lecture 11,12 date: 08,09-03-2020...the dynamic equation of gradually varied flow, presented...
TRANSCRIPT
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Lecture 11,12
Date: 08,09-03-2020
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FLOW PROFILES IN SERIAL ARRANGEMENT OF CHANNELS
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Flow profiles in a mild slope channel followed by a steep slope channel
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Flow profile in a steep slope channel followed by a mild slope channel
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Lecture 13
Date: 16-03-2020
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Example 6.1
A trapezoidal channel with b = 6m, n = 0.025, s
= 2 and S0 = 0.001 carries a discharge of 28 m3/s.
At a certain section A of the channel the depth of
flow is 1.30 m. (i) Determine the type of channel
slope. (ii) Determine the type of flow profile. (iii)
If at another section B, the depth of flow is 1.50 m,
state whether section B is located upstream or
downstream of section A.
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Example 6.1
A trapezoidal channel with b = 6m, n = 0.025, s = 2 and Sn = 0.001 carries a discharge of 28
m3/s. At a certain section A of the channel the depth of flow is 1.30 m. (i) Determine the type of
channel slope. (ii) Determine the type of flow profile. (iii) If at another section B, the depth of
flow is 1.50 m, state whether section B is located upstream or downstream of section A.
b= 6
s= 2
S0= 0.001
Q= 28
α= 1
n= 0.025
22
h A P R AR2/3
hn 1.91 1.91 18.76 14.54 1.29 22
9
h A B D Z = AD
hc 1.14 1.14 9.44 10.56 0.89 9
068.11001.0
14025.0
0
3/2
S
nQAR 068.11
001.0
14025.0
0
3/2
S
nQAR
/g
QZ c /g
QZ c
i) hn>hc Mild slope channel
ii) M1 flow profile.
iii) As, hA=1.30m, hB=1.50m
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Lecture 14
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Example 6.2: A rectangular channel b=10 m
wide and having = 1.10 and n = 0.025 has
three reaches arranged serially. The bottom
slopes of these reaches are S0=0.0040,
0.0065 and 0.0090, respectively. For a
discharge of Q=35 m3/s in this channel,
sketch the resulting flow profiles.
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PROBLEM
Solution The critical depth for the given conditions is obtained as
mgb
QhC 11.1
1081.9
3510.13
2
2
32
2
Since critical slope is the slope for which flow in the channel is both uniform and critical, hence
hn = hc = 1.11 m
Therefore, A = 10 1.11 = 11.12 m2, P = 10 + 2 1.11 = 12.22 m and R = A/P = 0.9 m, and
0070.091.012.11
35025.022
3/2
AR
nQSc
Thus, the bottom slopes of the three reaches are mild, steeper mild and steep, respectively. The resulting
flow profiles are M2, M2 and S2, as shown in the following figure. Mild (S0 < Sc)
Steep (S0 > Sc)
0.0040Mild
0.0065 Steeper mild
0.0090>0.0070->Steep
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COMPUTATION OF GRADUALLY VARIED FLOW PROFILES
)18.6.......()/(1
)/(10 M
c
N
n
hh
hhS
dx
dh
Dynamic equation of gradually varied flow
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COMPUTATION OF GRADUALLY VARIED FLOW PROFILES
The computation of the gradually varied flow basically involves the integration of
the dynamic equation of gradually varied flow, presented earlier.
This equation is a non linear ordinary differential equation of the first order and its
solution requires one boundary condition for depth, i.e. the depth at the section
where the computation begins must be given.
This equation can be easily integrated (i) for a wide channel, and (ii) for a
horizontal channel. For other channels, the integration of the gradually varied flow
equation has to be performed either graphically or numerically.
The computation of gradually varied flow profile must begin with the known depth
of flow at a control and proceed in the direction in which the control operates.
the computation of flow profiles is usually terminated at a section where the depth
of flow is about 5% greater or less than the normal depth.
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Information are generally required:
1. The discharge Q for which the flow profile is desired.
2. The depth of flow or stage at the control section where the computation
begins.
