Venturimeter and Pitot tube problems and solutions

1. 9C304.19 1
Department of Technical Education
Andhra Pradesh
Name : T. Subbarayudu
Designation : Lecturer
Branch : Civil Engineering
Institute : S.V. Govt. Polytechnic, Tirupati
Year/Semester : III Semester
Subject : Hydraulics
Subject Code : C304
Topic : Flow Of Liquid
Duration : 50 Minutes
Sub Topic : Practical application of bernoulli’s
theorem - venturi meter
Teaching Aids : PPT, Diagrams
Revised By : V. Srinivasa Rao, Lecturer, DAGPT,
Ongole

2. 9C304.19 2
Objectives
On completion of this session, you would be
able to
• Calculate discharge through the venturi meter.
• Design the diameter of venturimerter
• Find the Venturi head

3. 9C304.19 3
Recap
In the previous sessions, you have learnt in detail
about
• Applications of Bernoulli’s theorem
• Venturimeter and its applications
• Derivation of Discharge formula through venturi
meter

4. 9C304.19 4
We know that
• Discharge through the Venturimeter
Where Q = Discharge through venturimerter
Cd = Coefficient of discharge
= c/s area of inlet
= c/s. area of throat
h = Pressure difference between inlet
and throat
g = acceleration due to gravity
1
a
2
a
1 2
2 2
1 2
2
a a gh
Q Cd
a a



5. 9C304.19 5
Problems
1.A venturimerter 300x 100 mm is used fir measuring the
discharge of an oil flowing through a pipe. The difference
of pressure measured by a differential mercury
manometer is 150 mm . The specific gravity of oil is 0.8
and the venturi coefficient is 0.97 . Calculate the
discharge of oil in lt / sec.
1
2
2
2
1
Sol: Given data:
d 300mm 0.3m
d 100 mm 0.1m
π 0.3
a area atinlet 0.070 m
4
 
 

  

6. 9C304.19 6
m
0.15
mm
150
x
levels
mercury
between
difference
x
m
0.00785
4
0.1
π
throat
at
area
a 2
2
2








m
2.4
1
0.8
13.6
0.15
1
S
S
x
h m


















7. 9C304.19 7
1 2
2 2
1 2
2 2
3 3
discharge through the venturimeter
Cd a a 2gh
Q
a a
Cd 0.97
0.97 0.070 0.00785 2 9.81 2.4
0.070 0.00785
5.22 10 m /sec





    


 

8. 9C304.19 8
2.Find the discharge flowing through a venturimerter on inlet
dia. 30 cm and throat dia. 15 cm. The pressure difference
between inlet and throat as read with the help of differential
manometer is found to be 20 cm. of mercury. Specific gravity
of liquid 0.8 and that of mercury is 13.6
Cd = Coefficient of discharge venturimerter is 0.98

9. 9C304.19 9
2
1
2
2
2
2
Sol: Given data:
Diameter at inlet 30cm
Area at inlet a (30)
4
706.86 cm
Diameter at throat 15cm
Area at inlet a (15)
4
176.71 cm









10. 9C304.19 10
 
1 2
2 2
1 2
Let Q Disch arg eof oil through thepipe. Using the relation
Cd a a
Q = 2gh with usualnotation
a a


Specificgravityof oil 0.8
Differenceof pressurehead
h 20cmof mercury
13.6 0.8
20
0.8
320cmof oil
Coefficient of meter disch argeof venturimeterCd 0.98



 
  
 



11. 9C304.19 11
 
   
1 2
2 2
1 2
2 2
3
3
3
Cda a
Q= 2gh
a a
0.98 706.86 176.71
Q= 2 981 320
706.86 176.71
141718.07 cm /sec
141.718 lit /sec
0.141718 m /sec
Disch arge of oil through the venturimeter
Q = 0.141718 m /sec.

 
  
 

 
 





12. 9C304.19 12
3. A 20 cm x 10 cm venturimerter is mounted in a vertical
pipe carrying water the flow being upwards. The throat
section is 20 cm above the entrance section of the
venturimerter. For a certain flow through the meter, the
differential gauge between the throat and entrance
indicates a gauge deflection of 25 cm. Assuming the
Coefficient of venturimerter is 0.98. Find the discharge.

