Lecture 25 Design Example for a Channel Transition. I. Introduction

Transcription

1 Lecture 5 Design Example for a Channel Transition I. Introduction This example will be for a transition from a trapezoidal canal section to a rectangular flume section The objective of the transition design is to avoid backwater (GVF) profiles in the transition, and upstream & downstream of the transition We will specify a length for the transition, but the total net change in canal invert elevation across the transition will be defined as part of the solution The main design challenge will be to determine the shape (profile) of the canal invert across the transition II. Given Information The design flow rate is 15.0 m 3 /s The upstream trapezoidal section has 1:1 side slopes (m = 1) The bed slope of the upstream trapezoidal section is m/m The bed slope of the downstream rectangular flume is m/m The upstream and downstream channels are concrete-lined, as will be the transition In this example, the length of the transition is specified to be L = 8.0 m; in other cases the invert elevation change, z might be specified Both L and z cannot be specified beforehand because it would unnecessarily constrain the solution The base widths and uniform flow depths for the upstream and downstream channels are shown in the figure above; these were determined during the design procedures for the respective channels (canal & flume) These calculations can be confirmed by applying the Manning or Chezy equations The reduction in bottom width of the channel will be accomplished with a reverse parabola, from b =.5 m to b =.0 m The reduction in side slope from m = 1 to m = 0 will be done linearly across the length L of the transition BIE 5300/6300 Lectures 65 Gary P. Merkley

2 III. Confirm Subcritical Flow In the upstream channel, for uniform flow, the squared Froude number is: Q T Q (b + mh) (15) (.5 + (1)(1.87)) r F = = = = 0.6 (1) ga g h(b mh) (.5 (1)(1.87)) [ + ] [ + ] In the downstream channel (flume), for uniform flow, the squared Froude number is: QT Qb (15)(.0) r F = = = = () ga g hb 9.81 (.15)(.0) [ ] [ ] Therefore, F r < 1.0 for both the upstream canal and downstream flume Then, the flow regime in the transition should also be subcritical It would probably also be all right if the flow were supercritical in the flume, as long as the flow remained subcritical upstream; a hydraulic jump in the transition would cause a problem with our given design criterion IV. Energy Balance Across the Transition For uniform flow, the slope of the water surface equals the slope of the channel bed Then, the slope of the upstream water surface is , and for the downstream water surface it is Since the mean velocity is constant for uniform flow, the respective energy lines will have the same slopes as the hydraulic grade lines (HGL), upstream and downstream For our design criterion of no GVF profiles, we will make the slope of the energy line through the transition equal to the average of the US and DS energy line slopes: SEL = = (3) Gary P. Merkley 66 BIE 5300/6300 Lectures

3 This means that the total hydraulic energy loss across the transition will be: E = ( )(8.0) = m (4) where the length of the transition was given as L = 8.0 m The energy balance across the transition is: Q Q hu + + z = h + + E ga d u gad (5) where h u is the upstream depth (m); Q is the design flow rate (m 3 /s); A u is the upstream cross-sectional flow area (m ); z is the total net change in canal invert across the transition (m); h d is the downstream depth (m); A d is the downstream cross-sectional flow area (m ); and E is the hydraulic energy loss across the transition (m) The z value is unknown at this point, but the slope of the water surface across the transition should be equal to: S ws h + z h L = u d (6) where S ws is the (constant) slope of the water surface across the transition (m/m); and L is the length of the transition (m) Combining Eqs. 5 & 6: S ws Q E g Ad Au = (7) L For Q = 15 m 3 /s; A d = (.15)(.0) = 4.30 m ; A u = (1.87)(.5)+(1.0)(1.87) = 8.17 m ; E = m; and L = 8.0 m: (15) (9.81) + (4.3) (8.17) Sws = = (8) 8.0 Note that S ws S EL BIE 5300/6300 Lectures 67 Gary P. Merkley

