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Chapter 10: Problem 17

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?

Short Answer

Expert verified
The minimum height of the hump is 0.408 meters.

Step by step solution

01

Convert Units

Convert the depth from millimeters to meters. Given: Depth = 900 mm Convert: Depth = \( \frac{900}{1000} = 0.9 \) meters.
02

Calculate Flow Rate Using Manning's Equation

Use Manning's equation to calculate the flow rate. Manning's Equation:\[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \] Where:\(n = 0.015\) (Manning's roughness coefficient)\(A\) = cross-sectional area of flow = \( b \times d = 2.5 \times 0.9 = 2.25 \) square meters\( R = \frac{A}{P} \) (hydraulic radius, \( P \) is the wetted perimeter)\( P = b + 2d = 2.5 + 2 \times 0.9 = 4.3 \) meters\( R = \frac{2.25}{4.3} \approx 0.523 \) meters\( S = \frac{1}{1200} \approx 0.000833 \) (bed slope)Substitute values in Manning's equation and calculate the flow rate:
03

Continue Calculation of Flow Rate (Q)

\[ Q = \frac{1}{0.015} \times 2.25 \times (0.523)^{2/3} \times (0.000833)^{1/2} \] Calculate each part step by step: \[ (0.523)^{2/3} \approx 0.632 \] \[ (0.000833)^{1/2} \approx 0.0289 \] \[ Q = 66.67 \times 2.25 \times 0.632 \times 0.0289 \approx 2.723 \text{ cubic meters per second}\]
04

Determine Critical Depth

Use the critical flow condition to find the critical depth. For a rectangular channel, the critical depth \( d_c \) is given by: \[ d_c = \left( \frac{Q^2}{g b^2} \right)^{1/3} \] Where: \( g \) = 9.81 m/s² (acceleration due to gravity)\( Q = 2.723 \) cubic meters per second\( b = 2.5 \) metersSubstitute the values to find \( d_c \): \[ d_c = \left( \frac{(2.723)^2}{9.81 \times (2.5)^2} \right)^{1/3} \]
05

Calculate Critical Depth

Continue the calculation: \[ d_c = \left( \frac{7.413}{61.3125} \right)^{1/3} \approx \left(0.121 \right)^{1/3} \approx 0.492 \text{ meters} \] The critical depth is approximately 0.492 meters.
06

Calculate Minimum Height of Hump

Determine the minimum height of the hump by using the initial depth and the critical depth. Initial depth = 0.9 metersCritical depth = 0.492 metersMinimum height of the hump: \[ H = \text{Initial depth} - \text{Critical depth} = 0.9 - 0.492 = 0.408 \text{ meters} \]

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Key Concepts

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

Manning's equation
Manning's equation helps us calculate the flow rate in an open channel. This equation factors in the channel's roughness, cross-sectional area, hydraulic radius, and bed slope. The formula is as follows:


Here:
  • Q is the flow rate.
  • n is Manning's roughness coefficient.
  • A is the cross-sectional area of flow.
  • R is the hydraulic radius.
  • S is the bed slope.
In our example, we use this equation to find the flow rate given the channel dimensions, bed slope, and Manning's roughness coefficient. This step simplifies how water flows through a channel.
Critical Depth
Critical depth is the depth at which the flow velocity is at a critical state. At this point, the specific energy of the flow is at a minimum. In rectangular channels, the formula for critical depth is:




Here:
  • Q is the flow rate we calculated using Manning's equation.
  • g is the acceleration due to gravity, approximately 9.81 m/s².
  • b is the width of the channel.
In our calculation, we used this formula to derive the critical depth for the given flow rate. Identifying the critical depth allows us to understand when a flow will transition from subcritical to supercritical, which is essential for hydraulic engineering.
Hydraulic Radius
The hydraulic radius is a measure that describes how efficiently a channel can transport water. It is the ratio of the cross-sectional area of flow to the wetted perimeter. The formula is:

Here:
  • R is the hydraulic radius.

  • A is the cross-sectional area of flow.
  • P is the wetted perimeter.
For our channel, we calculated the hydraulic radius using the given depth and width. A higher hydraulic radius often indicates a more efficient channel, meaning less resistance to the flow. Therefore, understanding how the hydraulic radius impacts flow can significantly help in channel design and management.
Bed Slope
The bed slope is the gradient or incline along the flow direction of the channel. This property influences the flow velocity and kinetic energy of the water. It is usually given as a ratio, and in our case, the bed slope is 1 in 1200. This converts to:

  • The bed slope helps us use Manning's equation effectively. A steeper slope increases flow rate, but too steep a slope could lead to erosive conditions.
Calculating the bed slope allows us to predict and control the water flow characteristics accurately. This aids in designing channels that are efficient and safe.
Flow Rate Calculation
The flow rate calculation is crucial for understanding how much water a channel can convey. Using Manning's equation, we gather essential parameters like Manning's roughness coefficient, cross-sectional area, hydraulic radius, and bed slope to calculate the flow rate. In our example, this is:



This considers:
  • n as the roughness coefficient.
  • A the area.
  • R the hydraulic radius.
  • S the bed slope.
Calculating the flow rate is crucial for hydraulic analysis and channel design. This value helps ensure that our channels can handle expected water volumes without overflow.

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Most popular questions from this chapter

An open channel of trapezoidal section, \(2.5 \mathrm{~m}\) wide at the base and having sides inclined at \(60^{\circ}\) to the horizontal, has a bed slope of 1 in \(500 .\) It is found that when the rate of flow, is \(1.24 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the depth of water in the channel is \(350 \mathrm{~mm}\). Assuming the validity of Manning's formula, calculate the rate of flow when the depth is \(500 \mathrm{~mm}\).

Water flows at a depth of \(1.2 \mathrm{~m}\) in a rectangular prismatic channel \(2.7 \mathrm{~m}\) wide. Over a smooth hump \(200 \mathrm{~mm}\) high on the channel bed a drop of \(150 \mathrm{~mm}\) in the water surface is observed. Neglecting frictional effects, calculate the rate of flow.

The cross-section of a river \(30 \mathrm{~m}\) wide is rectangular. At a point where the bed is approximately horizontal the width is restricted to \(25 \mathrm{~m}\) by the piers of a bridge. If a flood of \(450 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) is to pass the bridge with the minimum upstream depth, describe the flow past the piers and calculate the upstream depth.

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 long horizontal channel has a base width of \(1 \mathrm{~m}\) and sides at \(60^{\circ}\) to the horizontal. When the flow in the channel is \(0.85 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the depth is \(500 \mathrm{~mm} .\) The discharge is suddenly reduced so that a surge wave of amplitude \(150 \mathrm{~mm}\) is propagated upstream. Determine the new rate of flow, the velocity of the wave and the Froude numbers before and after the wave.
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