National University
of Computer & Emerging Sciences
Fluid Mechanics
Week12&13: Flow Measurement:
Venturimeter, Orifices and Mouthpieces,
Pitot tube, Pitot static tube,
Weirs and notches, and Velocity methods.
Dr. Mohsin Siddique
Assistant Professor
NU-FAST Lahore
1
Date: 10/04/2013
17/04/2013
Flow Measurement
Pipes (pressure conduits)
Open channel (flumes, canals and
rivers etc)
1.
2.
3.
4.
5.
6.
1. Notches (Rectangular notch,V
notch)
2. Weirs
2
Venturimeter
Orifices
Orifice meter
Mouth pieces/tubes
Nozzle
Pitot static tube
Flow Measurement in Pipes
Venturimeter
3
Flow Measurements in Pipes
Venturimeter
Qact = Cd
A1 A2
A12 − A22
Figure shows a venturimeter in which
D1, A1,
discharge Q is flowing,
P1, Z1,V1
Let, D1 is diameter, A1 is cross-section
area, P1 is pressure, z1 is elevation head V1
is velocity at section 1. Similarly D2 , A2, P2,
z2 & V2 are corresponding values at
section 2
According to Bernoulli's Equation
between section 1 and 2 we can write;
v12 P2
v22
+ z1 +
= + z2 +
γ
2g γ
2g
P P 
2 g  1 − 2  + 2 g ( z1 − z 2 )
γ γ 
D2, A2,
P2, Z2,V2
Datum
P1
4
P P 
2 g  1 − 2  + 2 g ( z1 − z 2 ) = v22 − v12
γ γ 
Direction of flow
Flow Measurements in Pipes
Venturimeter
A1 A2
Qact = Cd
A12 − A22
P P 
2 g  1 − 2  + 2 g ( z1 − z 2 )
γ γ 
D1, A1,
P1, Z1,V1
 P1 P2 
Q2 Q2
2 g  −  + 2 g ( z1 − z 2 ) = 2 − 2
A2 A1
γ γ 
D2, A2,
P2, Z2,V2
Q Q = A1V1 = A2V2
 1
P P 
1 
2 g  1 − 2  + 2 g ( z1 − z 2 ) =  2 − 2 Q 2
γ γ 
 A2 A1 
 A12 − A22  2
 P1 P2 
2 g  −  + 2 g (z1 − z 2 ) =  2 2 Q
γ γ 
 A1 A2 
Qth =
5
A1 A2
A12 − A22
Datum
 P1 P2 
A12 A22
 −  + 2 g (z1 − z2 )
Q = 2
2
g
2
A1 − A2
γ γ 
2
P P 
2 g  1 − 2  + 2 g ( z1 − z2 )
γ γ 
Flow Measurements in Pipes
Venturimeter
Qact = Cd
A1 A2
A12 − A22
Since
D1, A1,
P1, Z1,V1
Qact = cd Qth
Qact = Cd
A1 A2
A12 − A22
 P1 P2 
2 g  −  + 2 g ( z1 − z 2 )
γ γ 
D2, A2,
P2, Z2,V2
P P 
2 g  1 − 2  + 2 g (z1 − z 2 )
γ γ 
Datum
Where Cd is coefficient of discharge and is defined as ratio of actual
discharge to theoretical discharge .
6
Flow Measurements in Pipes
Types of Venturimeter
a. Horizontal Venturimeter
b.Vertical Venturimeter
 P1 P2 
2 g  − 
γ γ 
A1 A2
Qact = Cd
A12 − A22
a. Horizontal Venturimeter
h
y
Figure shows a venturimeter
x
connected with a differential
manometer.
2
At section 1, diameter of pipe is D1,
1
and pressure is P1 and similar D2
and P2 are respective values at
According to gauge pressure equation
section 2.
P1
P
− x − Smh + y = 2
γ
γ
7
P1
γ
−
P2
γ
= S m h − ( y − x ) = S m h − ( h)
Flow Measurements in Pipes
Types of Venturimeter
a. Horizontal Venturimeter
b.Vertical Venturimeter
Qact = Cd
A12 − A22
a. Horizontal Venturimeter
A12 − A22
For horizontal venturimeter, ( z1 − z 2 ) = 0
Qact = Cd
