A broad crested weir with a crest height of 0.3m is located in a channel. With a measured head of 0.6m above the crest, the problem asks to calculate the rate of discharge per unit width, accounting for velocity of approach. Broad crested weirs follow the relationship that discharge per unit width (q) is proportional to the head (H) raised to the power of 3/2. Using this relationship and the given values of 0.3m for crest height and 0.6m for head, the problem is solved through trial and error to find the value of q.

1. Open Channel Hydraulics
1
Hydraulics
Dr. Mohsin Siddique
Assistant Professor

2. Steady Flow in Open Channels
Specific Energy and Critical Depth
Surface Profiles and Backwater Curves in Channels of
Uniform sections
Flow over Humps and through Constrictions
Hydraulics jump and its practical applications.
Broad Crested Weirs andVenturi Flumes

3. Flow over Humps and through
Constrictions

4. Flow Over Hump
Hump:
is a streamline construction provided at the bed of the channel.
It is locally raised bed.
Let’s examine the case of hump in a rectangular channel.
We will neglect the head loss.

5. Flow Over Hump
For frictionless two-dimensional flow,
sections 1 and 2 in Fig are related by
continuity and energy:
EliminatingV2 between these two
gives a cubic polynomial equation for
the water depth y2 over the hump.
1 1 2 2
2 2
1 2
1 2
2 2
v y v y
v v
y y Z
g g
=
+ = + +
2 2
3 2 1 1
2 2 2
2
1
2 1
0
2
2
v y
y E y
g
v
where E y Z
g
− + =
= + −
y2
y1
y3
Z
V1
V2
1 2 3
This equation has one negative and
two positive solutions if Z is not
too large.
It’s behavior is illustrated by E~y
Diagram and depends upon
whether condition 1 is Subcritical
(on the upper) or Supercritical
(lower leg) of the energy curve.
B1=B2

6. Flow Over Hump
The specific energy E2 is exactly Z
less than the approach energy E1, and
point 2 will lie on the same leg of the
curve as E1.
A sub-critical approach, Fr1 <1, will
cause the water level to decrease at
the bump. Supercritical approach
flow, Fr1>1, causes a water-level
increase over the bump.
If the hump height reaches
Zmax=Zc=E1-Ec, as illustrated in fig, the
flow at the crest will be exactly
critical (Fr =1).
If Z = Zmax, there are no physically
correct solutions to Eqn. i.e., a hump
too large will “choke” the channel
and cause frictional effects, typically a
hydraulic jump.
1 2
Super-Critical
Approach
These hump arguments are reversed if the channel has a depression (Z<0): Subcritical
approach flow will cause a water-level rise and supercritical flow a fall in depth. Point 2 will
be |Z| to the right of point 1, and critical flow cannot occur.
Z
Zmax

7. Flow Over Hump
Damming
Action
y1=yo, y2>yc, y3=yo
y1=yo, y2>yc, y3=yo
y1=yo, y2=yc, y3=yo y1>yo, y2=yc, y3<yo
yc
y1
y3
Z
Z=Zc
y1
Z<<Zc
y2 y3
Z
Z<Zc
y2
y1
y3
Z
Z>Zc
Afflux=y1-yo
y3
yc
yo
Z
y1

8. Flow Over Hump
As it is explained with the help of E~y Diagram, a hump of any height “Z”
would cause the lowering of the water surface over the hump in case of
subcritical flow in channel. It is also clear that a gradual increase in the
height of hump “Z” would cause a gradual reduction in y2 value. That height
of hump which is just causing the flow depth over hump equal to yc is know
as critical height of hump Zc.
Further increase in Z (>Zc) would cause the flow depth y2 remaining equal
yc thus causing the water surface over the hump to rise. This would further
cause an increase in the depth of water upstream of the hump which mean
that water surface upstream of the hump would rise beyond the previous
value i.e y1>yo. This phenomenon of rise in water surface upstream with
Z>Zc is called damming action and the resulting increase in depth upstream
of the hump i.e y1-yo is known as Afflux.

