Pressure measurement.

1. M.405.8 0
DEPARTMENT OF TECHNICAL EDUCATION
ANDHRA PRADESH
Name : Dr. P. Ugandhar
Designation : Sr. Lecturer in Mechanical Engg.
Branch : Mechanical Engineering
Institute : Govt. Polytechnic, SKLM
Year / Semester : IV
Subject : Hydraulics & Hydraulic Machinery
Subject Code : M-405
Topic :Fluid Properties
Duration :50 Minutes
Sub Topic : Problems on Manometers
Teaching Aids : PPT Animations

2. M.405.6 1
OBJECTIVES
On completion of this period, you would be able to
• Know the various pressure measuring devices
• Know the Difference between simple and differential manometers

3. M.405.6 2
Measurement of Pressure
Differential
(measure pressure
difference)
Fig. 1

4. M.405.6 3
Hydrostatic Law:
· The pressure at any depth ‘h’ from the free surface in a liquid of
density  is given by P = gh
· If ‘h’ is measured in upward direction P = -gh
The pressure of liquid if expressed in terms of the height (h) of
liquid column, it is known as pressure head of liquid.
Pressure head h = P/ g

5. M.405.6 4
Piezometer is used to measure low liquid pressures
What is Piezometer
Fig. 2 Piezometer

6. M.405.6 5
Manometer
• It is pressure measuring device
• It is used to measure difference of pressures between two
points.
• Its one column of liquid is balanced by another column of
liquid

7. M.405.6 6
Types of Manometers
• Simple U – tube Manometer
• Differential U – tube Manometer
• Simple U-tube manometer is used to measure
pressure at a point (gauge or vacuum)
• Differential U-tube manometer is used to measure
the difference of pressure between two points

8. M.405.6 7
How to find pressure at a point using U-tube Manometer
Fig 3 U-tube Manometer
• The Pressures at two
points P and Q must be equal
(being the same liquid)

9. M.405.6 8
U-tube Manometer:
• Equating pressures at P and Q, we have
P1+Ag(y+x) = patm +Bgx
P1-Patm = (B-A) gx - Agy
• Mercury is generally used as a manometric fluid because of its
high density and low vapour pressure

10. M.405.6 9
To find Vaccum Pressure:
• The Pressures at two points
P and Q must be equal.
• Equating pressures at P and
Q, we have
Fig 4 U-tube Manometer

11. M.405.6 10
Conversion of one fluid column into other fluid
column
• The pressure is given by P =  gh where h is the height
in terms of fluid of density 
• The same pressure can be expressed in terms of different
fluid columns.
i.e. P = 1gh1 = 2gh2
Where h1 and h2 are heights in terms of fluids of density
 1 and  2 respectively.
 1h1 = 2h2
Or
s1h1 = s2h2

12. M.405.6 11
SUMMARY
• In this class, we have discussed about
• Hydrostatic Law
• Piezometers
• Simple U-tube manometer to measure positive gauge pressures.
• Simple U-tube manometer to measure Vaccum Pressures

13. M.405.6 12
QUIZ
• A U-tube manometer with both limbs open to atmosphere
contains two immiscible liquids of densities 1 and 2 as shown
in figure. Under equilibrium ‘h’ is given by
(a) h =0
(b) h = L  2/ 1
(c) h = L(1- 1/ 2)
(d) h = L ( 2/ 1 –1)
Fig-5

14. M.405.6 13
• A U-tube manometer is connected to a pipeline conveying
water as shown in figure. Calculate the pressure head of water
in the pipeline is
Assignment
Fig-6

15. M.405.7 14
OBJECTIVES
On the completion of this period, you would be
able to
• Know the principle of differential manometer
• Working principle of Bourdon’s pressure gauge

16. M.405.7 15
Differential Manometer:
 
1 2
w w m
P y x g P y g x g
  
    
 
1 2 m w
P P gx
 
  
• The Pressures at two points P and
Q must be equal.
• Equating pressures at P and Q, we
have
Fig-1

17. M.405.7 16
Inverted Differential Manometer
• When a small difference of pressure between two
points is to be measured, then inverted differential manometer
is used
• A liquid of specific gravity less than that of the
flowing liquid is used

