Republic of the Philippines
TARLAC STATE UNIVERSITY
College of Engineering and Technology
CEP 372: FLUID MECHANICS LECTURE
MODULE 2: PRINCIPLES OF HYDROSTATIC PRESSURES
Prepared By: Engr. Karl Angelo G. Clarete
2.1 INTRODUCTION
a.) PRESSURE: refers to the effect of a pressure
acting against and distributed over a surface. They
may be created by a solid, liquid or gas. It is the
force per unit area exerted by a liquid or gas on a
body or surface with the force acting at right angles
to the surface uniformly in all directions.
The pressure at any point in a fluid at rest is equal to
its specific weight (γ) multiplied by the vertical
depth (h) of the point from the surface, or
𝒑 = 𝜸𝒉
where:
p – pressure (Pa, psi, psf)
𝛾 – specific or unit weight (N/m3, lb/ft3)
h – height or depth (m, ft)
If F represents the total force on an area A, and dF is
the elemental force on an elemental area dA, the
intensity of pressure is,
𝒑=
𝒅𝑭
𝒅𝑨
b.) PASCAL’S LAW: developed by French
mathematician Blaise Pascal. States that, “The
pressure on a fluid is equal in all directions and in
all parts of the container.”
Note: If the pressure is uniform over the area A, then
𝒑=
𝑭
𝑨
Otherwise this formula gives only the average
pressure
𝐬𝐢𝐧 𝜽 =
𝑨𝑨𝑩𝑬𝑭
𝑨𝑨𝑩𝑪𝑫
where:
𝐜𝐨𝐬 𝜽 =
𝑨𝑪𝑫𝑬𝑭
𝑨𝑨𝑩𝑪𝑫
p – pressure (Pa, psi, psf)
F – force (N, lb)
A – cross sectional area (m2, ft2)
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of mercury 1 inch in height at the standard
+
acceleration of gravity.
𝑭𝒙 = 𝟎
Millibar is 1/1000th of a bar and is the amount of
𝑭𝒙 − 𝑭𝒚 𝐬𝐢𝐧 𝜽 = 𝟎
𝑷𝒙 𝑨𝑨𝑩𝑬𝑭 − 𝑷𝒚 𝑨𝑨𝑩𝑪𝑫
𝑨𝑨𝑩𝑬𝑭
= 𝟎
𝑨𝑨𝑩𝑪𝑫
one centimeter, in one second. Millibar values used
in meteorology ranges from about 100 to 1050. At
sea level, standard air pressure in millibars is 1013.2.
𝑷𝒙 = 𝑷𝒚
+
force it takes to move an object weighing a gram,
2.3 GAGE PRESSURE (RELATIVE
𝑭𝒚 = 𝟎
PRESSURE)
a.) GAGE PRESSURE: are pressures above or below
𝑭𝒛 − 𝑭𝒚 𝐜𝐨𝐬 𝜽 = 𝟎
𝑷𝒛 𝑨𝑪𝑫𝑬𝑭 − 𝑷𝒚 𝑨𝑨𝑩𝑪𝑫
𝑨𝑪𝑫𝑬𝑭
= 𝟎
𝑨𝑨𝑩𝑪𝑫
𝑷𝒛 = 𝑷𝒚
∴ 𝑷𝒙 = 𝑷𝒚 = 𝑷𝒛
∴ The pressure on a fluid is equal in all directions
2.2 ATMOSPHERIC PRESSURE
a.) ATMOSPHERIC PRESSURE: is the pressure at
any one point on the earth’s surface from the weight
of the air above it. Refers to the prevailing pressure
in the air around us.
a.1.) STANDARD ATMOSPHERIC PRESSURE
𝑷𝒂𝒕𝒎 = 𝟏 𝒂𝒕𝒎
𝑷𝒂𝒕𝒎 = 𝟏𝟎𝟏. 𝟑𝟐𝟓 𝒌𝑷𝒂
𝑷𝒂𝒕𝒎 = 𝟏𝟒. 𝟕 𝒑𝒔𝒊
𝑷𝒂𝒕𝒎 = 𝟐𝟏𝟔𝟔 𝒍𝒃/𝒇𝒕𝟐
𝑷𝒂𝒕𝒎 = 𝟕𝟔 𝒄𝒎 𝒐𝒇 𝑯𝒈
𝑷𝒂𝒕𝒎 = 𝟐𝟗. 𝟗 𝒊𝒏 𝒐𝒇 𝑯𝒈
𝑷𝒂𝒕𝒎 = 𝟏𝟎𝟏𝟑 𝒎𝒊𝒍𝒍𝒊𝒃𝒂𝒓𝒔
Note:
Centimeters of Mercury is a small pressure unit
which represents the pressure pushing down due to
gravity of any volume of liquid mercury which is 1
the atmosphere and can be measured by pressure
gauges or manometers.
b.) VACUUM PRESSURE: A space that has all
matter removed from it. Can also be described as a
region of space where the pressure is less than the
normal atmospheric pressure of 760 mm (29.9 in) of
mercury. It is the pressure below the atmospheric
pressure and is used when the gage pressure is
negative.