9. Flow Through Contraction
When the width of the channel is reduced while the bed remains flat, the discharge
per unit width increases. If losses are negligible, the specific energy remains constant
and so for subcritical flow depth will decrease while for supercritical flow depth will
increase in as the channel narrows.
B1 B2
y2
yc
y1
( )
1 1 1 2 2 2
2 2
1 2
1 2
1 2
2 2 2 2 2 2
2 2
1 1
'
2 2
Using both equations, we get
2
Q=B y v =B y
1
Continuity Equation
B y v B y v
Bernoulli s Equation
v v
y y
g g
g y y
B y
B y
=
+ = +
 
 
− 
 
  −    

10. Flow Through Contraction
If the degree of contraction and the flow conditions are such that
upstream flow is subcritical and free surface passes through the critical
depth yc in the throat.
ycyc
y1
( )
3/2
2
2
sin
3
2 1
2
3 3
1.705
c c c c c c
c
c
Q B y v B y g E y
ce y E
Therefore Q B E g E
Q BE in SI Units
= = −
=
 
=  
 
=
B1 Bc
y2
yc
y1

11. Example # 11.3
In the accompanying figure, uniform flow
of water occurs at 0.75 m3/s in a 1.2m
wide rectangular flume at a depth of
0.6m.
(a). Is the flow sub-critical or super-
critical.
(b). If a hump height of Z=0.1 m is placed
in the bottom of flume, calculate the
water depth over the hump. Neglect the
head loss in flow over the hump.
(c). If the hump height is raised to
Z=0.2m, what then are the water depths
upstream and downstream of hump.
Neglect head loss over hump.
y1
y2 y3
Z
E.L
H.G.LV1
2/2g V2
2/2g
Q = 0.75 m3/sec
B = 1.2 m
y1 = 0.6 m
q = Q/B = 0.625 m3/sec/m

12. Example # 11.3
Solution
(a)
(b) First calculate Zc
Z<Zc therefore y2> yc
(C) Z>Zc therefore y2= yc
2 2
33
0.625
9.81
0.341
c
q
y
g
m y
Flow is subcritical
= =
= <
∴
2 2
1 22 2
1 2
2 2
2 12 2
2 1
2
Applying Bernoulli's Equation
2 2
2 2
0.46
q q
y Z y
gy gy
q q
y y Z
gy gy
y m
+ = + +
+ = + −
=
2 2
1 2 2
1
1
2 2
3 2 2
3
3
Applying Bernoulli's Equation
2 2
0.665
2 2
0.2
c
c
c
c
q q
y Z y
gy gy
y m
q q
y Z y
gy gy
y m
+ = + +
=
+ = + +
=
( ) ( )
cmZ
Z
g
Z
g
gy
q
yZ
gy
q
y
c
c
c
c
cc
2.14
512.0655.0
341.02
625.0
341.0
6.02
625.0
6.0
22
2
2
2
2
2
2
2
2
=
+=
++=+
++=+
(c).

13. Problem 11.54
A rectangular channel 1.2 m wide
carries 1.1 m3/sec of water in
uniform flow at a depth of 0.85m.
If a bridge pier 0.3m wide is placed
in the middle of this channel, find
the local change in water surface
elevation. What is the minimum
width of the constricted channel
which will not cause a rise in
water surface upstream.
Given that
Q = 1.1 m3/sec
B1 = 1.2 m
q1 = 0.92 m3/sec/m
yo = 0.85 m
B2 = B1-0.3 = 0.9 m
B1
0.3m
2 2
1 2
2
2
'
2 2
o
o
c c c
Bernouli s Equation
q q
y y
gy gy
Energy Equation
Q B y v Byv
+ = +
= =
B2=B1-0.3

14. Problem 11.54
2
1
sin 0.91
2
2
0.606
3
2.473 /sec
1.1
0.744
o
o
c
c c
c c c
c
q
ce E y m
gy
y E m
V gy m
Therefore
Q B y v
B m
= + =
= =
= =
= =
=
2 2
1 2
2
2
2
2
2
2
2
2 2
0.91
2
.38 & 0.785
o
o
q q
y y
gy gy
q
y
gy
y o m m
+ = +
+ =
=

15. Broad Crested Weirs and
Venturi Flumes

16. Broad Crested Weirs and Venturi Flumes
Flow Measurement in Open Channels
Temporary Devices
Floats
PitotTube
Current meter
SaltVelocity Method
Radio ActiveTracers
Permanent Devices
Sharp CrestedWeir/Notch
Broad CrestedWeir
Venture Flume
Ordinary Flume
Critical Depth Flume
Broad Crested Weirs and Venturi
Flumes are extensively used for
discharge measurement in open
channel.
Broad Crested Weirs and Critical
flumes are based and worked on
the principle of occurrence of
critical depth.

