Chapter 13: Problem 42
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
Step by step solution
Understand the Problem
Apply the Continuity Equation
Calculate Initial Cross-Sectional Area
Convert Velocity Units
Calculate Initial Flow Rate
Set Up Continuity Equation for Second Point
Solve for Depth
Conclude the Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fluid Dynamics
Cross-Sectional Area
- 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: \[A_2 = \text{width} \times \text{depth}\]
Flow Velocity
- 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 \(0.11 \, \text{m/s}\).
Conservation of Mass
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.