\) represents the second point where adjustments in depth or width could occur.
By holding the flow rate constant, the equation allows us to solve for unknown values like the water's depth at the second point. This understanding is crucial for maintaining efficient water systems, preventing overflow or insufficient supply, which could lead to broader environmental impacts.

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Most popular questions from this chapter

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 crosssectional area of the pipe is \(0.070 \mathrm{~m}^{2}\) and the magnitude of the fluid velocity is \(3.50 \mathrm{~m} / \mathrm{s}\). What is the fluid speed at points in the pipe where the cross-sectional area is (a) \(0.105 \mathrm{~m}^{2},\) (b) \(0.047 \mathrm{~m}^{2}\) ?

A sealed tank containing seawater to a height of \(11.0 \mathrm{~m}\) also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. Calculate the speed with which the water comes out of the tank.

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A shower head has 20 circular openings, each with radius \(1.0 \mathrm{~mm}\). The shower head is connected to a pipe with radius \(0.80 \mathrm{~cm} .\) If the speed of water in the pipe is \(3.0 \mathrm{~m} / \mathrm{s},\) what is its speed as it exits the shower-head openings?

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