17. Broad Crested Weir
A weir, of which the ordinary dam is
an example, is a channel obstruction
over which the flow must deflect.
For simple geometries the channel
discharge Q correlates with gravity
and with the blockage height H to
which the upstream flow is backed up
above the weir elevation.
Thus a weir is a simple but effective
open-channel flow-meter.
Figure shows two common weirs,
sharp-crested and broad-crested,
assumed. In both cases the flow
upstream is subcritical, accelerates to
critical near the top of the weir, and
spills over into a supercritical nappe.
For both weirs the discharge q per
unit width is proportional to g1/2H3/2
but with somewhat different
coefficients Cd.

18. Broad Crested Weir
1
act d
3/ 22
d
3/ 22
d
Velocity of approach =Q/By
H= Head over the crest
B= Width of Channel
Since Q =C
V
1.7C
2g
V
3.09C
2g
act
act
V
Q
Q B H in SI
Q B H in FPS
=
 
∴ = + 
 
 
= + 
 
L
22
2
2 22
2
1
2 3
3
2
3/ 22
Applying Energy Equation ignoring h
V
H+Z+
2g 2
For Critical flow
2 2
2V
H+
2g 2 2
V2
3 2g
:
V2
3 2g
V
1.7
2g
c
c
c c
c c
c
c c
c c c
V
Z y
g
V y
g
V V
g g
V g H
V BV
Since Q By V B V
g g
B
Q g H
g
Q B H i
= + +
=
∴ = +
 
= + 
 
= = =
  
 ∴ = + 
   
 
= + 
 
3/ 22
V
3.09
2g
n SI
Q B H in FPS
 
= + 
 
Z>Zc
Vc
y1

19. Broad Crested Weir
Coefficient of Discharge, Cd also called Weir Discharge Coefficient Cw
Cw depends upon Weber number
W, Reynolds number R and weir
geometry (Z/H, L, surface roughness,
sharpness of edges etc). It has been
found that Z/H is the most
important.
The Weber number W, which
accounts for surface tension, is
important only at low heads.
In the flow of water over weirs
the Reynolds number, R is
generally high, so viscous effects
are generally insignificant. For
Broad crested weirs Cw depends
on length for. Further, it is
considerably sensitive to surface
roughness of the crest.
Z>Zc
Vc

20. Venturi Flume
Ordinary Flume
An ordinary flume is the one in which a stream line contraction of width is provided
so that the water level at the throat is drawn down but the critical depth doesn’t
occur.
B1 B2
y2
yc
y1
1 1 1 2 2 2
2 2
1 2
1 2
2 2 2 2 2 2
2 2
1 1
'
2 2
Using both equations, we get
2
Q=B y v =B y
1
Continuity Equation
B y v B y v
Bernoulli s Equation
v v
y y
g g
gH
B y
B y
=
+ = +
 
 
 
 
  −    
H = y2-y1

21. Venturi Flume
Critical Depth Flume (Standing Wave Flume)
A critical depth flume is the one in which either the width is contracted to such
an extent that critical depth occurs at the throat or more common both a
hump/weir in bed & side contractions are provided to attain critical depth with
hydraulic jump occurrence at d/s of throat.
B1 B2
ycy1
Z
H
vc
1 1 1 2 2 2
22
1
2 c c
'
2 2
Using both equations, we get
Q=B y v
c
c
Continuity Equation
Q B y v B y v
Bernoulli s Equation
vv
Z H Z y
g g
= =
+ + = + +
V1

22. Problem: 12.66
A broad crested weir rises 0.3m above the bottom of channel. With a
measured head of 0.6m above the crest, what is rate of discharge per unit
width? Allow for velocity of approach.
1
3/ 22
3/22
3
0.3
0.6
???
As we know that;
1.7
2
1.7
2
1; using Trial and Error
= q =0.505 /sec/
act d
act d
act
Z m
H m
y Z H
q
V
Q C B H
g
Q
Q C B H
By g
Since B
Q m m
=
=
= +
=
 
