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11/11/2013
1
Non uniform flow in channels
By
Dr. Ajit Pratap Singh
Civil Engineering Department
Dynamic Equation of GVF� The basic differential equation for GVF
� When dy/dx = 0, So = Sf and the water surface profile is parallel to the channel bottom
� When dy/dx = +ve, water surface is rising and
� When dy/dx = -ve, water surface is falling
2
fo
Fr1
SS
dx
dy
−
−=
Dynamic Equation of GVF in Wide
Rectangular Channel� The differential equation for GVF in wide rectangular
channel
used isequation sChezy' if ,
y
y1
y
y1
Sdx
dy
used isequation sMannaing' if ,
y
y1
y
y1
Sdx
dy
3
c
3
n
o
3
c
10/3
n
o
−
−
×=
−
−
×=
Classification of Flow ProfilesClassification of Flow Profiles�� Bed slope SBed slope S00 is classified asis classified as
�� Steep : Steep : yynn < y< yc c or or ssoo>s>scc
�� Critical : Critical : yynn = y= yc c or or ssoo= s= scc
�� M ild : M ild : yynn > y> yc c oror ssoo< s< scc
�� Horizontal : SHorizontal : S00 = 0= 0
�� Adverse : SAdverse : S00 < 0< 0
�� Initial depth is given a zoneInitial depth is given a zone
�� Zone 1 : y > Zone 1 : y > yynn
•• The space above both critical and The space above both critical and normal depthnormal depth
�� Zone 2 : Zone 2 : yynn < y < y< y < ycc
•• The region lies between the normal The region lies between the normal and critical depthand critical depth
�� Zone 3 : y < yZone 3 : y < ycc
•• The lowest zone of space that lies The lowest zone of space that lies above the channel bed but below above the channel bed but below both critical and normal depth linesboth critical and normal depth lines
• 12 distinct configurations for surface profiles in GVF.
• It should be noted that a continuous flow profile usually occurs only in one zone
• Figure 16.4 shows various surface profiles which can be classified as backwater curves and drawdown curves depending on whether depth of flow increases or decreases in the direction of flow
• All the surface profiles with subscript 1 and 3 are backwater or rising curves while those with subscript 2 are drawdown or falling curves.
Non-uniform flow in Open Channel by Dr. Ajit Pratap Singh
MILD BACKWATER CURVES M1, M2 AND M3
M1: y1 >yn > yc
Again the case of constant bed slope S is considered. Recall that
M2: yn > y1 > yc
M3: yn > yc > y1
+
+=
−
−=
)(1
)(
12
1
1y
ySS
dx
dy fo
x Fr
Depth increases downstream,
decreases upstream
Depth increases downstream,
decreases upstream
Depth decreases downstream,
increases upstream
A bed slope is considered mild if yn > yc. This is the most common case in
alluvial rivers. There are three possible cases.
3
21/3
s
2rf3
2
ff3
2
2
gy
q
k
yα(y)Sor
gy
qC(y)S,
gy
q(y)Fr
−
−
===
+
−=
−
−=
)(1
)(
12
1
1y
ySS
dx
dy fo
x Fr
−
−=
−
−=
)(1
)(
12
1
1y
ySS
dx
dy fo
x Fr
11/11/2013
2
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
M1 CURVEM1 CURVEM1 CURVEM1 CURVE
� When y > yn > yc, what will be the limiting value of y??
side downstream on the y
side upstream on the yy n
∞→
→
o
n
Sdx
dy side, downstream on the y as and
0dx
dy side, upstream on the yy as and
vedx
dyget weHere
→∞→
→→
+→
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note that
normally line ingcorrespond meet the curves specified that theindicatesit ,dx
dyWhen
normally line ingcorrespond meet the curves specified that theindicatesit ,dx
dyWhen
horizontal be to tendscurves specified that theindicatesit ,Sdx
dyWhen
ally asymptotic NDL meet the curves specified that theindicatesit 0,dx
dyWhen
o
−∞→
∞→
→
→
Thus M1 curves meet the NDL asymptotically on the u/s andit tends to be horizontal on the d/s. M1 curve becomeshorizontal as the depth becomes larger. An example of sucha flow is a river entering a lake or reservoir or Flow behindan overflow weir
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh Non-uniform flow in Open Channel by Dr. Ajit Pratap Singh
M2 CURVEM2 CURVEM2 CURVEM2 CURVE
� When yn >y> yc, what will be the limiting value of y??
