03/03/2024
CIV1202 FLUID MECHANICS
Unit 2: Hydrostatics
Eng. Dr. Seith Mugume (PhD, MUIPE, REng)
Lecturer, Department of Civil and Environmental Engineering
College of Engineering, Design, Art and Technology
Makerere University
Email: seith.mugume@mak.ac.ug; smugume@gmail.com
Tel: +256 771 358 124
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Course Content
Unit 2: Hydrostatics
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Pressure in fluids
• Pressure, p, is defined as force per unit area:
𝐹𝑜𝑟𝑐𝑒
𝐹
• p = 𝐴𝑟𝑒𝑎 = 𝐴 [Units – Nm-2 or Pa]
• Atmospheric pressure (1 atm.) is equal to
101,325 Nm-2.
The
fluid
exerts
pressure
internally as well as on the same
container. The internal pressure is
the same in all directions
Unit 2: Hydrostatics
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Pressure in fluids
• In the presence of gravity, pressure
in a static fluid increases with
depth. This allows an upward
pressure force to balance the
downward gravitational force.
• This condition is hydrostatic
equilibrium.
• Incompressible fluids like liquids
have constant density; for them,
pressure as a function of depth h is
p = p0 + ρgh
where p0 = pressure Unit
at2:surface
Hydrostatics
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Pressure & Head
Vertical pressure relationship:
𝑑𝑝
= −𝜌𝑔
𝑑𝑧
Integrating gives:
𝑝 = −𝜌𝑔𝑧 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
If z is measured from the free surface,
𝑧 = −ℎ
 𝑝 = 𝜌𝑔ℎ + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
 The surface pressure is the atmospheric pressure, 𝑝𝑎𝑡𝑚
 𝑝𝑎𝑡𝑚 = constant
 𝑝 = 𝜌𝑔ℎ + 𝑃𝑎𝑡𝑚
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Pressure and head
• It is often convenient to take atmospheric pressure
as the datum. Pressure quoted this way is known as
gauge pressure
𝑃𝑔𝑢𝑎𝑔𝑒 = 𝜌𝑔ℎ
• Lower limit of any pressure is the pressure
measured in perfect vacuum
• Pressure measured above a perfect vacuum (zero)
is known as absolute pressure
𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝜌𝑔ℎ + 𝑃𝑎𝑡𝑚
=> Absolute pressure = (Gauge + Atmospheric)
pressure
Unit 2: Hydrostatics
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Gauge Pressure, p
• Gauge pressure can be given using the height of
any fluid.
𝑝 = 𝜌𝑔ℎ
• Vertical height, h corresponds to the head
• If pressure is quoted as head, then the density of
the fluid should be specified
Example:
What is pressure of 500 kN/m2 in terms of head of:
a)
b)
c)
Water of density, 𝜌 = 1,000 kg/m3
Mercury of density, 𝜌 = 13.6 x 103 kg/m3
A fluid with relative density, 𝛾 = 8.7
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Pressure measurement
• Manometers
– Single tube
– U-tube
• Piezometers
• Bourdon gauges
Unit 2: Hydrostatics
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Pressure measurement
• A manometer is applied to measure
pressure differences.
• Manometers use the relationship between
pressure head to measure pressure.
• Gauge pressure is a measure of pressure
relative to the ambient atmosphere:
p = patm + ρgh
=> Manometer measures gauge pressure
Unit 2: Hydrostatics
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Piezometric tube manometer
• Simplest manometer is an
open tube
• Attached to the top of the
container containing liquid at
pressure
• Tube open to atmosphere
• Measured pressure is relative
to atmospheric pressure =>
Gauge pressure
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Piezometric tube manometer
• Pressure at A, PA
𝑃𝐴 = 𝜌𝑔ℎ1
• Pressure at B, PB
𝑃𝐵 = 𝜌𝑔ℎ2
Problems with Piezometers
a) Can only be used for liquids
b) Pressure must be above atmospheric
c) Liquid height must not be too small or too
large
Unit 2: Hydrostatics
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Pressure measurement
Unit 2: Hydrostatics
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Exercise 2
What is the maximum gauge pressure of
water that can be measured by a piezometer
of height 1.5m? And if the liquid had a
relative density of 8.5, what would be the
maximum gauge pressure?
