# ME 3560 Fluid Mechanics Chapter II. Fluid Statics Summer 2016

```ME3560 – Fluid Mechanics
Summer 2016
ME 3560 Fluid Mechanics
Chapter II. Fluid Statics
1
Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
Examples of cases where forces on submerged surfaces must
be calculated.
• Fluid power cylinder.
• Storage tank.
• Aquarium observation
windows.
• Tank with curved
surface.
Chapter II. Fluid Statics
• Retaining wall.
• Fluid reservoir and
2
hatch.
Summer 2016
ME3560 – Fluid Mechanics
2.1 Pressure at a Point
• The term pressure is used to indicate the normal force per unit area at a
given point acting on a given plane within the fluid mass of interest.
ΣFy = p yδ xδ z − psδ xδ s sin θ = ρ
δ xδ yδ z
ay
2
δ xδ yδ z
δ xδ yδ z
ΣFz = p zδ xδ y − psδ xδ s cosθ − γ
=ρ
az
2
2 3
Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
• Realizing that
δ y = δ s cosθ ;
δ z = δ s sin θ
•Then the previous equations can be expressed as:
p y − ps =
δy
2
ρa y ;
p z − ps =
δz
2
( ρa z + γ )
•Since we are really interested in what is happening at a point, we take
the limit as δx, δy, and δz approach zero : p = p ;
p =p
y
s
z
s
p y = p z = ps
• The pressure at a point in a fluid at rest, or in motion, is independent
of direction as long as there are no shearing stresses present(Pascal's
law).
σyy
p=−
σ xx + σ yy + σ zz
3
Chapter II. Fluid Statics
σzz
σxx
σxx
σzz
σyy
4
Summer 2016
ME3560 – Fluid Mechanics
2.2 Basic Equation for Pressure Field
• Consider a small rectangular element of fluid.
•There are two types of forces acting on this element:
•surface forces due to the pressure
•body force equal to the weight of the element.
• Let p be the pressure at the
center of the element be
designated as p, then the
surface forces in the y direction
is:
∂p δy
)δxδz −
δF y = ( p −
∂y 2
∂p δ y
(p+
)δxδz
∂y 2
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
• Simplifying the previous expression we get:
∂p
δF y = − δ xδyδ z
∂y
•Following a similar procedure the surface forces for x and z are:
∂p
δFx = − δxδyδz
∂x
∂p
δFz = − δxδyδz
∂z
•The surface forces exerted on the element of fluid can be expressed as:
r
δFs = δFx iˆ + δFy ˆj + δFz kˆ =
r
∂p ˆ ∂p ˆ ∂p ˆ
δFs = − ( i +
j+
k )δxδyδz
∂x
∂y
∂z
r
∂ () ˆ ∂ () ˆ ∂ () ˆ
δFs = −∇ pδxδyδz ; ∇ () =
i+
j+
k
∂x
∂y
∂z 6
Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
• By considering the body forces, we can find that the weight of the
element is:
ˆ
ˆ
ˆ
− δWk = − ρgδxδyδzk = −γδxδyδzk
•Adding up body and surface forces, and applying Newton’s 2nd Law we
r
get:
ˆ
− (∇p + ρgk )δxδyδz = ρδxδyδz a
•Which after simplification becomes:
r
ˆ
− (∇p + ρgk ) = ρ a
•This is the general equation of motion for a fluid in which there are no
shearing stresses.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
2.3 Pressure Variation in a Fluid at Rest
r
ˆ
• For a fluid at rest, the equation: − (∇p + ρgk ) = ρ a
•Becomes: ∇p + ρgkˆ = 0
•Which can also be expressed as:
∂p
=0
∂x
∂p
=0
∂y
∂p
= − ρg
∂z
•These equations show that the pressure does not depend on x or y.
•Therefore p depends only on z (vertical direction – depth)
dp
= − ρg = −γ
dz
•This is the fundamental equation for fluids at rest and can be used to
determine how pressure changes with elevation.
•This equation expresses that the pressure decreases as we move upward
in a fluid at rest.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
2.3.1 Incompressible Fluid
• A fluid with constant density is called an incompressible fluid.