side downstream on the yy
side upstream on the yy
c
n
→
→
−∞→→
→→
−→
dx
dy side, downstream on the yy as and
0dx
dy side, upstream on the yy as and
vedx
dyget weHere
c
n
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note thatThus M2 curves meet the NDL asymptotically on the u/s andit meet the CDL normally on the d/s. An example of such aflow is over a free overfall. It is also obtained when a steepslope is preceded by a mild slope or a mild slope by a milderslope.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
M3 CURVEM3 CURVEM3 CURVEM3 CURVE
� When yn > yc >y, what will be the limiting value of y??
side downstream on the 0y
side upstream on the y c
→
→ y
undefined is dx
dy i.e.
dx
dy side, upstream on the 0y as and
dx
dy side, downstream on the yy as and
vedx
dyget weHere
c
∞→→
−∞→→
+→
11/11/2013
3
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note thatThus M3 curves meet the CDL and channel bottom linenormally. An example of such a flow is d/s of a sluice gate. Itis also obtained when the bottom slope changes from steepto mild.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
S1 CURVES1 CURVES1 CURVES1 CURVE
� When y>yc > yn, what will be the limiting value of y??
side downstream on the y
side upstream on the yy c
∞→
→
o
c
Sdx
dy side, downstream on the y as and
dx
dy side, upstream on the yy as and
vedx
dyget weHere
→∞→
∞→→
+→
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note that
Thus S1 curves meet the CDL normally on the u/s and ittends to be horizontal on the d/s. Example Flow behind anoverflow weir or profile formed behind a dam constructed ona steep channel
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
S2 CURVES2 CURVES2 CURVES2 CURVE
� When yc >y> yn, what will be the limiting value of y??
side downstream on the yy
side upstream on the yy
n
c
→
→
0dx
dy side, downstream on the yy as and
dx
dy side, upstream on the yy as and
vedx
dyget weHere
n
c
→→
∞→→
−→
11/11/2013
4
Remember to note thatThus S2 curves meet the CDL normally on the u/s and itmeet the NDL asymptotically on the d/s. An example of sucha flow is over a free overfall. It is also obtained when a steepslope is preceded by a mild slope or a mild slope by a milderslope.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
S3 CURVES3 CURVES3 CURVES3 CURVE
� When yc > yn >y, what will be the limiting value of y??
side upstream on the 0y
side downstream on the y n
→
→ y
undefined is dx
dy i.e.
dx
dy side, upstream on the 0y as and
0dx
dy side, downstream on the yy as and
vedx
dyget weHere
n
∞→→
→→
+→
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note thatThus S3 curves meets the channel bed normally and it isasymptotic to the normal depth line NDL. An example ofsuch a flow is d/s of a sluice gate. It is also obtained whenthe bottom slope changes from steep to mild.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
C1 CURVEC1 CURVEC1 CURVEC1 CURVE
� When y > yn = yc, what will be the limiting value of y??
side downstream on the y
side upstream on the yyy nc
∞→
=→
co
coc
SSdx
dy side, downstream on the y as and
SSdx
dy side, upstream on the yy as and
vedx
dyget weHere
=→∞→
=→→
+→
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note that horizontal be to tendscurves specified that theindicatesit ,S
dx
dyWhen
o→
Thus C1 curves will be more or less a horizontal line. For example flowbehind an overflow weir, flow behind a sluice gate.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
C3 CURVEC3 CURVEC3 CURVEC3 CURVE
� When y < yc = yn, what will be the limiting value of y??
side upstream on the 0y
side downstream on the yyy nc
→
=→
co
coc
SSdx
dy side, upstream on the 0y as and
SSdx
dy side, downstream on the yy as and
vedx
dyget weHere
=→→
=→→
+→
11/11/2013
5
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note that
horizontal be to tendscurves specified that theindicatesit ,Sdx
dyWhen
o→
Thus C3 curves will be more or less a horizontal line. Forexample back-water curve below a sluice gate provided in achannel with a critical slope is typical example of C3 profile.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
H2 CURVEH2 CURVEH2 CURVEH2 CURVE
� When yn (=∞)>y> yc, what will be the limiting value of y??
side downstream on the yy
side upstream on the y
c→
∞→
−∞→→
→∞→
−→
dx
dy side, downstream on the yy as and
0dx
dy side, upstream on the y as and
vedx
dyget weHere
c
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note that
Thus H2 curves meet the CDL normally at d/s end and at theu/s end it tends to approach horizontal line tangentially
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
H3 CURVEH3 CURVEH3 CURVEH3 CURVE
� When yn > yc >y, what will be the limiting value of y??
