Chapter 4:
Non uniform flow in open channels
Learning outcomes
• By the end of this lesson, students should be
able to:
– Relate the concept of specific energy and
momentum equations in the effect of change in bed
level - Broad Crested Weir
– Relate the concept of specific energy and
momentum equations in the effect of lateral
contraction of channel ( Venturi Flume)
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Introduction
• Analysis of steady non uniform flow in open
channels.
• Non uniform flow occurs in transitions where
there is change in cross section or obstruction
in channel.
• Analysis requires a different approach,
requiring the use of the energy equation in a
different form.
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Specific energy & alternative
depths of flow
• Specific energy, E,
v2
E  D
2g
(16.1)
• For a wide rectangular channel, mean velocity is,
Q
Q
q
v 

A BD D
• While the volume rate of flow per unit width,
Q
q
B
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• Substituting v & q into E,
q2
E  D
2 gD2
q2
D  ED 
0
2g
3
2
(16.2)
(16.3)
• This equation has 3 roots:
– 1 root is negative & unreal
– 2 roots are positive & real, which give 2 alternate
depths:
• Larger depth: deep slow flow (subcritical/ tranquil/
streaming flow).
• Smaller depth: shallow fast flow (supercritical/ shooting
flow)
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• Critical depth, DC:
– Depth at which the 2 roots coincide
– q = qmax
– E = Emin
• To find DC:
dE
2q 2
dD
 1
2 gD3
• When dE/dD = 0,
q
Criticaldepth,DC  
 g
2



1
3
 Q 
  2 
 gB 
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1
3
(16.4)
8
• Sub.
q 2  gDC
3
from (16.4) to (16.2),
3
gDC
3
E  DC 

DC
3
2
2 gDC
DC 
2
E
3
• Therefore, for rectangular channel,
• Differentiating (16.3) & assuming E is constant
q  D2 g E  D 
1
2
1 D 

1
dq
2

 2 g   E  D  2 
1
dD

E  D  2 

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(16.5)
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• When
becomes,
D  DC ,
qmax
dq
1
 0, E  DC   DC  0
dD
2
2  
2 
 E 2 g  E  E 
3  
3 
• Or qmax  gDC
3
1
2
, (16.5)
2 
2
 g  E
3 
1
3
2
(16.6)
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• Critical
velocity:
velocity
corresponding to critical depth.
• Sub. E  3 2 D , D  D into (16.1),
C
of
flow
C
2
v
3
DC  DC  C
2
2g
vC 
gDC 
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DC for non rectangular sections
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• For any shape and cross sectional area the E for any
D,
v2
E  D
2g
• Since v = Q/A,
Q2
E  D 2
2A g
(16.7)
• For flow at DC and vC, Emin, differentiating
(16.7),
(16.8)
2Q 2 A3 dA
1
2g
dD
0
• But a change in depth will produce a change in
cross-sectional area, therefore dA/dD=B.
Q2 B
1
3
Ag
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(16.9)
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• For critical flow,
Q  Ag 
vC   

A  B 
1
2
• From (16.9)
vC  g D
(16.10)
where D is the average depth.
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Froude Number
• Assume a surface wave
of
height
δZ
is
propagated from left to
right of observer.
• Wave is brought to rest
relative to observer by
imposing a velocity c
equal to wave velocity
on the observer, flow
will appear steady.
Mass per unit time(left of wave)  Mass per unit time(right of wave)
  BZ  Z u  u  c     B  Z  u  u  c 
Zu  uZ  Zu  cZ  0
c  u Z  Z  Z u
(5.32)
Hydrostatic forcedue toZ  Mass per unit time Change of velocity
gZBZ  BZu - c   u 
gZ
u 
c  u 
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• Sub. δu to (5.32),
c  u Z  gZ
Z  Z  c  u 
c  u 2  Z  Z g
 gZ
• If wave height δZ is small,
Velocityof thepropationof the wave relativeto thefluid  c - u
 gZ
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• Ratio of the stream velocity u to the
propagation velocity c-u is known as Froude
Number Fr.
u
u
(5.34)
Fr 

c u
gZ
• Fr can be used to determine the type of flow
for open channel.
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• For critical flow conditions, the Froude Number is,
Fr 
v

