Download Chapter 2: Pressure Distribution in a Fluid

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Professor Fred Stern
Chapter 2
1
Fall 2005
Chapter 2: Pressure Distribution in a Fluid
2.1) Pressure and pressure
gradient
In fluid statics, as well as in fluid
dynamics, the forces acting on a
portion of fluid (C.V.) bounded by
a C.S. are of two kinds: body forces and surface forces.
Body Forces: act on the entire body of the fluid (force
per unit volume).
Surface Forces: act at the C.S. and are due to the
surrounding medium (force/unit areastress).
In general the surface forces can be resolved into two
components: one normal and one tangential to the surface.
Considering then a cubical fluid element, we see that the
stress in a moving fluid comprises a 2nd order tensor.
y
σxy
σxz
x
z
Face
σxx
Direction
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⎡σ
σ = ⎢⎢σ
⎢⎣σ
ij
xx
yx
zx
σ
σ
σ
σ
σ
σ
xy
yy
zy
xz
yz
zz
⎤
⎥
⎥
⎥⎦
Since, by definition, a fluid cannot withstand a shear
stress without moving, (deformation) a stationary fluid
must necessarily be completely free of shear stress (σij=0,
i ≠ j). The only stress is the normal stress, which is
referred to as the pressure.
σii = -p
σn = -p, which is compressive, as it should be since fluid cannot
withstand tension. (Sign convention for p>0 is direction of n)
n
Or
(one value at a point,
independent of
direction, p is a scalar)
px = py = pz = pn = p
i.e. normal stress (pressure) is isotropic. This can be easily
seen by considering the equilibrium of a wedge shaped fluid
element
z
∑F
x : pn dA sin α − p x dA sin α = 0
∑F
pn = p x
pndA
dl
: − pn dA cosα + pz dA cosα − W = 0
pn = p z
1
Where W = ρgdA cos α dl sin α
2
= O ( dl )
pxdAsinα
α
dA=dldy
z
W=ρgV
pzdAcosα
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Note: For a fluid in motion, the normal stress is different
on each face and not equal to p.
σxx ≠ σyy ≠ σzz ≠ -p
By convention, p is defined as the average of the normal
stresses:
p = -1/3(σxx + σyy + σzz ) = -1/3 σii
The fluid element experiences a force on it as a result of
the fluid pressure distribution if it varies spatially.
Consider the net force in the x direction due to p(x,t).
∂p ⎞
⎛
dFx = pdydz − ⎜ p + dx ⎟dydz
∂x ⎠
⎝
∂p
= −
dxdydz
∂x
net
dy
pdydz
∂p ⎞
⎛
⎜ p + dx ⎟dydz
∂x ⎠
⎝
dz
dx
The result will be similar for dFy and dFz; consequently,
we conclude:
⎡ ∂p
∂p ˆ ∂p ˆ ⎤
dFpress = ⎢ − iˆ −
j − k ⎥ ∆∀
x
y
∂
∂
∂z ⎦
⎣
Or:
f = −∇p
force per unit volume due to p(x,t).
Note: if p=constant, f = 0 .
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2.2) Equilibrium of a fluid element
Consider now a fluid element which is acted upon by both
surface forces and a body force due to gravity
or f = ρ g (per unit volume)
dF = ρ g∀
grav
grav
Application of Newton’s law yields: ma = ∑ F
ρd∀ a = (∑ f )d∀
ρ a = ∑ f = f body + f
f body = ρ g = − ρg
f surface = f
f
pressure
f viscous
pressure
surface
k
+ f
viscous
(due to viscous stresses, since in general σ ij = − pσ ij + τ ij )
= −∇p
⎡ ∂ 2V ∂ 2V ∂ 2V ⎤
= µ ⎢ 2 + 2 + 2 ⎥ = µ∇ 2 V
∂z ⎦
∂y
⎣ ∂x
For ρ=constant, the viscous force will have this form (chapter 4).