= + 
 
 
= + 
 
=
Take Cd=0.62

23. Problem: 12.67
A broad crested weir of height 0.6m in a channel 1.5m wide has a flow over it
of 0.27m3/sec.What is water depth just upstream of weir?
1
3
3/ 2
2
1
3/ 2
2
1
1
1
0.6
0.6
1.5
0.27 /sec
0.62
As we know that;
1.7
2
0.27
0.27 1.7 0.62 1.5 0.62
1.5 2
Solving above equations reults
y 0.905
act d
Z m
H y
B m
Q m
Cd
Q
Q C B H
By g
x x y
y g
m
=
= −
=
=
=
 
= + 
 
 
= − + 
 
=

24. Hydraulic jump and its practical
applications.

25. Hydraulic jump
Hydraulic jump formed on a spillway model
for the Karna-fuli Dam in Bangladesh.
Rapid flow and hydraulic jump on a dam

26. Hydraulics Jump or Standing Wave
Hydraulics jump is local non-uniform flow phenomenon resulting from the
change in flow from super critical to sub critical. In such as case, the water
level passes through the critical depth and according to the theory
dy/dx=infinity or water surface profile should be vertical. This off course
physically cannot happen and the result is discontinuity in the surface
characterized by a steep upward slope of the profile accompanied by lot of
turbulence and eddies. The eddies cause energy loss and depth after the
jump is slightly less than the corresponding alternate depth. The depth
before and after the hydraulic jump are known as conjugate depths or
sequent depths.
y
y1
y2
y1
y2
y1 & y2 are called
conjugate depths

27. Classification of Hydraulic jump
Classification of hydraulic jumps:
(a) Fr =1.0 to 1.7: undular jumps;
(b) Fr =1.7 to 2.5: weak jump;
(c) Fr =2.5 to 4.5: oscillating jump;
(d) Fr =4.5 to 9.0: steady jump;
(e) Fr =9.0: strong jump.

28. Classification of Hydraulic jump
Fr1 <1.0: Jump impossible, violates second law of thermodynamics.
Fr1=1.0 to 1.7: Standing-wave, or undular, jump about 4y2 long; low
dissipation, less than 5 percent.
Fr1=1.7 to 2.5: Smooth surface rise with small rollers, known as a weak
jump; dissipation 5 to 15 percent.
Fr1=2.5 to 4.5: Unstable, oscillating jump; each irregular pulsation creates a
large wave which can travel downstream for miles, damaging earth banks
and other structures. Not recommended for design conditions. Dissipation
15 to 45 percent.
Fr1=4.5 to 9.0: Stable, well-balanced, steady jump; best performance and
action, insensitive to downstream conditions. Best design range. Dissipation
45 to 70 percent.
Fr1>9.0: Rough, somewhat intermittent strong jump, but good performance.
Dissipation 70 to 85 percent.

29. Uses of Hydraulic Jump
Hydraulic jump is used to dissipate or destroy the energy
of water where it is not needed otherwise it may cause
damage to hydraulic structures.
It may be used for mixing of certain chemicals like in case
of water treatment plants.
It may also be used as a discharge measuring device.

30. Equation for Conjugate Depths
1
2
F1 F2
y2y1
So~0
1 2 2 1
1
2
( )
Resistance
g f
f
Momentum Equation
F F F F Q V V
Where
F Force helping flow
F Force resisting flow
F Frictional
Fg Gravitational component of flow
ρ− + − = −
=
=
=
=
Assumptions:
1. If length is very small frictional resistance may be neglected. i.e (Ff=0)
2. Assume So=0; Fg=0
Note: Momentum equation may be stated as sum of all external forces is equal
to rate of change of momentum.
L

