007c (PPT) Pitot tube, Notches & Weirs.pdf

1. Flow measurement
Session 3
Pitot tube
&
Notches/Weirs
Dr. Vijay G. S.
Professor
Dept. of Mech. & Mfg. Engg.
MIT, Manipal
email: vijay.gs@manipal.edu
Mob.: 9980032104 1

2. 2
Pitot tube
• The Pitot tube (named after the French scientist Pitot) is one of the simplest and most
useful instruments ever devised for flow measurement.
• It is generally used to measure the local velocity of flow in pipes or open channels.
• The Pitot tube is a small tube bent at right angles
and is placed in the flow such that lower end
(called the probe), is bent through 90o and is
directed in the upstream direction.
• The probe is maintained such that the axis of its
hole (probe) is parallel to the velocity of the fluid
flowing.
• The liquid rises in the tube by a distance h due to
conversion of kinetic energy into potential energy.
• The velocity can be determined by measuring the
rise of liquid in the tube.
At point (1), Static pressure, p1 = ogz
At point (2), Stagnation pressure, p2 = og(z + h)
Free surface of liquid
Pitot
tube
Flow
o
Pitot tube used in an open channel
Tube dia should not
influence capillarity
Probe

3. 3
Pitot tube for a closed conduit (pipe)
Pitot tube used in a closed conduit (a pipe)
• In a closed conduit (like a pipe), there is some internal pressure, which results in the
static pressure head z at point (1), as indicated by the piezometer at (1).
• Due to stagnation of fluid at point (2), the stagnation pressure head indicated by the pitot
tube will be z+h.
• Thus the Dynamic pressure head,
h = (Stagnation pressure head –
Static pressure head)

4. 4
V2 = 0
• The streamline along the probe axis comes to a stop at (2). i.e., the flow stagnates at (2).
• This point is known as the “Stagnation point” where the fluid velocity is zero, i.e., V2 = 0.
• Since the flow is suddenly brought to rest, the pressure at that point rises sharply due to
the conversion of the kinetic head into pressure head.
• The pressure p2 at (2) is greater than p1 at (1) and is known as the “Stagnation pressure”.
Pitot tube – contd…
V1 = ?

5. 5
Pitot tube – Stagnation pressure
• The Stagnation pressure at a point in the fluid flow is defined as the total pressure which
would result if the fluid was suddenly brought to rest isentropically.
• The relationship between the Stagnation pressure (p2) and the Static pressure (p1) can be
deduced by applying the Bernoulli’s equation at points (1) and (2).
• Let p1, V1 and p2, V2 be the pressures and velocities at points (1) and (2) respectively.
2 2
1 1 2 2
1 2
2 2
 
    
o o
p V p V
z z
g g g g
• At the stagnation point (2), the velocity V2 = 0.
• Also z1 = z2
2
1 1 2
2
 
 
o o
p V p
g g g
2
1
2 1
2

  oV
p p

• Stagnation pressure (p2) = Static pressure (p1) + Dynamic pressure
2
1
2

 
 
 
oV
o

6. 6
Pitot tube – Stagnation pressure – contd…
• Stagnation pressure (p2) = Static pressure (p1) + Dynamic pressure
2
1
2

 
 
 
oV
• The stagnation pressure (p2) is more than the static pressure (p1) by an amount equal to
the dynamic pressure
2
1
2

 
 
 
oV
 The dynamic pressure head is
2
1
2
1
2
2
(2)


 
  
 


o
o
V
h g
V
g
Kinetic head lost at
(head = p/g)

7. 7
Pitot tube – Finding flow velocity
2
1
2 1
2

  oV
p p
If the stagnation pressure p2 and the static pressure
p1 are known, then the velocity of the flow V1 can
be determined.
2
1
2 1
2

 
oV
p p
 
2 1
2
1
2



o
p p
V
 
 
2 1 2 1
1 2 1
2
2 2 2
  
  
     
 
 
o o o
g p p p p
V g g h h gh
g g g
 
1 2

th
V gh
V1
Static
pressure
head (h1)
Stagnation
pressure
head (h2)

