6. A weir is a small overflow-type dam commonly
used to raise the level of a river or stream.
Weirs have been used to create mill ponds in
such places. Water flows over the top of a weir.
7. Some weirs have sluice gates which release
water at a level below the top of the weir.
The crest of an overflow spillway on a large dam
is often called a weir.
9. Sharp-crested weirs are used in the
measurement of irrigation water. The
sharp edge in the crest causes the
water to spring clear of the crest, and
thus accurate measurements can be
made.
10. Broad-crested weirs: are commonly
incorporated in hydraulic structures of various types
and, although sometimes used to measure water
flow, this is usually a secondary function.
11. In simples state weir consists of:
A bulkhead of timber,
Metal or concrete with an opening of fixed dimensions
cut in its top edge (weir notch)
Its bottom edge is the weir crest.
The depth of flow over the crest is called the head (H).
The overflowing sheet of water is known as the nappe.
14. Open-channel discharge measurement structures
(a) Side view of a sharp-crested weir, (b) Front view of a rectangular
weir and (c) Front view of a triangular (V-notch) weir.
Broad-crested weir
27. When the sides of the flow channel act as the
ends of a rectangular weir, no side contraction
exists, and the nappe does not contract from the
width of the channel
A Suppressed Weir is:
30. For a rectangular sharp-crested weir that takes
end contractions into account and neglecting the
velocity of approach, the discharge is determined
by:
Where:
L is the crest length (l), and
n is the number of end contractions
34. In some cases, the weir is made significantly
longer than the width of the river by forming it in a
'U' shape or running it diagonally, instead of the
short perpendicular path.
Weirs are used in conjunction with locks, to
render a river navigable and to provide even flow
for navigation.
Function
35. Weirs are used in conjunction with locks
A long weir is made to allow a lot of water with a
small increase in over flow depth.
36. Broad crested weirs are robust structures (
نشأم
هيلكه
قوي
) that
are generally constructed from reinforced concrete and
which usually span the full width of the channel. They are
used to measure the discharge of rivers, and are much
more suited for this purpose than the relatively flimsy sharp
crested weirs.
Additionally, by virtue of being a critical depth meter, the
broad crested weir has the advantage that it operates
effectively with higher downstream water levels than a
sharp crested weir.
Board Crested Weir
41. A duck bill weir is a type of long-crested weir that is
designed to control water levels.
As shown in the next slide, the design of a duck bill
weir involves a staggered weir crest that has the
appearance of teeth or duck bills. This staggering
effect increases the weir crest length while
minimizing the footprint of the weir.
The result is a smaller weir that is effective at
the upstream water depth.
What is a Duck-bill Weir?
45. Even though the water around weirs can often
appear relatively calm, they are dangerous places to
boat, swim or wade; the circulation patterns on the
downstream side can submerge a person indefinitely.
46. Weirs allow hydrologists and engineers a simple
method of measuring the rate of fluid flow in small to
medium-sized streams, or in industrial discharge
locations.
Since the geometry of the top of the weir is known,
and all water flows over the weir, the depth of water
behind the weir can be converted to a rate of flow.
47. The calculation relies on the fact that fluid will
pass through the critical depth of the flow regime
in the vicinity of the crest of the weir.
A weir will artificially reduce the upstream water
velocity, which can lead to an increase in siltation
(see next slide).
48.
49. The following general rules should be observed in the
construction and installation of weirs.
A weir should be set at right angles to the
direction of flow in a channel that is straight for a
distance upstream from the weir at least ten
times the length of the weir crest.
Construction and Placement
50. The crest and sides of the weir should be
straight and sharp-edged. The crest of the
rectangular and Cipolletti weirs should be level
and the sides should be constructed at exactly
the proper angle with the crest.
Each side of the V-notch weir should make a
45° angle with a vertical line through the vertex
of the notch.
51. Avoid restrictions in the channel below the weir that
would cause submergence. The crest must be
placed higher than the maximum downstream
water surface to allow air to enter below the nappe.
The channel upstream should be large enough to
allow the water to approach the weir in a smooth
stream, free from eddies, and with a mean velocity
not exceeding 0.3 foot per second.
59. Qact Actual discharge,
B Width of the weir,
H Head on the weir,
Cd Coefficient of discharge.