P1
γ
8
−
P2
γ
A1 A2
A −A
2
1
h
P P 
2 g  1 − 2  + 2 g ( z1 − z 2 )
γ γ 
A1 A2
Qact = Cd
2
2
P P 
2 g  1 − 2 
γ γ 
= S m h − ( y − x ) = S m h − ( h)
 P1 P2 
2 g  − 
γ γ 
A1 A2
y
x
2
1
According to gauge pressure equation
P1
− x − Smh + y =
P2
= S m h − ( y − x ) = S m h − ( h)
γ
P1
γ
−
γ
P2
γ
Flow Measurements in Pipes
Types of Venturimeter
Q = Cd
a. Horizontal Venturimeter act
b.Vertical Venturimeter
A1 A2
A12 − A22
 P1 P2 
2 g  −  + 2 g ( z1 − z 2 )
γ γ 
b.Vertical Venturimeter
Figure shows a venturimeter
connected with a differential
manometer.
According to gauge pressure equation
y
P1
γ
P1
P1
γ
9
γ
+ x − Sm h − y =
γ
−
P2
= Smh + y − x
−
P2
= S m h + ∆z − h
γ
γ
∆z
P2
h
Q x + ∆z = h + y
1
x
Datum
Flow Measurements in Pipes
Types of Venturimeter
Q = Cd
a. Horizontal Venturimeter act
b.Vertical Venturimeter
A1 A2
A12 − A22
 P1 P2 
2 g  −  + 2 g ( z1 − z 2 )
γ γ 
b.Vertical Venturimeter
Qact = Cd
P1
γ
−
P2
γ
A1 A2
A12 − A22
= S m h + ∆z − h
(z1 − z 2 ) = ∆z
10
 P1 P2 
2 g  −  + 2 g ( z1 − z 2 )
γ γ 
y
h
∆z
1
x
Datum
Q x + ∆z = h + y
Coefficient of Discharge for Venturimeter
11
Numerical Problem
Find the flow rate in venturimeter as shown in
figure if the mercury manometer reads h=10cm.
The pipe diameter is 20cm and throat diameter is
10 cm and ∆z =0.45m. Assume Cd=0.98 and
direction of flow is downward.
Qact = Cd
P1
γ
−
P2
γ
A1 A2
A12 − A22
P P 
2 g  1 − 2  + 2 g ( z1 − z 2 )
γ γ 
= S m h + ∆z − h
y
h
∆z
1
x
Datum
Q x + ∆z = h + y
12
Numerical problems
11.7.1-11.7.4
11.26-11.29 if you can
13
Orifice
An orifice is an opening (usually circular) in wall of a tank or in plate
normal to the axis of pipe, the plate being either at the end of the pipe or
in some intermediate location.
An orifice is characterized by the fact that the thickness of the wall or plate
is very small relative to the size of opening.
14
Orifice
A standard orifice is one with a sharp edge as in Fig (a) or an absolutely
square shoulder (Fig. b) so that there in only a line contact with the fluid
Those shown in Fig. c and d are not standard because the flow through
them is affected by the thickness of plate, the roughness of surface and
radius of curvature (Fig. d).
Hence such orifices should be calibrated if high accuracy is desired.
15
Classification of Orifice
According to size
1. Small orifice
2. Large orifice
An orifice is termed as small when
its size is small compared to head
causing flow. The velocity does not
vary appreciably from top to
bottom edge of the orifice and is
assumed to be uniform.
The orifice is large if the
dimensions are comparable with
the head causing flow. The variation
in the velocity from top to bottom
edge is considerable.
16
According to shape