13. 9C304.19 13
2
1
2
2
2
2
Sol: Given data:
Diameter at inlet 20cm
Area at inlet a (20)
4
314.16 cm
Diameter at throat 10cm
Area at inlet a (10)
4
78.54 cm









14. 9C304.19 14
Mercury manometer reading,
h 25cm of mercury
13.6 1
25
1
315cm of water
Cd 0.98


 
  
 


 
1 2
2 2
1 2
Let Q Disch argeof water through thepipe. Using the relation
Cda a
Q= 2gh with usualnotation
a a



15. 9C304.19 15
 
   
1 2
2 2
1 2
2 2
3
Cda a
Q= 2gh
a -a
0.98×314.16×78.54
Q= × 2× 981× 315
314.16 - 78.54
Q= 62490 cm /sec
Q= 62.490 lit /sec
 
 
 

16. 9C304.19 16
4. A venturimerter has an area ratio of 9 to 1, the larger
diameter being 30 cm. During the flow, the recorded
pressure head in the larger section is 6.5 m and that at
the throat 4.25 m. If meter coefficient , Cd = 0.99
Compute the discharge through the metre.

17. 9C304.19 17
1
2
2
1
2
2
2
Sol: Given data:
a
Ratioof areas, =9
a
Diameter at larger section = 0.3m
π
Area at inlet a = (0.3)
4
=0.07069 m
0.07069
Area at smallersection a =
9
=0.00785 m

18. 9C304.19 18
 
1 2
2 2
1 2
Let Q Dischargeof oilthrough thepipe. Using the relation
Cda a
Q= 2gh with usualnotation
a a


Pressure head at larger section
6.5 m
Pressure head at smaller section
4.25 m
Difference of pressure heads
h = 6.5 4.25 = 2.25 m of liquid
Coefficient of meter Cd= 0.99





19. 9C304.19 19
 
   
1 2
2 2
1 2
2 2
3
Cd a a
Q = 2gh
a -a
0.99×0.07069 ×0.00785
Q = × 2× 981× 2.25
0.07069 - 0.00785
Q = 0.052 m / sec
Q = 52.00 lit / sec
 
 
 

20. 9C304.19 20
5. A venturi meter consisting of 10 cm diameter at throat
in a 25 diameter pipe has coefficient of 0.99. The pipe
delivers oil specific gravity of 0.8. The pressure
difference on the two sides of the orifice plate is
measured by a mercury oil differential manometer. If the
differential gauge reads 80 cm of mercury, calculate the
rate of flow in litre / sec.

21. 9C304.19 21
2 2
1
2 2
2
Sol: Given data:
Diameter of venturimeter 10cm
π
Area a 10 78.59cm
4
Diameter of pipe 25cm
π
Area a 25 490.87cm
4
Venturi meter coefficient Cd 0.99
Specificgravity of oil 0.8

   

   



22. 9C304.19 22
m
0.15
mm
150
x
levels
mercury
between
difference
x
m
0.00785
4
0.1
π
throat
at
area
a 2
2
2








m
2.4
1
0.8
13.6
0.15
1
S
S
x
h m


















23. 9C304.19 23
1 2
2 2
1 2
2 2
3 3
Cd a a 2gh
Q
a a
Cd 0.99
0.99 0.070 0.00785 2 9.81 2.4
0.070 0.00785
55.31 10 m /sec





    


 
Discharge through the venturimerter

24. 9C304.19 24
6. Find the discharge flowing through a venturi meter on
inlet dia. 30 cm and throat dia. 15 cm. The pressure
difference between inlet and throat as read with the help
of differential manometer is found to be 20 cm. of
mercury. Specific gravity of liquid 0.8 and that of
mercury is 13.6
Cd = Coefficient of venturimerter is 0.97

25. 9C304.19 25
2
1
2
2
2
2
Sol: Given data:
Diameter at inlet 30cm
Area at inlet a (30)
4
706.86 cm
Diameter at throat 15cm
Area at inlet a (15)
4
176.71 cm









26. 9C304.19 26
Specific gravity of oil 0.8
Difference of pressure head
h 20 cm of mercury
13.6 0.8
20
0.8
320 cm of oil
Cd 0.97



 
  
 


 
1 2
2 2
1 2
Let Q Disch argeof oilthrough thepipe. Using the relation
Cda a
Q= 2gh with usualnotation
a a



27. 9C304.19 27
 
   
1 2
2 2
1 2
2 2
3
3
3
Cda a
Q= 2gh
a a
0.97 706.86 176.71
Q= 2 981 320
706.86 176.71
140270 cm /sec
140.27 lit /sec
0.1402 m /sec
Disch arge of oil through the venturimeter
Q = 0.1402 m /sec.

