Chapter 3: Channel Controls
A control is any channel feature, which fixes a relationship between depth and discharge in its
neighborhood. It may be natural or man-made. Also we may have:
Overflow structures-spillways, weirs, free falls
Underflow structures-sluice gates, gates
Why use need control structures?
1) Flow profile computation-provides the boundary condition.
2) To measure discharge; as an engineer we are concerned with the functioning of the
control itself, such as the ability of a spillway to discharge floodwaters at the required
rate.
We have rapid flow in the vicinity of control structures (sections)
Since the streamlines are highly curved near the control sections we can not solve the
equations analytical, Therefore we go to empirical relations based on experiments.
Flow through orifices and short tubes
Vena contract a is the section where area of the jet becomes constant
Typical examples of orifices and short tubes are shown. A knowledge of the laws of the flow
through them is necessary in determining the discharge through sluiceways and the entrances
to conduits.
If the entrance is not properly shaped, a contraction of the jet occurs as in sketches a, c and h,
and the area of the jet is not as great as the area of the orifice or tube. For properly rounded
approaches to orifices as in sketches b and e, and the constant diameter short tubes, the
diameter of the jet is equal to the area of the orifice or tube.
In case of short tubes without rounded entrances, the contraction does occur; but the jet
expands again, with certain exceptions, a partial vacuum occurring just inside the entrance
Let
H=
the head of water on the center line of a freely flowing orifice or tube, or the
difference in water level for a submerged orifice or tube
A=
the area of the orifice and tube
V=
the theoretical velocity corresponding to head H;
g=
the acceleration of gravity
Q=
the discharge
cc=
the coefficient of contraction
c'=
the coefficient of friction, the reduction in total head due frictional
cd=
the coefficient of discharge

A jet
Aorif
The general equation for the velocity of the spouting water is
V  2 gH
Considering the friction, the actual velocity due to the head H is
Vaction  c ' 2 gH
The discharge is equal to the product of the actual velocity and the area of the jet;
A j  cc A
Therefore Q  vact cc A  c ' cc A 2 gH
In experiments conducted to determine the discharge through orifices and tubes, the
coefficient of friction and the contraction coefficient are combined and the general equation is
Q  cd A 2 gH
Cd= discharge coefficient → it depends on the shape of the orifice and tube, it is not greatly
affected by the submergence. If we do not have submergence and enough friction contracted
jet does nor expend in tube it shoots out as contracted without touch the walls.
Dimensional Analysis
Let’s consider the following figure;
According to values of L, w and roundness, it can be a weir or sharp crested weir.
The discharge coefficient, cd, will be a function of:
cd   vo , H , w, L, g ,  ,  ,  
→ 8 variables
→ 3 basic dimensions, MLT
→ 5 dimensionless quantity
→ choose ρ, vo, H as primary (repeating variables);
Then dimensional analysis would give us:
H H

c d  f  , , Re , Fr , W 
W L

H H

c D  f  , Re , W 
W L

w

vo2 H
Sharp-Crested Weirs
A sharp crested weir normally consists of a vertical plate mounted at right angles to the flow
and having a sharp-edged crest, as shown in the figure.
Sharp-crested weirs are commonly used as a means of flow measurement. But it has an
importance in Open Channel Hydraulics, simply because, its theory forms a basis for the
design of spillway.
Because, the edge is shapp. → no boundary layer development only on the vertical surface on
which the velocities are small
Therefore no viscous effects → no energy dissipation
Let’s consider the simplest form of weir; consisting of
-
a plate set perpendicular to the flow in a rectangular channel,
-
its horizontal upper edge running the full width of the channel
-
2 dimensionality → no lateral contraction effects. Since the lateral contraction effects
are suppressed by the channel sidewalls, this type of weir is sometimes termed the
SUPPRESSED weir.
Let’s make an elementary analysis by assuming;
-
no contraction over the weir
-
pressure it atmospheric across the whole section AB.
The total head at a point c, then;
h
P