side upstream on the 0y
side downstream on the y c
→
→ y
undefined is dx
dy i.e.
dx
dy side, upstream on the 0y as and
dx
dy side, downstream on the yy as and
vedx
dyget weHere
c
∞→→
−∞→→
+→
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Remember to note that
Thus H3 curves meet the CDL and channel bottom linenormally.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
11/11/2013
6
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� An M1 profile occurs behind a dam or a sluice gate located on a mildchannel. The dam or a sluice gate piles up water behind it such that theflow depth is greater than the normal depth. Far away from the dam orsluice gate on the upstream side, the flow would be occurring underuniform conditions and the flow depth would be normal.
� In a similar manner, S1 and C1 profiles occur on the upstream side of asluice gate located on a channel with steep and critical slopes,respectively.
� An M2 profile occurs on the upstream side of a free over fall at thedownstream end of a mild channel since a critical depth occurs in thevicinity of a free over fall.
� Similarly, a H2 profile occurs on the upstream side of a free over fall atthe downstream end of a horizontal channel.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� Critical flow conditions occur at the entrance to a steepchannel from a lake or a reservoir. However, flow shouldtend towards uniform flow conditions far away from theentrance if the channel is long. Therefore, a S2 profile occursin steep channels, on the downstream side of the entrance.
Gradually Varied FlowGradually Varied Flow Gradually Varied FlowGradually Varied Flow
�� Typical OC system Typical OC system involves several involves several sections of different sections of different slopes, with transitionsslopes, with transitions
�� Overall surface profile is Overall surface profile is made up of individual made up of individual profiles described on profiles described on previous slidesprevious slides
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Steps to be followed to sketch the water
surface profiles
� Compute normal and critical depths for each reach of the channelsystem based on specified flow rate, roughness coefficient, slope ofthe reach,and the channel cross section.
� Plot the channel bed, the normal depth line (NDL) and the criticaldepth line (CDL) for each reach in the system.
� By comparing normal depth and critical depth, determinewhether the channel slope is mild, critical, steep, adverse orhorizontal
� Mark the control sections i.e., identify the sections where (i) theflow passes through a critical depth (ii) the flow is expected tooccur under uniform conditions, and (iii) there is a controlstructures such as a weir, a sluice gate, and a spillway. A controlsection is that at which for a given discharge the depth of flow isknown or it can be controlled to a required value
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� It may be noted that uniform flow conditions occur in longprismatic channels, far away from control sections. Criticaldepth occurs at (i) the free overfall, and (ii) the entrance to asteep channel from a lake, when the water level in the lake isabove the elevation of the CDL at the entrance. Criticaldepth also occurs when channel bed slope changes from mildto steep.
� Knowing the normal depth and depth at the control section,determine the surface profile
11/11/2013
7
SURFACE WATER PROFILE� In practice, often the length of the surface profile of the gradually
varied flow needs to be known
� From all the equations the derivations show the changes of waterdepth (y) for some distances/lengths (x). By using the integrationmethod in those equations, we can know,
� Distances/lengths from one point to another point when bothdepth are known. For example, if a weir is constructed across ariver having a mild slope then it may be required to estimate thedistance on the u/s side up to which the effect of resulting M1profile exists.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� Starting from each control point, sketch the appropriate water surfaceprofile depending on the zone in which the depth at the controlsection falls and the nature of the slope.
� Qualitatively locate the hydraulic jumps wherever the flow changesfrom supercritical to sub critical.
� For example, if there is a sluice gate at the downstream end of a steepchannel, the flow is sub critical on the upstream side of the gate.However, if the channel is long, flow is supercritical far away from thegate on the upstream side. Therefore, a hydraulic jump occurs in such achannel. Also, on the downstream side of a sluice gate on a long mildchannel, the flow is supercritical immediately downstream of the gate.However, far away from the gate on the downstream side, flow issubcritical.Therefore,a hydraulic jump occurs in such a case also.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
Real Life Cases of Water Surface Real Life Cases of Water Surface Real Life Cases of Water Surface Real Life Cases of Water Surface
ProfilesProfilesProfilesProfiles
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� There are several method to obtain surface water profile.
