Lecture 25
Design Example for a Channel Transition
I. Introduction
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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
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The design flow rate is 15.0 m3/s
The upstream trapezoidal section has 1:1 side slopes (m = 1)
The bed slope of the upstream trapezoidal section is 0.000516 m/m
The bed slope of the downstream rectangular flume is 0.00292 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 = 2.5 m to b = 2.0 m
The reduction in side slope from m = 1 to m = 0 will be done linearly across the
length L of the transition
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Gary P. Merkley
III. Confirm Subcritical Flow
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In the upstream channel, for uniform flow, the squared Froude number is:
Fr2 =
•
Q 2T
gA
3
=
g [h(b + mh)]
3
=
(15)2 (2.5 + 2(1)(1.87))
9.81 [1.87(2.5 + (1)(1.87))]
3
= 0.262
(1)
In the downstream channel (flume), for uniform flow, the squared Froude number
is:
Fr2
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Q2 (b + 2mh)
=
Q2 T
gA3
=
Q2b
g [hb]
3
=
(15)2 (2.0)
9.81 [(2.15)(2.0)]
3
= 0.577
(2)
Therefore, Fr2 < 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
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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 0.000516, and for the
downstream water surface it is 0.00292
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 =
Gary P. Merkley
0.000516 + 0.00292
= 0.001718
2
266
(3)
BIE 5300/6300 Lectures
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This means that the total hydraulic energy loss across the transition will be:
∆E = (0.001718)(8.0) = 0.0137 m
(4)
where the length of the transition was given as L = 8.0 m
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The energy balance across the transition is:
hu +
Q2
2gA u2
+ ∆z = hd +
Q2
2gA d2
+ ∆E
(5)
where hu is the upstream depth (m); Q is the design flow rate (m3/s); Au is the
upstream cross-sectional flow area (m2); ∆z is the total net change in canal invert
across the transition (m); hd is the downstream depth (m); Ad is the downstream
cross-sectional flow area (m2); and ∆E is the hydraulic energy loss across the
transition (m)
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The ∆z value is unknown at this point, but the slope of the water surface across
the transition should be equal to:
S ws =
hu + ∆z − hd
L
(6)
where Sws is the (constant) slope of the water surface across the transition
(m/m); and L is the length of the transition (m)
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Combining Eqs. 5 & 6:
S ws
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(7)
For Q = 15 m3/s; Ad = (2.15)(2.0) = 4.30 m2; Au = (1.87)(2.5)+(1.0)(1.87)2 = 8.172
m2; ∆E = 0.0137 m; and L = 8.0 m:
S ws
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Q2 ⎛ 1
1 ⎞
⎜⎜ 2 − 2 ⎟⎟ + ∆E
2g ⎝ A d A u ⎠
=
L
⎞
(15)2 ⎛ 1
1
−
⎜
⎟ + 0.0137
2(9.81) ⎝ (4.3)2 (8.172)2 ⎠
=
= 0.0578
8.0
(8)
Note that Sws ≠ SEL
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Gary P. Merkley
V. Change in Side Slope
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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:
m = 1 − 0.125 x
(9)
where 0 ≤ x ≤ 8 m
VI. Change in Bed Width
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The bed width decreases from 2.5 to 2.0 m over the length of the transition
This reduction is specified to be a reverse parabola, defined over L/2 = 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:
b = 2.5 −
x2
64
(10)
where 0 ≤ x ≤ 4 m
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For x > 4 m, the equation is:
(x − 8)2
b = 2.0 +
64
(11)
where 4 ≤ x ≤ 8 m
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You can also do this with a 3rd-degree polynomial:
b = Ax3 + Bx 2 + Cx + D
(12)
where A, B, C, D are fitted so that the slope is zero at x = 0 and at x = 8
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By quick inspection of Eq. 12, it is seen that b = 2.5 at x = 0, so D = 2.5
And, at x = 8, b = 2.0
The other two conditions are that the slope equal zero at the end points:
3Ax 2 + 2Bx + C = 0
(13)
where x = 0 and x = 8
Gary P. Merkley
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BIE 5300/6300 Lectures
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So, C must be equal to zero, and then A and B can be determined after a small
amount of algebra: A = 0.001953125, B = -0.0234375
The results are not identical, but very close (see the figure below)
2.50
2.45
