Topic 2 - Principles of Hydrostatics.pdf

TOPIC FOR HYDRAULICS, FIRST SEMESTER

1. CHAPTER 2
PRINCIPLES OF FLUID
STATICS
FLUID STATICS

2. FLUID STATICS
CHAPTER 2
PRINCIPLES OF FLUID
STATICS

3. FLUID STATICS
Fluid statics is the study of fluids in which there is no relative motion
between fluid particles.
If there is no relative motion, no shearing stresses exist, since velocity
gradients, such as dv/dy, are required for shearing stresses to be present.
The only stress that exists in a normal stress, the pressure, so it is the pressure
that is of primary interest in fluid statics.
UNIT PRESSURE OR PRESSURE, P

4. UNIT PRESSURE OR PRESSURE, P
Absolute pressure is referenced to regions such as outer
space, where the pressure is essentially zero because the
region is devoid of gas.
•The pressure in a perfect vacuum is called absolute
zero, and pressure measured relative to this zero
pressure is termed absolute pressure.
When the pressure is measured relative to prevailing local
atmospheric pressure, the pressure value is
called gage pressure.
When pressure is less than atmospheric, the pressure can
be described using vacuum pressure. Vacuum pressure is
defined as the difference between atmospheric pressure
and actual pressure. Vacuum pressure is a positive number
and equals the absolute value of gage pressure (which will
be negative ).
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FLUID STATICS
Fluid statics is the study of fluids in which there is no relative motion
between fluid particles.
If there is no relative motion, no shearing stresses exist, since velocity
gradients, such as dv/dy, are required for shearing stresses to be present.
The only stress that exists in a normal stress, the pressure, so it is the pressure
that is of primary interest in fluid statics.

5. UNIT PRESSURE OR PRESSURE, P
Absolute pressure is referenced to regions such as outer
space, where the pressure is essentially zero because the
region is devoid of gas.
•The pressure in a perfect vacuum is called absolute
zero, and pressure measured relative to this zero
pressure is termed absolute pressure.
When the pressure is measured relative to prevailing local
atmospheric pressure, the pressure value is
called gage pressure.
When pressure is less than atmospheric, the pressure can
be described using vacuum pressure. Vacuum pressure is
defined as the difference between atmospheric pressure
and actual pressure. Vacuum pressure is a positive number
and equals the absolute value of gage pressure (which will
be negative ).
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FLUID STATICS
Fluid statics is the study of fluids in which there is no relative motion
between fluid particles.
If there is no relative motion, no shearing stresses exist, since velocity
gradients, such as dv/dy, are required for shearing stresses to be present.
The only stress that exists in a normal stress, the pressure, so it is the pressure
that is of primary interest in fluid statics.
PRESSURE VARIATIONS

6. PRESSURE VARIATIONS
W=γV
W=γ(aL)
σF
x=0
F
2−F
1=Wsinθ
P
2a −P
1a= γaLsinθ
P
2−P
1=γLsinθ
P
2−P
1=γh
L
Wθ
θ
h
=
L
sin
θ
F
2
P
2
P
1
F
1
a

7. PRESSURE VARIATIONS
W=γV
W=γ(aL)
σF
x=0
F
2−F
1=Wsinθ
P
2a −P
1a= γaLsinθ
P
2−P
1=γLsinθ
P
2−P
1=γh
P
2=γh+P
1
L
Wθ
θ
h
=
L
sin
θ
F
2
P
2
P
1
F
1
a
PRESSURE BELOW LAYERS OF DIFFERENT LIQUIDS

8. PRESSURE BELOW LAYERS OF DIFFERENT LIQUIDS
LIQUID 1
LIQUID 2
LIQUID 3
AIR PRESSURE = Pa
PRESSURE VARIATIONS
W=γh
W=γ(aL)
σF
x=0
F
2−F
1=Wsinθ
P
2a −P
1a= γaLsinθ
P
2−P
1=γLsinθ
P
2−P
1=γh
P
2=γh+P
1
L
Wθ
θ
h
=
L
sin
θ
F
2
P
2
P
1
F
1
a

9. PRESSURE BELOW LAYERS OF DIFFERENT LIQUIDS
P
bottom=σγh+P=γ
1h
1+γ
2h
2+γ
3h
3+P
a
LIQUID 1
LIQUID 2
LIQUID 3
AIR PRESSURE = Pa
PRESSURE HEAD