1. Circular orifice
2. Rectangular orifice
3. Square orifice
4.Triangular orifice
According to shape of
upstream edge
1. Sharp-edged orifice
2. bell-mouthed orifice
According to discharge
condition
1. Free discharge orifice
2. Submerged orifice
Coefficients
Coefficient of contraction: It is the
ratio of area Ac of jet, to the area Ao of
the orifice or other opening.
Cc = Ac / Ao
Coefficient of velocity: It is ratio of
actual velocity to ideal velocity
V
Cv = act
Vth
Coefficient of discharge: It is the ratio
of actual discharge to ideal discharge.
Cd =
17
Qact Vact Aact
=
= C v Cc
Qth
Vth Ath
Vena-Contracta is section of
jet of minimum area. This section
is about 0.5Do from upstream
edge of the opening, where Do is
diameter of orifice
Orifice
Small orifice
Figure shows a tank having small orifice
at it bottom. Let the flow in tanks is
steady.
Let’s take section 1 (at the surface) and
2 just outside of tank near orifice.
inflow
1
H
Z1
According to Bernoulli’s equation
P1
v12 P2
v22
+ z1 +
= + z2 +
γ
2g γ
2g
2
v
0 + z1 + 0 = 0 + z 2 + 2
2g
2
v2
= z1 − z 2 = H
2g
vth = 2 gH
18
Where, H is depth of water above orifice
Outflow
2
Z2
Datum
Crosssectional area
Orifice
Small orifice
Qth = Avth = A 2 gH
Qact = Cd Avth = Cd A 2 gH
inflow
1
Q vth = 2 gH
Where, A is cross-sectional are of orifice
and Cd is coefficient of discharge.
H
Z1
Outflow
2
Z2
Datum
Cross-sectional
area, A
19
Orifice
Submerged orifice
According to Bernoulli’s Equation
inflow
1
H1
Z1
Z2
v12 P2
v22
+ z1 +
= + z2 +
γ
2g γ
2g
2
P
v
0 + z1 + 0 = 2 + z 2 + 2
γ
2g
2
v2
P
= z1 − z 2 − 2
2g
γ
v22
P
= ( z1 − z 2 ) − 2
2g
γ
v22
= H1 − H 2
2g
P1
Outflow
H2
2
P2
Datum
γ
= H1
vth = 2 g (H1 − H 2 )
According to steady flow
Inflow=outflow
20
Qth = Avth = A 2 g (H1 − H 2 )
Qact = Cd Avth = Cd A 2 g (H1 − H 2 )
Mouthpieces/tubes
A tube/mouth piece is a short pipe whose length is not more than
two or three diameters.
There is no sharp distinction between a tube and a thick walled
orifices.
A tube may be uniform diameter or it may diverge.
Figure: types and coefficients of tubes/mouthpieces
21
Nozzle
A nozzle is a tube of changing diameter, usually converging
as shown in figure if used for liquids.
Figure shows a nozzle. At section 1,
diameter of pipe is D1, and pressure
is P1 and similar D2 and P2 are
respective values at section 2.
v12 P2
v22
+ z1 +
= + z2 +
γ
2g γ
2g
P1
v12
v22
+0+
= 0+0+
γ
2g
2g
v22 v12 P1
−
=
2g 2g γ
P1
Q2
22
A22
−
Q2
A12
= 2g
P1
γ
1
2
According to continuity eq.
Q = Q1 = Q2
Q = A1V1 = A2V2
Nozzle
Jet: It is a stream issuing from a orifice, nozzle, or tube.
Q2
A22
−
Q2
A12
= 2g
P1
Jet
γ
 1
1 
P
Q 2  2 − 2  = 2 g 1
γ
 A2 A1 
 A A
1 2
Qth = 
 A2 − A 2
2
 1
Qact
23