Two parabolas
2.40
3rd-degree polynomial
Base width, b (m)
2.35
2.30
2.25
2.20
2.15
2.10
2.05
2.00
0
1
2
3
4
5
6
7
8
Distance, x (m)
VII. Change in Bed Elevation
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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 SEL = 0.001718 in the transition, and the slope of the water
surface is Sws = 0.0578
The velocity head can be described as follows:
V2
Q2
(15)2
=
+
x(S
−
S
)
=
+ 0.0561x
ws
EL
2
2g 2gA 2
2(9.81)(8.172)
(14)
or,
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Gary P. Merkley
V2
= 0.172 + 0.0561x
2g
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(15)
And, the cross-sectional area of flow, A, is equal to Q/V, which equals h(b+mh):
A=
Q
2g ( 0.172 + 0.0561x )
= h(b + mh)
(16)
where b and m are defined as functions of x in Eqs. 9, 10, 11; and 0 ≤ x ≤ 8 m
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Eq. 16 is quadratic in terms of h:
−b + b2 + 4mA
h=
2m
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(17)
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 2.0 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 ws x − z(x )
(18)
where 0 ≤ x ≤ 8 m; and z = 2.0 at x = 0
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Once h is known, use Eq. 18 to solve for z, then go to the next x value
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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+V2/2g, in the last column of the table (to the right)
decreases linearly at the rate of 0.001718 m per meter of distance, x, as we have
specified (see Eq. 3): the energy line has a constant slope
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Gary P. Merkley
270
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2.5
2.0
1.5
m (m/m)
b (m)
h (m)
z (m)
1.0
0.5
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
Distance (m)
4.050
9.0
Velocity head (m)
4.045
z+h (m)
Area (m2)
z+h+V2/2g
7.0
6.0
4.040
2
5.0
z+h+V /2g
2
Head (m) and Area (m )
8.0
4.035
4.0
4.030
3.0
2.0
4.025
1.0
0.0
4.020
0.0
2.0
4.0
6.0
8.0
Distance (m)
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Gary P. Merkley
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Note that the bed elevation, z, increases with x at first, then decreases to the final
value of 1.26 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 m3/s, but not for any other flow rate
Below are the transition design results using an arbitrary invert elevation of 2.00
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
272
BIE 5300/6300 Lectures
Transition Design Results
x
(m)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
6.2
6.4
6.6
6.8
7.0
7.2
7.4
7.6
7.8
8.0
m
(m/m)
1.000
0.975
0.950
0.925
0.900
0.875
0.850
0.825
0.800
0.775
0.750
0.725
0.700
0.675
0.650
0.625
0.600
0.575
0.550
0.525
0.500
0.475
0.450
0.425
0.400
0.375
0.350
0.325
0.300
0.275
0.250
0.225
0.200
0.175
0.150
0.125
0.100
0.075
0.050
0.025
0.000
BIE 5300/6300 Lectures
b
(m)
2.500
2.499
2.498
2.494
2.490
2.484
2.478
2.469
2.460
2.449
2.438
2.424
2.410
2.394
2.378
2.359
2.340
2.319
2.298
2.274
2.250
2.226
2.203
2.181
2.160
2.141
2.123
2.106
2.090
2.076
2.063
2.051
2.040
2.031
2.023
2.016
2.010
2.006
2.003
2.001
2.000
A
(m2)
8.165
7.911
7.680
7.467
7.272
7.091
6.922
6.766
6.619
6.482
6.352
6.230
6.115
6.007
5.903
5.805
5.712
5.623
5.538
5.456
5.379
5.304
5.233
5.164
5.098
5.034
4.973
4.914
4.857
4.802
4.748
4.697
4.647
4.599
4.552
4.506
4.462
4.419
4.378
4.337
4.298
h
(m)
1.87
1.84
1.82
1.80
1.78
1.76
1.75
1.73
1.72
1.72
1.71
1.70
1.70
1.70
1.70
1.70
1.70
1.70
1.71
1.72
1.73
1.74
1.75
1.76
1.78
1.79
1.81
1.82
1.84
1.86
1.88
1.90
1.92
1.94
1.96
1.99
2.02
2.05
2.08
2.11
2.15
273
2
V /2g
(m)
0.17200
0.18322
0.19444
0.20566
0.21688
0.22810
0.23932
0.25054
0.26176
0.27298
0.28420
0.29542
0.30664
0.31786
0.32908
0.34030
0.35152
0.36274
0.37396
0.38518
0.39640
0.40762
0.41884
0.43006
0.44128
0.45250
0.46372
0.47494
0.48616
0.49738
0.50860
0.51982
0.53104
0.54226
0.55348
0.56470
0.57592
0.58714
0.59836
0.60958
0.62080
2
z
z+h+V /2g
(m)
(m)
2.00
4.0420
2.02
4.0417
2.03
4.0413
2.04
4.0410
2.05
4.0406
2.05
4.0403
2.05
4.0400
2.05
4.0396
2.05
4.0393
2.05
4.0389
2.05
4.0386
2.04
4.0383
2.03
4.0379
2.02
4.0376
2.01
4.0372
2.00
4.0369
1.99
4.0366
1.97
4.0362
1.95
4.0359
1.93
4.0355
1.91
4.0352
1.89
4.0349
1.87
4.0345
1.84
4.0342
1.82
4.0338
1.79
4.0335
1.76
4.0332
1.74
4.0328
1.71
4.0325
1.68
4.0321
1.65
4.0318
1.62
4.0315
1.58
4.0311
1.55
4.0308
1.51
4.0304
1.48
4.0301
1.44
4.0298
1.40
4.0294
1.35
4.0291
1.31
4.0287
1.26
4.0284
Gary P. Merkley
Gary P. Merkley
274
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