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Part II: Non-uniform Flow
Changing conditions takesplace over a long distance
Gradually varied flowRapidly varied flow
Changing of conditionstakes place over a short
distance
Reference books1. Hydraulics in Civil and Environmental Engineering, Andrew Chadwick and John Morfett,
2nd Ed, E & FN Spon, 1995.
2. Introduction to Fluid Mechanics, Robert W. Fox, Alan T. McDonald, P. J. Pritchard, 6th
Ed, John Willey & Sons, 2004.
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Occurs when there is a sudden change in geometry,
Flow over a sharp-crested weirs,
Flow through regions of rapidly varied cross-section,e.g. venturi flumes and broad-crested weirs.
A sudden change in flow regime
Normally associated with hydraulic jump phenomena, i.e.
flow with high speed and low depth is rapidly changed to
low speed and high depth.
1. Rapidly varied flow:
length)sticcharacteri:(/ LgLVFr= (6.1)
In open channel flow, flow regime is defined by Froude number:
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1.1 Characteristics of rapidly varied flows:
The use of energy and momentum principles
Main task: to determine of water depth
Bernoulli equation and continuity equation
Specific energy E
Specific discharge q
Alternate depths
Water surface profile changes suddenly and has curvature,
Pressure distribution departs from hydrostatic distribution,
The assumption of parallel streamline does not apply.
1.2 Methods of analysis for rapidly varied flows:
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1.3 Energy considerations
Example: In a rectangular channel, fluid flows over a section inwhich the bed gradually rises by z. The upstream depth y1 and
discharge Q are known. What is the depth of the flow at position 2?
Assuming no energy loss between (1) and (2), apply the energy Eq:
y
V
g y
V
g z11
2
2
2
2
2 2+ = + + (6.2)
datum
(1) (2)
Figure 6.1
p2/=0p1/=0
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Apply the continuity equation between points (1) and (2):
q V y V y= =1 1 2 2 (6.3)
where q (m3/s/m) is the specific discharge (discharge per unit width).
yq
gyy
q
gyz
1
2
12 2
2
222 2
+ = + +
2 2 2 023
2
2
1
2
1
2
2gy y g z gy
q
y
q+ + =( )
(6.4)
3 solutions for y2. Which one(s) is(are) correct?
Substitute (6.3) into (6.2):
Cubic equation
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To answer the above question, we use the concept of specific energy
H z yV
gz y
Q
gAz E= + + = + + = +
2 2
22 2
E yV
gy
Q
gA= + = +
2 2
22 2
(6.5)
The energy at a location is given by:
Channel bedwhere
E: specific energy=energy between energy line (EL) and channel bed.
For a rectangular channel: Q/A = bq/by = q/y
y
z
Water surface
datum
gV 2/
2
EL
E yq
gy= +
2
22
(6.6)
Eq. (6.6) gives two asymptotes:
y E 0 y E y
(Please note the definition of E: E = local water depth + velocity head)
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Using Eq (6.6), Eq. (6.4) can be written as E E z1 2= + (6.7)
Point A is defined by E1 corresponds to location 1,
Location 2 in the channel have E2 and is on the specific energy curve.
Possible B1 and B2 whose depths are known as alternate depths To arrive at B2 from A requires that E2 < E1- z at some intermediate
points. This means energy loss in a frictionless system impossible.
Therefore the flow depth at (2) must be B1
on the specific energy curve.
alternate depths
Figure 6.2Eq. (6.6)
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Solution:
Q = 8 m3/s; b = 5 m;
q = Q/b = 8/5 = 1.6 m2/s
Example 6.1:
The discharge in a rectangular channel of width 5m is 8 m3/s. The
normal depth is 1.25 m. Determine the depth of flow above a sectionin which the bed gradually rises by 0.2 m. Use graphical methods to
find the solution.
zEE += 21
222 /13.0)2/( yygyqyE +=+=Specific Energy
00.20.40.60.8
11.21.41.61.8
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
E (m)
y
(m)myyE 33.1/13.0
2
111 =+=
From energy Eq.,
mzEE 13.112 ==
E1=1.33
z
y2
E2=1.13
From the figure we get: y2=1m
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Example 6.2
Discharge in a rectangular channel of width 5m is 8 m3/s. The normal
depth is 1.25 m. Determine the flow depth where the section contractsby 1.0 m. Use graphical and numerical methods to find the solution.
smAQV /28.1)25.15/(8/11
===
mgVyE 33.12/2111 =+=
2222 /2)4/(8/ yyAQV ===
222
2222 /204.02/ yygVyE +=+=
21 EE =From energy Eq. 0
0.5
1.0
1.5
2.0
0 0.5 1.0 1.5 2.0
y2=1.18m
E1=1.33m=E
2
E (m)
y(m)
we get y2=1.18m
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Example 6.3
The discharge in a rectangular channel of width 5m is 8 m3/s. The
normal depth is 1.25 m. Determine the depth of flow where the
section contracts by 1.0 m and the bed gradually rises by 0.2 m.
