Lecture 4
Pressure distribution in a fluid
Pressure and pressure gradient
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Chapter 2
•
•
•
•
Concept of pressure and pressure gradient
Hydrostatic pressure distribution
Hydrostatic forces
Buoyance and stability
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Forces in fluids
• Surface or volumetric (or mass)
• Surface forces can be normal (pressure) or
tangential (friction)
• Friction Forces are always parallel to velocity. This
is why they are often called “tangential forces”.
Their equation can be complex if velocity is not
parallel to any reference axis.
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Pressure
• Many fluid problems do not involve motion. They
concern the pressure distribution in a static fluid
and its effect on solid surfaces and on floating
and submerged bodies.
• Fluids at rest cannot support shear stress.
• Pressure is used to indicate the normal force per
unit of area at a given point acting on a given
plane within the fluid mass of interest.
• How the pressure at a point varies with the
orientation of the plane passing through the
point?
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Pressure at a point
• Small wedge (“pequena cunha”) of fluid at rest of size x by z by s and
depth b into the paper.
• There is no shear by definition, but we postulate that the pressures px, pz, and
pn may be different on each face
• The weight of the element also may be important.
• The element is assumed small, so the pressure is constant on each face.
• Summation of forces must equal zero (no acceleration) in both the x and z
directions.
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Pressure at a point
𝐹𝑥 = 0 = 𝑝𝑥 𝑏∆𝑧 − 𝑝𝑛 𝑏∆𝑠 sin 𝜃
1
2
𝐹𝑧 = 0 = 𝑝𝑧 𝑏∆𝑥 − 𝑝𝑛 𝑏∆𝑠 cos 𝜃 − 𝜌𝑔𝑏∆𝑥∆𝑧
By the geometry of the wedge:
∆𝑠 sin 𝜃 = ∆𝑧
∆𝑠 cos 𝜃 = ∆𝑥
𝑝𝑥 = 𝑝𝑛
1
𝑝𝑧 = 𝑝𝑛 − 2 𝜌𝑔∆𝑧
These relations illustrate two important principles of the hydrostatic, or shear-free, condition:
(1) there is no pressure change in the horizontal direction
(2) there is a vertical change in pressure proportional to the density, gravity, and depth change
In the limit as the fluid wedge shrinks to a “point”, ∆𝑧
𝑝𝑥 = 𝑝𝑦 = 𝑝𝑧 = 𝑝
0:
Since  is arbitrary, we conclude that the pressure at point (p) in a static fluid is a
point property, independent of the plane orientation.
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Pressure force in fluid element
• Pressure (or any other stress, for that matter) causes a
net force on a fluid element when it varies spatially.
• Consider the pressure acting on the two x faces in Fig2.2 and let
the pressure vary arbitrarily:
𝑝 = 𝑝 𝑥, 𝑦, 𝑧, 𝑡
• The net force in x direction on the element is:
𝜕𝑝
𝜕𝑝
𝑑𝐹𝑥 = 𝑝𝑑𝑦𝑑𝑧 − 𝑝 +
𝑑𝑥 𝑑𝑦𝑑𝑧 = − 𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑥
𝜕𝑥
• The total net-force vector on the
element due to pressure is:
𝑑𝐹𝑝𝑟𝑒𝑠𝑠 = −𝑖
𝜕𝑝
𝜕𝑝
𝜕𝑝
−𝑗
−𝑤
𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
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Equilibrium of fluid element
