00open channel
DESCRIPTION
HydraulicsTRANSCRIPT
![Page 1: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/1.jpg)
1
OPEN CHANNEL FLOW
![Page 2: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/2.jpg)
2
![Page 3: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/3.jpg)
3
![Page 4: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/4.jpg)
4
Question – What is the most obvious difference between pipe flow and open channel flow????????????? (in terms of flow conditions and energy situation)
Typical open channel shapes – Figure 14.1
![Page 5: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/5.jpg)
5
![Page 6: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/6.jpg)
6
Types of open channel flows – Steady flow – when discharge (Q) does not change with time. Uniform flow – when depth of fluid does not change for a selected length or section of the channel Uniform steady flow – when discharge does not change with time and depth remains constant for a selected section
- cross section should remain unchanged – referred to as a prismatic channel
Varied steady flow – when depth changes but discharge remains the same (how can this happen?) Varied unsteady flow – when both depth and discharge change along a channel length of interest. Rapidly varying flow – depth change is rapid Gradually varying flow – depth change is gradual
![Page 7: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/7.jpg)
7
Figure 14.3
Section 1 – rapidly varying flow Section 2 – gradually varying flow Section 3 – hydraulic jump Section 4 – weir and waterfall Section 5 – gradually varying Section 6 – hydraulic drop due to change in channel slope
![Page 8: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/8.jpg)
8
Hydraulic radius of open channel flow A parameter that is used often Ratio of flow cross sectional area (A) and wetted perimeter (WP)
R = A/ WP Hydraulic radius R for various channel shapes – Figure 14.1
![Page 9: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/9.jpg)
9
![Page 10: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/10.jpg)
10
Kinds (types) of open channel flow Reynolds number for pipe flow –
υvDN R =
Reynolds number for channel flow –
υvRNR =
For pipe flow – NR < 2000 – laminar NR > 4000 – turbulent For channel flow – NR < 500 – laminar NR > 2000 – turbulent
![Page 11: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/11.jpg)
11
Another “number” for channel flow! Froude Number [NF] (gravity versus inertial forces)
hF gy
vN =
Where yh is referred to as the hydraulic depth and given as –
yh = A/T where A is the area and T is the top width of the channel NF = 1.0 or when v = (gy)1/2 - critical flow NF < 1.0 – subcritical flow NF > 1.0 – super critical flow A combination of both the numbers is used to describe channel flow conditions.
![Page 12: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/12.jpg)
12
Uniform steady flow and Manning’s Equation When discharge remains the same and depth does not change then we have uniform steady flow. In this condition – The surface of water is parallel to the bed of the channel
Or S = Sw
Where S is the slope of the channel The slope of the channel can be expressed as –
- An angle = 1 degrees - As percent = 1% - Or as fraction = 0.01 or 1 in 100
![Page 13: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/13.jpg)
13
Velocity of flow (v) in a channel can be computed numerous empirical equations – One of them is Mannings equation –
2/13/20.1 SRn
v =
This the SI units form of the equation with v (meters/sec) and R (meters). Where n is the Manning’s coefficient (dimensionless) – values developed through experimentation Possible n values for various channel surfaces – Table 14.1
![Page 14: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/14.jpg)
14
![Page 15: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/15.jpg)
15
In English units the Manning’s equation form is –
2/13/249.1 SRn
v =
Where v is in feet/sec and the R value is in feet. If velocity is known, the discharge (Q) can then be computed as –
Q = A*v
2/13/20.1 SARn
Q =
Where Q is in m3/s For uniform flow, Q is referred to as Normal discharge
![Page 16: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/16.jpg)
16
The above equation can also be re-arranged such that –
2/13/2
SnQAR =
The left hand term is simply based on channel geometry.
![Page 17: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/17.jpg)
17
Problem 14.2 Determine normal discharge for a 200 mm inside diameter common clay drainage tile running half-full if the slope drops 1 m over 1000 m.