� They are� Direct Integration
� Numerical Integration
� Direct Step Method
� Graphical Integration
� Numerical/Computer Methods
Direct Step Method
� In general, a step method is characterized by dividing the channel into short reaches and carrying the computation step by step from one end of the reach to the other. The direct step method is a simple step method applicable to prismatic channel.
� Equating the total head at the two end section 1 and 2, the following may be written;
2 2
1 1 2 21 1 2 2
2 2L
p V p Vz z h
g ga a
g g+ + = + + +
2 2
1 21 2
2 2o f
V Vy S dx y S dx
g g+ + = + +
Turbulent flow (α ≅ 1)z - measured from horizontal datum
2 2
1 21 2
2 2o f
V Vy S dx y S dx
g g+ + = + +
1 2 0z z S dx− =
11/11/2013
8
Short reach (dx) of the channel
2
1V
2g2
2V
2g
oS dx
dx
f fh S dx=
energy grade line
hydraulic grade line
velocity head
water surfacewater surface
1 2
Where
E = specific energy at one point = y + v²/2g
1 2o fE S dx E S dx+ = +
2 1
0 f
E Edx
S S
−=
−solving for dx
2 2
4 /3f
n VS
R=
2
2f
VS
C R=
Manning Chezy
� Limitation: channel must be _________ (so that velocity is a function of depth only and not a function of x)
� Method� Find yn and yc. Determine the type of slope.� Identify type of profile (determines whether ∆y is + or -)� choose ∆y and thus yn+1
� Compute the area of flow section A, wetted perimeter P, hydraulic radius R and velocity at known value of depths of flow yn and yn+1
� Compute the mean velocity, the velocity head, the specfic energy E and energy line slope Sf at the channel sections where the depth flow is known i.e. at yn and yn+1
� calculate average friction slope � calculate dx
prismaticprismaticDirect Step Method
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� A rectangular channel conveys a discharge of 30 m3/s. It islaid at a slope of 0.0001. If at a section in this channel thedepth is 1.6 m, how far u/s or d/s from the section will thedepth be 2.0 m.Take Manning’s n = 0.015
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� A very wide rectangular channel conveys a discharge of 3.50m3/s per metre width at a depth of 2.50 m. The bed slope is1 in 5000. Due to a weir placed across the channel the waterlevel is raised by 1.50 m just on the upstream of it. Find atwhat distance upstream of the weir the depth of water will be3 m. Take C = 51. Use step method and take two steps. Alsoclassify the type of water surface profile.
Example� A wide rectangular channel carries water at 10 m3/s with channel
width = 8m, bed slope = 0.001 and n= 0.025.
� Find the length of back water which is formed due to a dam and obtained the 2 m water depth at the dam’s back.
� Find yn and yc.
yn
1.093703
yc
0.542064
yn > yc Mild slope channel
PROFILE M1 TYPE
11/11/2013
9
1. y = water depth (m)
2. R = A/P = hydraulic radius or y for very wide rectangular
3. v = q/y = flow velocity
4. v²/2g = kinetic energy
5. y + v²/2g = E = specific energy
6. E2 - E1 = ∆E = energy loss
7. sf = slope energy grade line = n2v2 = v2
R4/3 C2R
8. (sf1 + sf2)/2 = EGL slope average
9. (so - sf ) = slope difference
10. dx = reach = ∆E / (so - sf )
11. L = length of surface water profile which is to be calculated from dam
1. Find yo and yc2. Fill in the table
A/P q/y
1 + 4 (y+v2/2g) E2 -E1
n2v2
/R4/3 6/9
1 2 3 4 5 6 7 8 9 10 11
y A R v v2/2g E ∆E Sf ∆x
- - - -
fS 0 fS S−
Solution for example
(See EXCEL FILE)
A/P q/y
1 + 5
(y+v2/2
g) E2 -E1
n2v2
/R4/3
1 2 3 4 5 6 7 8 9 10 11
y A R v v2/2g E ∆E Sf Sfbar S0-Sfbar dx
2 16 1.33333 0.625 0.01991 2.01991 0.00017
0.12714 0.00018 0.000815746 155.8520063
1.87 14.96 1.27428 0.66845 0.02277 1.89277 0.0002
0.12647 0.00023 0.000774196 163.3564655
1.74 13.92 1.21254 0.71839 0.0263 1.7663 0.00025
0.12558 0.00028 0.00071855 174.7695148
1.61 12.88 1.14795 0.7764 0.03072 1.64072 0.00031
0.12437 0.00036 0.000642175 193.6630303
1.48 11.84 1.08029 0.84459 0.03636 1.51636 0.0004
0.12266 0.00047 0.000534278 229.5819527
1.35 10.8 1.00935 0.92593 0.0437 1.3937 0.00053
0.12019 0.00062 0.000376502 319.231189
1.22 9.76 0.93487 1.02459 0.05351 1.27351 0.00072
Σ (Σ (Σ (Σ (dx)))) 1236.454159
� The calculation must be from the dam to upstream until the water surface is 1% higher than the normal depth.
