=Lectures-9_Weirs_.pdf.dep of civil eng.

Lectures-9_Weirs_.pdf.

1. Lecture - 9
Flow over Weirs

2. Introduction
A structure, used to dam up a stream or river;
over which the water flows, is called a weir.
The conditions of flow, in the case of a weir are
practically the same, as those of a rectangular notch.
That is why, a notch is sometimes called as a weir
and vice versa.
The only difference between a notch and a weir is,
that the notch is of a small size; but the weir is of a
bigger one. Moreover, a notch is usually made in a
plate, whereas a weir is usually made of masonry or
concrete.

4. Types of Weirs
•Rectangular
•Cippoletti (Trapezoidal)
Shape
•Ordinary
•Submerged or drowned
Nature of
Discharge
•Narrow Crested
•Broad Crested
Width of Crest
•Sharp Crested
•Ogee
Nature of Crest

5. Discharge over a Rectangular Weir
Consider a rectangular weir, over which the water is flowing.






Let
H = Height of the water, above the crest of the weir,
L = Length of the weir, and
C
d = Coefficient of discharge.

6. Discharge over a Rectangular Weir
Let us consider a horizontal strip of water of thickness dh at a
depth h from the water surface as shown in.
Area of strip = L.dh
We know that theoretical velocity of water through the strip is,

 dq = C
d x Area of strip x Theoretical velocity

The total discharge, over the weir, may be found out by
integrating the above equation within the limits 0 and H.

gh2 ghdhLC
d 2..

7. 
2/3
0
2/3
0
2/3
0
2/1
0
2.
3
2
2.
3
2
2
3
2.
2.
2.
xHgLC
hgLCQ
h
gLC
dhhgLC
ghdhLCQ
d
H
d
H
d
H
d
H
d




















8. Discharge over a Rectangular Weir
Note:
Sometimes, the limit of integration to the above equation, are
from H
1 to H
2. (i.e,. the liquid level is at a height of H
1 above
the top of the weir and H
2 above the bottom of the weir)
instead of 0 and H; then the discharge over such a weir will be
given be the equation.
 
2/3
1
2/3
2 2
3
2
HHgLCQ
d 

9. Problem-1
A rectangular weir 4.5 meters long has a head of water 300mm,
Determine the discharge over the weir, if coefficient of
discharge is 0.60.
Solution:














slitressmxQ
mxxxQ
HgLC
Given
d
/ 1310/ 31.1164.0972.7
sec/3.081.925.46.0
3
2
values,Putting
2.
3
2
Q
r weir,rectangula over the Discharge
60.0C
0.3m 300mm H
4.5m L
:
3
32/3
2
3
d







10. Problem-2
A weir 8 m long is to be built across a rectangular channel to
discharge a flow of 9 m
3
/sec. If the maximum depth of water on
the upstream side of the weir is to be 2 m. what should be the
height of the weir? Adopt C
d = 0.62
Solution:













1.28m0.72-2 be should weir ofheight theTherefore
72.0
614.0
645.14
9
81.92862.0
3
2
9
values,Putting
2.
3
2
Q
waterof sill theabove water ofHeight HLet
62.0C & m 2 water ofDepth
/m 9 Q m 8 L
:
2
3
2
3
2
3
d
3







mH
HxHxxx
HgLC
s
Given
d

11. Problem-3
The daily record of rainfall over a catchment area is 0.2 million
cubic meters. It has been found that 80% of the rainwater
reaches the storage reservoir and then passes over a rectangular
weir. What should be the length of the weir, if the water is not to
rise more than 400mm above the crest?
Assume suitable value of coefficient of discharge for the weir as
0.61.
Solution:












61.0C
0.4m 400mm H
rain water of 80% reservoir in the Discharge
/0.2x10 Rainfall
:
d
36



 daym
Given

12. 
mL
LxxxLx
HgLC
smdaym
daym
d
06.4456.0/85.1
456.0)4.0(81.9261.0
3
2
85.1
values,Putting
2.
3
2
Q
r weir,rectangula over the Discharge
/85.1
60 x 60 x 24
10 x 0.16
/ 10 x 0.16
/)10 x (0.2 x 0.8 rain water of 80% Q
area,catchment thefrom
reservoir thereachesch water whiof volume that theknow We
2
3
2
3
3
6
36
36






13. Francis’s Formula for Discharge over a
Rectangular Weir (Effect of End Contractions)
Francis, after carrying out a series of experiments, proposed
an empirical formula for the discharge over a rectangular weir.
He found that the length of the stream of liquid, while flowing
over a weir, gets contracted at the ends of the sill as shown in
Fig.