v2
 v 2 gh
2g
The discharge per unit width then;
q
vo2 / 2 g  H
vo2

vdh 
/ 2g
vo2 / 2 g  H
vo2

2 ghdh
/ 2g
3/ 2
3/ 2

 vo2  
2  vo2
 
q  2 g 
 H   
3  2 g
2 g  




Here h is measured downwards from the T.E.L, not from the free surface.
Now, the effect of the flow contractio n may be expressed by a contraction cc, leading to the
result:
3/ 2
3/ 2
 v 2
  vo2  
2
o
 
q  cc 2 g 
 H  

3
 2 g
  2 g  
We can make this expression more compact by introducing a discharge coefficient cd;
q
2
cd 2 g H 3 / 2
3
3/ 2
3/ 2

 vo2  
vo2 
 

where; c d  cc 1 
 2 gH   , when
 2 gH 

 
vo2
we should expect both cc and the ratio of
to be dependent on the boundary geometry
2 gH
alone, in particular on the ratio of H/W;
vo2
when W is large vo is small 
 c d  cc
2 gH
REHBOCK found experimentally that
c d  0.611  0.08
H
1

viscous and surface tension effects, Neglect unless H is very
W 305H
small.
As W becomes very large → cd  0.611 and since Vo2 / 2 g becomes very small
cc  cd  0.611
This happens to be the numerical value of

 2
, shown last century by Kirchoff to be the
contraction coefficient of a jet issuing without energy loss, and with negligible deflection by
gravity from a long rectangular slot in a large tank, a two-dimensional problem,
This is the only fluid flow problem in ideal fluid for which the contraction coefficient is
obtained theoretically.
By many investigators the value 0.611 have been conformed for high weirs.
On the other hand, when W becomes small, H/W becomes large, and this formula cannot be
true for high values of H/W. In fact, experimental work has shown that it is true only for the
values of H/W up to approximately 5;
H
5
W
For 5
H
10 , cd begins to diverge from the value given by the formula, reaching a value of
W
1.135 when
H
 10 .
W
If W vanishes completely, so that H/W becomes infinite, we have the case of free over fall. In
this case H  yc  critical depth and q is determined accordingly.
In fact, it has been shown experimentally that critical flow also occurs just u/s of a very low
weir. In this case the weir is called a sill, i.e in the range H / W  20 .
yc  H  W
q c  y c vc
vc  gy c
q c  H  W  g H  W   g H  W 
3/ 2
 W
 1  
H

3/ 2
H 3/ 2
On the other
2
 W
q  c d 2 g H 3 / 2  g 1  
3
H

 W
c d  1.061  
H

The range 10
3/ 2
H 3/ 2
3/ 2
H
 20 has not yet been completely exploxed. But KANDA SWANY AND
W
ROUSE showed that
Max. cd occurs at
H
 10  border line for weir and sill.
W
For completely free overfull, w = 0, the eq;
3/ 2
3/ 2

 vo2  
vo2 
 

c d  cc 1 
 2 gH  
 2 gH 

 
cd  1.484cc
 W
c d  1.061  
H

3/ 2
 1.06  cc 
1.06
 0.715
1.404
Two important features of the pressure distribution down the vertical section ABC;
-
Pressure distribution is nonhydrostatic, because of distinct curvature of the streamlines
in the vertical plane,
-
Pressure is not atmosphere contrary assumption in the elementary analysis the press >
atmospheric pressure
Consistent with the contraction and acceleration d/s of AB clearly a pressure force is
necessary to cause this acceleration in the region of atmospheric pressure.
It has been assumed throughout that the pressure is atmospheric along the lower surface of the
jet, or “nappe” as well as upper surface if this jet is just confined between parallel walls
downstream of the weir, air will be trapped and this all will be gradually removed by the flow,
pressure in this region is reduced.
For a given head Q head → cavitation→ is necessary.
Any other type of weir will be 3D because of the lateral contraction from the sides, as well as
in the vertical plane. Typical examples are the “contracted” rectangular weir and the
triangular weir.
Uses of such weirs
-
For large tanks which could be thought as effectively infinite to measure the flow rate
-
If used for smaller tank, then calibration is needed.
Francis found experimentally that the amount of lateral contraction at each end of the
contracted weirs was equal to 0.10 H, provided that the length L of the weir was greater than
3H. On the basis of this result, it is commonly accepted that the discharge Q is given by;
Q
2
cc L  0.2 H  2 g H 3 / 2 , cc  0.611
3
The triangular weir, leading to the result;
Q
8