Use graphical methods to find the solution.
Plan view
smQV /28.1)25.15/(8/ 11 === mgVyE 33.12/2
111 =+=
2222 /2)4/(8/ yyAQV ===222
2222 /204.02/ yygVyE +=+=
or 1221 zEEzEE =+=From energy equation, we have:
(1)
Plot the specific energy curve for E2,
the water depth corresponding
to E2 is y2.
z =
y2y1
Elevation
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0
0.5
1.0
1.5
2.0
0 0.5 1.0 1.5 2.0
Eq (1)
E2
z=0.2
y2=0.83 m
E1=1.33m
E (m)
y(m)
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1.4 Subcritical, critical and supercritical flow
E y
q
gy= +
2
22 (6.6)
Eq (6.6) is plotted in (a) and Eq(6.8) is plotted in (b).
The specific volume flow rate q can be expressed as a function of y,From Eq. (6.6),
)(2 2 yEgyq = (6.8)
(q is constant)
(E is constant)
subcritical
supercritical
(a) (b)
Figure 6.3Eq(6.6)
Eq(6.8)Critical point
12
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Derivation of the critical depth for a rectangular channel
The critical depth yc can be evaluated from Eq(6.6) by setting thederivative of E with respect to y equal to zero:
01/3
2
==gy
qdydE
2
2
2gy
qyE +=
0Fr112
2
== gy
V
From , to let E have an extremum value,
Since q=Vy
For a rectangular channel, we have 3gy
q
gy
VFr ==
1Fr =
Thus, for critical depth, Fr = 1.
(6.9)
Solving for depth in Eq (6.9) we have: ( ) cygqy ==3/12/ (6.11);
and 2/3 cc yE = (6.12)
(6.10)
c cV gy= (6.11a)
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A generalized derivation of the critical depth
For a generalized cross-section, the specific energy can be writtenin terms of the total discharge Q and the cross-sectional area A:
2
22
22 gA
Q
yg
V
yE +=+=
013
2
==dy
dA
gA
Q
dy
dE
(6.13)
The minimum-energy is obtained by differentiating (6.13) with
respect to y, i.e.
(6.14)
For incremental changes in depth. The change in area is
BdydA = (6.15)
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Substituting (6.15) to (6.14), we get:
2 3
1 / 0Q B gA = (6.16)
The second term in Eq(6.16) is the Froude number squared,
2 3 /Fr // /
Q A VQ B gA
gA B gA B= = = (6.18)
The ratio A/B is the hydraulic depth (B is the surface width).
For a rectangular channel, A/B=y, Eq (6.18) is simplified to (6.10).
Therefore, at the critical depth, we have:
2 3
max / 1c cQ B gA =
At the critical depth, Q is maximum and E is minimum.
(6.17)
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1. For a given constant discharge (Figure 6.3a)
The specific energy curve has a minimum value Emin at yc,
For any other E, there are two possible depths called
alternate depths, one is called subcritical and the other is
supercritical.
For supercritical flow, y < yc(Fr > 1)
For subcritical flow, y > yc (Fr < 1)
2. For a given constant specific energy (Figure 6.3b)
The discharge is maximum at depth yc;
For all other discharges, there are two possible depths for
any particular value of E.
1.5 Some more discussion to Figure (6.3):
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Summary of critical, subcritical and supercritical flows
Critical Flow (requires )
requires Eminimum or qmaximum;
For rectangular channels:
23 / ,c y q g= min 3 / 2,cE y=
Subcritical Flow:
Fr < 1,
Water velocity < wave velocity
Disturbances travel upstream and downstream
Upstream water levels are affected by downstream control
c cV gy=
1/ == gyVFr
V: water velocity in channel
gy : celerity of long wave when there is a disturbance in channel
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Slope classification: mild, steep and critical slopesA mild slope: on which uniform flow is subcritical:
A steep slop: on which uniform flow is supercritical:
A critical slope: on which uniform flow is critical:
where y0
is the uniform depth.
