Fluid mechanics 244 - Chapter 3: Fluid statics
Prof. Ryno Laubscher
Stellenbosch University, South Africa
TERM 3 - 2022
Prof. Ryno Laubscher
Fluid mechanics 244 - Chapter 3: Fluid statics
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Pressure
Pressure measurement equipment
Introduction to fluid statics
Hydrostatic forces on submerged plane surfaces
Hydrostatic forces on curved surfaces
Multi-layered hydrostatic forces
Buoyancy and stability
Fluids in rigid body motion
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Pressure
Pressure is defined as the normal force exerted by a fluid per unit area.
Pressure P is typically associated with liquids and gases, and its
counter part in solid mechanics is stress σ.
Pressure can be defined as absolute or gage. Absolute pressure is
measured relative to a pure vacuum, whereas gage pressure is
measured relative to atmospheric pressure.
Pgage = Pabs − Patm
(1)
Vacuum pressure is typically measured as the difference between the
absolute pressure of the system and the atmospheric pressure.
Pvac = Patm − Pabs
(2)
For engineering analysis and fluid properties we always use absolute
pressure.
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Pressure
Pressure at a point
Pressure is a compressive force per unit area.
Pressure is a scalar quantity, because the pressure in each direction
has the same magnitude.
Let’s prove this. Consider a fluid wedge as seen below with a unit
meter depth in the y-direction (into the slide).
Figure: Fluid wedge
Prof. Ryno Laubscher
Force balance for stationary fluid:
!
!
Fx = max = 0,
Fz = maz = 0
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Pressure
!
!
Fx = 0 : P1 ∆y ∆z − P3 ∆y · l · sin θ = 0
Fz = 0 : P2 ∆y ∆x − P3 ∆y · l · cos θ − 0.5ρ · g · ∆x∆y ∆z = 0
The above simplifies to
P1 − P3 = 0
P2 − P3 − 0.5ρg ∆z = 0
As ∆z → 0: P1 = P2 = P3 = P, thus the pressure in all directions at a
point is equal, therefore, the compressive force has no direction and is a
scalar quantity.
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Pressure
Variations of pressure with depth
As the weight of a fluid column increases with depth, the pressure at
the bottom of the column increases.
Figure: Depiction of fluid pressure increase with depth
Let’s find a mathematical expression to calculate the change in
pressure with depth. Consider a small 1D fluid element.
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Pressure
Figure: 1D fluid element (∆x = ∆y = 1m)
Assume fluid element is at rest (acceleration = 0.0 m/s 2 ), thus the force
balance in the z-direction can be taken as (up is positive z direction)
!
Fz = 0 : P(z1 ) · ∆x − ρ · ∆x · ∆z · g − P(z2 ) · ∆x = 0
P(z1 ) · ∆x − ρ · ∆x · ∆z · g − P(z1 + ∆z) · ∆x = 0
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We can use a Taylor series expansion about the function P(z1 ), to find a
expression of pressure at P(z1 + ∆z):
P(z1 + ∆z) = P(z1 ) + ∆z ·
dP
+ ...
dz
Substituting the above into the force balance yields
"
#
dP
P(z1 ) · ∆x − ρ · ∆x · ∆z · g − P(z1 ) + ∆z ·
· ∆x = 0
dz
dP
= −ρg
dz
$ 2
$ 2
dP = −
ρgdz
1
1
$ 2
P2 − P1 = −
ρgdz
1
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Pressure
For liquids we assume the density is constant, therefore the above
integral evaluates to
P2 − P1 = −ρg (z2 − z1 )
For ideal gases (such as air), the following integral can be evaluated
dP
= −ρg = fracPRT g
dz
$ 2
$
dP
P2
−g 2 dz
= ln( ) =
P1
R 1 T
1 P
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Pressure
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Pressure measurement equipment
Barometer
Barometer is used to measure atmospheric pressure.
It uses the expression P2 = P1 + ρg (z1 − z2 ) (2 is below 1) to
calculate the atmospheric pressure.
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Pressure measurement equipment
Manometer
It is seen from the above derivations, that a change in height ∆z
corresponds to ∆P/ρg .
Thus this principle can be used, similar to a barometer, to measure
pressure differences.
A manometer is a device used to measure pressure differences, e.g.
∆P = Pdevice − Patm .
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Worked example - Manometer
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Intro to fluid statics
Fluid statics deals with problems where the fluids are at rest. E.g.
dams and truck containers.
For fluid statics we are only concerned with the normal stresses acting
on surfaces, thus only the fluid pressures.
Seeing as the fluid is at rest there are no viscous stresses between
fluid elements.
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Hydrostatic forces on submerged plane surfaces
For submerged planes, the pressure due to the weight of the fluid on
the surfaces results in a system of parallel forces.
