open channel the islamic university of gaza faculty of engineering civil engineering department...
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![Page 1: Open Channel The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322 Chapter 6](https://reader034.vdocument.in/reader034/viewer/2022052202/56649ed95503460f94be6f33/html5/thumbnails/1.jpg)
Open Channel
The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Hydraulics - ECIV 3322
Chapter 6Chapter 6
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2
Example 1
1.5m
3.0m
2
1
Open channel of width = 3m as shown, bed slope = 1:5000, d=1.5m find the flow rate using Manning equation, n=0.025.
sVAQ
V
P
AR
P
A
SRn
V
h
h
/m 84.49538.0
m/s 538.050001927.0
025.0
1
927.0708.9
9
9.708 35.132
m 95.1935.0
1
3
3
2
22
2
3
2
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Example 2Open channel as shown, bed slope = 69:1584, find the flow rate using Chezy equation, C=35.
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sVAQ
V
P
AR
P
A
SRCV
h
h
/m 84.11352.1627.0
m/s 7.01584
69.0917.035
917.018.177
52.162
m 177.18 04.552.28.166.38.115072.0
m 52.16215072.06.32
52.272.08.1652.2
2
04.552.2
3
2222
2
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Example 3Circular open channel as shown d=1.68m, bed slope = 1:5000, find the Max. flow rate & the Max. velocity using Chezy equation, C=70.
sVAQ
V
mP
AR
dP
ddA
SRCV
h
h
/m 496.117.269.0
m/s 69.05000
1485.070
485.05.4
17.2
m 4.5 68.1180
154
m 17.21542sin8
68.1
180154
4
68.12sin
84
3
22222
154Max. flow rate
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m/s 748.05000
157.070
57.03775.3
93.1
m 378.368.1180
75.128
m 93.175.1282sin8
68.1
18075.128
4
68.12sin
842
2222
V
mP
AR
dP
ddA
SRCV
h
h
75.128Max. Velocity
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Trapezoidal open channel as shown Q=10m3/s, velocity =1.5m/s, for most economic section. find wetted parameter, and the bed slope n=0.014.
Example 4
mD
DDDA
DnDBA
mV
QA
BD
DBD
nDBnD
78.1
667.6)2
36055.0(
667.65.1
10
6055.02
232
231
2
21
2
2
2
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mP
nDDP
nDBP
49.72
3178.12)78.1(6055.0
126055.0
12
2
2
2
To calculate bed Slope
6.1941:1
5.189.0014.0
1
89.049.7
667.6
m 49.7
m 667.6
1
3
2
2
3
2
S
SV
P
AR
P
A
SRn
V
h
h
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Example 5
13
2
gA
TQ
1
81.93
14.2
3
124.233.1
23
2
3
2
cc
c
cc
c
DD
D
gDnDB
nDBQ
m 31.0cD
Determine the critical depth if the flow is 1.33m3/s. the channel width is 2.4m
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Example 6
44
81.9
25
3
2
3
1
2
3
1
2
23
1
2
2
BBBgB
Qyc
0.006 2
D
0.016
125
2
P
AR
S R n
1V
32
c
32
BD
B
BD
BD
BD
cc
c
c
Rectangular channel , Q=25m3/s, bed slope =0.006, determine the channel width with critical flow using manning n=0.016
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mB
B
B
B
BB
BB
BB
3
0.006 8
4
0.016
1
4
25
0.006 42
4
0.016
14
25
32
35
31
32
32
32
32
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Example 7A 3-m wide rectangular channel carries 15 m3/s of water at a 0.7 m depth before entering a jump. Compute the downstrem water depth and the critical depth
72.27.081.9
14.7
/14.77.0
5
366.181.9
5
/s.mm 53
15
1
11
11
3
2
3
gd
VF
smd
qV
md
q
r
c
md
d
365.2
1)72.2(812
1
7.0
2
22
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Example 8
dn = Depth can calculated from manning equation
d1=d
n
d2
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calsupercriti is flow the142.108.181.9
63.4
/63.43
15
08.1
004.032
3
01.0
1
3
15
322
3
1
1
11
11
1
32
32
gd
VF
smd
V
mddD
D
D
D
DBDP
DBDA
SRnA
Q
r
n
h
d1=d
n
d2
a)
b)
md
d
7.1
1)42.1(812
1
08.1
2
22
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c)
d1=d
n
d2
mgg
E
smd
V
032.02
94.27.1
2
63.408.1
/94.27.13
15
3
15
22
22
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Example 9A trapezoidal channel with a bottom width of 5m, side slope of 1H: 1V, and a Manning n of 0.013 carries a discharge of 50m3/s at a slope of 0.0004. Compute by the direct step method the backwater profile created by a dam that backs up the water to a depth of 6 m immediately behind the dam. The upstream end of the profile is assumed at a depth equal to 1% greater than the normal depth.
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17
3
2
3
5
3
22
3
5
2
1
3
22
3
5
3
5
3
2
2
1
2
13
2
3
22
1
3
2
2
1
3
2
225
15
1125
15
0004.0
013.0*50
12
*
11
1
n
nn
n
nn
n
n
y
yy
y
yy
nyb
ynyb
P
A
S
nQ
S
P
A
nSR
nA
Q
SRn
V
By trial and error, Yn = 2.87 m
determine normal depth, Yn
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1
581.9
5250
12
3
2
3
2
3
2
3
2
cc
c
cc
c
YY
Y
gA
TQ
YbnYg
bnYQ
gA
TQ
determine critical depth, Yc
By trial and error, Yc = 1.90 mControl depth = 1.01*2.87 =2.90m
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1 2 3 4 5 6 7 8 9 10 11 12
y A P R V E ΔE Sf Sfm So-Sfm Δx L
m m2 m m m/s m m *104 *104 *104 m m
2.9 22.91 13.2 1.735 2.182 3.165 3.710
0.057 3.545 0.455 1250 1250
3.0 24.0 13.49 1.779 2.080 3.222 3.380
0.080 3.190 0.810 987 2237
3.1 25.11 13.77 1.824 1.990 3.303 3.000
0.082 2.830 1.170 700 2937
3.2 26.24 14.05 1.868 1.900 3.384 2.660
0.260 2.268 1.732 1500 4437
3.5 29.75 14.90 1.997 1.680 3.644 1.875
0.455 1.505 2.495 1750 6187
4.0 36.0 16.31 2.207 1.390 4.099 1.135
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1 2 3 4 5 6 7 8 9 10 11 12
y A P R V E ΔE Sf Sfm So-Sfm
Δx L
m m2 m m m/s m m *104 *104 *104 m m
4.0 36.0 16.31 2.207 1.390 4.099 1.135
0.472 0.925 3.075 1526 7713
4.5 42.75 17.73 2.411 1.170 4.570 0.715
0.481 0.592 3.408 1410 9123
5.0 50.0 19.14 2.612 1.000 5.051 0.469
0.487 0.394 3.606 1350 10473
5.5 57.75 20.56 2.808 0.865 5.538 0.318
0.491 0.271 3.729 1320 11793
6.0 66.0 21.97 3.005 0.758 6.029 0.223