52
Hydraulic Jump• A hydraulic jump occurs when flow changes from a supercritical flow
(unstable) to a sub-critical flow (stable).
• There is a sudden rise in water level at the point where the hydraulic
jump occurs.
• Rollers (eddies) of turbulent water form at this point. These rollers cause
dissipation of energy.
•A hydraulic jump occurs in practice at the toe of a dam or below a sluice gate
where the velocity is very high.
53
General Expression for HydraulicJump:
In the analysis of hydraulic jumps, the following assumptions are made:
(1) The length of hydraulic jump is small. Consequently, the loss of head
due to friction is negligible.
(2) The flow is uniform and pressure distribution is due to hydrostatic
before and after the jump.
(3) The slope of the bed of the channel is very small, so that the
component of the weight of the fluid in the direction of the flow is
neglected.
54
Location of hydraulic jump
Generally, a hydraulic jump occurs when the flow changes from
supercritical to subcritical flow.
The most typical cases for the location of hydraulic jump are:
1. Jump below a sluice gate.
2. Jump at the toe of a spillway.
3. Jump at a glacis.
(glacis is the name given to sloping floors provided in hydraulic structures.)
11/11/2013
10
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
MOMENTUM EQATION� Steady Flow
� Momentum is a vector quantity. The momentum equationcommonly used in most of the open channel flow problems is thelinear-momentum equation. This equation states that the algebraicsum of all external forces acting in a given direction on a fluidmass equals the time rate of change of linear-momentum of thefluid mass in that direction. In a steady flow the rate of change ofmomentum in a given direction will be equal to the net flux ofmomentum in that direction.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� Figure shows a control volume (a volume fixed in space) bounded by sections 1 and 2, the boundary and a surface lying above the free surface.
Non-uniform flow in Open Channel by Dr. Ajit
Pratap Singh
� The various forces acting on the control volume in the longitudinal direction are:
� (i) Pressure forces acting on the control surfaces, and
� (ii) Tangential force on the bed, ,
� (iii) Body force, i.e. the component of the weight of the fluid in the longitudinal direction, .
� By the linear-momentum equation in the longitudinal direction for a steady-flow discharge of ,
1F
2F
3F
4F
Q
124321MMFFFFF −=+−−=∑
� Momentum of the flow passing a channel section per unit time =
� Rate of change of momentum in the body of water flowing in a channel is equal to the resultant of all the forces that are acting on the body
� Where Ff total external force of frictional resistance acting in the direction opposite to the flow along the surface of contact between water and channel, W is the weight of water enclosed between the section, θ is angle of incination
g
QVwM =
( )f21
12 FWSinθPPg
VVQw−+−=
−
59
•The net force in the direction of flow = the rate of change of moment in that direction
1 2( )
w QV V
g= −
The net force in the direction of the flow, neglecting frictional resistance and the
component of weight of water in the direction of flow,
R = P1 - P2 .
Therefore, the impulse-moment yields
2 1 1 2( )
wQP P V V
g− = −
Where P1 and P2 are the pressure forces at section 1 and 2, respectively.
2 12 1 1 2( )wQ
wA z wA z V Vg
− = −
2
2 12 1
1 2
1 1( )
wQwA z wA z
g A A− = −
2 2
1 21 2
1 2
Q QA z A z
gA gA+ = +
z = the distance from the water surface to the centroid of the flow area60
Comments:
• This is the general equation governing the hydraulic jump for any shape
of channel.
• The sum of two terms is called specific force (F). So, the equation can
be written as:
M1 = M2
This equation shows that the specific force before the hydraulic jump is
equal to that after the jump.