0 0 c(or Fr1 or S >S )cy y 1,Water velocity < Wave velocity
Disturbances travel downstream
Upstream water levels are unaffected by downstream control
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Example 6.4
The discharge in a rectangular channel of width 5m is 8 m3/s. The
normal depth is 1.25 m. Determine the depth of flow where the
section contracts by 1.0 m and the bed gradually rises by 0.3 m.
Use graphical methods to find the solution.
z =
y2y1
z=0.3m
..
6.15/8/ 11 === bQq
curve)(plot33.1/1305.0)2/( 2112
1
2
111 myygyqyE =+=+=
2/ 22 == bQq curve)(plot/204.0)2/(2
22
2
2
2
222 yygyqyE +=+=
1.03m0.3-1.33or 1221 ===+= zEEzEE
From energy equation, we have:(1)
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Corresponding to E2=1.03m, there are no crossing points on the
specific energy curve. Therefore, there are no solutions. This result
suggests that the depth y2 should be the critical depth.
From the specific energy curve, we get:
ycritical = 0.741m and E2=Emin = 1.112m. (refer to Figure a)
From E1=E2+z E1new=1.412m (refer to Figure b)
From the specific energy curve at location (1), we get the new water
depth at location (1):
y1new=1.34m
Therefore, at the contraction, the water depth is y2 = 0.74m
And the original water depth is increased to y1new = 1.34m.
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0
0.5
1.0
1.5
2.0
0 0.5 1.0 1.5 2.0
(a)E
2=1.03
No crossing pointswith the energy curve
No solution.Therefore, y
2should
be at critical depth
z=0.3
E1=1.33m
E (m)
y(m)
E2~y2
E1~y1
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0
0.5
1.0
1.5
2.0
0 0.5 1.0 1.5 2.0
(b)
E1=1.412m
E2=1.03m
E1=1.33m
y1=1.2m (original depth)
y1=1.34m (new depth)
q1=1.6m
2/s
at location (1)
q2=2m
2/s
at location (2)
z=0.3
Move by hMove by h
Emin
=1.112m
Emin
=1.112m
z=0.3
ycritical
=0.735 m
E (m)
y(m)
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Example 6.5
The discharge in a trapezoidal channel is 30 m3/s. The channel has a
base width of 4 m and side slopes of 1:2. Determine the critical
water depth corresponding to the flow rate in the channel, n=0.022.
1
2
B
yc
4m
Solution:
(1) At critical water depth,
13
2
== gA
BQFr
cccc yyyBysmQ )24(2/)4(A,44B,/30With3
+=+=+==
1)24(81.9
)44(3033
2
=+
+
cc
c
yy
ymyc 4.1=
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(2) Critical slope:
2/3 1/ 2h c
AQ R S
n=Using Eq. (5.6),
At critical flow, y = yc
2/13/230 chc
c SRn
A=
(1)
(2)
Using n = 0.022, ,)24( ccc yyA +=c
cchc
y
yyR
524
)24(
+
+=
Substituting into (2) with yc = 1.4m
.......=cS
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Example 6.6
The discharge in a rectangular channel of width 5m is 8 m3/s. The
normal depth is 1.25 m. Determine the depth of flow immediately
prior to, above and after a section in which the bed gradually rises
by 0.5 m as shown.
Solution:
We define: y0 = original depthy1 = immediately prior to the hump
y2 = depth after the hump
y3 = depth above the hump
y3y0=y1
y2
(3)
z=
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Write the energy Eq between (0) and (3):
zEE += 30mzmysmBQq 5.0,25.1,/6.15/8/For 0
2=====
2 2
0 0 0/(2 ) 1.33 E y q gy m= + =
(1)
(2)mE 83.05.033.13 ==
From the figure, we can see that for hump higher that 0.37m, therewould be no crossing points (no solution) between E3 and specific
energy curve, i.e. the fluid does not have enough energy to pass over
a hump higher than 0.37m.
To pass over a hump higher than 0.37m, water dams up (known as
damming action or backwater) before the hump until it acquires
sufficient energy to pass over, after which the hump acts as a control.
The minimum energy required is the Emin (or Ec) at the critical point.