ρ a = −∇p + ρ g + µ∇ 2 V
with
motn. pres. grav. visc.
a=
∂V
+ V ⋅ ∇V
∂t
This is called the Navier-Stokes equation and will be
discussed further in Chapter 4. Consider solving the N-S
equation for p when a and V are known.
(
)
∇p = ρ g − a + µ∇V = B( x, t )
2
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This is simply a first order p.d.e. for p and can be solved
readily. For the general case (v and p unkown), one must
solve the N-S and continuity equations, which is a
formidable task since the N-S equations are a system of
2nd order nonlinear p.d.e.’s.
We know consider the following special cases :
1) Hydrostatics ( a = V = 0 )
2) Rigid body translation or rotation ( ∇ V = 0 )
2
3) Irrotational motion ( ∇ × V = 0 )
ifρ =cons tan t
∇ ×V = 0 ⇒
P
∇ 2V
= 0 ⇒ Euler equation ⇒
∫
⇒ Bernoulli eqn.
also,
∇ × V = 0 ⇒ V = ∇θ & if ρ = const. ⇒ ∇ 2θ = 0
∇ × (∇ × a ) = ∇(∇ ⋅ a ) − ∇ a
vector identity
2
2.3) Case (1) Hydrostatic Pressure Distribution
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^
∇p = ρ g = − ρg k
∂p ∂p
i.e.
=
=0
∂x ∂y
z
∂p
= −ρ g
∂z
and
g
dp = − ρgdz
2
or
2
2
1
1
p − p = − ∫ ρgdz = − g ∫ ρ ( z )dz
2
1
⎛r ⎞
g = go ⎜ o ⎟
⎝r⎠
≅ const. near
earth ' s surface ro
liquids Æ ρ = const. (for one liquid)
p = -ρgz + constant
gases
Æ ρ = ρ(p,t) which is known from the equation
of state: p = ρRT Æ ρ = p/RT
g dz
dp
=−
p
R T (z )
which can be integrated if T =T(z) is
known as it is for the atmosphere.
2.4) Manometry
Manometers are devices that use liquid columns for
measuring differences in pressure. A general procedure
may be followed in working all manometer problems:
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1.) Start at one end (or a meniscus if the circuit is
continuous) and write the pressure there in an appropriate
unit or symbol if it is unknown.
2.) Add to this the change in pressure (in the same unit)
from one meniscus to the next (plus if the next meniscus
is lower, minus if higher).
3.) Continue until the other end of the gage (or starting
meniscus) is reached and equate the expression to the
pressure at that point, known or unknown.
2.5) Hydrostatic forces on plane surfaces
The force on a body due to a pressure distribution is:
F = − ∫ p n dA
A
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Professor Fred Stern
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Chapter 2
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where for a plane surface n = constant and we need only
consider |F| noting that its direction is always towards the
surface: F = ∫ pn dA .
A
Consider a plane surface AB entirely submerged in a
liquid such that the plane of the surface intersects the freesurface with an angle α. The centroid of the surface is
denoted ( x, y ).
To find the line of action of the force which we call the
center of pressure (xcp,ycp) we equate the moment of
inertia of the resultant force to that of the distributed force
about any arbitrary axis.
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Professor Fred Stern
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ycp F = ∫ ydF
A
= γ sin α ∫ y 2 dA
A
∫ y dA = I → 2nd moment of Inertia about O − O
2
o
A
2
= y A+ I
= moment of inertia w.r.t horizontal centroidal axis
Æ F = pA = γ sin α yA
I
Æ
I
y = y+
yA
F = pA
and
cp
= δ sin α y A
and similarly for xcp
x F = ∫ xdF
cp
where
I = product of inertia
xy
A
I = I + x yA
I
x =
+x
yA
xy
xy
xy
cp
Note that the coordinate system in the text has its origin at the centroid
and is related to the one just used by:
x = x−x
text
and
y = −( y − y )
text
2.6) Hydrostatic Forces on Curved Surfaces
z
x
y
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In general,
Horizontal Components:
F = − ∫ p dA
F = − ∫ p n dA
y
A
y
Ay
^
^
F = F ⋅ i = − ∫ p n ⋅ i dA
dA
x
x
dAx = projection of n A onto a plane perp. to x direction
That is, the horizontal component of force acting on a
curved surface is equal to the force acting on a vertical
projection of that surface (which includes both magnitude
and line of action) and can be determined by the methods
developed for plane surfaces.
