DATE:24112005

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DATE24/11/2005
Dr. N. Balasubramanya
Professor of Civil Engineering
M.S.R.I.T. Bangalore - 54
Emailjhrumbaa_at_yahoo.co.in
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• A sharp crested rectangular weir 1.5m long and
  90cm high is installed in a rectangular channel
  1.5m wide. If the head on the weir is 30cm, find
  the discharge
• Neglecting velocity of approach
• Considering velocity of approach

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• A cipolletti weir of 40cm bottom width is
• Installed in a channel 75cm wide 45cm
• deep. If the head over the weir crest is
• 25cm, find the discharge over the weir.
• Neglecting vel. of approach
• Taking vel. of approach.

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A rectangular channel is 4.5m wide. Water flows
at a depth of 1.2m at a velocity of 90cm/s. A
sharp crested weir is constructed across the
channel and the depth in the channel rises upto
1.75m. What should be the height of the weir?
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Two 900 V notches one cipolletti weir are to
be used side by side to measure a discharge of
0.85m3/s through a channel. If the head should
not exceed 30cm, what should be the dimensions of
the weirs?
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Find the discharge through a trapezoidal notch
which is 1.2m wide at the top and 0.5m at the
bottom and is 40cm in height. The head of water
on the notch is 30cm. Cd for the rectangular
portion is 0.62 while for triangular portion is
0.60.
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A discharge of 1500m3/s is to pass over a
rectangular weir. The weir is divided into a
number of openings each of span 7.5m. If the
velocity of approach is 3m/s, find the number of
openings needed in order the head of water over
the crest is not to exceed 1.8m.
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A rectangular channel 1.5m wide has a discharge
of 200lps, which is measured by a right angled V
notch. Find the position of the apex of the
notch from the bed of the channel if the maximum
depth of water is not to exceed 1m. Take Cd0.62.
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Types of Nappe The equations derived for the
discharge over notches were under the assumption
that pressure under the nappe is atmosphere.
However, when the liquid if flowing over the
notch (suppressed), it touches the walls of the
channel and the air gets dissolved or entrained
in water, continuation of this process results in
a negative pressure i.e. partial vaccum under the
nappe. Finally the nappe gets deflected closer to
the weir wall. The pressure on the inner side of
the the nappe decides its type in the following
ways.
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• Free Nappe
• In this type, the stream of water passing over
  the weir the springs clear of the weir.
• b) Depressed Nappe
• In this type, a partial vaccum is created between
  the nappe and the weir. Discharge for such a flow
  situation is 8 to 10 greater than that with a
  free nappe.
• c) Clinging Nappe
• In this type the nappe totally adheres to the
  face of the weir. The discharge in this case
  would be 20 to 30 more than that in a fully
  aerated nappe.

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Ventilation of Weirs The nappe emerging from a
weir should be of a correct form, so that the
discharge equations derived for them are
valid. For accurate gauging of flow the nappe
should spring clear or it should be free. In
other words the space between the weir and nappe
should be maintained under atmospheric
conditions, particularly when the weir is
suppressed. In practice ventilation holes are
made on the weir walls so that air circulates
freely between the weir and the nappe.
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This is known as ventilation or ventilation of
weirs. Submerged Weir A weir is said to be
submerged when water level on both the upstream
and downstream sides are above the crest level of
the weir as shown in figure. H1 and H2 are the
heads over the weir on the upstream and
downstream sides. In the case of submerged weir,
it is necessary to derive the discharge equation
considering that the flow over the weir is a
combination of a free weir and a submerged
orifice.
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In other words, the flow Q1 between H1 and H2 is
considered as a free weir and Q2 between H2 and
the weir crest as a submerged orifice. For a
free weir For a submerged orifice.
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Cd10.58 and Cd2 0.80 are usually considered
for the weir and the orifice. As in the earlier
cases the head due to velocity of approach
haVa2/2g can also be considered. In such a
case In all the above equation Llength of
the notch or weir.
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• Problems
• A submerged weir 1m high spans the entire width
  of a rectangular channel 7m wide. Estimate the
  discharge when the depth of water is 1.8m on the
  upstream side and 1.25m on the downstream side of
  the weir. Assume Cd0.62 for the weir.
• Solution.
• Qsubmerged weir QweirQsubmerged orifice