(3)curve)(plot/1305.0)2/(Plot222
yygyqyE +=+=
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The depth of water at a control for a known flow rate is given by (6.11):
( )1/ 3 32 2 3/ 1.6 / 9.81 0.64c y q g m y= = = =The minimum energy for the flow to pass the hump is:
myE c 96.02/64.032/3min ===
(4)
(5)
Since , the flow prior to the hump is subcritical.cyy >0
The E1 prior to the hump is obtained by usingmzEE 46.15.096.0min1 =+=+=
From the figure, corresponding to E1=1.46m, we have y1 = 1.39m
or 0.34m. However, because the flow prior to the hump is subcritical,It can only increase its specific energy by increasing its depth since
there is no control point prior to the hump. Thus y1 = 1.39m.
(Note: y1 can also be obtained using )46.1)2/( 21211 =+= gyqyE
(6)
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Assuming no energy loss from (1) to (2), i.e. E1 = E2,
2
2 222
1.46 2
qy Egy= + =
From the figure, we get y2 = 1.39m or 0.34m.
Since a control usually causes the flow to change from subcritical to
supercritical or vice versa, thus the depth y2 = 0.34m is expected
downstream of the hump as shown in the figure.
(7)
Damming action
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0
0.5
1.0
1.5
2.0
0 0.5 1.0 1.5 2.0
0.34m
y1=1.39m
E1=1.46m = E2
z=0.5m
q=1.6m2/s
y0=1.25m
yc=0.64m
E3=0.83m(a)
Emin
=0.96m
No crossing pointsNo solution.
zmax
=0.37m
E0=1.33m
E (m)
y(m)
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1.6 Flow characteristics over humps and contractions
y
E
y1 y2
y2
y2
y2
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Example 6.7
The discharge in a 6 m wide rectangular concrete channel (n =
0.013) is 50 m3/s. Determine:
(a) The critical water depth and critical slope of the channel
(b) Types of flow if the flow at different sections of the channel has
water depths of 3.72, 1.92 and 1.55 m.(c) The Froude Number of the flow corresponding to flow depths of
3.72, 1.92 and 1.55 m
(d) The normal water depths and the type of slopes if the channelslopes are 0.001 and 0.04 respectively
Solution:
013.06,/50:Given3
=== m, nbsmQ
(a) 2 23 3/ (50 / 6) / 9.8 1.92 ,c y q g a m= = =
,
2/13/2
ch SRn
A
Q = ,byA c= )2/( cch ybbyR += 0258.0=cS
From Eq. (5.6),
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(b) At section of:
y=3.73m, since y > yc, subcritical flow
y=1.92m, since y = yc, critical flow
y=1.55m, since y < yc, supercritical flow
(c) The Froude number: , V=Q/(yb) =50/(yb)gyVFr /=At y=3.73m, V=2.24m/s, Fr = V/(gy)1/2 = 0.37
At y= 1.92m, V=4.34m/s, Fr = V/(gy)1/2 = 1
At y=1.55m, V=5.76m/s, Fr = V/(gy)1/2 = 1.38
(d) The normal water depths: Since
With A=by, , Q = 50m3/s, S0 = 0.001 and 0.004, )2/( ybbyRh +=
> Sc = 0.0258Steep slope0.76m < yc0.04
< Sc = 0.0258Mild slope2.71m > yc0.001
Slope typeNormal depth yS0
5.6)(Eq,2/1
0
3/2SR
nQ h=
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2. Hydraulic Jumps
A hydraulic jump is a steplike increase in fluid depth in an openchannel flow. Commonly seen below weirs and sluice gates.
Supercriticalflow (upstream)
Hydraulic
jump
Hydraulic jump
length
Subcriticalflow
C id i f ifi
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Consideration of specific energy:
Upstream:Supercritical flow, y1 < yc,
Specific energy: E1,
E
Es
y
y1
y2
yc
V1
V2
y1
y2
E2
Emin
E1
Downstream:Subcitical flow, y2 > ycSpecific energy: E2,
Energy difference: Es = E1 E2 > 0So after a hydraulic jump, there is energy loss
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Formed when transit from supercritical flow to subcritical flow
2.1 Hydraulic jump-formation and characteristics
Formation:
Transition from supercritical to subcritical flows is rapid,
Length of hydraulic jump: about 7 times downstream depth,
The transition involves large energy loss due to turbulence,
Energy loss can be found in terms of y1, y2 and upstream Fr,
Hydraulic jump can be used to dissipate water energy,
such as in a spillway of a dam to reduce damage to riverbed.