• For liquids the variation in density is usually negligible, even over
large vertical distances (ρ and therefore γ are constant):
∫
p2
p1
z2
dp = − ρ g ∫ dz
z1
p1 − p2 = ρ g ( z 2 − z1 )
p1 − p2 = ρ gh = γ h
• The pressure head can be
defined from the previous
equation as:
p1 − p2 p1 − p2
h=
=
ρg
γ
•If p0 is a reference
pressure, the pressure at a
depth h is p = γh + p
0
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Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
• As demonstrated, the pressure in a homogeneous, incompressible fluid
at rest depends on the depth of the fluid relative to some reference plane,
and it is not influenced by the size or shape of the tank or container in
which the fluid is held. The actual value of the pressure along AB
depends only on the depth, h, the surface pressure, p0, and the specific
weight, γ, of the liquid in the container.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
2.3.2 Compressible Fluid
• Gases such as air, oxygen, and nitrogen are compressible fluids.
ρ gas = f (T , p )
•γgas<< γliq
•Since γgas are comparatively small, then ∂p/∂z is small (even over
distances of several hundred feet the pressure will remain essentially
constant for a gas).
•Thus the effect of elevation changes on the pressure in gases can be
neglected in tanks, pipes.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
• When the variations in heights are large, on the order of thousands of
feet, the variation in γ must be taken into account.
p
• For an ideal (or perfect) gas the eqn. of state is: ρ =
which when
RT
combined with dp
= − ρg = −γ
dz
dp
pg
Results in the following relation:
=−
dz
RT
•This equation can be integrated to obtain:
∫
p2
p1
dp
p2
g z2 dz
= ln
=− ∫
p
p1
R z1 T
•Finally, it is necessary to specify the nature of the variation of
temperature with elevation. If the temperature has a constant value T0
over the range z2 to z1 (isothermal conditions), then
 g ( z2 − z1 ) 
p2 = p1 exp−

RT0 

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Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
2.5 Measurement of Pressure
•The pressure at a point within a fluid mass is designated as either an
absolute pressure (pabs) or a gage pressure (pgage).
•pabs is measured relative to a perfect vacuum (absolute zero pressure).
•pgage is measured relative to the local atmospheric pressure.
•Absolute pressures are always positive.
•Gage pressures can be positive or negative.
•A negative gage pressure is also referred to as a suction or vacuum
pressure.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
• Since p=F/A, its units are lb/ft2 (psf), lb/in2
(psi) and N/m2 (Pa)
•Pressure can also be expressed as the
height of a column of liquid.
•The units will refer to the height of the
column (in., ft, mm, m), also the liquid in
the column must be specified (H2O, Hg).
•The measurement patm is usually done with
a mercury barometer.
•The column of mercury will come to an
equilibrium position where its weight plus
the force due to the vapor pressure (which
develops in the space above the column)
balances the force due to the atmospheric
pressure.
patm = ρgh + pvapor = γh + pvapor
Chapter II. Fluid Statics
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Summer 2016
ME3560 – Fluid Mechanics
2.6 Manometry
•A standard technique for measuring pressure involves the use of liquid
columns in vertical or inclined tubes.
•Pressure measuring devices based on this technique are called
manometers.
•Three common types of manometers are: the piezometer tube, the Utube manometer, and the inclined-tube manometer.
2.6.1 Piezometer Tube
•The simplest type of manometer consists of a vertical tube, open at the
top, and attached to the container in which the pressure is desired.
•Manometers involve columns of fluids at rest, thus
p = γh + p0
gives the pressure at any elevation within a
homogeneous fluid in terms p0 and h
•Applying this equation to the piezometer tube
Chapter II. Fluid Statics
p A = γ 1h1
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Summer 2016
ME3560 – Fluid Mechanics
2.6.2 U–Tube Manometer
•The U–Tube manometer consists of a tube formed into the shape of a U.
•The fluid in the manometer is called the gage fluid.
•To find the pressure pA in terms of the various column heights, an
analysis is started at one end of the system towards the other end.
p A + γ 1h1 − γ 2 h2 = 0
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
•The U-tube manometer is also used to measure the difference in
pressure between two containers or two points in a given system.
•Consider a manometer connected
between containers A and B. The
difference in pressure between A and B
can be found by again starting at one end
of the system and working around to the
other end. Thus:
p A + γ 1h1 − γ 2 h2 − γ 3h3 = pB
•Therefore, the pressure difference is:
p A − pB = γ 2 h2 + γ 3h3 − γ 1h1
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
2.6.3 Inclined–Tube Manometer
•To measure small pressure changes, an inclined–tube manometer is
frequently used.
p A + γ 1h1 − γ 2l2 sin θ − γ 3h3 = pB
p A − pB = γ 2l2 sin θ + γ 3h3 − γ 1h1
18
Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
Examples of cases where forces on submerged surfaces must
be calculated.