Rectangular Weir Equation
60. A V-shaped notch is a vertical thin plate which is placed
perpendicular to the sides and bottom of a straight channel is
defined as a V-notch sharp-crested weir. The line which bisects
(
ينصف
) the angle of the notch should be vertical and at the same
distance from both sides of the channel .
The V-notch sharp-crested weir is one of the most precise (
دقة
)
discharge measuring devices suitable for a wide range of flow.
In international literature, the V-notch sharp-crested-weir is
frequently referred to as the ‘Thomson weir’.
Triangular or V-Notch Weir
70. A side-flow weir is a structure installed along the
side of a main channel or pipe.
They are used to divert flow during high flow
conditions.
These types of weirs are commonly seen in
irrigation and sewer systems.
71.
72.
73. A V-notch weir is to be designed to measure an
irrigation channel flow. For case in reading the
upstream water-level gage, a reading is desired for
the design flow rate of 150 m3/h. .
What is the appropriate angle for the V notch? Take
Cd = 0.62.
Worked Example
74. The designed flow rate
The actual flow rate for a sharp-crested triangular ( V
notch angle) is
H
76. The rectangular sharp-crested weir shown in the below figure is used to
maintain a relatively constant depth in the channel upstream of the weir.
I. How much deeper will the water be upstream of the weir during a
flood when the flowrate is 45 ft3/s compared to normal conditions
when the flowrate is 30 ft3/s? Assume the weir coefficient remains
constant at Cd = 0.62.
II.Repeat the calculations if the weir of part (a) is replaced by a
rectangular sharp-crested “duck bill” weir that is oriented at an
angle of 30o relative to the channel centerline as shown in Figure.
The weir coefficient remains the same.
Worked Example
77. For case (a):
Q = 30 ft3/s when B = 20 ft, Cwr = 0.62
……………(1)
B
Worked Example
78. If the discharge is increased to Q =45ft3/s
If the discharge is increased to Q =45ft3/s with keeping “B”, Cwr
…………………………….(2)
…………………………….(3)
Worked Example
80. If the discharge is increased to Q =45ft3/s
If the discharge is increased to Q =45ft3/s with keeping “B”, Cwr
…………………………….(2)
…………………………….(3)
Worked Example
97. 97
Specific Energy
“Specific energy: is the total mechanical energy with respect
to the local invert elevation of the channel”.
We can write 𝐸𝑆 = 𝑦 + 𝛼
𝑉2
2 𝑔
=
𝑦 + 𝛼
𝑄 𝐴 2
2 𝑔
≅ 𝑦 +
𝑄 𝐴 2
2 𝑔
(since 𝑉 = 𝑄 𝐴 )
98. For a given cross section, the flow area, A is a function of y, therefore the
specific energy is a function of Q and y,
We can thus study the variation of
98
Specific Energy
𝐸𝑆 ≅ 𝑦 +
𝑄 𝐴 2
2 𝑔
𝑦 𝑎𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸𝑆
(Specific energy curve)
𝑦 𝑎𝑠 𝑎𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑄
(Specific Discharge curve)
𝐸𝑆 ≅ 𝑦 +
𝑄 𝐴 2
2 𝑔
For Q =
const
For y =
const.
99. 99
Let us consider a specific energy for a steady flow
𝒁 +
𝑷
𝝆 𝒈
+
𝑽𝟐
𝟐 𝒈 𝟏
= 𝒁 +
𝑷
𝝆 𝒈
+
𝑽𝟐
𝟐 𝒈 𝟐
+ 𝒉𝑳 𝟏 →𝟐
Note that since: 𝑄 = 𝐴. 𝑉 , we can write
𝐸𝑆 = 𝑦 + 𝛼 ∙
𝑉2
2 𝑔
= 𝑦 + 𝛼 ∙
𝑄2
2 𝑔∙ 𝐴2 ≅ 𝑦 +
𝑄2
2 𝑔∙ 𝐴2
Concept of Specific Energy
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔
Specific
energy
Specific energy:
Is the total mechanic energy
with the local invert elevation
of the channel.
100. 100
For a given cross sectional, the flow area ”A” is a function of
”y”. Therefore, the specific energy is a function of “A” & “y”
𝑬𝑺 ≅ 𝒚 +
𝑸𝟐
𝟐 𝒈 ∙ 𝑨𝟐
We can thus study the function of:
𝒚 as a function of 𝑬𝑺
{Specific energy curve}
𝒚 as a function of 𝑸
{Specific discharge curve}
Concept of Specific Energy
for Q constant
for Es constant
101. 101
We wish to plot the specific energy curve ( i.e. ”y” as a function of “Es“ for
constant “Q”) .