10. PRESSURE HEAD
P
h
z
PRESSURE BELOW LAYERS OF DIFFERENT LIQUIDS
P
bottom=σγh+P=γ
1h
1+γ
2h
2+γ
3h
3+P
a
LIQUID 1
LIQUID 2
LIQUID 3
AIR PRESSURE = Pa

11. PRESSURE HEAD
ℎ=
??????
??????
P
h
z DEVICES FOR MEASURING PRESSURE

12. DEVICES FOR MEASURING PRESSURE
BAROMETER
▪An instrument that is used to measure atmospheric
pressure
▪The most common types are mercury barometer and
the aneroid barometer.
oA mercury barometer is made by inverting a
mercury-filled tube in a container of mercury. The
pressure at the top of the mercury barometer will
be the vapor pressure of mercury, which is very
small: Pv = 2.4x10−6 atm at 20 °C.
oAn aneroid barometer works mechanically. An
aneroid is an elastic bellows that has been tightly
sealed after some air was removed.
PRESSURE HEAD
ℎ=
??????
??????
P
h
z

13. DEVICES FOR MEASURING PRESSURE
BAROMETER
▪An instrument that is used to measure atmospheric
pressure
▪The most common types are mercury barometer and
the aneroid barometer.
oA mercury barometer is made by inverting a
mercury-filled tube in a container of mercury. The
pressure at the top of the mercury barometer will
be the vapor pressure of mercury, which is very
small: Pv = 2.4x10−6 atm at 20 °C.
oAn aneroid barometer works mechanically. An
aneroid is an elastic bellows that has been tightly
sealed after some air was removed.
BOURDON-TUBE GAGE
▪A Bourdon-tube gage measures pressure by sensing
the deflection of a coiled tube. The tube has an
elliptical cross section and is bent into a circular arc.
When atmospheric pressure (zero gage pressure)
prevails, the tube is undeflected, and for this
condition the gage pointer is calibrated to read zero
pressure. The Bourdon-tube gage is common
because it is low cost, reliable, easy to install, and
available in many different pressure ranges. There are
disadvantages: dynamic pressures are difficult to
read accurately; accuracy of the gage can be
lower than other instruments; and the gage can be
damaged by excessive pressure pulsations.

14. DEVICES FOR MEASURING PRESSURE
BOURDON-TUBE GAGE
▪A Bourdon-tube gage measures pressure by sensing
the deflection of a coiled tube. The tube has an
elliptical cross section and is bent into a circular arc.
When atmospheric pressure (zero gage pressure)
prevails, the tube is undeflected, and for this
condition the gage pointer is calibrated to read zero
pressure. The Bourdon-tube gage is common
because it is low cost, reliable, easy to install, and
available in many different pressure ranges. There are
disadvantages: dynamic pressures are difficult to
read accurately; accuracy of the gage can be
lower than other instruments; and the gage can be
damaged by excessive pressure pulsations.
BAROMETER
▪An instrument that is used to measure atmospheric
pressure
▪The most common types are mercury barometer and
the aneroid barometer.
oA mercury barometer is made by inverting a
mercury-filled tube in a container of mercury. The
pressure at the top of the mercury barometer will
be the vapor pressure of mercury, which is very
small: Pv = 2.4x10−6 atm at 20 °C.
oAn aneroid barometer works mechanically. An
aneroid is an elastic bellows that has been tightly
sealed after some air was removed.
PIEZOMETER
▪A piezometer is a vertical tube, usually transparent, in
which a liquid rise in response to a positive gage pressure.
Pressure in the pipe pushes the water column to a height
h, and the gage pressure at the center of the pipe is P=γh.
Advantages
✓Simplicity direct measurement (no need for calibration)
✓Accuracy
Disadvantages
✓Cannot easily be used for measuring pressure in a gas
✓Limited to low pressure because the column height
becomes too large at high pressures