8. (V1)act < (V1)th  (V1)act = Cv (V1)th
where, Cv is known as “Coefficient of velocity”
(V1)act = Cv (V1)th
This equation gives the actual local velocity of flow measured by a pitot tube.
Cv  0.98
8
Pitot tube – Finding flow velocity – contd…

v
Actual velocity
C
Theoretical velocity
 
1 2
v
act
V C gh

 
1 2

th
V gh

9. Pitot-Prandtl probe OR Pitot-Static probe
Probe opening for sensing
stagnation pressure p2
Connections
to differential
manometer
9
Holes for
sensing static
pressure p1


Vmax
V1

10. 10
Central
velocity V1
Pitot-Prandtl probe OR Pitot-Static probe – contd…
Importance of positioning the pitot-static probe (for a pipe flow)

11. 11
Pitot-Prandtl probe OR Pitot-Static probe – contd…
1 1
m m
o o
S
h x x
S


   
   
   
   
1 1
m m
o o
S
h x x
S


   
   
   
   
If m > o OR Sm > So then
an upright differential
manometer is used.
If m < o OR Sm < So then
an inverted differential
manometer is used.
m
m
o
o
Computing the dynamic pressure head from manometer reading

12. 12

13. PROBLEMS ON PITOT TUBE
13

14. Problem 1: Find the velocity of the flow of an oil through a pipe, when the
difference of mercury level in a differential U-tube manometer connected to the
two tappings of the pitot-tube is 100 mm. Take co-efficient of pitot-tube = 0.98
and specific gravity of oil = 0.8.
o = 800 kg/m3 (Flowing oil)
m = 13600 kg/m3 (Manometric fluid)
x = 100 mm = 0.1 m of mercury
Cv = 0.98
2
0.98 2 9.81 1.6
5.49 /

   

act v
V C gh
m s
14
1
13600
0.1 1 1.6
800


 
 
 
 
 
   
 
 
m
o
h x
h m of oil

15. Problem 2: A pitot-static tube is used to measure the velocity of water in a pipe. The
stagnation pressure head is 6 m of water and static pressure head is 5 m of water.
Calculate the velocity of flow assuming that the coefficient of pitot tube is equal to 0.98.
15
Stagnation pressure head, h2 = 6 m of water (Flowing water)
Static pressure head, h1 = 5 m of water (Flowing water)
Cv = 0.98
Dynamic pressure head, h = h2 – h1 = 6 – 5 = 1 m of water
2
0.98 2 9.81 1
4.34 /

   

act v
V C gh
m s

16. Problem 3: A sub-marine moves horizontally in sea and has its axis 15 m below
the free surface of water. A pitot-tube properly placed just in front of the sub-
marine and along its axis is connected to the two limbs of a U-tube containing
mercury. The difference of mercury level is found to be 170 mm. Find the speed
of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of
sea water is 1.026 with respect to that of fresh water. Take Cv = 0.98.
16
h1 = 15 m (Static pressure head)
o = 1026 kg/m3 (Flowing sea water)
m = 13600 kg/m3 (Manometric fluid)
x = 170 mm = 0.17 m of mercury
Cv = 0.98
1
13600
0.17 1 2.083
1026


 
 
 
 
 
   
 
 
m
o
h x
h m of sea water
2
0.98 2 9.81 2.083
6.265 /

   

act v
V C gh
m s
Vact = 6.265 × 60 × 60/1000
= 22.55 km/hr

17. Problem 4: A pitot-static tube placed in the center of a 300 mm pipe line has one
orifice pointing upstream and other perpendicular to it. The mean velocity in the
pipe is 80% of the central velocity. Find the discharge through the pipe if the
pressure difference between the two orifices is 60 mm of water. Take the
coefficient of pitot tube as 0.98.
17
d1 = 300 mm = 0.3 m (pipe dia)  a1 = (d1
2/4) = 0.0707 m2
Vavg = 0.8×V1 (Mean velocity, V1 = Central velocity)
o = 1000 kg/m3 (Flowing water)
h = 60 mm = 0.06 m of water (Dynamic pressure head)
Cv = 0.98
Q = ?  
1 2
0.98 2 9.81 0.06
1.063 /

   

v
act
V C gh
m s

18. 18
Problem 4: contd…
Vavg = 0.8×V1
= 0.8×1.063
= 0.851 m/s
Discharge through the pipe,
Q = a1× Vavg = 0.0707 × 0.851 = 0.06 m3/s
Q = 60 lps
Central
velocity V1