18. M.405.7 17
Fig-2 Inverted differential manometer
Pressure at D = Pressure at C
PA -  1gb- mgc = PB - 2g(a+b+c)
PB-PA =  2g(a+b+c) -  1gb -  mg c

19. M.405.7 18
Bourdon’s Pressure Gauge:
∙ It is a mechanical gauge and is used for measuring pressure in
steam boilers. This gauge has an elastic metallic tube of elliptical
section bent into circular arc
Fig-3 Bourdon´s Pressure Guage

20. M.405.7 19
Bourdon’s Pressure Gauge:
•When fluid enters the bent tube at one end, it tends to straighten
the tube.
• This causes the other end to move, causing pointer to move on a
graduated scale giving pressure reading
As the tube is surrounded by atmospheric pressure, the movement
of the tube is against atmospheric pressure and hence this gauge
measures pressure above or below atmospheric pressure (gauge
pressure)

21. M.405.7 20
SUMMARY
In this class, we have discussed about
• Differential Manometers
• Differential inverted U-tube Manometers
• Bourdon’s Pressure Gauge

22. M.405.7 21
QUIZ
1. The manometer shown in figure connects
two pipes carrying oil and water respectively.
The difference between oil and water pressure
is
(a) ( Water+  oil)gH
(b) ( Water / oil)gH
(c) ( Water -  oil)gH
(d)  Oil gH
Fig-4

23. M.405.7 22
2. The manometer shown in figure connects two pipes carrying
oil and water respectively. From this figure it can be concluded
that
(a) pressure in pipes are equal
(b) the pressure in oil pipe is higher
(c) the pressure in water pipe is
higher
(d) The pressure in two pipes can not
be compared (for want of Sufficient
data)
QUIZ
Fig-5

24. M.405.7 23
Frequently Asked Questions
1. What is differential manometer?
2. Why is mercury used as a manometric liquid?
3. How does Bourdon’s pressure gauge work?

25. M.405.8 24
RECAP
In the previous class, we have discussed about
• Principle of manometers
• Finding pressure at a point by using simple manometer
• Finding pressure difference between two points by using differential
manometer

26. M.405.8 25
OBJECTIVES
• On the completion of this period, you would be able to
•solve simple problems on manometers

27. M.405.8 26
Problem –1:
• Find the pressure of a fluid of specific gravity 0.9 at the
point A as shown in figure.
Fig-1

28. M.405.8 27
Solution
• Let x-x be the datum line
 Density of oil (0) = 0.9 x 1000 = 900 kg/m3
• Pressure in the left limb at x-x = Pressure in the right limb at XX
0
20 12 20
0
100 100
m
g g
 

   
 
   
   
• PA+

29. M.405.8 28
Solution
3
8 20
900 9.81 13.6 10 9.81
100 100
A
P
   
    
   
   
• PA = 13.6  103 9.81  0.2 – 900 9.81  0.08
• PA = 25976.88 N/m2
• PA+

30. M.405.8 29
• Problem – 2
A differential manometer is connected as shown in Fig.
Find the pressure difference between A and B in N/m2 and in
cm of water.
• Let us choose xx as datum line.
• Pressure in left limb at xx = Pressure in right limb at xx
Fig-2

31. M.405.8 30
Solution
2 2
0
38 23
(0.53) (0.23)
100
A H o B H o
P g g P g
  

 
      
 
 
2 0
0.61 0.53 (0.23)
A B H o
P P g g
 
   
2 0
(0.08) (0.23)
A B H o
P P g  
 
  
 
 
9.81 1000(0.08) 810(0.23)
A B
P P
  
 
3
0.81 1000 810 /
o kg m
   
2
1042.8 /
A B
P P N m
  
.
.
.
.
.
.
.

32. M.405.8 31
• Pressure difference in terms of water
P = gh


• h = -0.1063 m of water
or h = -0.1063 100 cm
 h = -10.63 cm of H20
2
A B
P = P P 1042.8 /
N m
   
.
.

33. M.405.8 32
ASSIGNMENT
1. The pressure at certain point is observed to be 40 m of oil of
specific gravity 0.6. Find the corresponding height in terms of
mercury of specific gravity 13.6 at that point.
2. Pressures have been observed at four different point in different
units of measurements of follows:
(a) 150 kpa (b) 1800milli bar
(c) 20m of water (d) 1240mm of mercury
Arrange the points in the descending order of magnitude of
pressure