4 V. Change in Side Slope The side slope will change linearly from 1 to 0 over the length of the transition The equation for m, with x = 0 at the upstream end of the transition, is: where 0 x 8 m VI. Change in Bed Width m = x (9) The bed width decreases from.5 to.0 m over the length of the transition This reduction is specified to be a reverse parabola, defined over L/ = 4.0 m Specific criteria could be used to define the shape of the parabola, but a reduction of 0.5 m in bed width over an 8.0-m distance can be accomplished in a simpler way Define the bed width, b, for the first half of the transition as follows: x b =.5 (10) 64 where 0 x 4 m For x > 4 m, the equation is: where 4 x 8 m You can also do this with a 3 rd -degree polynomial: (x 8) b =.0+ (11) 64 3 b = Ax + Bx + Cx+ D (1) where A, B, C, D are fitted so that the slope is zero at x = 0 and at x = 8 By quick inspection of Eq. 1, it is seen that b =.5 at x = 0, so D =.5 And, at x = 8, b =.0 The other two conditions are that the slope equal zero at the end points: where x = 0 and x = 8 3Ax + Bx + C = 0 (13) Gary P. Merkley 68 BIE 5300/6300 Lectures

5 So, C must be equal to zero, and then A and B can be determined after a small amount of algebra: A = , B = The results are not identical, but very close (see the figure below) Two parabolas 3rd-degree polynomial.35 Base width, b (m) Distance, x (m) VII. Change in Bed Elevation The change in bed elevation can be determined by setting up and solving a differential equation, or by the known change in velocity head across the transition Setting up and solving the differential equation can be done, but it is easier to apply the velocity head, which is the difference between the energy line (EL) and the hydraulic grade line (HGL) The slope of the EL is S EL = in the transition, and the slope of the water surface is S ws = The velocity head can be described as follows: or, V Q (15) = + x(s ws S EL) = x (14) g ga (9.81)(8.17) BIE 5300/6300 Lectures 69 Gary P. Merkley

6 V x g = + (15) And, the cross-sectional area of flow, A, is equal to Q/V, which equals h(b+mh): Q A = = h(b + mh) (16) g x ( + ) where b and m are defined as functions of x in Eqs. 9, 10, 11; and 0 x 8 m Eq. 16 is quadratic in terms of h: h b+ b + 4mA = (17) m Use Eq. 16 to calculate A as a function of x, then insert A into Eq. 17 and solve for h at each x value Using an arbitrary invert elevation of.0 m at the transition inlet, the relationship between depth of water, h, and canal bed elevation, z, across the 8-m transition is: h= 3.87 S x z(x where 0 x 8 m; and z =.0 at x = 0 ) (18) Once h is known, use Eq. 18 to solve for z, then go to the next x value ws The graph below shows the results of calculations using the above equations The numerical results are shown in the table below Note that the sum z+h decreases linearly through the transition (the water surface has a constant slope) Note that the velocity head increases linearly through the transition Note that the summation, z+h+v /g, in the last column of the table (to the right) decreases linearly at the rate of m per meter of distance, x, as we have specified (see Eq. 3): the energy line has a constant slope Gary P. Merkley 70 BIE 5300/6300 Lectures

7 m (m/m) b (m) h (m) z (m) Distance (m) Head (m) and Area (m ) Velocity head (m) z+h (m) Area (m) z+h+v/g z+h+v /g Distance (m) BIE 5300/6300 Lectures 71 Gary P. Merkley

8 Note that the bed elevation, z, increases with x at first, then decreases to the final value of 1.6 m Note that the cross-sectional area decreases non-linearly from 0 to 8 m, but the inverse of the area squared increases linearly, which is why the velocity head also increases at a linear rate This transition design will produce a smooth water surface for the design flow rate of 15 m 3 /s, but not for any other flow rate Below are the transition design results using an arbitrary invert elevation of.00 m at the inlet to the transition Why would you want to have a smooth water surface for the design flow rate in such a transition? Gary P. Merkley 7 BIE 5300/6300 Lectures

9 Transition Design Results x m b A h V /g z z+h+v /g (m) (m/m) (m) (m) (m) (m) (m) (m) BIE 5300/6300 Lectures 73 Gary P. Merkley

10 Gary P. Merkley 74 BIE 5300/6300 Lectures