Solution:
a) Water
𝑃𝑔𝑎𝑢𝑔𝑒 = 𝜌𝑔ℎ = 1000 x 9.81 x 1.5 = 14,715 N/m2 (14.715 kN/m2)
b) Liquid with relative density of 8.5
𝑃𝑔𝑎𝑢𝑔𝑒 = 𝜌𝑔ℎ = 8.5 x 1000 x 9.81 x 1.5 = 125,077.5 N/m2
(125.077 kN/m2)
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Equality of pressure
• Consider a horizontal cylindrical element
• Cross sectional area = A; Mass density = 𝜌; Left end pressure
= pl; Right end pressure = pr
• For equilibrium, the sum of forces in the x direction is zero
• 𝒑𝒍 𝑨 = 𝒑𝒓 𝑨 ⇒ 𝒑𝒍 = 𝒑𝒓 (Horizontal direction  Constant
pressure)
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The “U”-Tube Manometer
•
Enables the pressure of both liquids and gases to be measured
•
“U” tube is connected as shown and filled with the manometric fluid
•
Manometric fluid density should be greater than density of the fluid
being measured 𝜌𝑚𝑎𝑛 > 𝜌
•
The two fluids should not be able to mix  immiscible
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The “U”-tube Manometer
• Pressure in a continuous static fluid is the
same at any horizontal level
• Pressure at B, pB = Pressure at C, pC
• For the left arm;
pB = pA+ pressure of height of liquid measured
𝑝𝐵 = 𝑝𝐴 + 𝜌𝑔ℎ1
• For the right arm;
𝑝𝐶 = 𝑝𝑎𝑡𝑚 + 𝜌𝑚𝑎𝑛 𝑔ℎ2 (omit patm)
𝑝𝐵 = 𝑝𝐶 => 𝑝𝐴 + 𝜌𝑔ℎ1 = 𝜌𝑚𝑎𝑛 𝑔ℎ2
=> 𝒑𝑨 = 𝝆𝒎𝒂𝒏 𝒈𝒉𝟐 − 𝝆𝒈𝒉𝟏
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Gas pressure measurement
• The manometer works in the same way for
gases
• However, the manometric fluid is liquid
(i.e. mercury, oil or water)
• Liquid density is much greater than gas
=> 𝜌𝑚𝑎𝑛 ≫ 𝜌
𝜌𝑔ℎ1 can be neglected
Gauge pressure = 𝝆𝒎𝒂𝒏 𝒈𝒉𝟐
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Exercise 3
Using a “U”-tube manometer to measure
gauge pressure of fluid density 𝜌 =
700 𝑘𝑔/𝑚3 , and the manometric fluid is
mercury, with relatively density of 13.6. What
is the gauge pressure if:
a) h1 = 0.4 m and h2 = 0.9 m ?
b) h1 = 0.4 m and h2 = -0.1 m ?