𝑬𝑺 ≅ 𝒚 +
𝑸𝟐
𝟐 𝒈∙ 𝑨𝟐
One immediately sees that the curve has two asymptotes
For 𝑦 → 0, we have 𝐴 → 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸𝑆 → ∞
𝐹𝑜𝑟 𝑦 → ∞, we have 𝐴 → ∞, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸𝑆 → 𝑦
Specific Energy Curve
102. 102
In addition, for a given discharge “Q”, the curve has a minimum value 𝑦𝑐𝑟.
We will see about this minimum later in details.
Some observations imposed:
For a given 𝐸𝑆, there are always ( except when 𝑦𝑐𝑟 = 𝑦) two depths 𝑦1
& 𝑦2. They are called alternate depths.
The depth corresponding to minimum specific energy 𝐸𝑆 𝑚𝑖𝑛. is called
critical depth 𝑦𝑐𝑟.
Minimum specific energy 𝐸𝑆 𝑚𝑖𝑛. increases with increase discharge Q.
There are three possible flow regimes:
1. Sub-critical (𝑦 > 𝑦𝑐𝑟),
2. Critical (𝑦 ≥ 𝑦𝑐𝑟), and
3. Super-critical (𝑦 < 𝑦𝑐𝑟).
109. 109
Critical Depth and its Importance
Critical depth in a channel “𝑦𝑐𝑟” is the flow depth at which:
The specific energy at which “𝐸𝑆 𝑚𝑖𝑛." for a given discharge “Q”.
The discharge is maximum “𝑄𝑚" for a give specific energy “𝐸𝑆 ".
𝐸𝑆 − ℎ𝑐𝑟 =
𝑦ℎ
2
Recall that:
𝑄𝑚 = 2 𝑔 ∙ 𝐸𝑆 − ℎ𝑐𝑟 & 𝑄𝑚 = 2 𝑔 ∙ 𝑦ℎ
The average velocity corresponding to the critical depth
𝑉
𝑐𝑟 = 2 𝑔 ∙ 𝑦ℎ &
𝑉𝑐𝑟
2
2 𝑔
=
𝑦𝑐𝑟
2
112. 112
Chezy and Manning Coefficients
Attention Dimensional Coefficients
Chezy Equation
Manning-Strickler Equation
𝐶 𝐿1 2
∙ 𝑆−1
𝑛 𝐿−1 3
∙ 𝑆−1
Tables are available for various
surfaces
Tables are available for various
surfaces
113. 113
Depth of flow for a given discharge, where the specific
energy is at a minimum,
Occurs when 𝑑𝐸𝑆 𝑑𝑦 = 0 and 𝐹𝑟= 1,
It is important to calculate 𝑦𝑐𝑟 in order to determine if the
flow in the channel will be sub-critical or super-critical,
Can be found through specific energy diagram
Critical Depth
116. 116
Best Hydraulic Section (Most Efficient Cross-section)
What is best hydraulic section?
A section that gives the largest flow area for the smallest
perimeter.
For a rectangular channel,
𝐴 = 𝐵 𝑦 & 𝑃 = 𝐵 + 2 𝑦 ∴ 𝑃 =
𝐴
𝑦
+ 2 𝑦
Let us keep “A” as constant “P” is only a function of “y”. Let
us vary “y” to minimize the perimeter:
𝑑𝑃
𝑑𝑦
= −
𝐴
𝑦2 + 2 = 0 →
𝐴
𝑦2 = 2
→
𝐵 𝑦
𝑦2
= 2 → ∴ 𝐵 = 2𝑦
To obtain best hydraulic section the width must equal
twice the depth.
117. 117
y
B = 2 y
T = 2 r
r
Rectangular Channel Semi-circle Channel
118. 118
The triangular drainage ditch shown in the figure, has a side slope of Z
= 2, Find:
The critical depth, ycr, for a discharge of Q = 0.35 m3/s and the
corresponding minimum specific energy.