14. This end contraction, of the stream of liquid, is known as
lateral contraction or side contraction.
Francis also found that the amount of the end contractions
depend upon the conditions of sides of the channel and top of
the sill, as well as velocity of liquid.
But an approximate value of end contraction, at each end, is
1/10 of the height of the liquid above the sill of the weir. Thus
if, there are two end contractions only (as in the case of simple
rectangular weir) the effective length of the weir is(L – 0.2 H).
Substituting this value of length in equation for discharge,
 
2
3
22.0
3
2
HxgHLCQ
d
Francis’s Formula for Discharge over a
Rectangular Weir (Effect of End Contractions)

15. Sometimes, the total length of a weir is divided into a number
of bays or spans by vertical posts as shown in Fig. In such a
case, the number of end contractions will be twice the number
of bays, or spans, into which the weir is divided. Thus, in
general, we may write the empirical formula proposed by
Francis as:

Where, n = No. of end contractions.
Now substituting C
d = 0.623 and g = 9.81 m/sec
2
in the
general equation for discharge,

Note: When the end contractions are suppressed, the value of
n in the above equation, is taken as zero.


 
2
3
21.0
3
2
xHgnHLCQ
d  2
3
1.084.1 nHLQ 
Francis’s Formula for Discharge over a
Rectangular Weir (Effect of End Contraction)

16. Problem-4
A 30 meters long weir if divided into 10 equal bays by vertical
posts each 0.6 meter wide. Using Francis' formula, calculate the
discharge over the weir, under an effective head of 1 meter.
Solution:












m 24.6 0.6) x (9-30 L
weir, theoflength effective theand
ns)contractio end twohasbay Each ( 20 2 x 10n
ns,contractio end of no. that theknow We
1mH
0.6m post each ofWidth
10 bays of No.
30m weir oflength Total
:







Given

17.  
 
s
x
HnHLQ
/m 41.6
)1(1) x 20 x (0.1-24.6 1.84
1.084.1
formula) Fransis(by weir over the Discharge
3
2/3
2
3




18. Problem-5
A reservoir has a catchment area of 25 square kilometers. The
maximum rainfall over the area is 25mm per hour, 40% of
which flows to the reservoir over a weir. Using Francis'
formula, find the length of the weir. The head of water over the
water should not exceed 0.8m.
Solution:












0.8m H
rainfall totalof 40% reservoir theinto Discharge
m/hr 0.025 mm 25 =hourper rainfall Maximum
m 10 x 25 = km square 25 = areaCatchment
:Given
26




19.  
 
52.74m L
8.08.021.084.14.69
1.084.1
formula) Fransis(by weir over the Discharge
/4.69
6060
1025.0

/1025.0025.0)10 x (25 x 0.4
Rainfall x Area x 0.4 rainfall of 40% Q
area,catchment thefrom
reservoir thereaches which water,of volume theand
weir)simple of( 2 n
:nscontractio end of no. that theknow We
2/3
3
6
366
2/3







xxxL
HnHLQ
sm
x
x
hrmxx


20. Bazin’s formula for discharge over a
Rectangular Weir
Bazin, after carrying out series of experiments, proposed an
empirical formula for the discharge over rectangular weir.
He found that the value if coefficient of discharge varies with
the height of water over the sill of the weir. Thus, he proposed
an amendment in formula for rectangular weir.
We know that discharge over rectangular weir,


Bazin proposed that the discharge over a weir,


Where m = 2/3 C
d
2/3
2
3
2
HxgLCQ
d 2/3
2 HxgLmQ

21. Bazin’s formula for discharge over a
Rectangular Weir
He found that the value of m varies with the head of
water, whose value may be obtained from the
relation:




He found the above relation by experiments in which
he avoided the effect of end contractions. metersin water ofHeight H Where
003.0
405.0