^
γ∀
Fz = − ∫ pn ⋅ k dA = − ∫ p dAz = γ ∫ h dAz = δ∀
Az
Az
That is, the vertical component of force acting on a
curved surface is equal to the net weight of the total
column of fluid directly above the curved surface and has
a line of action through the centroid of the fluid volume.
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Professor Fred Stern
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Chapter 2
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2.7) Hydrostatic Forces in Layered Fluids
2.8) Buoyancy and Stability
Archimedes Principle
FB = FV ( 2 ) − FV (1)
= fluid wt. above 2ABC –
fluid wt. above 1ADC
= wt. of fluid equivalent to
body volume
In general, FB = ρg ∀ ( ∀ = submerged volume).
The line of action is through the centroid of the displaced
volume, which is called the center of buoyancy.
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Stability: Immersed Bodies
Condition for static equilibrium: (1) ∑Fv=0 and (2) ∑M=0
Condition (2) is met only when C and G coincide,
otherwise we can have either a righting moment (stable)
or a heeling moment (unstable) when the body is heeled.
For a floating body the situation is slightly more
complicated since the center of buoyancy will generally
shift when the body is rotated, depending upon the shape
of the body and the position in which it is floating.
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Professor Fred Stern
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Chapter 2
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The center of buoyancy (centroid of the displaced
volume) shifts laterally to the right for the case shown
because part of the original buoyant volume AOB is
transferred to a new buoyant volume EOD.
The point of intersection of the lines of action of the
buoyant force before and after heel is called the
metacenter M and the distance GM is called the
metacentric height. If GM is positive, that is, if M is
above G, then the ship is stable; however, if GM is
negative, then the ship is unstable.
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Professor Fred Stern
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Consider a ship which has taken a small angle of heel α
1. evaluate the lateral displacement of the center of
buoyancy, CC ′
2. then from trigonometry, we can solve for GM and
evaluate the stability of the ship
Recall that the center of buoyancy is at the centroid of the
displaced volume of fluid (moment of volume about yaxis – ship centerplane)
x∀ = ∫ x d∀ = ∑ x ∆∀
i
i
This can be evaluated conveniently as follows:
x∀ =
moment of ∀ before heel (goes to zero due to
symmetry of original buoyant volume AKKD
about centerplane)
- moment of ∀ AOB
+ moment of ∀ EOD
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x∀ = − ∫ (− x ) d∀ + ∫ ( x) d∀
AOB
tan α = y
DOE
d∀ = y dA = x tan α dA
x
x∀ = ∫ x tan α dA + ∫ x tan α dA
2
2
OA
OD
= tan α ∫ x dA
2
ship waterplane area
(moment of inertia of ship waterplane = I00, i.e, Izz)
x∀ = I 00 tan α
CC ' = x =
I 00 tan α
∀
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CC ' = CM tan α (from Section View)
CM =
I 00
∀
GM = CM − CG
GM =
I 00
∀
− CG
This equation is used to determine the stability of floating
bodies:
• If GM is positive, the body is stable
• If GM is negative, the body is unstable
Roll:
The rotation of a ship about the longitudinal axis
through the center of gravity.
Consider symmetrical ship heeled to a very small angle θ.
Solve for the subsequent motion due only to hyrdostatic
and gravitational
forces.