2.4 ABSOLUTE PRESSURE
a.) ABSOLUTE PRESSURE: Is the pressure above
absolute zero. Absolute pressure can never be
negative and absolute zero is obtained if all air is
removed. It is the lowest possible pressure
attainable.
𝑷𝒂𝒃𝒔 = 𝑷𝒈𝒂𝒈𝒆 + 𝑷𝒂𝒕𝒎
where:
𝑃
– absolute pressure (Pa, psi, psf)
𝑃
– gage pressure (Pa, psi, psf)
𝑃
– atmospheric pressure (Pa, psi, psf)
cm high.
Inch of mercury is a non-SI unit of measurement for
pressure. It is used for barometric pressure in
weather reports, refrigeration and aviation in the
United States. It is the pressure exerted by a column
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2.5 EQUIPMENT FOR PRESSURE
MEASUREMENT
a.) MERCURY BAROMETER: is an accurate and
relatively simple way to measure changes in
c.) MANOMETER: is a tube, usually bent in a form
of U, containing a liquid of known specific gravity,
the surface of which moves proportionally to
changes of pressure. It is used to measure pressure.
atmospheric pressure. At sea level, the weight of the
atmosphere forces mercury 760 mm (29.9 in) goes up
a calibrated glass tube. Higher elevations yield
lower readings because the atmosphere is less dense
there, and the thinner air exerts less pressure on the
mercury.
2.6 TYPES OF MANOMETER
a.) OPEN TYPE: has an atmospheric surface in one
leg or two legs and is capable of measuring gage
pressures.
b.) ANEROID BAROMETER: In an aneroid
barometer, a partially evacuated metal drum
expands or contracts in response to changes in air
pressure. A series of levers and springs translates
the up and down movement of the drum top into
the circular motion of the pointers along the aneroid
barometer’s face.
b.) DIFFERENTIAL TYPE: without an atmospheric
surface and capable of measuring only differences of
pressure.
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c.) PIEZOMETER: the simplest form of open
at rest, all forces acting upon it must be in
manometer. It is a tube tapped into a wall of a
equilibrium.
container or conduit for the purpose of measuring
FLS
pressure. The fluid in the container or conduit rises
in this tube to form a free surface.
c.1.) LIMITATIONS OF PIEZOMETER


Note:
Large pressures in the lighter liquids require
FREE LIQUID SURFACE (FLS): refers to a liquid
long tubes.
surface subject to zero gage pressure or with
Gas pressures cannot be measured because
atmospheric pressure only.
gas cannot form a free surface.
2.7 STEPS IN SOLVING MANOMETER
PROBLEMS
1.) Decide on the fluid in feet or meter, of which the
heads are to be expressed, (water is the most
With reference to illustration shown,
𝑾 = 𝜸𝑽
𝑾 = 𝜸(𝒂𝑳)
+
𝑭𝒙 = 𝟎
advisable base fluid).
2.) Starting from an end point, number in order, the
interface of different fluids.
3.) Identify points of equal pressure (taking into
account that for a homogeneous fluid at rest, the
pressure along the same horizontal plane are equal).
Label these points with the same number.
4.) Proceed from level to level, adding (if going
down) or subtracting (if going up) pressure heads as
the elevation decreases or increases, respectively
with due regard of the specific gravity of the fluids.