cc tan
2g H 5 / 2
15
2
The most commonly used value of the notch angle α is 90o, for this case cc is found to be
around 0.585, somewhat less than for the rectangular weir.
Although the value of cc is near to 0.59 in general, it is affected by viscosity, surface tension,
and weir plate roughness;
A comprehensive study of triangular weir flow has been made by lenz, He used many liquids
in order to discover the effects of viscosity and surface tension on weir coefficients, thus
extending the utility of the triangular weir as a reliable measuring decree.
For α = 90o, lenz proposed that
C c  0.56 
0.70
R W 0.170
0.165
Applicable to all liquids providing that the failing sheet of liquid does not ding to the weir
plate and that H > 0.06 m, Re 300
,
W 300
R & W Cc
Formula given by British Standard Spec fractions for all three of these weirs are:
For suppressed weir q 
Contracted weir;
Q
2
H

1.5
2 g  0.604  0.081 H  0.0034
3
W

2
L
2 g 0.615L  0.1H H 1 / 5
2
3
H
For V-notch Q  250H 2 / 50
Experiments show that Q  2.48H 2.48 is more accurate
For H  23 cm
H 9 ''
P 31 cm
P12' '
L122 cm
L 4 '
For 23cm H  46cm
9 '' H 18 ''
P 46cm
P18
L183cm
L 6 '
These values are taken from the British Standard Specification: Since the offer … guidance as
to the exact form of the weir edge we will see them.
Within the above limitations and using   90 o
g  32.2 ft / sec 2
The formula given for the suppressed weir is:
q
2
H

1.5
2 g  0.604  0.081 H  0.0034
3
W

L in foot second units
Virtually identical with that of Rehbock’s
The stated margin of error is ± 1.5%
For contracted rectangular weir; the formula is the equivalent of
Q
2
2 g x0.615L  01H H 1.5
3
 2%
L/ H  2
Q  2.50H 2.50
Although precise measurements have tended to indicate that the equation
Q  2.48H 2.48 is more precise
3'' H 15 ''
Half V-notch, that is α = 45o
Q  1.24H 2.48
 1.5%
The finish of the edge and upstream surface of a weir is important, since the roughness of the
surface or rounding of the edge tends to suppers the lateral components of flow, increase cc
and hence increase the discharge.
It should be measured 3-4 it upstream of weir.
CONTRACTED WEIRS
A contracted weir involves a 3-D flow problem, because of:
- the contraction from the sides as well as in the vertical plane.
Typical examples:
-
contracted rectangular weirs
-
triangular weir
Such weirs are commonly used for flow measurement, normally in tanks large enough to be
effectively infinite, so that cc has its minimum value. In this case, general equations are
available relating discharge to head; but for small tanks weirs should be calibrated for each
particular problem.
a) Contracted Rectangular Weirs
Francis found experimentally that the amount of lateral contraction at each end of the
contracted rectangular weir was equal to
Leff  L  0.2 H
1
H
10
L 3 H
On the basis of this result, it is commonly accepted that the discharge Q is given bey
Q
2
cc L  0.2 H  2 g H 3 / 2
3
cc  0.611
b) The triangular weir (V-Notch)
The triangular weir can be analyzed in the same elementary way as the suppressed
rectangular weir leading to the result:
Q
8

cc tan
2g H 5 / 2
15
2
The most commonly used value of the notch angle α is 90o. For this case cc=0.585
→somewhat less than for rectangular weir.