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Pressure difference
measurement using a “U”-tube
manometer
𝑝𝐶 = 𝑝𝐷
𝑝𝐶 = 𝑝𝐴 + 𝜌𝑔ℎ𝑎
𝑝𝐷 = 𝑝𝐵 + 𝜌𝑔 ℎ𝑏 − ℎ + 𝜌𝑚𝑎𝑛 𝑔ℎ
𝑝𝐴 − 𝑝𝐵
= 𝑝𝑐 − 𝜌𝑔ℎ𝐴 − 𝑝𝐷 + 𝜌𝑔 ℎ𝐵 − ℎ
+ 𝜌𝑚𝑎𝑛 𝑔ℎ
= 𝝆𝒈 𝒉𝑩 − 𝒉𝑨 + (𝝆𝒎𝒂𝒏 − 𝝆)𝒈𝒉
For gases,
𝜌𝑚𝑎𝑛 ≫ 𝜌
𝒑𝑨 − 𝒑𝑩 = 𝝆𝒎𝒂𝒏 𝒈𝒉
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Pressure forces on submerged
bodies
Key static fluid principles already introduced
• Hydrostatic vertical pressure distribution
𝑑𝑝
= −𝜌𝑔
𝑑𝑧
• Pressure at any equal depths in a continuous fluid
are equal
• In an uninterrupted fluid continuum, pressure at
any point acts equally in all directions (Pascal’s
law)
• Forces from a fluid on a boundary act at right
angles to that boundary
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Pressure on plane surfaces
• Consider a solid object
submerged in a liquid
– Liquid applies a pressure force
which is perpendicular to the
surface of the object at all
points
– All other pressure components
are also in equilibrium (unless
if there is relative motion
between the liquid and the
object)
Pressure at a point in a liquid
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Pressure on plane surfaces
Pressure force on a plate (a) Horizontal plate (b) Vertical plate
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(a) Horizontal plate
• Consider a flat plate with width, B, length,
L, suspended in water horizontally at
depth y.
⇒ 𝑝 = 𝜌𝑔𝑦
▪ For y = 2 m, B = 1m, and L = 2.5 m; Calculate
the pressure at depth of 2 m
▪ Calculate the total force, F on the top surface
of the plate
Solution (p = 19,620 N/m2; F = 49,050 N)
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(a) Horizontal plate
• In most cases, engineers find it more
convenient to treat the total (or resultant)
force as if it was concentrated at a point
(as opposed to a distributed load)
• Point => Centre of Pressure or Point of
action
• The resultant force will act through the
centre of pressure, acting a right angles
to the plane through the centre of pressure
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(b) Vertical plate
• Pressure, p increases with depth
• At the water surface, gauge pressure = 0
• At the bottom of the plate, pressure, 𝑝 = 𝜌𝑔𝑦 =
1,000 × 9.81 × 2.5 = 24,525 𝑁
• For linear pressure variation with y,
F = Average pressure x Area
24,525
𝐹=
× 1 × 2.5 = 30,656 𝑁
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Note: Locating the point of action of F is not so easy
as for the horizontal plate problem. A more general
approach is needed
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(b) Vertical plate
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(b) Vertical plate
• Consider a small element 𝛿𝐴. The force, 𝛿𝐹 acting on that
area due to pressure of the liquid, is 𝑝𝛿𝐴.
– 𝑝 is not constant
– For an element at a distance, y below the free surface of the
liquid;
𝑝 = 𝜌𝑔𝑦
𝛿𝐹 = 𝜌𝑔𝑦𝛿𝐴
Therefore, the total force, 𝐹 = 𝜌𝑔 ‫𝐴𝛿𝑦 ׬‬
For 𝑦 = 𝑙𝑠𝑖𝑛𝜃;
𝐹 = 𝜌𝑔 න 𝑙𝑠𝑖𝑛𝜃𝛿𝐴 = 𝜌𝑔(𝑠𝑖𝑛𝜃) න 𝑙𝑑𝐴
‫ >= 𝐴𝑑𝑙 ׬‬geometric characteristic of the shape: First moment of
Area
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(b) Vertical plate
• ‫ = 𝐴𝑑𝑙 ׬‬product of 𝐴𝑙 ҧ
Where A is the area of the plane surface and 𝑙 ҧ is the distance from the
origin O to the centroid of the plane.
