Section 2

SECTION 2: Forces in Static Fluids
Objectives
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Introduce the concept of pressure;
Prove it has a unique value at any particular elevation;
Show how it varies with depth according to the hydrostatic equation and
Show how pressure can be expressed in terms of head of fluid.
2.1. Fluids statics
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•
a static fluid can have no shearing force acting on it, and that
any force between the fluid and the boundary must be acting at right angles to the
boundary.
Fig. 2.1: Pressure force normal to the boundary
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For an element of fluid at rest, the element will be in equilibrium - the sum of the
components of forces in any direction will be zero.
The sum of the moments of forces on the element about any point must also be zero.
It is common to test equilibrium by resolving forces along three mutually perpendicular axes and
also by taking moments in three mutually perpendicular planes and to equate these to zero.
2.2. Pressure
If the force exerted on each unit area of a boundary is the same, the pressure is said to be
uniform.
pressure =
Force
Area over which the force is applied
(2.1)
F
A
Units: Newton's per square metre, N m-2, kg m-1 s-2.
(The same unit is also known as a Pascal, Pa, i.e. 1Pa = 1 N m-2)
(Also frequently used is the alternative SI unit the bar, where 1 bar = 105 N m-2)
Dimensions: M L-1 T-2.
p=
2.3. Pascal's Law for Pressure at a Point
Fig. 2.2: Triangular prismatic element of fluid
ps acts perpendicular to surface ABCD,
px acts perpendicular to surface ABFE and
py acts perpendicular to surface FECD.
And, as the fluid is at rest, in equilibrium, the sum of the forces in any direction is zero.
Pressure at any point is the same in all directions.
This is known as Pascal's Law and applies to fluids at rest.
2.4. Variation of Pressure Vertically in a Fluid under
Gravity
Fig. 2.3: Vertical elemental cylinder of fluid
Force due to p1 on A (upward) = p1A
Force due to p2 on A (downward) = p2A
Force due to p2 on A (downward) = mg
= mass density x volume = ρg (z2 –z1)
Taking upward as positive, in equilibrium we have
p1A - p2A – ρgA (z2 – z1) = 0
p1 – p2 = – ρg (z2 – z1)
(2.2)
Under gravity, pressure decreases with increase in height z = (z2 – z1).
2.5. Equality of Pressure at the same Level in a Static Fluid
Fig. 2.4: Horizontal elemental cylinder of fluid
The fluid is at equilibrium so the sum of the forces acting in the x direction is zero.
plA = prA
pl = pr
Pressure in the horizontal direction is constant.
Two connected tanks:
Fig 2.5: Two tanks of different cross-section connected by a pipe
With pl = pr:
and
so
pl = pp + ρgz
pr = pq + ρgz
(2.3)
pp + ρgz = pq + ρgz
pp = pq
The pressures at the two equal levels, P and Q are the same.
2.6 Pressure and Head
In a static fluid of constant density we have the relationship dp/dz = - ρg. This can be integrated
to give
p = -ρgz + const
In a liquid with a free surface the pressure at any depth z measured from the free surface so that
z = -h (see figure 2.6)
Fig. 2.6: Fluid head measurement in a tank
p = ρgh + const
At the surface of fluids, the pressure is the atmospheric pressure, patmospheric. So
p = ρgh + patmospheric
Gauge pressure
Pgauge = ρgh
Absolute pressure is
Pabsolute = ρ g h + patmospheric
Absolute pressure = Gauge pressure + Atmospheric pressure
This vertical height is known as head of fluid.
p = ρgh
Note: If pressure is quoted in head, the density of the fluid must also be given.
(2.4)
(2.5)
Example 2.1:
We can quote a pressure of 500 kN m-2 in terms of the height of a column of water of density,
ρ = 1000 kg m-3. Using p = ρgh,
Solution:
h=
500 x 10 3
p
=
= 50.95 m of water
ρg 1000 x 9.81
And in terms of Mercury with density, ρ = 13.6 x 103
500 x 10 3
h=
= 3.75 m of Mercury
13.6 x 10 3 x 9.81
2.7 APPLICATIONS OF THE HYDROSTATIC EQUATION
2.7.1 Manometers
Fig. 2.7 A simple U-tube manometer
p1 – p2 = ρm g (z2 – z1)
that is,
p1 – p2 = ρm g ∆h
ρm : density of the manometer fluid
2.7.2 Barometers
Fig. 2.8 Mercury barometer as devised by Torricelli.
pa = ρ g ∆h = absolute pressure
When the barometer fluid is mercury, ∆h = 760 mm and when it is water, ∆h = 10.33 m.
2.7.3 Vertical Walls of Constant Width
ρgh
Fig. 2.9
Vertical wall with water pressure acting on one side, atmospheric pressure on the other. The
graph on the left is a plot of the pressure as a function of depth.