19. Problem 5: A pitot static probe is inserted in a pipe of 300 mm diameter. The
static pressure in the pipe is 100 mm of mercury (vacuum). The stagnation
pressure at the center of the pipe, recorded by pitot-tube is 0.981 N/cm2.
Calculate the rate of flow of water through pipe, if the mean velocity of flow is
0.85 times the central velocity. Take Cv = 0.98.
19
d1 = 300 mm = 0.3 m (pipe dia)  a1 = (d1
2/4) = 0.0707 m2
Vavg = 0.85×V1 (Mean velocity, V1 = Central velocity)
o = 1000 kg/m3 (Flowing water)
h1 = 100 mm (vacuum) = -100 mm = - 0.1 m of Hg (Static pressure head)
p2 = 0.981 N/cm2 = 0.981×104 N/m2 (Stagnation pressure)
Cv = 0.98
Q = ?
p1 = Hg× g × h1 = 13600×9.81×-0.1 = -13341.6 N/m2

20. 20
 
2 1
4
0 981 10 13341 6
1000 9 81
2 36
. .
.
.

 

  
 
  



o
p p
h
g
m of water
 
1 2
0.98 2 9.81 2.36
6.668 /

   

v
act
V C gh
m s
Vavg = 0.85×V1
= 0.85×6.668
= 5.668 m/s
Discharge through the pipe,
Q = a1× Vavg = 0.0707 × 5.668 = 0.4 m3/s
Q = 400 lps
Problem 5: contd…

21. Notches/Weirs
Measurement of discharge through open channels
21

22. 22
Notches/Weirs
Rectangular weir/notch
Triangular weir/notch
A Notch or a Weir is used for measuring the rate of flow of liquid through
open channels or rivers.

23. Notches/Weirs – contd…
23
Notch Weir
• It is a structure which obstructs the flow
from a tank or reservoir
• It is a structure which obstructs the flow
in an open channel or river
• It is an opening provided on one side of
the tank or reservoir, with free surface of
liquid present below the top edge of the
opening
• It is an opening provided on the
obstructing structure, such that flow
occurs on the top edge of the weir
• Notch is normally made of a metallic
plate
• Weir is made of concrete or masonry
• Comparatively smaller in size • Comparatively larger in size
• It is used to measure discharge in small
channels or in laboratory reservoirs.
• It is used to measure discharge in larger
bodies like rivers and large channels.

24. 24
Rate of flow through a rectangular notch/weir
Crest or Sill
Top edge of notch

25. 25
Rate of flow through a rectangular notch/weir – contd…
H = Head of water over the crest
L = Length of the notch or weir
• Consider a horizontal strip of water
of thickness dh and length L at a
depth h below the free surface of
water
• The area of strip = L  dh
• Theoretical velocity of water flowing through strip = 2gh
• The discharge dQ, through strip is
dQ = Cd  Area of strip  Theoretical velocity
2
d
dQ C L dh gh
   
where Cd = Co-efficient of discharge

26. 26
Rate of flow through a rectangular notch/weir – contd…
• The total discharge, Q, for the whole notch or weir is determined by integrating the above
equation between the limits 0 and H.
0 0
2
H H
d
Q dQ C L dh gh
    
 
 
0
3/2
0
3/2
2
2
3/ 2
2
2
3
   
 
    
 
   

H
d
H
d
d
Q C L g h dh
h
C L g
C L g H
 
3/2
2
2
3
d
Q C L g H
   


27. 27
Rate of flow through a triangular notch/weir
Apex
Top edge of notch
• Let H = head of water above the apex of
the V- notch or triangular weir.
•  = angle of notch
• Consider a horizontal strip of water of
thickness dh at a depth of h from the free
surface of water

28. 28
Rate of flow through a triangular notch/weir – contd…
tan
2
AC AC
OC H h

 
 
 

 
 tan
2
AC H h

 
    
 
Width of the strip AB =  
2 2 tan
2
AC H h

 
   
 
Area of the strip =  
2 tan
2
H h dh

 
 
 
 
Theoretical velocity of water flowing through strip = 2gh
The discharge dQ, through strip is
dQ = Cd  Area of strip  Theoretical velocity
 
2 tan 2
2
d
dQ C H h dh gh

 
    
 
 
where Cd = Co-efficient of discharge

29. 29
Rate of flow through a triangular notch/weir – contd…
The total discharge, Q, for the whole notch or weir is determined by integrating the above
equation between the limits 0 and H.
 