• Fluid power cylinder.
• Storage tank.
• Aquarium observation
windows.
• Tank with curved
surface.
Chapter II. Fluid Statics
• Retaining wall.
• Fluid reservoir and
hatch. 19
ME3560 – Fluid Mechanics
Summer 2016
2.8 Hydrostatic Force on a Plane Surface
•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.
•For fluids at rest we know that the force must be perpendicular to the
surface since there are no shearing stresses present.
•For incompressible fluids, the pressure varies linearly with depth.
•For horizontal surfaces, magnitude of
the resultant force is FR=pA.
•If patm acts on both sides of the
bottom, the FR on the bottom is only
due to the liquid in the tank.
•Since the pressure is constant and
uniformly distributed over the bottom,
FR acts through the centroid of the
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area.
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•When a submerged plane surface is inclined and assuming that the fluid
surface is open to the atmosphere.
•Let the plane in which the
surface lies intersect the free
surface at 0 and make an
angle θ with this surface.
•The x–y coordinate system is
defined so that 0 is the origin
and y = 0 (i.e., the x-axis) is
directed along the surface as
shown.
•The objective is to determine
the direction, location, and
magnitude of the resultant
force acting on one side of the
submerged area.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
•At any given depth, h, the force acting on dA is dF = h dA and is
perpendicular to the surface.
•Thus, the magnitude of the
resultant force can be found
by summing these differential
forces over the entire surface.
In equation form:
FR = ∫ γ hdA
A
FR = ∫ γ y sin θdA
A
•Since γ and θ are constant:
FR = γ sin θ ∫ ydA
A
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
•The integral appearing in the previous equation is the first moment of
the area with respect to the x axis
∫
A
ydA = yc A
•yc is the y coordinate of the centroid of area A measured from the x axis.
The equation for FR can be written as
FR = γ Ayc sin θ = γ hc A
•hc is the vertical distance from the fluid surface to the centroid of the
area. The magnitude of the force depends only on the γ, A, and hc (it is
independent of θ).
•The magnitude of the resultant fluid force is equal to the pressure
acting at the centroid of the area multiplied by the total area.
•The y coordinate, yR, of the resultant force can be determined by
summation of moments around the x axis. That is, the moment of the
resultant force must equal the moment of the distributed pressure force:
FR yR = ∫ ydF = ∫ γ sin θ y dA
2
Chapter II. Fluid Statics
A
A
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ME3560 – Fluid Mechanics
•Therefore, since FR= γAyc sinθ:
yR
∫
=
A
2
y dA
Summer 2016
yc A
•The integral in the numerator is the moment of inertia, Ix, with respect to
an axis formed by the intersection of the plane containing the surface and
the free surface (x axis). Thus
Ix
yR =
yc A
•By using the parallel axis theorem to express Ix: I x = I xc + Ayc
•Ixc is the moment of inertia of the area with respect to an axis passing
through its centroid and parallel to the x axis:
I xc
2
yR =
∫
=
•In a similar analysis
xR
Chapter II. Fluid Statics
A
xydA
yc A
=
yc A
+ yc
I xy
yc A c
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ME3560 – Fluid Mechanics
•In a similar analysis:
xR
∫
=
A
xydA
yc A
Summer 2016
=
I xy
yc A c
•Ixy is the product of inertia with respect to the x and y axes. Again, using
the parallel axis theorem
I xyc
xR =
yc Ac
+ xc
•Ixyc is the product of inertia with respect to an orthogonal coordinate
system passing through the centroid of the area and formed by a
translation of the x–y coordinate system.
•If the submerged area is symmetrical with respect to an axis passing
through the centroid and parallel to either the x or y axes, the resultant
force must lie along the line x =xc.
•The point through which FR acts is called the center of pressure.
•As yc increases the center of pressure gets closer to the centroid of the
area.
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Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
2.10 Hydrostatic Force on a Curved Surface
•To determine the resultant force exerted by a fluid on a curved surface,
it is necessary to determine independently the vertical and horizontal
components of FR
26
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•The horizontal force, FH, is determined by projecting the curved surface
onto the vertical plane (y–z) and calculating the force that the fluid would
exert on this imaginary vertical surface.
•The point of application of FH corresponds to the location where the
force would act on the imaginary (projection) vertical surface.
27
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•The vertical force, FV, is the weight of the fluid above the curved
surface up to the free surface.
•The line of action of FV passes through the centroid of the volume of
fluid above the curved surface.