• Forces acting on the fluid element: surface forces due to the pressure,
body force equal to weight of the element and friction
𝜏𝑦𝑥
𝜏𝑦𝑥
𝑦
≈ 𝜇
𝑑𝑉
𝑑𝑦
𝑦
𝑑𝑉
𝑑𝑓 = lim 𝜇
𝑑𝑦 0
𝑑𝑦
𝑑𝑉
𝑑𝑦
𝑦+𝑑𝑦
Weight=g(dxdydz)
−𝜇
𝑦+𝑑𝑦
𝑑𝑉
𝑑𝑦
By volume unit and considering  constante:
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𝑦+𝑑𝑦
≈ 𝜇
=
𝑦
𝑑
𝑑𝑉
𝜇
𝑑𝑥𝑑𝑦𝑑𝑧
𝑑𝑦
𝑑𝑦
𝑓𝑠 = 𝜇
𝑑2 𝑉
𝑑𝑥 2
+
𝑑2 𝑉
𝑑𝑦2
+
𝑑2 𝑉
𝑑𝑧 2
= 𝜇𝛻 2 𝑉
Equilibrium of fluid element
• 𝑓𝑝𝑟𝑒𝑠𝑠 = −𝛻𝑝 (net pressure force per unit volume)
• 𝑓𝑔𝑟𝑎𝑣 = 𝜌𝑔 (net gravity force per unit volume)
• 𝑓 = 𝜇𝛻 2 𝑉(net viscous force per unit volume)
• By Newton’s law, the sum of these per-unit-volume
forces equals the mass per unit volume (density)
times the acceleration a of the fluid element:
𝑓 = 𝑓𝑝𝑟𝑒𝑠𝑠 + 𝑓𝑔𝑟𝑎𝑣 + 𝑓 = −𝛻𝑝 + 𝜌𝑔 + 𝜇𝛻 2 𝑉 = 𝜌𝑎
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General equation
𝜌𝑎 = −𝛻𝑝 + 𝜌𝑔 + 𝜇𝛻 2 𝑉
𝑑𝑉
𝜌
𝑑𝑡
= −𝛻𝑝 + 𝜌𝑔 + 𝜇𝛻 2 𝑉
• If the fluid is at rest or at constant velocity, a=0 and
𝛻 2 𝑉 = 0 the equation for the pressure distribution
reduces to
𝛻𝑝 = 𝜌𝑔
This is a hydrostatic distribution and is correct for all fluids at
rest, regardless of their viscosity, because the viscous term
vanishes identically.
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Hydrostatic pressure
• In our customary coordinate system z is “up”. Thus the
local-gravity vector for small-scale problems is:
𝑔 = −𝑔𝑘
where g is the magnitude of local gravity, for
example, 9.807 m/s2.
For these coordinates, equation 𝛻𝑝 = 𝜌𝑔
has the
components
𝜕𝑝
𝜕𝑝
𝜕𝑝
=0
=0
= −𝜌𝑔 = −𝛾
𝜕𝑥
𝜕𝑦
𝜕𝑧
• These equations show that pressure does not depend on x and y.
Thus, as we move point to point in horizontal plane, pressure do
not change. Since p depends only on z, 𝜕𝑝 𝜕𝑧 can be replaced by
the total derivative
𝑑𝑝
𝑑𝑧
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= −𝜌𝑔 = −𝛾
Hydrostatic pressure
𝑑𝑝
𝑑𝑧
= −𝛾
or
𝑝2 −𝑝1 =
2
−𝛾
1
𝑑𝑧
• These equation is the solution to the hydrostatic problem. The
integration requires an assumption about the density and gravity
distribution. Gases and liquids are usually treated differently.
• Pressure in a continuously distributed uniform static fluid varies only
with vertical distance and is independent of the shape of the
container.
• The pressure is the same at all points on a given horizontal plane in
the fluid.
• The pressure increases with depth in the fluid.
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Efect of variable gravity
• For a spherical planet of uniform density, the acceleration
of gravity varies inversely as the square of the radius from
its center
𝑟0 2
𝑔 = 𝑔0
𝑟
where r0 is the planet radius and g0 is the surface value of
g. For earth, r0  3960 statute mi  6400 km.
In typical engineering problems the deviation from r0
extends from the deepest ocean, about 11 km, to the
atmospheric height of supersonic transport operation,
about 20 km. This gives a maximum variation in g of
(6400/6420)2, or 0.6 percent. We therefore neglect the
variation of g in most problems.