Calculate the discharge if the flow depth is y = 0.60 m
(The channel has a Manning’s coefficient of n = 0.025 m-1/3/ S and
a bed slope of So = 0.001.
y
Z
1
T
119. Solution
Froude number for a triangular channel is given by :
𝐹𝑟 =
𝑉
𝑔 ∙ 𝑦ℎ
When the flow is critical, we have: 𝐹𝑟
2 = 1.0 =
𝑄2∙ 𝑇
𝑔 ∙ 𝐴3
𝑦= 𝑦𝑐𝑟
∴
𝑄2
𝑔
=
𝐴3
𝑇 𝑦= 𝑦𝑐𝑟
For a triangular channel, we have
𝐴 = 𝑍 ∙ 𝑦2
& 𝑦ℎ =
𝐴
𝑇
=
𝑍 𝑦2
2 𝑍 𝑦
=
𝑦
2
Substituting into Eq. (1) gives:
𝑄2
𝑔
=
𝐴3
𝑇 𝑦= 𝑦𝑐𝑟
=
𝑍 𝑦2 3
2 𝑍 𝑦
𝑦= 𝑦𝑐𝑟
=
𝑦𝑐𝑟
5
2
∴ 𝑦𝑐𝑟 =
5 2 𝑄2
𝑔
=
5 2 ×0.352
9.81
= 0.362 𝑚
119
121. Solution
The geometric relationships for a trapezoidal channel can be calculated follows:
𝑇 = 𝐵 + 2 𝑍 𝑦 = 2.0 + 2 × 1.5 × 0.56 = 3.68𝑚
𝐴 = 𝑦 𝐵 + 𝑍 𝑦 = 0.56 (2.0 + 1.5 × 0.56) = 1. 59 𝑚2
𝑃 = 𝐵 + 2 𝑦 𝑍2 + 1 = 2.0 + 2 × 0.56 1.5 2 + 1 = 4.02 𝑚
𝑅ℎ = 𝐴 𝑃 = 1.59 4.02 = 2.0 + 2 × 1.5 × 0.56 = 0.396 𝑚
𝑦ℎ = 𝐴 𝑇 = 𝐴 𝑃 = 1.59 3.65 = 0.43 𝑚
121
A trapezoidal channel having a bottom width of B = 2.0 m and side slope of Z =
1.5 carries a uniform flow with a depth of y = 0.56 m. The channel has a bed slope
of So = 0.005 and the coefficient of Manning is 0.03.
What is the discharge of a uniform flow?
What is the regime of flow?
122. 1.5
1.0
y = 0.56 m
T = 0.56 m
B = 2.0 m
The discharge using Manning equation 𝑄 =
𝑘
𝑛
∙ 𝑅ℎ
2 3
∙ 𝑆𝑜
1 2
∙ 𝐴
or
𝑄 =
𝑘
𝑛
∙ 𝑅ℎ
2 3
∙ 𝑆𝑜
1 2
∙ 𝐴
=
1
0.03
× 0.3962 3
× 0.0051 2
× 1.59 = 2.02 𝑚3
/𝑆
To determine the regime of the flow, Froude number for a trapezoidal
channel is given by: 𝐹𝑟 =
𝑉
𝑔×𝑦=𝑦ℎ
=
𝑄
𝐴. 𝑔× 0.435
=
2.02
1.59 × 9.81×0.43
= 0.62
Since 𝐹𝑟 = 0.62 < 1.0, the uniform flow is sub-critical
123. 123
Depth of flow for a given discharge, where the specific
energy is at a minimum,
Occurs when 𝑑𝐸𝑆 𝑑𝑦 = 0 and 𝐹𝑟= 1,
It is important to calculate 𝑦𝑐𝑟 in order to determine if the
flow in the channel will be sub-critical or super-critical,
Can be found through specific energy diagram
Computation of Critical Depth
124. A rectangular laboratory channel has a width of B=2.0 m wide and a
Manning coefficient n = 0.020 m-1/3/ s.
What should be the bed slope to achieve a critical uniform flow in this
channel for a discharge of Q = 3.0 m3/s?