In fluid statics, we are concerned with the resultant force and its point
of application. The point of application is called the centre of
pressure.
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Hydrostatic forces on plane surfaces
A mathematical expression to calculate the depth to any point on the
tilted plane is
h = y · sin θ
where y is the length measured from the origin at the fluid surface level
parallel to the plane. The expression for the hydrostatic pressure at any
point on the plane is thus
P(y ) = P0 + ρ · g · y · sin θ
The resultant force is then the value equal to the pressure integrated over
the plane surface
$
$
FR =
P(y )dA =
(P0 + ρ · g · y sin θ) dA
A
A
$
FR = P0 A + ρ · g · sin θ ydA
A
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Hydrostatic forces on plane surfaces
%
In the equation above A ydA is the first moment of inertia and is related
to the area centroid y-coordinate yc by
$
1
yc =
ydA
A A
Substituting the above into the expression for the resultant force FR gives
FR = (P0 + ρ · g · yc sin θ) · A
FR = (P0 + ρ · g · hc ) · A = Pc A
where hc = yc sin θ and Pc = P0 + ρ · hc · g . Now that we have the
resultant force, we need to find the line of action. The line of action does
not pass through the area centroid but rather the centre of pressure.
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Hydrostatic forces on plane surfaces
To find the centre of pressure yP , by taking the moment about the origin
point O.
$
yP F R =
y · PdA
A
$
yP F R =
y (P0 + ρ · g · y sin θ) dA
A$
$
yP FR = P0 ydA + ρg sin θ y 2 dA
A
%
A
yP FR = P0 yC A + ρg sin θIxx,0
where Ixx,0 = A y 2 dA is the second moment of area. Typically second
moments of area are given aligned with the area centroid, not the point at
which we calculate the moments.
Ixx,O = Ixx,C + yC2 · A
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Hydrostatic forces on plane surfaces
Thus to find the vertical distance to the centre of pressure
Ixx,C
'
yP = yC + &
0
yC + ρgPsin
θ A
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Hydrostatic forces on plane surfaces
Special cases: Rectangular plates
Let’s consider completely submerged rectangular plate of height b and
width a and tilted at an angle θ from the horizontal plane.
The force acts on the centre of pressure, located beneath the centroid
of the area as
b
ab 3 /12
'
yP = s + + &
2
0
s + b/2 + ρgPsin
θ ab
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Fluid mechanics 244 - Chapter 3: Fluid statics
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Hydrostatic forces on plane surfaces
The resultant force acting at the centre of pressure yP is calculated as
"
"
#
#
b
FR = PC A = P0 + ρg s +
sin θ ab
2
For a vertical plate (θ = 900 ) the resultant force is calculated as
"
"
##
b
FR = P0 + ρg s +
ab
2
For the vertical plate yP = s + b/2.
For a horizontal plate (θ = 1800 ) yP = h. The resultant force is
calculated as
FR = (P0 + ρgh) ab
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Worked example - Submerged plate
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Worked example - Submerged plate
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Hydrostatic forces on curved planes
To determine the hydrostatic forces acting on a 2D curved surface,
we consider the horizontal and vertical force components of the
resultant force vector separately.
This is done by analysing a liquid block enclosing the curved surface
as seen below.
The resultant force is then calculated as FR =
Prof. Ryno Laubscher
Fluid mechanics 244 - Chapter 3: Fluid statics
(
FV2 + FH2 .
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Hydrostatic forces on curved planes
The vertical and horizontal force components are calculated as
"
"
##
b
FH = Fx = P0 + ρg s +
ab
2
FV = Fy ± W = (P0 + ρgh) ab ± ρVblock
where W = ρVblock is the weight of the liquid block.
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Multi-layered hydrostatic forces
For multilayered fluids of different densities, the separate resultant
forces are calculated for each liquid layer and then summed as shown
below
!
!
FR =
FR,i =
PC ,i Ai
The centre of pressure can be found by performing a moment balance
using the resultant force about any point and the individual moments
exerted by the individual liquid layer forces.
!
F R yP =
FR,i yPi
(3)
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Worked example - Hydrostatic forces on curved surfaces
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Worked example - Hydrostatic forces on curved surfaces
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Buoyancy and stability
Buoyancy
When objects are placed in a liquid it experiences an upward force,
called the buoyancy force.
E.g. wood floating on water.
The buoyancy force is due to the increase in pressure with depth. The
difference between the downward and upward fluid pressures creates
the buoyant force.
Consider a submerged horizontal plate, as seen below.
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Buoyancy and stability
A force balance in the vertical direction yields
ρf · g · s · A + ρs Vbody · g = ρf · g · (s + h) · A
ρs Vbody g = ρf g Vsubmerged
In the above equation FB = ρf g Vsubmerged is the buoyant force
experience by the object.