𝑭𝟐 − 𝑭𝟏 = 𝑾𝒔𝒊𝒏𝜽
𝑷𝟐 𝒂 − 𝑷𝟏 𝒂 = 𝜸(𝒂𝑳)𝒔𝒊𝒏𝜽
𝑷𝟐 − 𝑷𝟏 = 𝜸𝑳𝒔𝒊𝒏𝜽
𝒃𝒖𝒕 𝑳𝒔𝒊𝒏𝜽 = 𝒉
𝑷𝟐 − 𝑷𝟏 = 𝜸𝒉
b.) PRESSURE POINTS LIE ON THE SAME
ELEVATION: Considering that two points (1 and 2)
lie on the same elevation, such that ℎ = 0, then:
FLS
2.8 VARIATIONS IN PRESSURE
a.) PRESSURE POINTS DO NOT LIE ON THE
SAME ELEVATION: Considering any two points (1
and 2), whose difference in elevation is ℎ, to lie in
the ends of an elementary prism having a cross
sectional area 𝑎 and a length of 𝐿. Since the prism is
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With reference to illustration shown,
𝛾 – specific or unit weight (N/m3, lb/ft3)
a.1.) TO CONVERT PRESSURE HEAD OF
+
LIQUID A TO LIQUID B:
𝒔𝑨
𝝆𝑨
𝜸𝑨
𝒉𝑩 = 𝒉𝑨
𝑜𝑟 𝒉𝑩 = 𝒉𝑨
𝑜𝑟 𝒉𝑩 = 𝒉𝑨
𝒔𝑩
𝝆𝑩
𝜸𝑩
𝑭𝒙 = 𝟎
𝑭𝟐 − 𝑭𝟏 = 𝟎
𝑷𝟐 𝒂 − 𝑷 𝟏 𝒂 = 𝟎
(𝒂)(𝑷𝟐 − 𝑷𝟏 ) = 𝟎
𝑷𝟐 − 𝑷𝟏 = 𝟎
𝑷𝟐 = 𝑷𝟏
c.) PRESSURE BELOW LAYERS OF DIFFERENT
LIQUIDS: Consider the tank shown to be filled with
liquids of different densities and with air at the top
under a gage pressure of 𝑝 , the pressure at the
bottom of the tank is:
a.2.) TO CONVERT PRESSURE HEAD OF ANY
LIQUID TO WATER:
𝒉𝒘𝒂𝒕𝒆𝒓 𝒔𝒘𝒂𝒕𝒆𝒓 = 𝒉𝒍𝒊𝒒𝒖𝒊𝒅 𝒔𝒍𝒊𝒒𝒖𝒊𝒅
𝒔𝒘𝒂𝒕𝒆𝒓 = 𝟏
𝒉𝒘𝒂𝒕𝒆𝒓 = 𝒉𝒍𝒊𝒒𝒖𝒊𝒅 𝒔𝒍𝒊𝒒𝒖𝒊𝒅
2.10 LIQUID PRESSURE IN A
HYDRAULIC SYSTEM
In a hydraulic system, fluid is confined to two
chambers. Each chamber has a piston that is free to
move.
𝑷𝒃𝒐𝒕𝒕𝒐𝒎 =
𝑷
𝑷=
𝑷𝒃𝒐𝒕𝒕𝒐𝒎 = 𝑷𝒂𝒊𝒓 + 𝜸𝟏 𝒉𝟏 + 𝜸𝟐 𝒉𝟐 + 𝜸𝟑 𝒉𝟑
2.9 PRESSURE HEAD
a.) PRESSURE HEAD: Is the height ℎ of a column of
homogenous liquid of unit weight γ that will
produce an intensity of pressure 𝑃.
From the formula for pressure,
𝑷 = 𝜸𝒉
𝑷
𝒉=
𝜸
where:
𝑷𝟏 =
𝑭
𝑨
𝑭𝟏
𝑭𝟐
𝑎𝑛𝑑 𝑷𝟐 =
𝑨𝟏
𝑨𝟐
According to Pascal’s Law:
𝑷𝟏 = 𝑷𝟐
𝑭𝟏
𝑭𝟐
=
𝑨𝟏
𝑨𝟐
where:
𝑃𝑖𝑠𝑡𝑜𝑛 1 − 𝑀𝐴𝑆𝑇𝐸𝑅 𝐶𝑌𝐿𝐼𝑁𝐷𝐸𝑅
𝑃𝑖𝑠𝑡𝑜𝑛 2 − 𝑆𝐿𝐴𝑉𝐸 𝐶𝑌𝐿𝐼𝑁𝐷𝐸𝑅
h – pressure head (m, ft)
p – pressure (Pa, psi, psf)
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2.11 SAMPLE PROBLEMS
EXAMPLE #1: In a depth of liquid of 1 m causes a
pressure of 7 kPa, what is the specific gravity of the
liquid?
EXAMPLE #7: In the figure shown, if the
atmospheric pressure is 101.03 kPa and the absolute
pressure at the bottom of the tank is 231.3 kPa, what
is the specific gravity of olive oil?
EXAMPLE #2: What is the pressure in kPa 12.5 m
below the ocean? Use specific gravity = 1.03 for
ocean water.
EXAMPLE #3: If the pressure 23 meter below a
liquid is 338.445 kPa, determine its (a) unit weight in
kN/m3, (b) mass density in kg/m3, and (c) specific
gravity.
EXAMPLE #4: If the pressure at a point in the ocean
is 60 kPa, what is the pressure in kPa 27 meters
below this point? Use specific gravity = 1.03 for
ocean water.
EXAMPLE #5: 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 in kPa 2 m below
the surface?
"Motivation is what gets you started.
Habit is what keeps you going." – Jim Ryun
EXAMPLE #6: 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 in kN/m3.
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