gD
vC
1
gDC
– If v < vc – subcritical flow
– If v > vc – supercritical flow
• Important difference,
– Subcritical : disturbances can travel upstream and
downstream thus enabling downstream conditions to
determine the behavior of the flow.
– Supercritical: disturbances cannot travel upstream and thus
downstream cannot control the behavior of the flow.
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Tranquil
Critical
Shooting
Subcritical
Critical
Supercritical
Depth
D > DC
D = DC
D < DC
Velocity
v < vC
v = vC
v > vC
Fr
Fr < 1
Fr = 1
Fr > 1
Channel slope
Mild
Critical
Steep
Control
Downstream
-
Upstream
Disturbance
Wave can travel
upstream
Standing waves
Waves cannot
travel upstream
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• Figure 4.1 shows that when flow is in the region of
• Dc , small changes of E and q results in relatively
large changes in D.
• Small surface waves are therefore easily formed but
since velocity of propagation vp = vc, these waves
will be stationary or standing waves.
• Their presence therefore, an indication of critical flow
conditions
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Figure 4.1a: Plot of D vs q for a constant E
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• Line OA from figure below can be drawn at 45°
through the origin.
• If scales for E and D are the same, horizontal
distances from vertical axis to line OA will be equal
to D, and the distance from OA to the specific energy
curve will be v2/2g.
• If q is constant,
– Tranquil flow:
• D increases, v increases, E curve is asymptotic to OA
– Shooting flow:
• D decreases, v increases, E curve will be asymptotic to the E axis.
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• Which of the two alternate depths for a given E will
occur at a cross section depends on the slope of the
channel.
• Critical slope, sc, is defined as the slope of the
channel which will maintain flow at critical depth,
DC.
– Uniform tranquil: s < sc – mild slope
– Uniform shooting flow: s > sc – steep slope
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Example 4.1
A rectangular channel 8 m wide conveys water at a
rate of 15 m3/s. If the velocity in the channel is 1.5
m/s, determine;
a) E
b) DC
c) vc
d) Emin
e) Type of flow
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Example 4.2
Determine the critical depth in the trapezoidal
channel shown below if the discharge in the channel
is 0.34 m3/s. The channel has side slopes with a
vertical to horizontal ratio of 1:1.
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Example 4.3
Determine the critical depth in a channel of triangular
cross section conveying water at a velocity of 2.75
m/s and at a depth of 1.25 m. The channel has side
slopes of 1:2.
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Exercise
A channel has a trapezoidal cross-section with a
base width of 0.6 m and sides sloping at 450.
When the flow along the channel is 20 m3 min-1,
determine the critical depth. (0.27 m)
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Control sections
• Control sections – cross sections at which the flow
passes through the critical depth.
• Such sections are limiting factor in the design of
channel. and some of the cases in which they occur
are:
–
–
–
–
Transition from tranquil to shooting flow
Entrance to a channel of steep slope from a reservoir
Free outfall from a channel with a mild slope
Change in bed level or change in width of channel
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Transition from tranquil to shooting flow
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• May occur where there is a change of bed slope s.
• Upstream slope is mild and s is less than the critical
slope sc.
• Over s considerable distance the depth will change
smoothly from D1 to D2.
• At the break in the slope, the depth will pass through
DC forming a control section which regulates the
upstream depth.
• At the tail end, the reverse transition from shooting to
tranquil flow occurs suddenly by means of a
hydraulic jump.
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Entrance to a channel of steep slope from a reservoir
• If depth of flow in the channel is less than DC for the
channel, water surface must pass through DC in the
vicinity of the entrance, since conditions in the
reservoir correspond to tranquil flow.
Free Outfall from a Channel with a Mild Slope
• If slope s of the channel is less than sc the upstream
flow will be tranquil.
• At the outfall, theoretically, the depth will be critical,
DC.
• In practice, gravitational acceleration will cause an
increase of velocity at the brink so that D < DC.
• While experiments indicate that depending on the
slope upstream:
– DC occurs at distance of between 3DC to 10DC from the
brink.
– D at the brink is approximately 0.7DC.
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Change in bed level or
channel width
Flow over a
broad-crested
weir
Effect of lateral
contraction of a
channel
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Flow over a broad-crested weir
• Broad-crested weir is an obstruction in the form of a
raised portion of the bed extending across the full
width of the channel with a flat upper surface or crest
sufficiently broad in the direction of flow for the
surface of the liquid to become parallel to the crest.
• Upstream edge is rounded to avoid separation losses
that occur at a sharp edged.
• The flow upstream of the weir is tranquil and the
conditions downstream of the weir allow a free fall
over the weir.
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• The discharge over the weir will be, therefore, be the
maximum possible and flow over the weir will take
place at DC.
• For a rectangular channel,
Q 
DC   2 
 gB 
2