(Hint: critical uniform flow is achieved when uniform flow depth for a
given discharge is equal to critical depth for the same discharge
Froude number for a rectangular channel is given by:
𝐹𝑟 =
𝑉
𝑔 ∙ 𝑦ℎ=𝑦
or 𝐹𝑟
2
= 1.0 =
𝑄2∙ 𝑇
𝑔 ∙ 𝐴3
𝑦= 𝑦𝑐𝑟
=
𝑄2∙𝐵
𝑔 ∙ 𝐵∙𝑦 3
𝑦= 𝑦𝑐𝑟
𝑦𝑐𝑟 =
3 𝑄2
𝑔 ∙ 𝐵2 𝑦𝑐𝑟 =
3 32
9.81×22 =𝟎. 𝟔𝟏𝟐 𝑚
124
Solution
126. Characteristics:
Unstable surface
Series of standing waves
Occurrence
Broad crested weir (and other weirs),
Channel controls (rapid changes in cross-
section),
Over falls,
Change in channel sloe from mild to steep
slope,
Used for flow measurements. 126
Critical Depth 𝑑𝐸𝑠 𝑑𝑦 = 0
128. Specific Energy : Sluice Gate
𝐸𝑆,1 = 𝐸𝑆,2
𝐸𝑆
Given: downstream depth and discharge
Find : Upstream depth ?
𝑦1& 𝑦2 are alternate depths (same specific energy)
Why not use momentum conservation to find 𝑦1????
129. The Hazen–Williams equation is used in the design of water
pipe systems such as fire sprinkler systems, water supply
networks, and irrigation systems. It is named after Allen
Hazen and Gardner Stewart Williams.
The Hazen-Williams equation has the advantage that:
the coefficient C is not a function of the Reynolds number,
but it has the disadvantage that it is only valid for water.
130. Also, it does not account for the temperature or viscosity
of the water.
The Hazen-Williams equation is limited to the flow of
water in pipe larger than 2.0 in. and smaller than 6.0
ft in diameter.
Velocity of flow should not exceed 10.0 ft/s
131.
132. Typical C factors used in design, which take into account
some increase in roughness as pipe ages are as follows:
Material C factor range
Asbestos-cement
Cast iron 10 years
Cast iron 20 years
Cast iron 30 years
Cast iron 40 years
Cast iron new
Cement-Mortar Lined Ductile Iron Pipe
Concrete
Copper
Fiber-reinforced plastic
Galvanized iron
Polyethylene
Polyvinyl chloride (PVC)
Steel
140
107 - 113
89 - 100
75 - 90
64 - 83
130
140
100 - 140
130 - 140
150
120
140
150
90 - 110
133. The general form can be specialized for full pipe flows. Taking
the general form,
Exponentiating each side by 1/0.54 gives (rounding exponents
to 3-4 decimals)
Rearranging gives:
and the discharge Q = A , V
148. A tank of 5 m2 plan area is fitted with a sharp-edged orifice of 5cm
diameter in its base. The coefficient of discharge of the orifice is 0.62.
Calculate:
The time taken for the level in the tank to fall from 2 meters depth to
0.5 meters depth.
If water is now admitted to the tank at a rate of 0.01 m3/s:
calculate the rate at which the surface will be rising when the depth
in the tank is 1 meter and the depth in the tank when the level
becomes steady.
149. An oil (S. G. = 0.92) flows through a vertical tube of 8 in diameter. The
flow is measured by a Venturi tube of 4 in. diameter throat with a U-tube
manometer, containing mercury, as shown in the figure. If Cd = 0.98,
what is:o
(a) the flow, for a manometer reading of 9 in ?
(b) the manometer reading, for a flow of 2 ft3/sec ?
150. A flow nozzle of 6 in diameter is placed in an 18 in diameter pipe. During
calibration the pressure differential was measured to be 10 psi for a flow
of 7.2 cfs and 17 psi for a flow of 9.3 cfs. Determine:
The discharge coefficient of the measuring nozzle (see Figure).
151. A 4-in. by 1-in. nozzle, shown in the figure, is attached to the
end of a 4-in hose line. The velocity of the water leaving the nozzles is 96
fps, the coefficient of velocity, Cv, is 0.96 and the coefficient of contraction,
Cc, is 0.80. Determine:
The necessary pressure at the base of the nozzle.
152. An orifice of area A0 and velocity coefficient Cv = 0.8 is installed in a pipe
of area Ap = 2A0. The pipe is attached to a dam as shown and the water
level in the dam is 100 ft. higher than the outlet from the reservoir. The
contraction coefficient of the orifice is unity. There are no other losses in
the pipe nor between the end of the orifice and the pipe. The velocity
leaving the pipe Vp is to be determined.