Example 2.2 Force acting on the vertical wall
At the free surface, the pressure is atmospheric and below the surface, the pressure increases
linearly with depth to a maximum value of pa + ρgh. What is the resultant force due to the water
pressure acting over the area of the reservoir wall, and where does it act?
Solution
We can use absolute pressures. From the hydrostatic equation, the pressure at any depth is given
by
p – pa = ρg(h - z)
(z is positive going up). On the side of the wall where the water acts, the pressure p acts on
element dA = w dz to produce a force dFx that acts to the right. So
dFx = (pressure) x (area) = pdA =pwdz = [pa+ ρg(h - z)] w dz
The resultant force, dF, acting on the area w dz is given by
dF = dFx – pa w dz.
and it acts to the right. That is,
dF = [pa + ρg(h - z)]w dz – pa w dz = ρg(h - z)w dz
(2.6)
Integrating equation 2.6 from z = 0 to z = h gives the resultant force
h
F = ∫ ρ g (h − z) w dz
0
so that
F = ½ ρ g w h2
(2.7)
Moment Balance
Where should the resultant force F be placed so that it exerts the same moment as the distributed
force due to pressure?
dM0 = z × dF
z: moment arm of the force dF, about the axis through z = 0, measured perpendicular to the line
of action of the force. This moment is clockwise about the origin. Substituting for dF from
equations 2.6, we obtain
dM0 = z × dF = pg(h - z)wz dz,
Integrating from z = 0 to z = h gives the resultant moment M0 = 1/6 ρgwh3.
By definition, the resultant force times its moment arm must be equal to the total moment
exerted by the pressure acting on the wall, so that
M0 = z × F
(2.8)
where z is the moment arm for the resultant force, as measured from the axis through z = 0.
Finally,
z
= 1/3 h
When atmospheric pressure acts everywhere, the resultant force exerted by a fluid of depth h
acting on a vertical wall of constant width is given by
F = ½ ρgwh2
and it acts a distance h/3 from the bottom of the reservoir.
EXAMPLE 2.3 Forces on a Sloping Plate
A water bassin has on the sloping site a rectangular trap. Calculate the resultant force F on this
trap and the position of the acting point xF.
h =3m
b =4m
ρW = 1000 kg/m3
α = 60°
Solution:
We can neglect the atmospheric pressure because it acts from both sites on the plate.
Introducing the coordinate x from xF = 0 in the direction of xF the differential force is
dF = ρg (2h – x) sin α b dx
and
h
F = ρ g b sin α ∫ (2h − x) dx
x =0
F = 3/2 ρ g b sinα h2
The moment of the resultant force F· xF about the origin x = 0 must be the same like the sum
of the moments of all differential forces and their attached moment arms.
F x F = ∫ dF ⋅ x
1
2 h
4
x F = ∫ dF ⋅ x = 2 ∫ (2 hx − x 2 ) dx = h
F
3 h x =0
9
2.8 Archimedes Principle
Two forces act on the can in fig. 2.10:
The mass force
y
z
p0
x
m·g
ρ
and the force resulting from the pressure distribution
FB
ρW g A ∆h = ρW g V
mg
Fig. 2.10: Bouyant Force on a Can in Water
VρW g is just the weight of the displaced water. An object is buoyed up by a force equal to the
weight of the displaced water.
Fig. 2.11 A V-shaped boat floating in water.
The both vertical components of the force due to water pressure act up, and since they are of the
same magnitude, they add.
W=2
ρgwh 2
2 tan Θ
(In the limit, we see that when Θ → π/2 , the weight that can be supported by the water
pressure goes to zero, which is the correct limiting behavior.)
Note that the volume of the displaced fluid is wh2/tanΘ, and its weight W is given by
ρgwh2/tanΘ.
Archimedes principle:
The buoyancy force on a solid is equal to the weight of liquid it has displaced, and it acts
through the centroid of the displaced volume.
(2.9)
2.9 Rigid-Body Rotation
If the rate of rotation is ω [rad/s], the equation of motion in the r-direction becomes
∂p
V2
=ρ
= ρ r ω2
r
∂r
(2.10)
For the z-direction (where z is positive down), we get the usual hydrostatic variation
∂p
= ρg
∂z
(2.11)
Fig. 2.12 Fluid in rigid-body motion under rotation.
Integrating equation 2.10 with respect to r we obtain
p=
1 2 2
ρ r ω + f ′(z ) + C 3
2
(2.12)
and integrating equation 2.11 with respect to z we obtain
p = ρ g z + g ′(x ) + C 4
(2.13)
From equations 2.12 and 2.13
p = ρgz +
1 2 2
ρ r ω + C′
2
2.14)
To find C´, we need to know the pressure at a given point. At the free surface, the pressure is
equal to atmospheric pressure, so that p = pa at r = z = 0, and
z=
p − pa r 2ω2
−
ρg
2g
(2.15)

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