^
^
⎛
⎞
Fb = ⎜ cos θ j − sin θ i ⎟ ρg∀
⎝
⎠
r
Fb
( ρg∀ = ∆)
M = r×F
g
b
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M g = ⎛⎜ − GC j + CC ' i ⎞⎟ × ∆⎛⎜ cos θ j − sin θ i ⎞⎟
⎠
⎠ ⎝
⎝
^
^
^
^
= (− GC sin θ + CC ' cos θ )∆ k
^
= (− GC + CM ) sin θ ∆
Note: recall that M = F ⋅ d ,
= GM sin θ ∆
tan θ =
o
where d is the perpendicular
distance from O to the line of
action of F .
CC ' sin θ
=
CM cos θ
M = Gz ∆
..
∑ M = −I θ
G
= GM sin θ ∆
G
I = mass moment of inertia about long axis through G
θ = angular acceleration
..
..
I θ + ∆GM sin θ = 0
.. ∆GM
θ =0
for small θ : θ +
I
∆GM ρg∀GM mgGM
=
+
I
I
I
k=
k =I
2
I
m
m
definition of radius of gyration
mk = I
2
∆GM gGM
=
I
k
2
O
d
F
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The solution to this equation is,
0 for no initial
velocity
.
θ (t ) = θ cos ω t +
o
n
θ
sin ω t
ω
o
n
n
θ = the initial heel angle
where
ω
o
n
= natural frequency
=
gGm
k
2
=
gGM
k
Simple (undamped) harmonic oscillation:
The period of the motion is
T=
T=
2π
ω
n
2πk
gGM
Note that large GM decreases the period of roll,
which would make for an uncomfortable boat ride (high
frequency oscillation).
Earlier we found that GM should be positive if a ship
is to have transverse stability and, generally speaking, the
stability is increased for larger positive GM. However,
the present example shows that one encounters a “design
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tradeoff” since large GM decreases the period of roll,
which makes for an uncomfortable ride.
2.9) Rigid Body Translation or Rotation
In rigid body motion, all particles are in combined
translation and/or rotation and there is no relative motion
between particles; consequently, there are no strains or
strain rates and the viscous term drops out of the N-S
2
equation (µ∇ V = 0) .
∇p = ρ (g − a )
from which we see that ∇p acts in the direction of (g − a ),
and lines of constant pressure must be perpendicular to
this direction (by definition, ∇f is perpendicular to f =
constant).
Rigid body of
fluid translating
or rotating
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Chapter 2
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The general case of rigid body translation/rotation is as
shown. If the center of rotation is at O where V = V 0 , the
velocity of any arbitrary point P is:
V = V 0 + Ω × r0
where Ω = the angular velocity vector
and the acceleration is:
dV0
dΩ
dV
=a=
+ Ω × (Ω × r 0 ) +
× r0
dt
dt
dt
N
2
1
3
1
=
acceleration of O
2
=
centripetal acceleration of P relative to O
3
=
linear acceleration of P due to Ω
Usually, all these terms are not present. In fact, fluids can
rarely move in rigid body motion unless restrained by
confining walls for a long time.
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Professor Fred Stern
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1.) Uniform Linear Acceleration
For uniform rigid body acceleration, a has same
magnitude and direction for all particles. Vector sum of g
and –a gives direction of pressure gradient ∇p , which is
the direction of greatest rate of increase of p.
∇p = ρ (g − a ) = constant
[
= − ρ (g + a ) k + a i
^
z
^
x
]
Consider for example specific one dimensional cases:
dp
= − ρ ax , if a is positive, dp is negative; this means
x
dx
dx
acceleration occurs with decreasing pressure in x
direction, and vice-versa, deceleration occurs with
increasing pressure in x direction.