Characteristics:
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2.2 Downstream or sequent depth for a rectangular channel
The basic equations involved are:
Continuity equation
Momentum equation
Assumptions:
Ignore boundary friction (due to a short length of hydraulic jump),
The channel bed slope is very small,
Pressure distributions at (1) and (2): hydrostatic
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Momentum equation: F Mx =
F F M M 1 2 2 1 =
(6.19)
F M F M 1 1 2 2+ = + = (6.20)
where M, the pressure force plus momentum flux, is constant for
any given flow rate q.y
y1
y2
p2
p1 F +MIgnore friction
Initial depth
Sequent depth
(1)
(2)
A i h d t ti t (1) d (2) h
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Assuming hydrostatic pressure at (1) and (2), we have:
2
1 1 / 2,F gy= F gy22
2 2=
/
1 1 ,qV= M qV2 2=
qgy
y qgy
y
2
1
1
2 2
2
2
2
2 2+ = + (6.21)
= +q
gy
y2 2
2(for rectangular channel) (6.22)
Momentum function
( )1/ 2
22 11
11 8 1 ,2
yFry
= +
( )
[ ]
y
y
Fr1
2
2
21 21
2
1 8 1= + /
(6.23)
(6.24)
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2.3 Energy loss
h E E E yq
gyy
q
gyL= = = + +
1 2 1
2
2 2
2
22 2
1 2
( ) ( )
Since FrV
gy
q
gy
2
2 2
3= = And use Eqs. (6.23) and (6.24):
(6.25)
h Ey y
y yL= =
( )2 13
1 24
(6.26)
Energy loss increases dramatically as the relative height of the jump
increases.
Example 6 8
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Example 6.8
Determine the depth y4 immediately upstream of the hydraulic jump
that will form in the situation shown below if q = 1.6 m2/s.
From Example 6.6, we have:
y1 = 1.39m, y2 = 0.34m, yc = y3 = 0.64m
y0 = 1.25m
Damming action
y5=y0
y4
y3=yc
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37.025.181.9
25.1/6.1/
2
5
5
52 =
===
gy
yq
gy
VFr
( )1/ 2
242
5
11 8 1 ........
2
yFr
y
= + =
y4 can be calculatedAs y5 = 1.25m,
Energy loss: .......4
)(
54
345 ==
yy
yyhL
Assuming y5 = y0 = 1.25m,
2 4 Classification of hydraulic jumps:
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2.4 Classification of hydraulic jumps:
The actual structure of a hydraulic jump depends on Fr
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2.5 Some further discussion on hydraulic jumps
The relations for hydraulic jumps developed in the present
unit are based on a rectangular and horizontal channel,
Hydraulic jumps in other channel configurations are similar
to those for rectangular channels,
The expression of sequent depth and energy loss in other
channel configurations are somewhat different from jumps
in rectangular channels.
3 Analysis of Gradually Varied Flows
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3. Analysis of Gradually Varied Flows
Tasks
Deduce the trend of water surface change (classification ofsurface profiles)
Calculate water levels and velocity along the course of the
channel (quantitative evaluation) Analysis method
Energy equation
Continuity (mass conservation) equation
Characteristics of gradually varied flow Water depth and velocity change gradually,
Flow is non-uniform,
Water surface changes smoothly and continuously, Friction loss along the channel is not negligible.
3 1 The equations for gradually varied flow
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3.1 The equations for gradually varied flow
Assuming the change in the total energy is due to frictional lossesover some distance x, the energy grade line (EGL) and hydraulic
grade line (HGL) can be drawn as shown below.
datum
Friction slope Sf
Conservation of the energy gives:
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Conservation of the energy gives:
S x y
V
g S x y
V
gf0 11
2
2
2
2
2 2 + + = + +
S x S x y yV
g
V
gf0 2 12
2
1
2
2 2 = + +
(7.1)
(7.2)
The differential form by choosing x small dx:
S dx S dx dy d V
gf0
2
2= + + ( )
(7.3)
ddy
Vg
ddy
QgA
QgA
dAdy
Q BgA
( ) ( )
2 2
2
2
3
2
32 2
= = = (7.4)Since ,
A: the cross-sectional area; Q: discharge (m3/s).
(7.5)320
/1 gABQ
SS
dx
dy f
=
Substitute (7.4) into (7.3) and rearrange:
E (7 5) i lid f i h
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(7.6)
Eq. (7.5) allows the water surface profiles of gradually varied flow
to be deduced, y is measured vertically from the channel bottom,
the slope of the water surface dy/dx is relative to channel bottom.