15. DEVICES FOR MEASURING PRESSURE
PIEZOMETER
▪A piezometer is a vertical tube, usually transparent, in
which a liquid rise in response to a positive gage pressure.
Pressure in the pipe pushes the water column to a height
h, and the gage pressure at the center of the pipe is P=γh.
Advantages
✓Simplicity direct measurement (no need for calibration)
✓Accuracy
Disadvantages
✓Cannot easily be used for measuring pressure in a gas
✓Limited to low pressure because the column height
becomes too large at high pressures
BOURDON-TUBE GAGE
▪A Bourdon-tube gage measures pressure by sensing
the deflection of a coiled tube. The tube has an
elliptical cross section and is bent into a circular arc.
When atmospheric pressure (zero gage pressure)
prevails, the tube is undeflected, and for this
condition the gage pointer is calibrated to read zero
pressure. The Bourdon-tube gage is common
because it is low cost, reliable, easy to install, and
available in many different pressure ranges. There are
disadvantages: dynamic pressures are difficult to
read accurately; accuracy of the gage can be
lower than other instruments; and the gage can be
damaged by excessive pressure pulsations.
MANOMETER
▪A manometer, often shaped like the letter “U”, is a
device for measuring pressure by raising or lowering a
column or liquid.
▪Figure shows that positive gage pressure in the pipe
pushes the manometer liquid up at height Δh. To use
a manometer, engineers relate the height of the
liquid in the manometer to pressure.

16. DEVICES FOR MEASURING PRESSURE
MANOMETER
▪A manometer, often shaped like the letter “U”, is a
device for measuring pressure by raising or lowering a
column or liquid.
▪Figure shows that positive gage pressure in the pipe
pushes the manometer liquid up at height Δh. To use
a manometer, engineers relate the height of the
liquid in the manometer to pressure.
PIEZOMETER
▪A piezometer is a vertical tube, usually transparent, in
which a liquid rise in response to a positive gage pressure.
Pressure in the pipe pushes the water column to a height
h, and the gage pressure at the center of the pipe is P=γh.
Advantages
✓Simplicity direct measurement (no need for calibration)
✓Accuracy
Disadvantages
✓Cannot easily be used for measuring pressure in a gas
✓Limited to low pressure because the column height
becomes too large at high pressures
SAMPLE PROBLEM 1
If the pressure in the air space above an oil (s = 0.75) surface in a closed
tank is 115 kPa absolute, what is the gage pressure 2 m below the surface?

17. SAMPLE PROBLEM 1
If the pressure in the air space above an oil (s = 0.75) surface in a closed
tank is 115 kPa absolute, what is the gage pressure 2 m below the surface?
P=P
surface+γh P
surface=115 −101.325
P=13.675+(9.81)(0.75)(2)
DEVICES FOR MEASURING PRESSURE
MANOMETER
▪A manometer, often shaped like the letter “U”, is a
device for measuring pressure by raising or lowering a
column or liquid.
▪Figure shows that positive gage pressure in the pipe
pushes the manometer liquid up at height Δh. To use
a manometer, engineers relate the height of the
liquid in the manometer to pressure.
P
surface=13.675 kPa gage

18. SAMPLE PROBLEM 1
If the pressure in the air space above an oil (s = 0.75) surface in a closed
tank is 115 kPa absolute, what is the gage pressure 2 m below the surface?
P=P
surface+γh P
surface=115 −101.325
P
surface=13.675 kPa gageP=13.675+(9.81)(0.75)(2)
P=28.39 kPa
SAMPLE PROBLEM 2
A pressure gage 6 m above the bottom of the tank containing a liquid
reads 90 kPa. Another gage height 4 m reads 103 kPa. Determine the
specific weight of the liquid.

19. SAMPLE PROBLEM 2
A pressure gage 6 m above the bottom of the tank containing a liquid
reads 90 kPa. Another gage height 4 m reads 103 kPa. Determine the
specific weight of the liquid.
P
2−P
1=γh
2 m
4 m
90 kPa
103 kPa
1
2
103−90=γ(2)

20. SAMPLE PROBLEM 2
A pressure gage 6 m above the bottom of the tank containing a liquid
reads 90 kPa. Another gage height 4 m reads 103 kPa. Determine the
specific weight of the liquid.
P
2−P
1=γh
2 m
4 m
90 kPa
103 kPa
1
2
103−90=γ(2)
γ=6.5 kN/m
3
SAMPLE PROBLEM 3
An open tank contains 5.8 m of water covered with 3.2 m of kerosene (γ=
8 kN/m
3
). Find the pressure at the interface and at the bottom of the tank.

