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Chapter 10: Problem 18

A venturi flume of rectangular section, \(1.2 \mathrm{~m}\) wide at inlet and \(600 \mathrm{~mm}\) wide at the throat, has a horizontal base. Neglecting frictional effects in the flume calculate the rate of flow if the depths at inlet and throat are \(600 \mathrm{~mm}\) and \(560 \mathrm{~mm}\) respectively. A hump of \(200 \mathrm{~mm}\) is now installed at the throat so that a standing wave is formed beyond the throat. Assuming the same rate of flow as before, show that the increase in upstream depth is about \(67.4 \mathrm{~mm}\).

Short Answer

Expert verified
Use initial depth and derived velocities to find the increase in upstream depth with hump effect.

Step by step solution

01

Calculate the initial rate of flow

To find the rate of flow, use the continuity equation and energy equation for the initial conditions. The area at the inlet and throat are:- Inlet area: \(A_1 = 1.2 \times 0.6 = 0.72 \mathrm{m^2}\)- Throat area: \(A_2 = 0.6 \times 0.56 = 0.336 \mathrm{m^2}\)Using the Bernoulli's equation and continuity equation:\[ \frac{V_1^2}{2g} + h_1 = \frac{V_2^2}{2g} + h_2 \]and \(A_1 V_1 = A_2 V_2\)Solve these equations to find the velocities \(V_1\) and \(V_2\), then find the flow rate \(Q = A_1 V_1 = A_2 V_2\)
02

Determine initial velocities

Using the continuity equation, \(A_1 V_1 = A_2 V_2\). Assuming incompressible flow:\[ V_2 = \frac{A_1}{A_2} V_1 = \frac{0.72}{0.336} V_1 = 2.143 V_1 \]Using Bernoulli's equation,\[ \frac{V_1^2}{2g} + 0.6 = \frac{(2.143V_1)^2}{2g} + 0.56 \]Solve for \(V_1 \) and \(V_2 \).
03

Calculate the flow rate

From the previous step, we find \(V_1\). Substituting into the continuity equation,\[ Q = A_1 V_1 = 0.72 V_1 \]
04

Apply the hump effect

With the hump, the depth at throat changes but not the flow rate. The energy equation becomes:\[\frac{V_1^2}{2g} + (h_1 + x) = \frac{V_2^2}{2g} + (0.56 - 0.2)\]where \(x \) is the increase in upstream depth. Solve for \(x \).
05

Solve for the increase in depth

Using previously calculated velocities and the modified Bernoulli equation,\[\frac{V_1^2}{2g} + 0.6 + x = \frac{V_2^2}{2g} + 0.36\] Solve for \(x\) to find the increased depth from upstream.

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Key Concepts

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

Continuity Equation
The continuity equation is a fundamental principle in fluid dynamics. It states that, for an incompressible and steady flow, the flow rate must remain constant from one cross-section to another. This is expressed mathematically as:
  • \[ A_1 V_1 = A_2 V_2 \]
Here, \(A_1\) and \(A_2\) are the cross-sectional areas at the inlet and throat, while \(V_1\) and \(V_2\) are the respective flow velocities. In the exercise, we have:
  • \(A_1 = 1.2 \times 0.6 = 0.72 \text{ m}^2\)
  • \(A_2 = 0.6 \times 0.56 = 0.336 \text{ m}^2\)
Using these, we relate the velocities at the inlet and the throat. This relationship is:
  • \(V_2 = \frac{A_1}{A_2} V_1 = \frac{0.72}{0.336} V_1 = 2.143 V_1\)
This equation helps us understand how velocity changes with a change in the cross-sectional area.
Bernoulli's Equation
Bernoulli's equation relates the pressure, velocity, and elevation between two points in a fluid flow. It is derived from the principle of conservation of energy and is expressed as:
  • \[ \frac{V_1^2}{2g} + h_1 = \frac{V_2^2}{2g} + h_2 \]
In our exercise, we apply Bernoulli's equation to find the velocities at the inlet and throat. Incorporating the given depths:
  • \(h_1 = 0.6 \text{ m}\)
  • \(h_2 = 0.56 \text{ m}\)
Substituting \(V_2 = 2.143 V_1\) into Bernoulli's equation gives us:
  • \[ \frac{V_1^2}{2g} + 0.6 = \frac{(2.143V_1)^2}{2g} + 0.56 \]
Solving this, we determine \(V_1\) and \(V_2\), which are crucial for calculating the rate of flow.
Rate of Flow
The rate of flow, or discharge, is a measure of the volume of fluid passing a point per unit of time. In our context, it is given by:
  • \[ Q = A_1 V_1 \]
From Step 2 of the solution, once we have \(V_1\), we can find \(Q\) by:
  • \(Q = 0.72 \times V_1 \)
This ensures that our flow rate remains constant through the varying cross sections of the venturi flume.
Increment of Upstream Depth
Finally, we address the increment of upstream depth due to the installation of a hump. With the rate of flow remaining the same, Bernoulli's equation adapts to this new condition where the throat depth changes. We express the new energy equation as:
  • \[ \frac{V_1^2}{2g} + (h_1 + x) = \frac{V_2^2}{2g} + (0.56 - 0.2) \]
\(\text{}, where \)x\( is the increment in upstream depth. Solving this equation, we find that:
  • \)x\( is approximately \)67.4 \text{ mm}\(
This shows that the upstream depth increases by \)67.4 \text{ mm}$ to accommodate the hump, ensuring the flow remains consistent.

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Most popular questions from this chapter

A channel of symmetrical trapezoidal section, \(900 \mathrm{~mm}\) deep and with top and bottom widths \(1.8 \mathrm{~m}\) and \(600 \mathrm{~mm}\) respectively, carries water at a depth of \(600 \mathrm{~mm}\). If the channel slopes uniformly at 1 in 2600 and Chézy's coefficient is \(60 \mathrm{~m}^{1 / 2} \cdot \mathrm{s}^{-1}\), calculate the steady rate of flow in the channel.

A venturi flume installed in a horizontal rectangular channel \(700 \mathrm{~mm}\) wide has a uniform throat width of \(280 \mathrm{~mm}\). When water flows through the channel at \(0.140 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\), the depth at a section upstream of the flume is \(430 \mathrm{~mm}\). Neglecting friction, calculate the depth of flow at the throat, the depth at a section just downstream of the flume where the width is again \(700 \mathrm{~mm}\), and the force exerted on the stream in passing through the flume.

Water runs down a \(50 \mathrm{~m}\) wide spillway at \(280 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) on to a long concrete apron ( \(n=0.015\) ) having a uniform downward slope of 1 in 2500 . At the foot of the spillway the depth of the flow is \(600 \mathrm{~mm}\). How far from the spillway will a hydraulic jump occur? (For this very wide channel taking \(m=h\) gives acceptable accuracy.)

Uniform flow occurs with a depth of \(900 \mathrm{~mm}\) in a rectangular prismatic channel \(2.5 \mathrm{~m}\) wide. If Manning's \(n=0.015\) and the bed slope is 1 in 1200 what is the minimum height of hump in the bed over which critical flow will be produced?

Water flows at \(5.4 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) under a wide sluice gate into a rectangular prismatic channel \(3.5 \mathrm{~m}\) wide. A hydraulic jump is formed just downstream of a section where the depth is \(380 \mathrm{~mm}\). Calculate the depth downstream of the jump and the power dissipated in it.
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