Post on 04-Apr-2018
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Characteristics of Gradually Varying Flow (GVF) Flow depths vary gradually along the channel.
Changes in flow depth occurs over a long distance.
Water surface nearly horizontal
2 conditions:
(a) Backwater (b) Drawdown
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Analysis and derivation of GVF Equation
Head loss is similar to uniform flow
and
Bed slope is small
Channel is prismatic
Constant roughness along the channel reach
oRSACQ oSARnQ3
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Differential Equation of GVF
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(5.1)
Differentiate with respect to x:
(5.2)
Where;
and
g
v
dx
d
dx
dy
dx
dz
dx
dE
dx
dz
dx
dH
2
2
2
2vH z E z y
g
;fS
dx
dH oS
dx
dz
dx
dy
dy
dA
gA
Q
dx
dy
gA
Q
dx
d
g
v
dx
d3
2
2
22
22
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Since dA/dy = T
Rewrite Eqn. (5.2),
Rearrange the above Eqn.,
(5.3)
3
2
1gA
TQ
SS
dx
dy fo
dx
dy
gA
TQ
dx
dy
g
Q
dx
d3
22
2
dx
dy
gA
TQ
dx
dSS of
3
2
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If K = conveyance at any depth y and Ko = conveyancecorresponding to normal depth yo, then:
So,
;
f
QK
S
2
2
K
K
S
So
o
f
o
o
QK
S
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Rearrange the above Eqn.,
(5.4)
Eqn. (5.4) is useful in developing direct integrationtechniques and it is applicable for channel with any
geometrical shape
2
0
2
3
1
1
o
KS
Kdy
Q TdxgA
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Differential Equation for Rectangular Channel
Substitute relevant parameters and simplifying Eqn. (5.4)resulted;
(5.5)
Where;
and
3
2
1
1
y
y
K
K
Sdx
dy
c
o
o
n
ARK
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oo
QK
S
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Differential Equation for Very Wide RectangularChannel (R = y)
Substitute relevant parameters and simplifying Eqn. (5.5)
resulted;
;for Mannings n (5.6)
3
310
1
1
y
y
y
y
Sdx
dy
c
o
o
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Substitute relevant parameters and simplifying Eqn. (5.5)resulted;
;for Chezy (5.7)
3
3
1
1
o
o
c
y
ydyS
dx y
y
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Computations of GVF
Computation of GVF involves basically solution of dynamiceqn. of GVF
Several methods of computation:
(a) Direct Integration method
(b) Numerical Integration method
(c) Graphical integration method
(d) Standard step method Numerical Integration method will be discussed
The GVF differential equation is written in finite differenceform
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Numerical Integration Method
The unknown length of channel to be determined is dividedinto several portions called segment
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Referring to the segment above:
= length of the segment
= changes of flow depth between segments, whichfixed in the calculation =
= average flow depth in each segment =y
x
y
1
2
1
ii yy
ab yy 2
1
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Rectangular Channel (Eqn. 5.5)
Substitute relevant parameters and simplifying Eqn. (5.4)resulted;
(5.8)
Where;
and
3
2
1
1
y
y
K
K
Sx
y
c
o
o
;32
g
qyc
o
ooo
S
Q
n
RAK
32 3
23
2
2
yB
yB
n
yB
n
RAK
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Very Wide Rectangular Channel (Eqn. 5.6)
Substitute relevant parameters and simplifying Eqn. (5.6)resulted;
For Manning; (5.9a)
Where;
3
310
1
1
y
y
y
y
Sx
y
c
o
o
3
2
gqyc
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Very Wide Rectangular Channel (Eqn. 5.6)
Substitute relevant parameters and simplifying Eqn. (5.7)resulted;
For Chezy; (5.9b)
Where;
3
3
1
1
o
o
c
y
yy Sx y
y
32
gqyc
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Differential Equation of GVF
Y
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A very wide rectangular channel with bed slope
0.001, Mannings n 0.025 and discharge 2.5
m3/s.m. Determine the backwater curve created by
a low dam that has water depth of 2.0 m
immediately upstream of the dam. The upstream
computations may be carried out to a depth 1%
greater than the normal depth. (Use numericalintegration method in your computations and divide
the range of depth into 4 equal parts).
Example 1
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3
10
3
1
1
yy
yy
Syx
o
c
o
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Solution
Find yo, so from Manning Eqn.
R = y
n
SAR
Qo
32
n
SyBy
Qooo
32
n
Syq
oo3
5
53
1000
025.05.2
oy
q=Q/B
myo 5.1
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Solution
Find yc, for numerical integration method
32
g
qy
c
3
2
81.9
)5.2(
myc 86.0
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Solution
1% above yo, 1% of 5.101.0 oy
my 52.1
m02.0015.0 02.05.1 y
?L
oy my 52.1 my 0.2
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m52.1 m0.2m88.1m76.1m64.1
4x 3x 2x 1x
4 3 2 1y y y y
divide the range of depth into 4 equal parts
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Solution
From Eqn (5.9a)
where
310
3
1
1
yy
yy
S
y
xo
c
o
my 12.04
52.12
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Solution
Or
Where and
3
10
3
1
1
001.0
12.0
y
y
y
y
x
o
c
BAx 120
3
1
yy
A c3
10
1
yy
B o
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Solution
Or
Where and
3
10
3
1
1
001.0
12.0
y
y
y
y
x
o
c
BAx 120
3
1
yy
A c3
10
1
yy
B o
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xy A B
2.00-1.88
1.88-1.76
1.76-1.64
1.64-1.52
L=
Create Table!!!
The length of backwater, L = m
y
ix
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xy A B
2.00-1.88 1.94 0.912 0.576 190.1
1.88-1.76 1.82 0.895 0.475 226.1
1.76-1.64 1.70 0.871 0.341 306.9
1.64-1.52 1.58 0.839 0.159 632.9
So, L= 1355.3m
Create Table!!!
The length of backwater, L = 1355m
y
ix
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xy A B
2.00-1.88
1.88-1.76
1.76-1.64
1.64-1.52
So, L=
Create Table!!!
The length of backwater, L = 1718m
y
ix
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