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2) The upstream and downstream water surfaces are
150mm and 75m above the crest of a drowned weir.
If the length of the weir is 2.5m, find the
discharge, the coefficients of discharge for the
free and drowned portions may be taken as 0.58
and 0.8 respectively. Allow for velocity of
approach. Solution. H11500mm15m,
H275mm0.075m L2.5m, Cd10.58, Cd20.8
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Velocity of approach
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OGEE WEIR When the weir is suppressed and its
height is large, the nappe emerging out may be
subjected to the problems of ventilation. Hence,
in such cases the weir profile downstream is
constructed conforming to the shape of the lower
side of the nappe. Such a weir is known as a
spillway or ogee weir. The cross section of an
ogee weir will be shown. The coordinates of the
spillway profile can be worked out for the head H
using the equation.
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The u/s face of the spillway is generally kept
vertical. The discharge equation for an ogee weir
will be Same as that for a suppressed
rectangular notch.
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Problem Calculate the discharge over an ogee weir
of 8.5m length, when the head over the crest is
2.15m and Cd0.61. Solution. L8.5m, H2.15m,
Cd0.61
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Broad Crested Weir A weir is said to broad
crested when its width (parallel to flow) b is
greater than 0.5xmaximum head acting on it.
Let Llength of the weir HHead of water u/s of
the weir w.r.t. the crest hDepth of water over
the weir crest VVel. Of flow over the
weir Applying Bernoullis equation between (1)
and (2) with the crest of the weir as datum
neglecting losses (hL)
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Discharge over the weir Qarea of flow over the
weir x vel. of flow over the weir. i.e. Actual
discharge
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From Equation (2), we see that Qact is a function
of h for a given value of H.
is maximum when
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Substituting the value of h in eq(2) and
simplifying.
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Problems Determine the discharge over a broad
crested weir 26m long, the upstream level of
water is measured as 0.5m above the crest level.
The height of the weir is 0.6m and the width of
the approach channel is 36m. Take
Cd0.9. Solution. For a broad crested weir.
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Since, the width of the channel, we have to
consider the velocity of approach Va
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A reservoir discharges water at 60,000 m3/day
over a broad crested weir, the head of length of
the weir, if Cd0.65. Solution. Q60,000m3/day60
,000/24x60x600.694m3/s H500mm0.5m Cd?, L?
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A channel of 45m2 cross sectional area,
discharging 50 cumecs of water is to be provided
with a broad crested weir. If the crest of the
weir is 1.6m below the upstream water surface,
find the length of the weir, if
Cd0.85. Solution.
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or
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Submerged Orifice A fully submerged orifice is
one in which the entire outlet side or the
downstream side is completely under the liquid.
It is also known as a drowned orifice. Consider
points (1) and (2) situated upstream of orifice
and at the Vena contracta respectively. H1Height
of water above the top of the orifice on the
upstream side. H2Height of water above the
bottom level of the orifice.
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HDifference in water level bwidth of
orifice CaCoefficient of discharge. Height of
water above the centre of orifice on upstream
side Height of water above the centre of the
orifice on the downstream side
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Applying Bernoullis equation between (1) and (2)
with the horizontal passing through (A) (B) as
datum and neglecting losses. (hL)
(negligible)
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or
From continuity equation Discharge QactCdxarea
of orifice x velocity
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Large Rectangular Orifice An orifice is said to
be large when the head acting on it is five times
the depth of the orifice. Unlike in the case of a
small orifice, the discharge cannot be calculated
from the equation for the reason that the
velocity is not constant over the entire cross
section of the jet.
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Consider an elementary horizontal strip of depth
dh at a depth h below the free surface of the
liquid as shown. Area of the strip Theoretical
velocity through the strip Discharge through the
elementary strip Therefore through the entire
orifice is obtained by integrating the above
equation between the limits H1 and H2.
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Problems 1. Find the discharge through a fully
submerged orifice of width 2m if the difference
of water levels on both the sides of the orifice
be 800mm. The height of water from the top and
bottom of the orifice are 2.5m and 3m
respectively. Take Cd0.6 Solution. For a
submerged orifice.
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Where, Cd0.6, b2m, H23m, H12.5m, H800mm
0.8m
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2. Find the discharge through a totally drowned
orifice 1.5m wide and 1m deep, if the difference
of water levels on both the sides of the orifice
is 2.5m, Take Cd0.62. Solution. b1.5m, d1m,
H2.5m, Cd0.62
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3. Find the discharge through a rectangular
orifice 3m wide and 2m deep fitted to a water
tank. The water level in the tank is 4m above the
top edge of the orifice. Take Cd0.62. Solution.
For a rectangular orifice where B3m, Cd0.62,
H2(42)6m, H14m.
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4. A rectangular orifice 1m wide and 1.5m deep is
discharging water from a vessel. The top edge of
the orifice is 0.8m below the water surface in
the vessel. Calculate the discharge through the
orifice if Cd0.6. Also calculate the percentage
error if the orifice is treated as
small. Solution. For a rectangular orifice.
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