Q  B gDC
DC 
1
3
3

1
2
2
E
3
8


Q  B g  E 3 
27 

1
2
 1.705BE
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2
(16.11)
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• The specific energy, E, measured above the crest of the weir
will be (assuming no losses),
v2
EH
2g
H is the height of the upstream water level above the crest and
v is the mean velocity at a point upstream where flow is
uniform.
• If the upstream depth is large compared with the depth over
the weir, (v2/2g) is negligible, therefore,
EH
• Rewriting (16.11),
Q  1.705 BH
3
2
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(16.12)
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• A single measurement of the head, H above the
crest of the weir would then be sufficient to
determine Q.
• Since D  Q / gB  , the depth over the crest of
the weir is fixed, irrespective of its height.
• Any increase in the weir height will not change
DC but will cause an increase in the depth of
the flow upstream.
• Therefore, maximum height of the weir,
2
2
1
3
C
z  D1  D2
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• If the level of the flow downstream is raised, the
surface level will be drawn down over the hump, but
the depth may not fall to the critical depth.
• The rate of flow can be calculated by applying
Bernoulli’s s Equation and continuity equation and
depends on the difference in surface level upstream
and over the weir.
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Example 4.4
A broad crested weir 500 mm high is used to measure
the discharge in a rectangular channel. The width of
the channel is 20 m and the height of the channel
upstream of the weir is 1.25 m. What is the discharge
in the channel if water falls freely over the weir?
Assume that the velocity upstream is very small.
Determine the difference in water level between
upstream and over the top of the weir.
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Effect of lateral contraction of a channel
• When width of a channel is reduced while bed
remains flat, q increases.
• As channel narrows - neglecting losses, E remains
constant – for tranquil flow, depth will decrease while
for shooting flow depth will increases.
Free surface
does not pass
through DC
Lateral
contraction
Free surface
passes through
DC
With hump
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Free surface does not pass through DC
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• Arrangement forms a venturi flume (venturi meter).
• Applying energy equation between upstream & throat
and ignoring losses,
2
2
v
v
D1  1  D2  2
2g
2g
2
2
2
v2  B2 D2 
1  2 2   D1  D2  h
2 g 
B1 D1 
v2 
2 gh
B D 
1   2 2 
 B1 D1 
Q  B2 D2 v2
Q  B2 D2
(applyingQ1  Q2 )
2
2 gh
B D 
1   2 2 
 B1 D1 
2
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• Actual discharge,
Q  Cd B2 D2
2 gh
 B2 D2 

1  
 B1 D1 
2
Cd is a coefficient of discharge – 0.95 to 0.99
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Free surface passes through DC
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• The flowrate is given by,
Q  B2 D2 v2
 B2 D2 2 g E  DC 
2
1
 B2  E 2 g  E
3
3
 1.705BE
3
2
• Assuming that the upstream velocity head is
negligible,
3
Q  1.705 BH 2
(16.15)
where H is the of the upstream free surface above bed
level at the throat.
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Lateral contraction with hump
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• Height of upstream water level above the hump, H =
D1 – Z
• When upstream conditions are tranquil and the bed
slope is the same downstream as upstream,
impossible for shooting flow to be maintained for any
great distance from the throat.
• Revert to tranquil flow downstream by means of a
hydraulic jump or standing wave.
• Venturi flume operating in this mode is known as
standing wave flume.
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Example 4.5
A venturi flume is constructed in a channel
which is 3.5 m wide. If the throat width in the
flume is 1.2 m and the depth upstream from
the constriction is 1.25 m , calculate the
discharge in the channel when the depth at the
throat is 1.2 m. If the conditions are such that a
standing wave is formed, what is the
discharge?
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Example 4.6
• A Venturi flume is 2.5m wide and 1.4m deep
upstream with a throat width of 1.3m.
Assuming that a standing wave form
downstream, calculate the rate of flow of water
if the discharge coefficient is 0.94. Do not
ignore the velocity of approach.
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Review of past semesters’
questions
APR 2010
• A 10 m wide channel conveys 25 m3/s of
water at a depth of 1.6 m. Determine :
i) specific energy of the flowing water
ii) critical depth, critical velocity and
minimum specific energy
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APR 2010
• A venturi flume is 1.40 m wide at the entrance
and 0.7 m wide at the throat. Determine the
flow if the depths at the entrance and at the
throat is 0.8 m and 0.6 m respectively. Neglect
all losses.
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