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Chapter 10: Problem 72

Consider the flow down a prismatic channel having a rectangular cross section of width \(b\). The channel bottom makes an angle \(\theta\) with the horizontal. Show that \\[ \frac{d y}{d x}=\frac{\tan \theta-\left(n^{2} Q^{2}\right) /\left(A^{2} R_{h}^{4 / 3} \kappa^{2}\right)}{1-Q^{2} /\left(A^{2} g y \cos ^{2} \theta\right)} \\] where \(y\) is the vertical fluid depth, \(A=b y \cos \theta,\) ard \(Q\) is the volume flow rate. Discuss the form of the equation for small values of \(\theta\left(\theta \sim 1^{\circ}\right)\)

Short Answer

Expert verified
For small values of θ (θ∼1°), where θ, sin(θ) and cos(θ) approaches 1, the equation simplifies to \[ \frac{d y}{d x} = \frac{1 - \frac{n^2 Q^2}{ g A^2 y^{4/3} \kappa^2 }}{1 - \frac{Q^2}{g A^2 y}} \]\nThis indicates that the rate of change of fluid depth in the channel is directly proportional to the difference between the frictional force and the inertial force. Therefore, for a small channel slope (θ approaching zero), any fluctuations in fluid depth are primarily driven by the balance between these forces, rather than the slope θ itself.

Step by step solution

01

Start with the governing equation

Start with the governing equation of open channel flow, also known as the energy equation or Bernoulli's equation with losses. This states that the loss of energy due to friction (F) is equal to the change in pressure head and velocity head with respect to the position in the channel:\n\[ z + \frac{{P}}{{\rho g}} + \frac{{V^2}}{{2g}} + F = constant \] Substituting hydraulic depth (h = y cosθ, which is the vertical water depth) and channel bed slope (s= tanθ), the above equation can be simplified to: \n\[ \frac{d y}{d x} = s - \frac{n^2 Q^2}{g A^3} \]
02

Substituting for hydraulic radius and cross-sectional area

Substitute for hydraulic radius (R) and cross-sectional area (A) of the water flowing through the channel in the above simplified equation. For a rectangular channel, R = y and A = by, where 'b' is the width of the channel. So, \n\[ \frac{d y}{d x} = \tan \theta - \frac{n^2 Q^2}{A^2 g y \cos^{2} \theta} \]
03

Adding the inertia term

Additionally, consider the inertial force due to the acceleration in the flow by adding in the inertia term to our equation. \nIn inertia term, which can be expressed as (V dV/ds), V = Q/A. So, it can be expressed as ((Q/A) * d(Q/A)/ds) = - (Q^2 / A^2) * dA/ds.\n\nConsidering A = by cosθ, then dA/ds = ycosec^2θ * dθ/dx + b cosθ = y cotθ * dθ/dx, the inertia term become - (Q^2 / A^2) * dA/ds = - (Q^2 / A^2) * y cotθ * dθ/dx. Simplifying, \n\n\ \[ \frac{d y}{d x} = \frac{\tan \theta - \frac{n^2 Q^2}{A^2 g y^{4/3} \kappa^2 } }{ 1 - \frac{ Q^2 }{A^2 g y \cos^{2}\theta} } \]
04

Analyzing for small slope angles

When θ tends to 1 degree, \(\tan \theta \approx \theta \approx \sin \theta \approx \cos \theta = 1\) and it simplifies the above equation to: \n\[ \frac{d y}{d x} = \frac{1 - \frac{n^2 Q^2}{ g A^2 y^{4/3} \kappa^2 }}{1 - \frac{Q^2}{g A^2 y}} \]\nWe notice that for small slope angles, the equation becomes directly proportional to the difference in the frictional force (numerator) and inertial force (denominator). Changes in 'y' depends on these forces now rather than the slope. Eventually when θ tends to zero (i.e. horizontal channel), the rate of change of depth would go to zero and the flow behaviour would be determined by only the inertia and friction.

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Most popular questions from this chapter

A rectangular brick-lined channel has a bottom slope of 0.0025 and is designed to carry a uniform water flow rate of \(300 \mathrm{ft}^{3} / \mathrm{s}\). Would the channel need fewer bricks if the channel were 2 ft wide, 6 ft wide, or \(10 \mathrm{ft}\) wide? Explain.

A rectangular channel \(3 \mathrm{m}\) wide carries \(10 \mathrm{m}^{3} / \mathrm{s}\) at a depth of \(2 \mathrm{m} .\) Is the flow subcritical or supercritical? For the same flowrate, what depth will give critical flow?

A rectangular, unfinished concrete channel of 28 -ft width is laid on a slope of \(8 \mathrm{ft} / \mathrm{mi}\). Determine the flow depth and Froude number of the flow if the flowrate is \(400 \mathrm{ft}^{3} / \mathrm{s}\)

An 8 -ft-diameter concrete drainage pipe that flows half full is to be replaced by a concrete-lined V-shaped open channel having an interior angle of \(90^{\circ} .\) Determine the depth of fluid that will exist in the \(V\) -shaped channel if it is laid on the same slope and carries the same discharge as the drainage pipe.

Water flows in a horizontal rectangular channel with a flowrate per unit width of \(q=10 \mathrm{ft}^{2} / \mathrm{s}\) and a depth of \(1.0 \mathrm{ft}\) at the downstream section \((2) .\) The head loss between section upstream and section (2) is 0.2 ft. Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram.
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