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Professor Fred Stern
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Similarly in z:
dp
= − ρ ( g − az )
dz
Which means pressure increases with depth unless az>g
For 2-D cases, unit vector in the direction of ∇p (i.e.
direction of greatest rate of increase of p) is given by
^ ∇p
s=
^
| ∇p |
==
^
(g + a z ) k + a x i
[(g + a )
2
z
+ ax
]
1
2 2
Lines of constant pressure are perpendicular to ∇p .
Unit vector in direction of p = constant is given by
^
^
^
a k − (g + a z ) i
n = sˆ × ˆj = x
1
(Where ĵ - unit normal to paper)
2 2
2
a x + (g + a z )
^
angle between n and x axes:
[
]
θ = tan
−1
a
(g + a )
x
z
In general the pressure variation with depth is greater than
in ordinary hydrostatics; that is:
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^
dp
= ∇p ⋅ s = ρ [a + ( g + a ) ]
ds
G
2
2
x
1
2
which is > ρg
z
p = ρGs + cons.
= ρGs
gage pressure
2.) Rigid Body Rotation
Consider a cylindrical tank of liquid rotating at a constant
^
rate Ω = Ω k :
a = Ω × (Ω × r )
0
^
= − rΩ e
2
r
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∇p = ρ ( g − a )
^
^
= − ρg k + ρrΩ e
2
r
∂p
= ρr Ω 2
∂r
i.e.
Integrating, we get
ρ 2 2
p =
2
and
r Ω + f ( z ) + c and
∂p
= − ρg
∂z
p = − ρgz + f (r ) + c
ρ
2 2
Hence f (r ) = 2 r Ω and f ( z ) = − ρgz
Thus,
p=
ρ
2
r Ω − ρgz + const .
2
2
The constant is determined by specifying the pressure at
one point; say, p = p0 at (r,z) = (0,0).
1
p = p0 − ρ gz + r 2Ω2
2
(Note: Pressure is linear in z and parabolic in r)
Curves of constant pressure are given by:
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p −p rΩ
z=
+
= a + br
ρg
2g
2
2
0
2
which are paraboloids of revolution, concave upward,
with their minimum points on the axis of rotation.
The position of the free surface is found, as it is for linear
acceleration, by conserving the volume of fluid.
The unit vector in the direction of ∇p is:
^
^
s=
See Fig 2.23
on page 85
tan θ =
^
− ρg k + ρrΩ e
[( ρg )
2
r
2
+ ( ρrΩ )
2
dz
= −g
rΩ
dr
2
z
]
1
2
^
2
slope of s
r
θ
^
⎛ Ω z⎞
i.e. r = C exp⎜ −
⎟
g
⎝
⎠
2
1
s
equation of ∇p surfaces
2.10) Pressure Distribution in Irrotational Flow
If the flow is irrotational,
∇ ×V = 0
(
i.e.
)
∇p = ρ g − a + µ∇ 2 V
ω=0
(vorticity)
incompressible flow
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now use the vector identity to show that the viscous terms
dropout automatically for incompressible irrotational flow
∇ 2 Vv = ∇ (N
∇ ⋅ Vv) − ∇ × (∇ × Vv)
=0
=0
ρ = const.
ω =0
^
∇p = − ρg k − ρ a
a = −∇ ⎛⎜ p + gz ⎞⎟
ρ
⎝
⎠
piezometric
pressure
Or
Euler’s equation
DV ∂V
=
+ V
⋅ ∇
V = −∇⎛⎜ p + gz ⎞⎟
⎝ ρ
⎠
dt
∂t
Or
1
V ⋅ ∇V = ∇ VN
⋅ V − V × (∇ × V )
2
V2
=0
vector identity
Lastly, for steady flow:
⎛1
⎞
∇⎜ V 2 + p + gz ⎟ = 0
ρ
⎝2
⎠
Or
p+
ρ
2
V 2 + ρgz = constant
(Bernoulli eq.)
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=
B
For steady rotational flow, it is also possible to derive a
Bernoulli equation; however, in this case
B = B(ψ)
that is, the Bernoulli constant varies from streamline to
streamline.