3. The channel shape at various channel sections.
4. The bottom slope S0 of the channel.
5. The energy coefficient .
6. The Manning’s n or Chezy’s C.
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There are many methods for computing gradually varied flow profiles.
However, these methods can be broadly classified into the following two categories:
i) Methods used for computing flow profiles in prismatic channels.
ii) Methods used for computing flow profiles in non-prismatic channels.
The methods used for computing flow profiles in prismatic or regular or uniform
channels compute a longitudinal distance x for a given h explicitly without involving
any trial. The direct step method and the direct integration method fall in this
category.
On the other hand, the methods used for computing flow profiles in non-prismatic or
irregular or non-uniform channels compute h from a given x. In this case, a trial-
and-error procedure is necessary. The standard step method falls in this second
category.
Methods for computing GVF
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Computation of Flow Profiles in Prismatic Channels
(a) Direct Integration Method
Direct integration of the gradually varied flow equation for computing flow profiles in a
wide channel and in a horizontal channel is simple and considered here.
(i) Flow Profile a Wide Channel: Breese Method
Equation (6.18) can be integrated exactly for a wide rectangular channel with the
conveyance expressed in terms of the Chezy equation. For this case, M = N = 3 and
3
3
0)/(1
)/(1
hh
hhS
dx
dh
c
n
Putting u = h/hn, so that du = dh/hn in above Eq. and rearranging yields
duuh
h
S
hdx
n
cn
33
3
0 1
111
which on integration gives 13
3
0
1 Ch
hu
S
hx
n
cn
)18.6.......()/(1
)/(10 M
c
N
n
hh
hhS
dx
dh
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where is the Breese function given by
u
uu
uu
u
du
0
1
2
2
3 12
3tan
3
1
1
1ln
6
1
1
and C1 is a constant of integration.
This integration was first performed by J.A. Ch. Breese in 1860.
A determination of the flow profile by this solution is widely known as the Breese
method.
For a wide channel the critical depth hc is given by 3
2
g
qhc
and, using the Chezy formula, the normal depth hn is given by2
32
0
n
qh
C S
where C is Chezy’s C and q is the discharge per unit width, and hence
g
SC
h
h
n
c 0
23
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The length of the flow profile between two consecutive sections of depth h1 and h2 is
123
3
12
0
12 1 n
cn
h
huu
S
hxxL
13
3
0
1 Ch
hu
S
hx
n
cn
3
1 1 1 13
0
1n c
n
h hx u C
S h
3
2 2 2 13
0
1n c
n
h hx u C
S h
where 1 and 2 are the values of corresponding to u1 = h1/hn and u2 = h2/hn, respectively
211 1
1 2
11
212 2
2 2
21
11 1 3ln tan
6 2 131
11 1 3ln tan
6 2 131
u u
uu
u u
uu
h2h1
L
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Example 6.3
A wide rectangular channel with Chezy’s C = 47 m1/2/s and S0 = 0.0001
carries a discharge of 2 m2/s. A dam raises the water level by 0.50 m above
the normal depth at the dam site. Compute the length of the resulting flow
profile between the dam site and the location where the depth is 2.90 m.
0.50 h2=2.90
h1=hn+0.50
L
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mSC
qh
mg
qh
n
c
626.20001.00.47
2
742.081.9
2
32
2
3
0
2
2
3
2
3
2
Since hn > hc, the channel slope is mild.
Now, h1 = 2.626 + 0.50 = 3.126 m, h2 = 2.90 m, Since h1 or h2 > hn > hc, the
profile is M1.
So, u1 = h1/hn = 1.190
u2 = h2/hn = 1.104.
6687.02858.09545.0
12
3tan
3
1
1
1ln
6
1
4933.02734.07667.012
3tan
3
1
1
1ln
6
1
2
1
2
1
2
2
22
1
1
2
1
1
2
11
uu
uu
uu
uu
Hence, the length of the profile is obtained using Eq. (6.33) as
3
2 1 2 13
0
3
3
1
0.7422.6261.104 1.190 1 0.6687 0.4933 6760.06
0.0001 2.626
n c
n
h hL u u
S h
m
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Flow Profile in a Horizontal Channel
*By yourself