H
m

22. Problem-6
Find the discharge over a rectangular weir 4.5m long under
head of 600mm by using Bazin’s formula.
Solution:












s
x
HxgLmQ
H
m
mmmH
mL
/m 3.8
0.6 9.81 x 2 x 4.5 x 0.41
2
formula) sBazin'(By weir over the discharge theand
41.0
6.0
003.0
405.0
003.0
405.0
:relation sBazin' From
6.0600
5.4
:Given
3
3/2
2/3







23. Problem-7
A rectangular weir 6m long is discharging water under a head
of 300mm. Calculate the discharge over the weir by using
i) Francis formula
ii) Bazin’s formula
Solution:













 
 
s
x
HnHLQ
mmmH
mL
/m 1.796
)3.0(0.3) x 2 x (0.1-6 1.84
1.084.1
Q weir,over the Discharge
2n ns,contractio end of No.
formula sFranci'by weir over the Discharge i)
3.0300
5.4
:Given
3
2/3
2
3







24. s
x
HxgLmQ
H
m
/m 1.18
0.6 9.81 x 2 x 4.5 x 0.41
2
formula) sBazin'(By weir over the discharge theand
415.0
3.0
003.0
405.0
003.0
405.0
:relation sBazin' From
formula, sBazin' usingby weir over the Discharge ii)
3
3/2
2/3





25. Velocity of Approach:
It is defined as:
“The velocity with which water approaches or reaches a weir
before it flows over it is called velocity of approach.”
If V
a is the velocity of approach then additional head h
a equals
to V
2
/2g due to velocity of approach is acting over the water
flowing over the weir.
The initial height of water becomes H + h
a and the final height
becomes h
a.
It is determined by finding discharge over the weir neglecting
velocity of approach. Then dividing discharge by the cross
sectional area of the channel on the upstream side of the weir,
the velocity of approach is obtained.

26. Velocity of Approach:
Mathematically,

This velocity of approach is used to find an additional head:


Again the discharge is calculated and above process is
repeated for more accurate discharge.
Discharge over the rectangular weir with velocity of approach
is: Channel of Area
Q
V
a 2g
2
a
a
V
h ])[( 2
3
2 2/32/3
aad hhHxgLCQ 

27. Problem-8
A weir 2.0m long has 0.6m head of water over the crest. Using Francis’
formula, find the discharge over the weir and, if the channel
approaching the weir is 6m wide and 1.2m deep.
Also determine the new discharge, considering the velocity of approach.
Solution:













 
 
s
x
HnHLQ
mH
mL
/m 1.95
)6.0(0.6) x 2 x (0.1-2.4 1.84
1.084.1
2n ns,contractio end of No.
:approach of velocity thegconsiderinout weir withover the Discharge i)
1.2m depth and 6m channel ofWidth
6.0
0.2
:Given
3
2/3
2/3








28. smQ
xxxQ
hHxnHLQ
m
m
xg
sm
a
/ 01.2
])015.0()615.0[()]615.021.0(4.2[84.1
][ )1.0(84.1
615.0015.06.0hHH Head Total
values,Putting
015.0
81.92
54.0
2
v
h
approach, of velocity todue head and
/54.0
6.3
95.1
A
Q
vapproach, ofVelocity
3.6m 0.6 x 6 A
channel, in the flowing water of area sectional Cross
approach of velocity thegconsiderin weir over the Discharge ii)
3
2/32/3
2/32/3
11
a1
22
a
3








29. Flow through Mouthpieces
We know that discharge through an orifice depends upon its
coefficient of discharge. It was felt by the engineers that the
discharge through an orifice is too less (due to low value of
coefficient of discharge).
It was found after conducting series of experiments by
engineers that if a short pipe be fitted to an orifice, it will
increase the value of coefficient of discharge and of course
discharge too.
Such a pipe whose length is generally more than 2 times the
diameter of the orifice and is fitted (externally or internally) to
the orifice is known as mouthpiece.

30. Flow through Mouthpieces

31. Flow through Nozzles
A nozzle is a tapering mouthpiece, which is fitted to the outlet
end of a pipe.
A nozzle is generally, used to have a high velocity of water,
as it converts pressure head into kinetic head at its outlet.
A high velocity of water is required in fire fighting, service
station, mining power development etc.