che354 immersed
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Flow Around Immersed Objects
Incompressible Flow
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Goals
Describe forces that act on a particle in a fluid.
Define and quantify the drag coefficient for
spherical and non-spherical objects in a flow
field.
Define Stokes and Newtons Laws for flowaround spheres.
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Flow Around Objects
There are many processes that involve flow through aporous medium such as a suspension of particles:
Packed Bed Chemical Reactor
Food Industry
Oil Reservoirs
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Forces
Dynamic
Fk results from the relative motion
of the object and the fluid (shear
stress)
Static
Fs results from external pressure
gradient (Fp) and gravity (Fg).
pgk FFFFMM 12
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Dynamic Forces
A drag force (friction) arises in situations in which
moving fluids are in contact with a solid surface (recall
pipe flow.
Fwis the kinetic force exerted on the solid wall or
particle, A is the wetted surface area, K is the kinetic
energy per unit volume, and fis the friction factor.
K
AFf w
2
2
V
w
headvelocitydensity
shearwallf
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Dynamic Forces
For flow around a submerged object a drag coefficientCd is defined and the equation becomes:
2
2
0uAC
F dk
U0 is the approach velocity (far from the object), is
the density of the fluid, A is the projected area of the
particle, and Cd is the drag coefficient analogous tothe friction factor in pipe flow (keep this in mind).
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Projected Area
The projected area used in the Fkis the area seen bythe fluid.
Spherical Particle
A 4
2
2 DR
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Projected Area
Cylinder
For objects having shapes other than spherical, it is
necessary to specify the size, geometry and orientationrelative to the direction of flow.
Axis perpendicular to flow Rectangle LDA
Axis parallel to flow Circle
4
2D
A
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Drag Coefficient
The drag coefficient, like the friction factor in pipes
depends on the Reynolds number
0DuRe
D is particle diameter or a characteristic length and
and are fluid properties.
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Drag Coefficient
For slow flow around a sphere and Re
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Drag CoefficientReCRe d 2410 44.01000 dCRe
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Why Different Regions?As the flow rate increases wake drag becomes an important
factor. The streamline pattern becomes mixed at the rear ofthe particle and at very high Reynolds numbers completely
separate in the wake. This causes a greater pressure at the
front of the particle and thus an extra force term due to
pressure difference.
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Static Forces
Static forces exist in the absence of fluid motion. They
include the downward force of gravity and the upwardforce of buoyancy that results from the gravity induced
pressure gradient in the z-direction
gVgmF pppg 1P
ghPP f 12
gV
ghAPPAF
fp
fb
12
bF
gF
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Total Force
The gravity and buoyancy forces on an object immersedin liquids do not generally balance each other and the
object will be in motion.
bgkt FFFF
bF
gF
What is the direction of Fk?
It is always opposite to the direction ofparticle motion
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Equilibrium
When a particle whose density is greater than that of the
fluid begins to fall in response to the force imbalance, it
begins to accelerate (F=ma). As the velocity increases
the viscous drag force also increases until all forces are in
balance. At this point the particle reaches terminal
velocity.
bgk FFF 0
gVgmuACF fppt
fpd 20
2
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Terminal Velocity
If the particle has a uniform density , the particle
mass is Vpp and
gVuAC gfptfpd 2
0
2
fd
pfp
t
C
Dgu
3
4
Use: Falling ball viscometer to measure viscosity
fdp
pfp
t
C
mgu
3
2
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Settling VelocityThe settling (terminal) velocity of small particles is
often low enough that the Reynolds number is lessthan unity (Cd = 24/Re).
18
2gD
ufpp
t
1Re
Between 1000
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Criterion for Settling Regime
Reynolds number is a poor criteria for determining theproper regime for settling. We can derive a value K that
depends solely on the physical parameters
312
fpfp
g
DK
K < 2.6 Stokes Law
Newtons LawK > 68.9
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Example
A cylindrical bridge pier 1 meter in diameter is submerged
to a depth of 10m in a river at 20C. Water is flowing pastat a velocity of 1.2 m/s. Calculate the force in Newtons on
the pier.
smkgx
mkg
water
water
3
3
10005.1
2.998
smu 2.10
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Example
2
2
0uACF dk
6
3
3
0 10192.1
10005.1
12.12.998
smkg
msmmkgDuRe
Fig. 7.3 gives Cd 0.35
Projected Area = DL = 10 m2
Ns
m
m
kgmFk 515,22.12.99810
2
35.02
22
3
2
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ExampleEstimate the terminal velocity of limestone particles
(Dp = 0.15 mm, = 2800 kg/m3) in water @ 20C.
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Example
9.3
10005.1
2.99828002.99898.9
105.1
31
2
3
3324
sm
kgm
kg
m
kg
s
m
mK
Guess Re = 4 Cd = 9.0
s
m
m
kg
mm
kg
s
m
ut 02.0
2.9980.93
105.12.99828008.94
3
4
32
310005.1
02.02.998105.13
34
smkg
smmkgmRe
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Example
Guess Re = 2 Cd
= 15
U0 = 0.015 m/s Re = 2.3
Guess Re = 2.5 Cd = 12
U0 = 0.017 m/s Re = 2.6
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