Stability - self-study
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Worked example - Buoyancy
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Worked example - Buoyancy
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Fluids in rigid body motion
In this subsection, we will consider changes in fluid pressure due to
movement of the fluid container.
E.g. pressure variations of fuel in a truck container, you running with
a glass of milk.
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Fluids in rigid body motion
Rigid body assumption ← assume no relative movement between fluid
particles, therefore, no viscous stresses exist. All fluid particles will have
the same acceleration.
Let’s consider a small differential rectangular element and associated
surface pressure forces and weight-induced body force. Assume the
pressure inside the entire fluid element is constant and is evaluated at
its centre as P(x, z).
From Newton’s 2nd law we know that
!
F̄ = m · ā
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Fluids in rigid body motion
The forces acting on the fluid element can be divided into pressure
and body forces
!
!
!
F̄ =
F̄S +
F̄B
)
)
)
)
)
)
where
F̄S =
FS,x i¯ + FS,z k̄ and
F̄B =
FB,x i¯ + FB,z k̄.
Let’s look at the surface forces first. The forces acting in the
x-direction are the two compressive pressure forces at x − δx/2 and
x + δx/2
!
FS,x = P(x −
δx
δx
) · δz − P(x + ) · δz
2
2
Using Taylor series expansion about function P(x, z) we can evaluate
the pressure at the left and right faces as
!
Prof. Ryno Laubscher
FS,x = (P −
∂P δx
∂P δx
) · δz − (P +
)·δ
∂x 2
∂x 2
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Fluids in rigid body motion
The above simplifies to
!
FS,x = −
!
FS,z = −
∂P
δxδz
∂x
Similarly in the z-direction, the surfaces forces simplifies to
∂P
δxδz
∂z
Therefore, the total surface force vector for the entire fluid element
can be written as
"
#
!
∂P ¯ ∂P
F̄S = −
i+
k̄ δxδz
∂x
∂z
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Fluids in rigid body motion
For the body forces, there is no force in the x-direction, seeing as the
gravitational force is selected in the z-direction, thus
!
!
F̄B =
FB,z k̄
!
FB,z = −ρg δxδz
The total force vector acting on the fluid element is
"
#
!
∂P ¯
∂P
F̄ = −
i +(
+ ρg )k̄ δxδz
∂x
∂z
or in shorthand vector notation
!
F̄ = −(∇P + ρg k̄)δxδz
∂ ¯
Note: ∇ = ( ∂x
i+
∂ ¯
∂
∂y j + ∂z k̄)
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Fluids in rigid body motion
Inserting the above force vector into the equation of motion yields
−(∇P + ρg k̄)δxδz = ρδxδz · ā
or in expanded vector notion
∂P ¯ ∂P
i+
k̄ + ρg k̄ = −ρ(ax i¯ + az k̄)
∂x
∂z
thus
∂P
∂x
= −ρax and
Prof. Ryno Laubscher
∂P
∂z
= −ρ(az + g )
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Fluids in rigid body motion
Special cases: rigid body motion
Fluids at rest:
∂P
∂P
∂P
= 0,
= 0,
= −ρg
∂x
∂y
∂z
Free falling fluids:
∂P
∂P
∂P
= 0,
= 0,
=0
∂x
∂y
∂z
thus P constant everywhere in the fluid.
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Fluids in rigid body motion
Acceleration along a straight path
For acceleration along a straight path the pressure gradients in each
direction are
∂P
= −ρax
∂x
∂P
= −ρay
∂y
∂P
= −ρ(az + g )
∂z
In there is no movement in the y-direction the container moves in a
2D plane and ∂P
∂y = 0.
Thus, pressure is only a function of x, z, P = P(x, z). The derivative
of dP is then (using chain rule)
∂P
∂P
dP =
· dx +
· dz
∂x
∂z
which give
dP = −ρax · dx − ρ(az + g ) · dz
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Fluid mechanics 244 - Chapter 3: Fluid statics
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Fluids in rigid body motion
Integrating the above expression while assuming ρ = constant, yields
P2 − P1 = −ρax (x2 − x1 ) − ρ(g + az )(z2 − z1 )
Using the above equation the pressure difference between any two
points in a fluid container can be calculated.
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Fluid mechanics 244 - Chapter 3: Fluid statics
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Fluids in rigid body motion
To find the vertical rise or drop of the fluid free surface, the above
equation is used, by picking P1 and P2 to be at the free surface.
z2 − z1 = −
ax
· (x2 − x1 )
g + ag
From the above expression the slope of the liquid level can be
determined as
Slope =
z2 − z1
dz
−ax
=
= − tan θ =
x2 − x1
dx
g + az
Rotation in cylindrical container - self study.
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Worked example - Rigid body motion
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Worked example - Rigid body motion
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