2 2
1 21 2
1 2
Q QA z A z
gA gA+ = +
11/11/2013
11
61
Hydraulic Jump in Rectangular Channels
1 1A By= 1
1
2
yz =
2 2A By= 2
2
2
yz =
2 2
1 2
1 2
1 2
( )( ) ( )( )2 2
y yQ QBy By
gBy gBy+ = +
2 2
1 21 2
1 2
Q QA z A z
gA gA+ = +
B=
2 22
2 1 2 1
1 22
y y y yq
g y y
− −=
2
1 2 2 1
2( )
qy y y y
g= +
2
2 2
2 1 2 1
20
qy y y y
g+ − =
using
we get
2
1 2 1 2
2( )
qy y y y
g= +
62
2 2
1 12
1
2
2 2
y y qy
gy
= − + +
2 2
2 21
2
2
2 2
y y qy
g y
= − + +
This is a quadratic equation, the solution of which may be written as:
where y1 is the initial depth and y2 is called the conjugate depth. Both are called
conjugate depths.
These equations can be used to get the various characteristics of hydraulic jump.
2
2
3
1 1
811 1
2
qy
y g y
= − + +
2
1
3
2 2
811 1
2
qy
y gy
= − + +
63
2
3
c
qy
g=
3
2
1 1
11 1 8
2
cyy
y y
= − + +
3
1
2 2
11 1 8
2
cyy
y y
= − + +
( )221
1
11 1 8
2
yFr
y= − + +
( )212
2
11 1 8
2
yFr
y= − + +
1
1
1
VF
gy=
2
2
2
VF
gy=
But for rectangular channels, we have
Therefore,
These equations can also be written in terms of Froude’s number as:
64
1 2E E E∆ = −
2
22
s
qE y
gy= +
( )
2 22 2 2
2 1
1 2 2 122 3
1 2 1 2
( )2 2 2
y yq q qE y y y y
gy gy g y y
−∆ = + − + = − −
Due to the turbulent flow in hydraulic jump, a dissipation (loss) of energy occurs:
Where, E = specific energy
For rectangular channels:
hence,
Head Loss in a hydraulic jump (HL):
After simplifying, we obtain
3
2 1
1 2
( )
4
y yE
y y
−∆ =
65
2 1jh y y= −
6j jL h≅
Height of hydraulic jump (hj):
The difference of depths before and after the jump is known as the
height of the jump,
Length of hydraulic jump (Lj):
The distance between the front face of the jump to a point on the
downstream where the rollers (eddies) terminate and the flow becomes
uniform is known as the length of the hydraulic jump. The length of the
jump varies from 5 to 7 times its height. An average value is usually
taken:
Dam at Hiram Falls on the Saco River near Hiram, Maine, USA
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� A spillway is designed to discharge 5m3/sec/m length. After flowing over the spillway, water flows on to a horizontal concrete apron (Manning’s rugosity coefficient as 0.015). The velocity of flow at toe is observed to be 15 m/sec and tail water depth is limited to 3.0 m. Calculate the minimum length of apron to contain the jump on the apron and consequent energy lost.
Surges in Open Channel� A surge or surge wave is a moving wave front which brings
about an abrupt change in depth of flow
� It is also referred to as moving hydraulic jump and is caused by sudden increase or decrease of depth of flow, such as that caused by sudden opening or closing of a gate fixed in the channel.
� Positive surges: which results in an increase in depth of flow
� Negative surges: causes a decrease in depth of flow
� Positive Surge: Type A which is Advancing D/S and Vw = C+V1
� Positive Surge: Type B which is Advancing U/S and Vw = C-V1
� Negative Surge: Type C which is retreating D/S and Vw = C+V1
� Negative Surge: Type D which is retreating U/S and Vw = C-V1
� Where C is the celerity of the wave which is defined as the velocity of wave relative the velocity of flow and Vw is the absolute velocity of the wave
� Applying continuity equation to the control volume of fig., if ρ= density of water; A2 = flow area behind the wave and A1 = flow area ahead of the wave. Since ρ is a constant
� Applying momentum equation to the control volume of fig., if ρ= density of water; A2 = flow area behind the wave and A1 = flow area ahead of the wave. Since ρ is a constant
� The channel is prismatic, horizontal and frictionless. Therefore, the only force acting on the control volume is pressure force. Pressure force acts in the positive x - direction at the inlet section and in the negative x - direction at the outlet section. Above Equation can be written as
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� For Rectangular channels
Problem� A wide rectangular channel is carrying a flow of 3 m3/sec
per meter width of channel at a flow depth of 1.5m. What should be the increase in discharge at the upstream end to cause a surge of 0.5m? What is the corresponding surge velocity?
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