 2 g P1

γ

 A A
1 2
= Cd 
 A2 − A 2
2
 1

 2 g P1

γ

1
2
According to continuity eq.
Q = Q1 = Q2
Q = A1V1 = A2V2
Nozzle
Vena-contracta is section of jet of
minimum area. This section is about
0.5Do from upstream edge of the
opening, where Do is diameter of orifice
 A A 
1 2

Qact = Cd 
 A2 − A 2 
2 
 1
 A C A
1
c o
Qact = Cd 
 A2 − C 2 A 2
c o
 1
P
Qact = K 2 g 1
(
)
(
2g
)
P1
γ

 2 g P1

γ

(
)
(
)
2
A2 = Cc Ao
γ
 A C A
1
c o
K = Cd 
 A2 − C 2 A2
c o
 1
1




Ao= cross-section area at nozzle
Where, K is coefficient of nozzle
24
Pressure, P1 is then measured with the help of piezometer or manometer
Nozzle
According to gauge pressure equation
P1
− x − Smh = 0
P1
= x + Smh
γ
γ
25
h
1
2
Nozzle
Head loss in nozzle
According to energy equation
v12 P1 v12
+ z1 +
= +
=H
γ
2g γ 2g
P1
v12 P2
v22
+ z1 +
= + z2 +
+ HL
γ
2g γ
2g
P1
v12
v22
+0+
= 0+0
+ HL
γ
2g
2g
P1
 P1 v12  v22
v22
 −
H L =  +
= (H ) −
2g
 γ 2g  2g
H is total head at section 1
26
1
2
v22 v22
+ z2 +
=
γ
2g 2g
P2
Nozzle
Head loss in nozzle
However, for ideal case, HL=0
vth2 2
=0
H L = (H ) −
2g
v12 P1 v12
+ z1 +
= +
=H
γ
2g γ 2g
P1
v2th = 2 gH
v22
v = C v 2 gH ⇒
= C v2 H
2g
2
2
2
Substituting back in eq (1), we get
H L = (H ) − Cv2 H
(
H L = H 1 − Cv2
27
)
1
2
v22 v22
+ z2 +
=
γ
2g 2g
P2
Nozzle
Efficiency of nozzle
According to energy equation
v12 P1 v12
+ z1 +
= +
=H
γ
2g γ 2g
P1
input
v12 P2
v22
+ z1 +
= + z2 +
+ HL
γ
2g γ
2g
P1
v22
v22
output
2g
2g
η=
=
=
P1 v12
input
H
+
γ 2g
28
η=
2
Cv H
2
= Cv
H
1
output
( )
Cv 2 H
2
η% =
= Cv 100
H
2
v22 v22
+ z2 +
=
γ
2g 2g
P2
v22
Q
= C v2 H
2g
Calibration and Calibration Curves
Calibration : Determine coefficients of flow measuring
devices, e.g.,
Cd, Cc, Cv, etc
Calibration curve: Plotting calibration curve
e.g., h1/2 Vs Qact
h3/2 Vs Qact
29
Numerical Problems
Discharge and headloss in nozzle are
20L/s and 0.5 m respectively. If dia of
pipe is 10cm and dia of nozzle is 4cm,
determine the manometric reading.
Manometric fluid is mercury.
h
1
v12 P2
v22
+ z1 +
= + z2 +
+ HL
γ
2g γ
2g
P1
30
2
P1
γ
= x + Smh
Numerical Problem
A jet discharges from an orifice in a vertical plane under a head of 3.65m.
The diameter of orifice is 3.75 cm and measured discharge is 6m/s. The coordinates of centerline of jet are 3.46m horizontally from the venacontracta and 0.9m below the center of orifice.
Find the coefficient of discharge, velocity and contraction.
Qact = Cd Avth = Cd A 2 gH
(
Cd = Qact / A 2 gH
inflow
)
1
Cv =
vact
=
vth
Cc = C d / C v
31
Outflow
H
gx 2
2y
x=3.46m
2 gH
2
x = vact t
1
y = gt 2
2
y=0.9m
Vact = gx 2 / 2 y
Numerical Problems
11.6.1-11.6.4
11.17-11.18
32
Part II
33
Bernoulli’s Equation
V2
+z+
=H
γ
2g
P
Pressure head + Elevation head + Velocity head = Total Head
Multiplying with unit weight,γ,
V2
P + ρgz + ρ
= contt
2
Static Pressure : P
Dynamic pressure : ρV 2 / 2
Hydrostatic Pressure: ρgZ
Stagnation Pressure: Static pressure + dynamic Pressure
2
P+ρ
34
V
= Pstag
2
Pitot Tube and Pitot Static Tube
Pitot Tube: It measures sum of velocity
head and pressure head
Piezoemeter: It measures pressure
head
Pitot-Static tube: It is combination of
piezometer and pitot tube. It can
measure velocity head.
35
Pitot Tube and Pitot Static Tube
Consider the following closed channel flow (neglect friction):
open
open
Pitot tube
piezometer
tube
Pitot static
tube
V2
2g
Uniform
velocity profile
P V2
+
γ 2g
P
γ
V
1
z
2
Stagnation point
V 2 V 2 P  P
= 
+  −
2g  2g γ  γ
 Pstag P 
Vth = 2 g 
− 
γ
 γ
36
Remember !!
Theoretical/ideal flow
velocity at elevation z in
pipe.
V2
= Pstag
P+ρ
2
P V 2 Pstag
+
=
γ 2g 2g
Pitot Static Tube
In reality, directional velocity
fluctuations increase pitot-tube
readings so that we must multiply
Vth with factor C varying from
0.98 to 0.995 to give true (actual)
velocity
 Pstag P 
− 
Vact = C 2 g 
γ
 γ
However, piezometer holes are rarely located in precisely
correct position to indicate true value of P/γ, we modify above
equation as;
 Pstag P 
Vact = C1 2 g 
− 
γ
 γ
Where C1 is coefficient of instrument to account for discrepancy.
37
Notches and Weirs
38
Notches and Weirs
39
Notches and Weirs
Notch. A notch may be defined as an opening in the side of a tank or
vessel such that the liquid surface in the tank below the top edge of the
opening.
A notch may be regarded as an orifice with the water surface below its