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Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
2.11 Buoyancy, Flotation and Stability
2.11.1 Archimedes’ Principle
•If a stationary body is completely or partially (floating) submerged in a
fluid, the force exerted by the fluid on the body called buoyant force.
•A net upward vertical force results because pressure increases with
depth and the pressure forces acting from below are larger than the
pressure forces acting from above.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
•Consider a flat plate of thickness h submerged in a liquid of density ρf
parallel to the free surface.
•The area of the top and bottom surface of the plate is A and its distance
to the free surface is s.
•The pressure at the top and bottom surfaces are ρf gs and ρf g(s+h).
•The hydrostatic force acting on the top surface is Ftop=ρf gsA, the larger
force Fbottom=ρfg(s+h)A acts upward on the bottom surface of the plate.
•The difference between these two forces is a net upward force
(buoyancy force)
FB = Fbottom − Ftop
FB = ρ f g ( s + h) A − ρ f gsA
FB = ρ f ghA = ρ f gV
30
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•The relation ρf gV is the weight of the liquid whose volume is equal to
the volume of the plate.
•The previous relation is developed for a simple geometry, but it is valid
for any body regardless of its shape.
•This can be shown mathematically by a force balance or simply by this
argument:
•Consider an arbitrary shaped solid body submerged in a fluid at rest and
compare it to a body of fluid of the same shape indicated by dotted lines
at the same distance from the free surface.
•The buoyant forces acting on these two
bodies, which depend only on depth, are
the same at the boundaries of both.
•The imaginary fluid body is in static
equilibrium, and thus, the net force and
net moment acting on it are zero.
31
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•Thus, the upward buoyant force must be equal to the weight of the
imaginary fluid body, whose volume is equal to the volume of the solid
body.
•Further, the weight and the buoyant force must have the same line of
action to have a zero moment.
•This is known as Archimedes’ Principle:
•The buoyant force acting on a body immersed in a fluid is equal to
the weight of the fluid displaced by the body, and it acts upward
through the centroid of the displaced volume.
2.11.2 Stability
•A body is in a stable equilibrium position if, when
displaced, it returns to its equilibrium position.
•It is in an unstable equilibrium if, when displaced
(even slightly), it goes to a new equilibrium position.
32
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•Stability considerations are particularly
important for submerged or floating bodies
since the centers of buoyancy and gravity
do not necessarily coincide.
•A small rotation can result in either a
restoring or overturning couple.
•For a completely submerged body with
CG below the center of buoyancy.
•A rotation from its equilibrium position will create a restoring couple
formed by the weight, W, and the buoyant force, FB, causing the body to
rotate back to its original position. Thus, for this configuration the body
is stable.
•The body will be stable as long as the center of gravity falls below
the center of buoyancy.
33
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•For the completely submerged body shown, which has a center of
gravity below the center of buoyancy, a rotation from its equilibrium
position will create a restoring couple formed by the weight, W, and the
buoyant force, FB, which causes the body to rotate back to its original
position. Thus, for this configuration the body is stable. It is to be noted
that the body will be stable as long as the center of gravity falls below
the center of buoyancy, thus the body is in a stable equilibrium position
with respect to small rotations. However, as is illustrated in Fig. 2.26, if
the center of gravity of the completely submerged body is above the
center of buoyancy, the resulting couple formed by the weight and the
buoyant force will cause the body to overturn and move to a new
equilibrium position. Thus, a completely submerged body with its center
of gravity above its center of buoyancy is in an unstable equilibrium
position.
34
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•If the center of gravity of the completely
submerged body is above the center of
buoyancy, the resulting couple formed by
the weight and the buoyant force will cause
the body to overturn and move to a new
equilibrium position.
•A completely submerged body with its
center of gravity above its center of
buoyancy is in an unstable equilibrium
position..
35
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•For floating bodies the stability problem is more complicated.
•As the body rotates the location of the center of buoyancy (which passes
through the centroid of the displaced volume) may change.
•A floating body such as a barge that rides low in the water can be stable
even though the center of gravity lies above the center of buoyancy.
•This is true since as the body rotates FB shifts to pass through the
centroid of the newly formed displaced volume and combines with W to
form a couple which will cause the body to return to its original
equilibrium position.
36
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
•However, for a relatively tall, slender body, a small rotational
displacement can cause the buoyant force and the weight to form an
overturning couple as illustrated.