31. Equation for Conjugate Depths
Let the height of jump = y2-y1
Length of hydraulic jump = Lj
2 1
1 1 2 2 2 1
1 1 1 2 2 2
1 2 ( )
( )
Depth to centriod as measured
from upper WS
.1
Eq. 1 stated that the momentum flow rate
plus hydrostatic force is the same at both
c c
c
c c
F F Q V V
g
h A h A Q V V
g
h
QV h A QV h A eq
g g
γ
γ
γ γ
γ γ
γ γ
− = −
− = −
=
+ = + ⇒
2 2
1 1 2 2
1 2
sections 1 and 2.
Dividing Equation 1 by and
changing V to Q/A
.2c c m
Q Q
A h A h F eq
A g A g
γ
+ = + = ⇒
2
;
Specific Force=
: Specific force remains same at section
at start of hydraulic jump and at end of hydraulic
jump which means at two conjugate depths the
specific force is constant.
m
Where
Q
F Ahc
Ag
Note
= +
( )
( )( )
2 2 2 2
1 2
1 2
1 2
2 22 2
1 2
1 2
2
2 2
2 1
1 2
2
2 1
2 1 2 1
1 2
Now lets consider a rectangular channel
2 2
.3
2 2
1 1 1
2
1
2
y yq B q B
By By
By g By g
y yq q
eq
y g y g
q
y y
g y y
or
y yq
y y y y
g y y
∴ + = +
+ = + ⇒
 
− = − 
 
 −
= − + 
 

32. Equation for Conjugate Depths
2
2 1
1 2
1 1 2 2
2 2
1 1 2 1
1 2
3
1
2
2
1 2 2
1 1 1
2
22 2
1
1 1
2
1
.4
2
Eq. 4 shows that hydraulic jumps can
be used as discharge measuring device.
Since
2
2
0 2 N
y yq
y y eq
g
q V y V y
V y y y
y y
g
by y
V y y
gy y y
y y
F
y y
y
y
+ 
= ⇒ 
 
= =
+ 
∴ =  
 
÷
 
= +  
 
 
= + − 
 
( )
2
1
21
2 1
1 1 4(1)(2)
2(1)
1 1 8
2
N
N
F
y
y F
− ± +
=
= − ± +
( )
( )
21
2 1
22
1 2
Practically -Ve depth is not possible
1 1 8 .5
2
1 1 8 .5
2
N
N
y
y F eq
Similarly
y
y F eq a
∴ = − + + ⇒
= − + + ⇒

33. Location of Hydraulic Jumps
Change of Slope from Steep to Mild
Hydraulic Jump may take place
1. D/S of the Break point in slope y1>yo1
2. The Break in point y1=yo1
3. The U/S of the break in slope y1<yo1
So1>Sc
So2<Sc
yo1
y2
yc
Hydraulic Jump
M3
y1

34. Location of Hydraulic Jumps
Flow Under a Sluice Gate
So<Sc
yo
yc
ys
y1 y2=yo
L Lj
Location of hydraulic jump where it starts is
L=(Es-E1)/(S-So)
Condition for Hydraulic Jump to occur
ys<y1<yc<y2
Flow becomes uniform at a distance L+Lj from sluice gate where
Length of Hydraulic jump = Lj = 5y2 or 7(y2-y1)

35. Problem
35

36. Problem 11. 87
A hydraulic Jump occurs in a triangular
flume having side slopes 1:1.The flow
rate is 0.45 m3/sec and depth before
jump is 0.3m. Find the depth after the
jump and power loss in jump?
Solution
Q= 0.45 m3/sec
y1=0.3m
y2=?
2 2
1 1 2 2
1 2
2
1 2
/3
0.858
0.679
2.997
Q Q
Ahc A hc
A g A g
hc y
y m
E E E
E
Power Loss Q E
Power Loss Kwatt
γ
+ = +
=
=
∆ = −
∆ =
= ∆
=
1:1
T=2y
hc=y/3

37. Problem 11. 89
A very wide rectangular channel with bed slope = 0.0003 and roughness n =
0.020 carries a steady flow of 5 m3/s/m. If a sluice gates is so adjusted as to
produce a minimum depth of 0.45m in the channel, determine whether a hydraulic
jump will form downstream, and if so, find (using one reach) the distance from the
gate to the jump.
Solution
So<Sc
yo
ycys=0.45m y1 y2=yo
L Lj

38. Problem 11. 89
1/32
2/3 1/2
2
1 2
2 2
4/3
1
1.366 0.45
( )
c
o
o
m
m
s
o
q
y Super Critical Flow
g
A
Q R S
n
y y
y f y
V n
S
R
E E
L
S S
 
= = > ⇒ 
 
=
≈
=
=
−
=
−

39. Thank you
Questions….
Feel free to contact:
39