upper edge. It is generally made of metallic plate. It is used for measuring
the rate of flow of a liquid through a small channel of tank.
Weir: It may be defined as any regular obstruction in an open stream over
which the flow takes place. It is made of masonry or concrete. The
condition of flow, in the case of a weir are practically same as those of a
rectangular notch.
Nappe: The sheep of water flowing through a notch or over a weir
Sill or crest. The top of the weir over which the water flows is known as
sill or crest.
Note: The main difference between notch and weir is that the notch is
smaller in size compared to weir.
40
Classification of Notches/Weirs
Classification of Notches
1. Rectangular notch
2. Triangular notch
3.Trapezoidal Notch
4. Stepped notch
41
Classification of Weirs
According to shape
1. Rectangular weir
2. Cippoletti weir
According to nature of
discharge
1. Ordinary weir
2. Submerged weir
According to width of weir
1. Narrow crested weir
2. Broad crested weir
According to nature of crest
1. Sharp crested weir
2. Ogee weir
Discharge over Rectangular Notch/Weir
Consider a rectangular notch or weir provided in channel carrying
water as shown in figure.
Figure: flow over rectangular notch/weir
42
H=height of water above crest of
notch/weir
P =height of notch/weir
L is length of notch/weir
dh=height of strip
h= height of liquid above strip
L(dh)=area of strip
Vo = Approach velocity
Theoretical velocity of strip neglecting
approach velocity =
2 gh
Thus,
discharge passing through strips
Area × velocity
=
Discharge over Rectangular Notch/Weir
Therefore, discharge of strip
(
dQ = Ldh 2 gh
)
vstrip = 2 gh
Astrip = Ldh
In order to obtain discharge over
whole area we must integrate above
equation from h=0 to h=H,
therefore;
H
Q = 2 g L ∫ h dh
0
2
2 g LH 3 / 2
3
2
= Cd
2 g LH 3 / 2
3
Q=
Qact
Where, Cd = Coefficient of discharge
Note: the expression for rectangular weir and sharp crested weirs are same.
43
Numerical Problems
A rectangular notch 2m wide has a constant head of 500mm. Find
the discharge over the notch if coefficient of discharge for the notch
is 0.62.
44
Numerical Problems
A rectangular notch has a discharge of 0.24m3/s, when head of water
is 800mm. Fine the length of notch. Assume Cd=0.6
45
Discharge over Rectangular Notch/Weir
It is more convenient to express the discharge equation in the
following form;
Qact = Cd
2
2 g LH 3 / 2 = Cw LH 3 / 2
3
Where Cw is coefficient of weir. Using the value of Cd as 0.62, we
can further write as;
Qact = 3.32 LH 3 / 2 in BG units
Qact = 1.18 LH 3 / 2
in SI units
The above equation give good results if H/P <0.4, which is well
within the usual operating range.
46
Discharge over Rectangular Notch/Weir
End Contractions: When length L of a
rectangular weir is less than the width of
the channel, the nappe will have end
contractions so that its width is less than L.
Experiment indicate that under the
condition depicted in figure, the effect of
each side contraction is to reduce the
effective width of the nappe by 0.1H.
Hence in such situations the flow rate may be computed by
substituting (L-0.1nH) instead of L, where n is number of end
contraction
Qact = Cd
47
2
2 g (L − 0.1nH )H 3 / 2 = Cw (L − 0.1nH )H 3 / 2
3
Rehbook Formulae for Cd
Rehbook of the karlsruhe hydraulic laboratory Germany obtained
following formulae for wide variety of variation in H and P
48
Numerical Problems
Solve example 11.7
11.11.1-11.11.5
11.41-11.45
49
Discharge over Triangular Notch (V-Notch)
In figure, V notch is shown.
Lets take a small strip of width, dh, at
any height, h, from water surface level.
Discharge, dQ, through strip is
(
dQ = dh(2 x ) 2 gh
)
From triangles
x
= tan (θ / 2 )
H −h
x = (H − h ) tan (θ / 2)
thus
(
dQ = dh(2(H − h ) tan (θ / 2 )) 2 gh
50
)
H=height of water above apex of notch
x is width of strip from center of notch
θ=Angle of notch
dh=height of strip
h= height of liquid above strip
L(2x)=Area of strip
Vo = Approach velocity
Theoretical velocity of strip neglecting
approach velocity =
2 gh
Discharge over Triangular Notch (V-Notch)
In order to obtain discharge over
whole area we must integrate
above equation from h=0 to h=H,
therefore;
(
H
Q = ∫ dh(2(H − h ) tan (θ / 2 )) 2 gh
)
0
H
Q = 2 2 g tan (θ / 2 )∫ (H − h ) h dh
0
H
(
)
Q = 2 2 g tan (θ / 2 )∫ Hh1/ 2 − h 3 / 2 dh
0
4