21. SAMPLE PROBLEM 3
An open tank contains 5.8 m of water covered with 3.2 m of kerosene (γ=
8 kN/m
3
). Find the pressure at the interface and at the bottom of the tank.
P
A= γ
kh
k
A
B
3.2 m
Kerosene
γ=8 kN/m
3
5.8 m
Water
P
A=(8)(3.2)
P
A=25.6 kPa
P
B=σγh
P
B=9.815.8+(8)(3.2)
SAMPLE PROBLEM 2
A pressure gage 6 m above the bottom of the tank containing a liquid
reads 90 kPa. Another gage height 4 m reads 103 kPa. Determine the
specific weight of the liquid.
P
2−P
1=γh
2 m
4 m
90 kPa
103 kPa
1
2
103−90=γ(2)
γ=6.5 kN/m
3

22. SAMPLE PROBLEM 3
An open tank contains 5.8 m of water covered with 3.2 m of kerosene (γ=
8 kN/m
3
). Find the pressure at the interface and at the bottom of the tank.
P
A= γ
kh
k
A
B
3.2 m
Kerosene
γ=8 kN/m
3
5.8 m
Water
P
A=(8)(3.2)
P
A=25.6 kPa
P
B=σγh
P
B=9.815.8+(8)(3.2)
P
B=82.498 kPa
SAMPLE PROBLEM 4
A barometer reads 760 mmHg and a pressure gage attached to a tank
reads 850 cm of oil (sg = 0.80). What is the absolute pressure in the tank in
kPa?

23. SAMPLE PROBLEM 4
A barometer reads 760 mmHg and a pressure gage attached to a tank
reads 850 cm of oil (sg = 0.80). What is the absolute pressure in the tank in
kPa?
P
abs=P
atm+P
gage
P
abs=9.8113.60.76+(9.81)(0.8)(8.5)
SAMPLE PROBLEM 3
An open tank contains 5.8 m of water covered with 3.2 m of kerosene (γ=
8 kN/m
3
). Find the pressure at the interface and at the bottom of the tank.
P
A= γ
kh
k
A
B
3.2 m
Kerosene
γ=8 kN/m
3
5.8 m
Water
P
A=(8)(3.2)
P
A=25.6 kPa
P
B=σγh
P
B=9.815.8+(8)(3.2)
P
B=82.498 kPa

24. SAMPLE PROBLEM 4
A barometer reads 760 mmHg and a pressure gage attached to a tank
reads 850 cm of oil (sg = 0.80). What is the absolute pressure in the tank in
kPa?
P
abs=P
atm+P
gage
P
abs=9.8113.60.76+(9.81)(0.8)(8.5)
P
abs=168.1 kPa abs
SAMPLE PROBLEM 5
Piston A has a cross-section of 1200 sq. cm. while that of piston B is 950 sq.
cm. with the latter higher than piston A by 1.75 m. If the intervening
passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B?

25. SAMPLE PROBLEM 5
Piston A has a cross-section of 1200 sq. cm. while that of piston B is 950 sq.
cm. with the latter higher than piston A by 1.75 m. If the intervening
passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B?
1200 sq. cm 950 sq. cm
SAMPLE PROBLEM 4
A barometer reads 760 mmHg and a pressure gage attached to a tank
reads 850 cm of oil (sg = 0.80). What is the absolute pressure in the tank in
kPa?
P
abs=P
atm+P
gage
P
abs=9.8113.60.76+(9.81)(0.8)(8.5)
P
abs=168.1 kPa abs

26. SAMPLE PROBLEM 5
Piston A has a cross-section of 1200 sq. cm. while that of piston B is 950 sq.
cm. with the latter higher than piston A by 1.75 m. If the intervening
passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B?
B
A
1.75 m
1200 sq. cm 950 sq. cm

27. SAMPLE PROBLEM 5
Piston A has a cross-section of 1200 sq. cm. while that of piston B is 950 sq.
cm. with the latter higher than piston A by 1.75 m. If the intervening
passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B?
P
A−P
B=γ
oh
o
P
A−P
B=(9810)(0.8)(1.75)
B
A
1.75 m
Oil
sg = 0.8
1200 sq. cm 950 sq. cm