0 0
2 tan 2
2
H H
d
Q dQ C H h dh gh

 
     
 
 
 
   
1/2 3/2
0 0
2 2 tan 2 2 tan
2 2
 
   
   
   
   
 
H H
d d
Q C g H h h dh C g Hh h dh
3/2 5/2 5/2 5/2
0
2 2
2 2 tan 2 2 tan
2 3/ 2 5/ 2 2 3 5
H
d d
Hh h H H
Q C g C g
 
   
   
   
   
   
   
   
5/2
8
2 tan
15 2
d
Q C g H

 
   
 
 
5/2
8 90
0.6 2 9.81 tan
15 2
o
Q H
 
     
 
 
For a right angled notch (weir) ( = 90o) and Cd = 0.6,
5/2
1.417
Q H
 

30. PROBLEMS ON NOTCHES/WEIRS
30

31. Problem 1: Find the discharge of water flowing over a rectangular notch of 2 m
length when the constant head over the notch is 300 mm. Take Cd = 0.60.
31
L = 2 m (length of notch)
H = 300 mm = 0.3 m (head over the notch)
Cd = 0.60
 
 
3/2
3/2
3
2
2
3
2
0.6 2 2 9.81 0.3
3
0.582 /
   
    

d
Q C L g H
m s
Q = 582 lps

32. Problem 2: The head of water over a rectangular notch is 900 mm. The
discharge is 300 liters/s. Find the length of the notch, when Cd is 0.62.
32
H = 900 mm = 0.9 m (head over the notch)
Q = 300 lps = 0.3 m3/s
Cd = 0.62
L = ? (length of notch)
 
 
3/2
3/2
2
2
3
2
0.3 0.62 2 9.81 0.9
3
0.192
   
    

d
Q C L g H
L
L m

33. Problem 3: Determine the height of a rectangular weir of length 6 m to be built
across a rectangular channel. The maximum depth of water on the upstream side
of the weir is 1.8 m and discharge is 2000 litres/s. Take Cd = 0.6 and neglect end
contractions.
33
L = 6 m
H1 = 1.8 m (depth of water on upstream)
Q = 2000 lps = 2 m3/s
Cd = 0.6
H2 = ? (Height of weir)
Let H2 = height of weir
H2 = H1 – H
H2 = 1.8 – 0.328
= 1.472 m
 
 
3/2
3/2
2
2
3
2
2 0.6 6 2 9.81
3
0.328
   
    

d
Q C L g H
H
H m

34. Problem 4: Find the discharge over a triangular notch of angle 60o when the
head over the V-notch is 0.3 m. Assume Cd = 0.6.
34
 = 60o (V notch angle)
H = 0.3 m (head of water above apex)
Cd = 0.6
Q = ? (Flow rate)
Q = 40 lps

35. Problem 5: Water flows over a rectangular notch 1 m wide at a depth of 150
mm and afterwards passes through a triangular right-angled notch. Taking Cd for
the rectangular and triangular weir as 0.62 and 0.59 respectively, find the depth
over the triangular weir.
35
HR = 150 mm = 0.15 m (head over the rectangular notch)
L = 1 m (length of rectangular notch)
Cd = 0.62 (rectangular notch)
Cd = 0.59 (triangular notch)
 = 90o (V notch angle)
HT = ? (head over the triangular notch)
HT
HR
90o
Q
QRectangular = QTriangular
 
3/2 5/2
2 8 90
2 2 tan
3 15 2
o
d R R dT T
C L g H C g H
 
       
 
 

36. Problem 5: Contd…
36
2
3
2
dR
C L g
    
3/2 8
R
H 
4
15
2
5
dT
C g
  5/2
90
tan
2
o
T
H
 
 
 
 
 
 
3/2
5/2 5
4 tan 45
dR R
T o
dT
C L H
H
C
  

 
 
 
3/2
5/2
2/5
5 0.62 1 0.15
4 0.59 tan 45
0.0763
0.0763 0.3573
o
T
T
H
H m
  

 

 