S = 1/1000 = 0.001 A = (1/2) * (π D2/4) = 0.5*π*(0.2)2/4 = 0.0157 m2 WP = (1/2) * (π D) = 0.5*π*0.2 = 0.3141 m R = 0.05 m From Table 14.1 n for clay tile = 0.013 Substitute these values in the equation –
![Page 18: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/18.jpg)
18
2/13/20.1 SARn
Q = And we get
2/13/2 )001.0()05.0(*0157.0*013.00.1
=Q Q = 5.18 x 10-3 m3/s
![Page 19: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/19.jpg)
19
Problem 14.3 Calculate slope of channel below If Q = 50 ft3/s Formed unfinished concrete channel
Equation that you will use
2/13/20.1 SARn
Q =
Or
![Page 20: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/20.jpg)
20
3/22/1
49.1 ARQnS =
Compute A = 12 ft2 WP = 9.66 ft R = A/WP = 12/9.66 = 1.24 ft Manning’s n for concrete channel = 0.017 Substitute And S = 0.00169 Drop 1.69 ft for every 1000 ft.
![Page 21: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/21.jpg)
21
Problem 14.4 Design rectangular channel in formed unfinished concrete Q = 5.75 m3/s S = 1.2% Normal depth = ½ of the width of the channel Since we have to design the channel – the equation that should be used –
2/13/2
SnQAR =
RHS is known. RHS = 0.017*5.75/(0.012)1/2 = 0.0892 Now we know that y = b/2 Express Area and the hydraulic radius in terms of b. A = by = b2/2
![Page 22: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/22.jpg)
22
WP = b+ 2y = 2b R = A/WP = b/4 Therefore, LHS = AR2/3 = b2/2 * (b/4) 2/3 = RHS = 0.892 B = 1.76 m y = 1.76/2 m
![Page 23: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/23.jpg)
23
Problem 14.5 In the problem above the final width was set at = 2m and the maximum Q = 12 m3/s; find the normal depth for this maximum discharge. OK again,
2/13/2
SnQAR =
RHS = 0.017*12/(0.012)1/2 = 1.86 B = 2m A = 2y WP = 2+2y R = 2y/(2+2y) Therefore LHS =
![Page 24: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/24.jpg)
24
86.122
223/2
=⎟⎟⎠
⎞⎜⎜⎝
⎛+ y
yy
Cannot solve this directly, will have to do trial and error. Set up a Table and compare y (m) A (m2) WP
(m) R (m) R2/3 AR2/3 Required
change in y
2.0 4.0 6.0 0.667 0.763 3.05 Make y lower
1.5 3.0 5.0 0.600 0.711 2.13 Make y lower
1.35 2.7 4.7 0.574 0.691 1.86 OK
![Page 25: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/25.jpg)
25
Conveyance and most efficient channel shapes Look at the RHS of the equation
2/13/20.1 SARn
Q =
Other than the S term, all other terms are related to channel cross section and its features. These terms together are referred to as the Conveyance (K) of the channel
3/20.1 ARn
K =
OR
2/1SKQ =
![Page 26: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/26.jpg)
26
K is maximum when WP is the least for a give area
this is also the most efficient cross section for conveying flow
For circular section – half full flow is the most efficient For other shapes – see Table 14.3 from the text.
![Page 27: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/27.jpg)
27
![Page 28: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/28.jpg)
28
Compound Sections When channel shape changes with flow depth – typical in natural stream sections during flooding During floods – water spills over the flood plain You need to know Q at various depths or vice-versa – so that you can design channels or determine channel safety for various flood magnitudes
![Page 29: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/29.jpg)
29
Cazenovia Creek in Buffalo during “normal” flow conditions
![Page 30: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/30.jpg)
30
Cazenovia Creek during flood!
![Page 31: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/31.jpg)
31
Problem 14.21E Figure 14.21 – natural channel with levees
Channel – earth with grass cover, n = 0.04 S = 0.00015 Determine normal Q for depth = 3 and 6 ft.