For a rectangular channel, since Froude number is defined as
FrV
gy=
Eq. (7.5) can then be rearranged as
Eq. (7.5) is valid for any cross section shapes
2
0
2
0
1/1 r
ff
F
SS
gyV
SS
dx
dy
=
=
3 1 1 A note on the friction slope
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Strictly speaking, the slope term in Mannings Eq (5.6) is the frictionslope Sf, since for steady uniform flow, S0 = Sf.
Rearranging Mannings equation to find Sfgives:
Sn Q P
Af=
2 2 4 3
10 3
/
/(7.8)
Eq. (7.8) implies that Sfdecreases as the water depth y increases,
which in combination with (7.6) means that
y y S Sf< >0 0
y y S Sf>
3 2 Calculation of Gradually Varied Flow Profile
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Task: calculate the distance between two cross sections
Resources: channel slope S0, channel properties, water depth y1 at
section 1 and water depth y2 at section 2 are known
S x y
V
g S x y
V
gf0 11
2
2
2
2
2 2 + + = + +
S x E S x E f0 1 2 + = +
x
E E
S S
E
Sf=
=
1 2
0
(7.11)
Sf can be calculated using Mannings equation:
S
n Q P
Af=
2 2 4 3
10 3
/
/
(7.12)
From Eq. (7.1)
3.2 Calculation of Gradually Varied Flow Profile
3.2.1 Direct step method:
How?
Example 7.1
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The water depth of a gradually varying flow ( q = 1.6 m2/s) in a
wide rectangular channel (n = 0.015) is 0.34 m right behind a humpand 0.38 m at a location further downstream. What would be the
distance between the two locations?
(7.11)
Sf: Slope of water surface (Eq. (7.12)
S0: Slope of channel bed (Eq. (5.6)
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q2
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3.2.2 Numerical methods:
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Task: calculate yi (i=1 to n)
Resources: channel properties;
Method: An iterative procedure
3
2
2
3/20
2
0
11'
gA
bQ
AR
QnS
F
SS
dx
dy
yh
r
f
=
== (7.13)
Using Taylors expansion:
xyyxdx
dyyy iii +=+=+ '1 (7.14)
From Eq.(7.5), we can get
y0
y1 y2 yi yi+1x x
. . .
where and)''(2
1' 1++= ii yyy
=
3
22
3/201/'
ihii
igA
bQ
RA
QnSy
3.2.2 Numerical methods:
The Iterative Procedure:
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1. Use Eq. (7.13) to calculate y'i where yi is known either as initial
point or from a previous cycle of this calculation
2. Set y 'i+1 = y ' i as a first estimate
3. Use current values of y ' i and y ' i+1 to calculate yi+1 from Eq.
(7.14) for a selected x
4. Find a revised estimate of y ' i+1 from Eq. (7.13) using the yi+1value from step 3
5. If the new y ' i+1 value is not close enough to the previouslycalculated value, repeat step 3, 4 and 5 using the latest estimate
of yi+1 found in step 4
6. Once the iteration procedure has yielded successive estimates ofy ' i+1 and yi+1 within acceptable limits of accuracy, proceed to
the next section of channel and repeat the process
7. Terminate the process when the desired reach has been covered
3 3 Water surface profiles and their classifications
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3.3 Water surface profiles and their classifications
Take a channel with mild slope as an example. The surface profile may
occupy the three regions shown below and the sign of dy/dx can be
found for each region:
y0 yc
(a) Region 1 for the mild slope:
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(a) Region 1 for the mild slope:
Since y>y0 > yc, we have Sf< S0 from Eq. (7.8)
Since y > yc, we have Fr2 < 1 from Eq. (7.10)
From Eq. (3.2.6) we get dy/dx is positive.
The asymptotic behaviour of the free surface M1:
00Frand,As Sdy/dxSy f
hence the water surface is asymptotic to a horizontal line.
0,As 00 dy/dxSSyy f
hence the water surface is asymptotic to the line y = yn.
(b)Region 2 on a mild slope:
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For the M2 profile, dxdyyy c /,as
This is physically impossible. This is because as the fluid entersa region of rapidly varying flow and Eq. (7.6) is no longer valid.
y y S Sf 0 0dy dx/ 012 rc Fyy
(c)Region 3 on a mild slope:
positive.is/1Frand, 200 dxdySSyyy fc >>>>
For the M3 profile, dxdyyy c /,as
This is physically impossible. In practice, an hydraulic jump willform before y = yc.
Note: above discussion is for mild slope. For steep, critical, horizontal
and adverse slopes, the surface profiles are given in next page.
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