28. SAMPLE PROBLEM 5
Piston A has a cross-section of 1200 sq. cm. while that of piston B is 950 sq.
cm. with the latter higher than piston A by 1.75 m. If the intervening
passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B?
P
A−P
B=γ
oh
o
P
A−P
B=(9810)(0.8)(1.75)
P
A−P
B=13734 Pa
B
A
1.75 m
Oil
sg = 0.8
1200 sq. cm 950 sq. cm
SAMPLE PROBLEM 6
Determine the pressure gage at A in the given open-type manometer

29. SAMPLE PROBLEM 6
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Determine the pressure gage at A in the given open-type manometer
P
A= σγh
0+13.69.811.6+13.69.81??????−13.69.81x
-9.810.3
2 m
1.6 m
0.7 m
water
mercury
SAMPLE PROBLEM 7
Determine the value of y in the manometer shown in the Figure.
SAMPLE PROBLEM 5
Piston A has a cross-section of 1200 sq. cm. while that of piston B is 950 sq.
cm. with the latter higher than piston A by 1.75 m. If the intervening
passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B?
P
A−P
B=γ
oh
o
P
A−P
B=(9810)(0.8)(1.75)
P
A−P
B=13734 Pa
B
A
1.75 m
Oil
sg = 0.8
1200 sq. cm 950 sq. cm

30. SAMPLE PROBLEM 7
Determine the value of y in the manometer shown in the Figure.
SAMPLE PROBLEM 6
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�=���.�� ??????????????????
Determine the pressure gage at A in the given open-type manometer
P
A= σγh
0+13.69.811.6+13.69.81??????−(13.6)(9.81)(x)
2 m
1.6 m
0.7 m
water
mercury

31. SAMPLE PROBLEM 7
0=γ
w sg
hg h − γ
w h −γ
w sg
o h−P
0=9.8113.6y−9.811.5−9.810.83−5
Oil
sg = 0.8
Water
Air, 5 kPa1 m
3 m
1 m
0.5 m y
B
A
Determine the value of y in the manometer shown in the Figure.

32. SAMPLE PROBLEM 7
Oil
sg = 0.8
Water
Air, 5 kPa1 m
3 m
1 m
0.5 m y
B
A
Determine the value of y in the manometer shown in the Figure.
0=γ
w sg
hg h − γ
w h −γ
w sg
o h−P
0=9.8113.6y−9.811.5−9.810.83−5
??????=�.��� ??????
SAMPLE PROBLEM 8
For the manometer shown, determine the pressure at the center of the
pipe.

33. SAMPLE PROBLEM 8
P
B=
0+9.8113.551+9.81(0.8)(1.5)
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�=���.�� ??????????????????
For the manometer shown, determine the pressure at the center of the
pipe.
0+γ
w sg
Hg h+γ
w sg
oil h
1 m
1.5 m
Mercury
sg = 13.55
Oil
sg = 0.80
SAMPLE PROBLEM 7
Oil
sg = 0.8
Water
Air, 5 kPa1 m
3 m
1 m
0.5 m y
B
A
Determine the value of y in the manometer shown in the Figure.
0=γ
w sg
hg h − γ
w h −γ
w sg
o h−P
0=9.8113.6y−9.811.5−9.810.83−5
??????=�.��� ??????
SAMPLE PROBLEM 9
A closed cylindrical tank contains 2 m of water, 3 m of oil (s = 0.82) and the
air above the oil has a pressure of 30 kPa. If an open mercury manometer
at the bottom of the tank has 1 m of water determine the deflection of
mercury.

34. SAMPLE PROBLEM 9
P=
30+9.810.823+9.812+9.811
−9.8113.6y=0
A closed cylindrical tank contains 2 m of water, 3 m of oil (s = 0.82) and the
air above the oil has a pressure of 30 kPa. If an open mercury manometer
at the bottom of the tank has 1 m of water determine the deflection of
mercury.
P
a+γ
w sg
oil h+γ
w h
1+γ
w h
2−γ
w sg
Hg y
??????=�.��� ??????
2 m
3 m
Oil
s = 0.82
Pa = 30 kPa
1 m
y
Water
SAMPLE PROBLEM 8
P
B=
0+9.8113.551+9.81(0.8)(1.5)
??????
�=���.�� ??????????????????
For the manometer shown, determine the pressure at the center of the
pipe.
0+γ
w sg
Hg h+γ
w sg
oil h
1 m
1.5 m
Mercury
sg = 13.55
Oil
sg = 0.80