37
Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
2.12 Pressure Variation in a Fluid with Rigid–
Body Motion
r
ˆ
•The general equation of motion: − ∇p − γ k = ρ a
•was developed for both fluids at rest and fluids in motion, for situations
where there were no shearing stresses present. This equation can be
expressed as ∂p
∂p
∂p
−
∂x
= ρa x
−
∂y
= ρa y
−
∂z
= ρa z + γ
•Problems involving fluid motion with no shearing stresses appear when
a mass of fluid undergoes rigid-body motion.
•For example, if a container of fluid accelerates along a straight path, the
fluid will move as a rigid mass (after the initial sloshing motion has died
out) with each particle having the same acceleration. Since there is no
deformation, there will be no shearing stresses. Thus the previous
equations apply.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
2.12.1 Linear Motion
•Consider an open container of a liquid that is translating along a straight
path with a constant acceleration a (ax = 0), therefore ∂p/∂x = 0.
•In the y and z directions:
∂p
−
= ρa y
∂y
∂p
−
= ρ (a z + g )
∂z
•The change in pressure between two closely spaced points located at y,
z, and y + dy, z + dz can be expressed as
∂p
∂p
dp =
dy +
∂y
dp = − ρa y dy − ρ ( g + a z ) dz
•Or in terms of the problem studied:
Chapter II. Fluid Statics
∂z
dz
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Summer 2016
ME3560 – Fluid Mechanics
•Along a line of constant pressure (for example
the free surface), dp = 0, then the slope of this line
ay
dz
is:
dy
=−
g + az
•Thus, the free surface of the accelerating
mass will be inclined if ay ≠ 0.
•In addition, all lines of constant pressure
will be parallel to the free surface.
•For the case when ay = 0, az ≠ 0 (the mass
of fluid accelerating in the vertical
direction), the fluid surface will be
horizontal with pressure distribution
different than hydrostatic, and given by dp
dz
= − ρ ( g + az )
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
2.12.2 Rigid–Body Rotation
•After an initial “start-up” transient, a fluid contained in a tank that
rotates with a constant angular velocity ω about an axis will rotate with
the tank as a rigid body.
•The acceleration of a fluid particle located at a distance r from the axis
of rotation is equal in magnitude to rω2, and the direction of the
acceleration is toward the axis of rotation.
•The pressure gradient ∇p can be expressed in cylindrical polar
coordinates r, θ, and z as:
∂p
∂p
1 ∂p
∇p =
eˆr +
eˆθ +
eˆz
∂r
r ∂θ
∂z
2
a
=
−
r
ω
eˆr ; aθ = 0; a z = 0
•In terms of this coordinate system: r
•It follows from
r
ˆ
− ∇p − γ k = ρ a
∂p
•That the eqns. are:
∂r
Chapter II. Fluid Statics
= ρ rω ;
2
∂p
∂θ
= 0;
∂p
∂z
= −γ
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Summer 2016
ME3560 – Fluid Mechanics
•Thus, for this type of rigid-body rotation, p = p(r, z), therefore the
differential pressure is
∂p
∂p
2
dp =
∂r
dr +
∂z
dz = ρ rω dr − γdz
•On a horizontal plane (dz = 0), then dp/dr = ρ ω2r, (greater than zero).
That is: due to the centrifugal acc., p increases in the radial direction.
42
Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
•Along a surface of constant pressure, such as the free surface, dp = 0,
dp = ρ rω dr − γdz
2
dz rω
=
dr
g
2
•Integration of this result gives the equation for surfaces of constant
2 2
pressure as
rω
z=
2g
+ constant
•This expression reveals that these surfaces of constant pressure are
parabolic.
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Chapter II. Fluid Statics
Summer 2016
ME3560 – Fluid Mechanics
•Integration of
•Yields
dp = ρ rω 2 dr − γdz
∫ dp = ρ ω ∫ rdr − γ ∫ dz
p=
2
ρω r
2 2
2
− γ z + constant
•The constant of integration can be
expressed in terms of a specified
pressure at some arbitrary point r0, z0.
•This result shows that the pressure
varies with the distance from the axis
of rotation, but at a fixed radius, the
pressure varies hydrostatically in the
vertical direction.
Chapter II. Fluid Statics
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Summer 2016
ME3560 – Fluid Mechanics
2.3.2 Compressible Fluid
2.6.3 Inclined–Tube Manometer
2.9. Pressure Prism
45
Chapter II. Fluid Statics
ME3560 – Fluid Mechanics
Summer 2016
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Chapter II. Fluid Statics
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