![Page 32: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/32.jpg)
32
![Page 33: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/33.jpg)
33
Assignment # 9
- 14.3E - 14.9M - 14.10M - 14.14M
![Page 34: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/34.jpg)
34
Compound section – More realistic situation – channel roughness n may be different for floodplain than the main channel Why?????
In that case
- determine velocity for each sub section - and then sum up the discharges for the sections
![Page 35: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/35.jpg)
35
2/13/2
49.1 SPA
nv
i
i
ii ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
i
n
ii AVQ ∑
=
=1
![Page 36: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/36.jpg)
36
Example Problem –
Compute discharge for depth of 8 feet S = 0.5% n for bank = 0.06 n for main channel = 0.03 A1 = 80*4 = 320 A2 = 50*8 = 400 A3 = 100*5 = 500 P1 = 80+4 = 84 P2 = 4+50+3 = 57 P3 = 100+5 = 105
![Page 37: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/37.jpg)
37
2/13/2
49.1 SPA
nv
i
i
ii ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
i
n
ii AVQ ∑
=
=1
⎥⎦
⎤⎢⎣
⎡++=
06.0500)105/500(
03.0400)57/400(
06.0320)84/320()005.0(49.1
3/23/23/22/1Q
Q = 9010 cfs.
![Page 38: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/38.jpg)
38
Energy Principles for Open Channel flow Energy at a particular point in the channel = PE + KE
gvyE2
2
+=
Where y is the depth of flow and v is the velocity Note – no pressure term! This is energy with respect to the channel bottom – Specific Energy When energy is measured with respect to another fixed datum – Total energy
gvzyE2
2
++=
Where z is the height of the channel bottom from the datum In terms of Q the specific energy can be expressed as –
![Page 39: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/39.jpg)
39
2
2
2gAQyE +=
Where Q is the discharge and A is the cross-sectional flow area
![Page 40: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/40.jpg)
40
Example Problem Channel width (rectangular) = 2m Depth = 1m Q = 4.0 m3/s Height above datum = 2m Compute specific and total energy A = by = 2.0*1.0 = 2 m2 Specific energy =
2
2
2gAQyE +=
2
2
2*81.9*241 +=E
E = 1.20 m Total energy = = Datum height + specific energy = 2.0 + 1.20 = 3.20 m
![Page 41: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/41.jpg)
41
Specific Energy Diagram The specific energy can be plotted graphically as a function of depth of flow.
2
2
2gAQyE +=
E = Es + Ek Es = y (static energy) Ek = Q2/2gA2 (kinetic energy) Relationship between y and Es & Ek
![Page 42: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/42.jpg)
42
Combining the two relationships – specific energy diagram
Key points from the diagram –
1. the diagram applies for a given cross section and discharge
2. as the depth of flow increases, the static energy increases, and the kinetic energy decreases
![Page 43: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/43.jpg)
43
3. the total energy curve approaches the static energy curve for high depths and the kinetic energy curve for small depths
4. The specific energy is minimum (Emin) for a particular
depth – this depth happens to be the critical depth – Depth for which the Froude’s number = 1.0. velocity = Vc.
5. Emin – only energy value with a singular depth!
6. Depths less than the critical depths – supercritical flow.
Froude Number > 1.0. V > Vc.
7. Depths greater than the critical depths – subcritical flow. Froude Number < 1.0. V < Vc.
8. For all other energy values – there are two depth
associated – one greater than the critical depth and one less than the critical depth.