 
  
 

 
 





28. 9C304.19 28
7.A horizontal venturi meter 16 cm x 8 cm is used to
measure the flow of an oil specific gravity 0.8 . Determine
the deflection of the oil mercury gauge, if the discharge
of a oil is 50 lt /sec. Take Cd =.95
2 2
2 2
3
π
Sol: Area of the inlet end = 16 =201.06 cm
4
π
Area of the inlet throat = ×8 =50.265cm
4
specific gravity of oil =0.8
Q = 50 lit /sec or 50x1000 = 50000 cm /sec

29. 9C304.19 29
h
f
S
h x 1
S
13.6-0.8
x
0.8
h 16x
 
  
 
 
 
   
 

1 2
2 2
1 2
2 2
a a 2gh
Q Cd
a a
0.95 201.06 50.26
50000 2 981 16x
201.06 50.26
x 32.74cm
differencesof liquid columns 32.74 cmof mercury


 
   




30. 9C304.19 30
8.A horizontal venturi meter 30 cm x 15 cm is used to
measure the flow of water. Determine the deflection of
the water mercury gauge, if the discharge of a water is
110 lt /sec. Assume Cd =0.96
2 2
2 2
3
Sol:
π
Area of the inlet end = 30 = 706.85 cm
4
π
Area of the throat = ×15 = 176.714cm
4
Q = 110 lit /sec or 110x1000 = 110000 cm /sec


31. 9C304.19 31
h
f
S
h x 1
S
13.6 - 1
x
1
h 12.6 x
 
  
 
 
 
   
 

1 2
2 2
1 2
2 2
a a 2gh
Q Cd
a a
0.96 706.85 176.714
110000 2 981 12.6x
706.85 176.714
x 15.944 cm of mercury
differencesof liquidcolumns 15.944 cmof mercury


 
   




32. 9C304.19 32
Summary
In this session, you have learnt about solving
problems on
• Finding discharge through venturimerter
• Designing the diameter at throat of venturi meter
• Finding the venturi head.

33. 9C304.19 33
Quiz
1. For measuring discharge through closed conduits
accurately the device used is
a) venturi meter
b) orifice meter
c) pitot tube
d) elbow meter

34. 9C304.19 34
2. For measuring velocity through streams accurately the
device used is
a) venturi meter
b) orifice meter
c) pitot tube
d) elbow meter
Quiz

35. 9C304.19 35
3. Energy conversion in the inlet of a venturi meter is
a) pressure energy increases
b) velocity energy decreases
c) pressure energy remains unaltered
d) velocity energy increases
Quiz

36. 9C304.19 36
Frequently Asked Questions
1. A venturimerter 200x 100 mm is used for measuring the
discharge of an oil flowing through a pipe. The difference
of pressure measured by a differential mercury manometer is
220 mm . The specific gravity of oil is 0.9and the Cd = 0.97 .
Calculate the discharge of oil in lt / sec
2 A horizontal venturi meter 18 cm x 9 cm is used to measure
the flow of an oil specific gravity 0.8 . Determine the
deflection of the oil mercury gauge, if the discharge of a oil is
90 lt /sec. Assume Cd =.95

37. 9C304.19 37
3. A 20 cm x 12 cm venturimerter is mounted in a vertical pipe
carrying water the flow being upwards. The throat section is
20 cm above the entrance section of the venturimerter. For a
certain flow through the meter, the differential gauge between
the throat and entrance indicates a gauge deflection of 28 cm.
Assuming the Coefficient of discharge of venturimerter is
0.98. Find the discharge.
4. A horizontal venturi meter 30 cm x 16 cm is used to measure
the flow of water. Determine the deflection of the water
mercury gauge, if the discharge of a water is 105 lt /sec.
Assume Cd =0.96
Frequently Asked Questions