 𝐹 = 𝜌𝑔 𝑠𝑖𝑛𝜃 𝐴𝑙 ҧ
• In addition to the magnitude of the force, F, it is also necessary to the
precise location and angle of its line of action
•
– Pressure is distributed over the whole immersed body
– Nonetheless, a concentrated point load is considered
– 𝛿𝐹 produces a moment 𝛿𝐹𝑙 about the origin
– 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝛿𝐹𝑙 = 𝜌𝑔𝑦𝛿𝐴𝑙 = 𝜌𝑔𝑙 𝑠𝑖𝑛𝜃 𝛿𝐴𝑙 = 𝜌𝑔 𝑠𝑖𝑛𝜃 𝑙2 𝛿𝐴
In the limit, 𝑑𝐹𝑙 = 𝜌𝑔 𝑠𝑖𝑛𝜃 𝑙 2 𝑑𝐴
•
Considering the whole surface,
𝑚𝑜𝑚𝑒𝑛𝑡 = 𝐹𝑙 = 𝜌𝑔 𝑠𝑖𝑛𝜃 න 𝑙2 𝑑𝐴
Quantity ‫ 𝑙 ׬‬2 𝑑𝐴 is another geometric characteristic => Second moment of area, I
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• Distance from the origin to the point of action,
• 𝑙′ =
𝑚𝑜𝑚𝑒𝑛𝑡
𝜌𝑔 𝑠𝑖𝑛𝜃 𝐼
𝐼
=
= ҧ
𝑓𝑜𝑟𝑐𝑒
𝜌𝑔 𝑠𝑖𝑛𝜃 𝐴𝑙 ҧ
𝐴𝑙
• I is evaluated using the Parallel axes theorem
𝐼 = 𝐼𝑜 + 𝐴𝑙 2ҧ
Where:
o I is the second moment of area of a plane surface about
origin O
o Io is the second moment of area of the surface about an
axis through its centroid
The axis is parallel to the axis at O. Io is a function of
geometric shape. 𝐴𝑙 2ҧ is the proportion of I due to the
distance from O to the centroid of the surface.
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2nd Moment of Area about a line through
the centroid of some common shapes
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Exercise
A rock fill dam is designed to have a cross section as represented in the figure
below. The reservoir design depth is to be 10 m. Estimate
(a) The forces on the dam per unit width
(b) The location of the centre of pressure
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Pressure forces on Curved
Surfaces
• Not all immersed
structures are flat
• Some dams have an
upstream face that may
be curved in both
vertical and horizontal
directions
• Other types of hydraulic
control structures such
as gates may also be
curved
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Pressure forces on curved
surfaces
• Pressure force is acts perpendicular to the immersed
surface
• Pressure on a curved surface is unevenly distributed
• The force is resolved into horizontal (𝛿𝐹𝑥 ) and vertical
(𝛿𝐹𝑦 )components
• 𝛿𝐹𝑦 = total weight of the volume of the fluid above the
surface => Acts through the centre of gravity
• 𝛿𝐹𝑥 = pressure force on vertical plane surface equal in
height to the projected height of the curved surface =>
Acts at distance 𝑙′ below the free surface of the fluid
• Resultant force acts through the point of intersection of
lines of action W and Fx
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Pressure forces on curved
surfaces
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Exercise 1: Hydrostatic Force on
a Quadrant gate
Determine the magnitude and direction of
the resultant force of water on the quadrant
gate below. The principal dimensions are,
Radius of gate = 1.2 m, Width of gate = 3.5
m, Density of water = 1000 kg/m3. The
position of the centre of gravity is 4R/3π
horizontally from the origin
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Pressure on a quadrant gate
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Exercise 2: Hydraulic forces on a
rockfill dam
The Government of Uganda intends to construct a rockfill
dam across a valley for water supply and irrigation
purposes. The dam has been designed with a cross
section as shown below and reservoir depth is 12 m.
(i)
Calculate the hydraulic forces on the dam.
(ii)
Determine the location of the centre of pressure
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References
Bansal, R. K. (2005). Fluid Mechanics and Hydraulic Machines. New Delhi: Laxmi
Publications Ltd.
Chadwick, A., Morfett, J., & Borthwick, M. (2013). Hydraulics in Civil and
Environmental Engineering (5th Ed.). London and New York: CRC Pres,
Taylor & Francis Group.
Pasche, E. (2007). Fundamentals of Fluid Mechanics. Hamburg: Hamburg
University of Technology.
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