Civil Engineering Hydraulics. Backwater Profile

Transcription

1 Civil Engineering Hydraulics God put me on this earth to accomplish a certain number of things. Right now I am so far behind that I will never die. Last class, we derived an expression for the change in depth as we moved along a channel carrying a gradually varying flow. S0 Sf ( 1 Fr ) To be consistent with Dr. Janna s text, I m going to return to using z as the depth of flow in the channel. 1

2 This is a non-linear differential equation and so we will have to utilize numerical methods to approximate a solution S0 Sf ( 1 Fr 3 ) To Fr do this, we can substitute for S0, Sf, and S0 Sf ( 1 Fr )

3 S0 is the slope of the channel so it is equal to tanθ. For small angles, tanθ sinθ θ S0 Sf (1 Fr ) sin θ Sf (1 Fr ) 5 Sf represents the rate ofenergy lost due to the flow being in contact with the channel surface. It is a function of the material, the hydraulic radius, and the velocity S0 Sf (1 Fr ) sin θ Sf (1 Fr ) 6 3

4 Sf can be expressed as sin θ Sf (1 Fr ) Sf sin θ (1 Fr ) 7 The Froude number is The subscript on the z denotes that the mean depth is used This is the ratio of the top width to the area of the flow For a rectangular channel it is equal to z 8 sin θ (1 Fr ) Fr v gzm sin θ v 1 gz m

5 To utilize this method for any other type of channel, substitute the ratio of the flow area divided by the top width Dr. Janna does this to arrive at equations 7.39b and 7.39c 9 sin θ (1 Fr ) Fr v gzm sin θ v 1 gz m He also uses cosθ in his expressions. For typical channels that have small slopes, the 1 is appropriate because cosθ 1 As channel slopes increase, using the cosθ may be more appropriate. 10 sin θ (1 Fr ) Fr v gzm sin θ v 1 gz m 5

6 To utilize the expression since so many of the variables are a function of z, we have to make an approximation. sin θ v 1 gz m sin θ Δz Δx v 1 gz m 11 We will start at some point where we know the z and work from there To do this, we will set the value for Δz. sin θ v 1 gz m sin θ Δz Δx v 1 gz m 1 6

7 The city of Memphis, Tennessee, is subject to heavy rainfalls during certain times of the year. Drainage ditches have been dug throughout the city. Most are concrete-lined and all drain ultimately into the Mississippi River. Consider one such rectangular, concretelined channel dug out to a width of.5 m and inclined at a slope of The channel is very long, and at one end there is a dam that partially restricts the flow. At the dam, the water depth is 3.5 m. Determine the variation of depth with upstream distance if the volume flow rate is 8 m3/s. sin θ Δz Δx v 1 gz m

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14 7 Since the value for critical depth in the channel is less than the depth given, the flow is subcritical and the expressions for gradually varied flow hold true. 8 1

15 The flow upstream of the obstruction was at some point normal depth we assume. We need to calculate what would be normal depth in the channel

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18 This value is slightly different than the one that Dr. Janna has in the text. I did not use any tables so that may be the reason for the difference. 35 We know that the elevation at the dam is 3.5 m and that at some point upstream the elevation is 1. m. These are the points we will work from. The idea is to use steps and locate how far up stream each elevation change is located

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24 7 Homework 8-1 Consider a rectangular brick-lined channel of width 10 ft and inclined at a slope of The channel is long and contains a dam at one end. The water depth just before the dam is 6.5 ft. The flow rate in the channel is 30 ft3/s. Determine the shape of the backwater curve 8