Q = 2 2 g tan (θ / 2 ) H 5 / 2 
15

Q=
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[
8
2 g tan (θ / 2 ) H 5 / 2
15
]
Qact =
[
8
Cd 2 g tan (θ / 2 ) H 5 / 2
15
]
Numerical Problems
Find the discharge over a triangular notch of angle 60o, when head
over triangular notch is 0.2m. Assume Cd=0.6
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Numerical Problems
During an experiment in a laboratory, 0.05m of water flowing over a right
angled notch was collected in one minute. If the head of sill is 50mm
calculate the coefficient of discharge of notch.
Solution:
Discharge=0.05m3/min=0.000833m3/s
Angle of notch, θ=90o
Head of water=H=50mm=0.05m
Cd=?
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Numerical Problems
A rectangular channel 1.5m wide has a discharge of 0.2m3/s, which is
measured in right-angled V notch, Find position of the apex of the notch
from the bed of the channel. Maximum depth of water is not to exceed 1m.
Assume Cd=0.62
Width of rectangular channel, L=1.5m
Discharge=Q=0.2m3/s
Depth of water in channel=1m
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Coefficient of discharge=0.62
Angle of notch= 90o
Height of apex of notch from bed=Depth of water in channel-height of
water over V-notch
=1-0.45= 0.55m
Discharge over Trapezoidal Notch
Assignment for you.
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Separation
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Cavitation
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Methods of Velocity Measurement
Assignment for you.
Reference: page 500 of Fluid mechanics with engineering
application by Finnermore and Franzini
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Thank you
Questions….
Feel free to contact:
mohsin.siddique@nu.edu.pk
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