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Hydrostatic pressure in liquids
• Liquids are so nearly incompressible that we can
neglect their density variation in hydrostatics. Thus we
assume constant density in liquid hydrostatic
calculations, for which hydrostatic conditions
integrates to
𝑝2 − 𝑝1 = −𝛾 𝑧2 − 𝑧1 or 𝑧1 − 𝑧2 =
𝑝2
𝛾
𝑝1
−
𝛾
We use the first form in most problems. The quantity  is
called the specific weight (“peso específico”) of the fluid,
with dimensions of weight per unit volume; The quantity
𝑝 𝛾 is a length called the pressure head (“carga de
pressão”) of the fluid.
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Hidrostatic pressure distribution
• For lakes and oceans, the coordinate system is usually
chosen as in Figure, with z=0 at the free surface, where p
equals the surface atmospheric pressure pa.
𝑝1 , 𝑧1 = 𝑝𝑎 , 0
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𝑝 = 𝑝𝑎 − 𝛾𝑧
Hydrostatic pressure in gases
• Gases are compressible, with density nearly proportional to
pressure. Thus density must be considered as a variable in the
hydrostatic pressure equation. It is sufficiently accurate to
introduce the perfect-gas law 𝑝 = 𝜌𝑅𝑇 in the hydrostatic
equation:
𝑑𝑝
𝑝
= −𝜌𝑔 = −
𝑔
𝑑𝑧
𝑅𝑇
Separate the variables and integrate between points 1 and 2:
2
𝑑𝑝
𝑝2
𝑔 2 𝑑𝑧
= ln = −
𝑝1
𝑅 1 𝑇
1 𝑝
The integral over z requires an assumption about the
temperature variation T(z). One common approximation is the
isothermal atmosphere, where T= T0:
𝑔 𝑧2 − 𝑧1
𝑝2 = 𝑝1 𝑒𝑥𝑝 −
𝑅𝑇0
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Hydrostatic forces on plane surfaces
• Objective: compute the hydrostatic force over flat and
curve surfaces and the application point (center of
pressure).
• When a surface is submerged in a fluid, forces develop
on the surface due to the fluid. The determination of
these forces is important in the design of storage
tanks, ships, dams, and other hydraulic structures.
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Hydrostatic forces on plane surfaces
• Plane panel of arbitrary shape completely submerged in a liquid
• Panel plane makes an arbitrary angle  with the horizontal free
surface, so that the depth varies over the panel surface.
• If h is the depth to any
element area dA of the
plate, from hydrostatic
pressure
equation
in
liquids the pressure there
is p=pa+h.
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Hydrostatic forces on plane surfaces
• To derive formulas involving the plate shape, establish an xy
coordinate system in the plane of the plate with the origin at its
centroid, plus a dummy coordinate  down from the surface in the
plane of the plate. A is the area of the plane surface.
• The total hydrostatic force on one side of the plate is given by
From the figure ℎ = sin 𝜃 and, by
definition, the centroidal slant distance
from the surface to the plate is
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• Therefore, since  is constant along the plate, becomes
• Finally, unravel this by noticing that
, the depth
straight down from the surface to the plate centroid. Thus
• The force on one side of any plane submerged surface in a uniform
fluid equals the pressure at the plate centroid times the plate area,
independent of the shape of the plate or the angle  at which it is
slanted.
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Center of pressure
• However, to balance the bending-moment portion of the stress, the
resultant force F acts not through the centroid but below it toward
the high-pressure side. Its line of action passes through the center
of pressure CP of the plate (see figure).
• To find the coordinates (xCP, yCP), we sum moments of the elemental
force p dA about the centroid and equate to the moment of the
resultant F. To compute yCP, we equate
Vanishes by def. of centroidal axes
Ixx is the area moment of inertia of the plate area about its centroidal x axis
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Center of pressure
• The negative sign shows that yCP is below
the centroid at a deeper level (below the
gravity center) and, unlike F, depends on
angle .
• The determination of xCP is exactly similar:
Ixythe product of inertia of the plate, again computed in the plane of the plate
• If Ixy = 0, usually implying symmetry, xCP=
0 and the center of pressure lies directly
below the centroid on the y axis.
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• In most cases the ambient pressure pa is neglected
because it acts on both sides of the plate; for
example, the other side of the plate is inside a ship
or on the dry side of a gate or dam. In this case
pCG=hCG, and the center of pressure becomes
independent of specific weight:
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Resume
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