9. The two depths associated with the same energy values are
referred to as – Alternate depths
10. As discharge increases, the specific energy curves move to the upper right portion of the chart.
![Page 44: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/44.jpg)
44
![Page 45: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/45.jpg)
45
Additional Equations Specific energy equation –
2
2
2gAQyE +=
Taking a derivative and equating it to zero (critical flow conditions when energy is minimum) We get – Condition at critical flow
13
2
=gA
BQ
Solving these further, for a rectangular channel (A = By), we get –
13
2
=cgy
q
![Page 46: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/46.jpg)
46
Or
3
2
gqyc =
Critical depth can be determined explicitly Also, for rectangular channel -
cyE23
min = Explicit equations that can quickly give you the critical depth and minimum specific energy for a rectangular channel – no need to interpolate from graph
![Page 47: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/47.jpg)
47
Example Problem: • Rectangular channel • Width = 4 m • Q = 12.0 m3/s • Depth of flow = 2.5 m
• Draw specific energy diagram • Find critical and alternate depth
2
2
2gAQyE +=
2
2
2gyqyE +=
Specific discharge – discharge per unit width = q = Q/B The advantage of using specific discharge is that we avoid using B and relate q directly to y q = 12/4 = 3m2/s
![Page 48: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/48.jpg)
48
Set up a table and compute the specific energy for every 0.2m depth increment.
2
2
2gyqyE +=
y KE total E
0.20 11.47 11.67 0.40 2.87 3.27 0.60 1.27 1.87 0.80 0.72 1.52 1.00 0.46 1.46 1.20 0.32 1.52 1.40 0.23 1.63 1.60 0.18 1.78 1.80 0.14 1.94 2.00 0.11 2.11 2.20 0.09 2.29 2.40 0.08 2.48 2.60 0.07 2.67 2.80 0.06 2.86 3.00 0.05 3.05 3.20 0.04 3.24 3.40 0.04 3.44 3.60 0.04 3.64 3.80 0.03 3.83 4.00 0.03 4.03
![Page 49: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/49.jpg)
49
00.5
11.5
22.5
33.5
44.5
5
0 1 2 3 4 5
E
y
00.5
11.5
22.5
33.5
44.5
5
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
E
y
![Page 50: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/50.jpg)
50
Explicit computation –
13
2
=cgy
q
Or
3
2
gqyc =
971.081.9
33
2
==cy
cyE23
min =
457.1971.0*23
min ==E
![Page 51: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/51.jpg)
51
Since given depth 2.5 m > 0.971 – the given depth is subcritical and the other depth should be supercritical Now determining alternate depths – Energy at 2.5 m =
2
2
2gyqyE +=
57.25.2*81.9*2
35.2 2
2
=+=E
This energy value is the same for the other alternate (supercritical) depth, so –
2
2
*81.9*2357.2
yy +=
![Page 52: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/52.jpg)
52
Solve equation by trail and error y E
0.40 3.27 0.41 3.14 0.42 3.02 0.43 2.91 0.44 2.81 0.45 2.72 0.46 2.63 0.47 2.55 0.48 2.47 0.49 2.40 0.50 2.33 0.51 2.27 0.52 2.22 0.53 2.16 0.54 2.11
y = 0.467 m – supercritical alternate depth.
![Page 53: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/53.jpg)
53
Problem 14.41E from text GIVEN – • Triangular channel with side slopes having ratio of 1:1.5
• Q = 0.68 ft3/s
• Channel – clean, excavated earth
CALCULATE –
a. critical depth b. Emin c. Plot specific energy curve d. Determine energy for 0.25 ft and alternate depth e. Velocity of flow and Froude number f. Calculate required slopes if depths from d are to be normal
depths for given flow
![Page 54: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/54.jpg)
54
Solution –
y, A = zy2, v = Q/A, T =2zy, yh=A/T ,
![Page 55: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/55.jpg)
55
![Page 56: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/56.jpg)
56
y, A = zy2, v = Q/A, T =2zy, yh=A/T ,
hF gy
vN =
2
2
2gAQyE +=
![Page 57: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/57.jpg)
57
![Page 58: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/58.jpg)
58
![Page 59: 00Open Channel](https://reader034.vdocument.in/reader034/viewer/2022042820/55cf9981550346d0339db592/html5/thumbnails/59.jpg)
59
Assignment # 10 - Specific Energy of Channel Flow • 14.39M • 14.42E