Module # 4
Flow Past of Immersed
Bodies
1 1
Incompressible Flow
2 2
Flow Around Objects
3 3
FLUID FLOW ABOUT IMMERSED BODIES
Drag due to surface stresses composed
of normal (pressure) and tangential
(viscous) stresses.
p7 p8
p6 5 6 p9
7
p5 4 8 p10
U DRAG
p4 3 p11
9
p3 2 p12
10
p2 1 ……
p13
p1 p…
Fluid Resistance
The transmission of energy from an object passing through a
fluid to the fluid is known as fluid resistance.
The resistance of an object passing through a fluid increases
as the speed of the object increases and as the viscosity of the
fluid increases.
Drag
◦ Is the
resistance an
airplane
experiences
in moving
forward
through the
air
Fn P A P : pressure
At any point on surface:
Ft A : shear stress
Integrate pressure and shear stress distributions around body
surface
Drag FD - component of resultant force in direction of flow
Lift FL - component of resultant force perpendicular to
direction of flow
Concept of Drag
Drag is the retarding force exerted on a moving body
in a fluid medium
It does not attempt to turn the object, simply to slow
it down
It is a function of the speed of the body, the size (and
shape) of the body, and the fluid through which it is
moving
Drag Force Due to Air
The drag force due to wind (air) acting on an object
can be found by:
FD = ½ ρ CDV2A
where: FD = drag force (N)
CD = drag coefficient (no units)
V = velocity of object (m/s)
A = projected area (m2)
ρ = density of air (kg/m3) {1.2 kg/m3}
Surface and Form Drag
Surface drag is a result of the friction between the
surface and the fluid.
The fluid closest to the object (boundary layer) rubs
against the object creating friction.
Form drag occurs when air is driven past an
object and is diverted outward creating a low
pressure region behind the object.
low pressure
high pressure
Form Drag
Low form drag
The orientation of the
object will affect the
frontal area and will
play an important role
in the amount of form
drag.
High form drag
Lift and Drag
shear stress and pressure integrated over the surface of a
body create force
drag: force component in the direction of upstream
velocity
lift: force normal to upstream velocity (might have 2
components in general case)
D
D dFx p cos dA w sin dA CD 1 2
2 U A
L
L dFy p sin dA w cos dA CL 1 2
2 U A
CD = FD/(1/2 U2A) = CD,pressure + CD,friction
Projected Area
The projected area used in the FD is the area “seen” by the
fluid.
Spherical Particle
2
2 D
A R
4
Projected Area
For objects having shapes other than spherical, it is
necessary to specify the size, geometry and orientation
relative to the direction of flow.
Cylinder
Axis perpendicular to flow Rectangle A LD
2
D
Axis parallel to flow Circle A
4
Resultant
FR
Lift FL
C
A B
drag D
FD
Drag force due
to pressure difference
low pressure region
motion of air
high pressure region motion of object
Drag force due
to pressure difference
flow speed (high) vair + v
reduced pressure
v
vair (vball)
v flow speed (low) vair - v
increased pressure
Boundary layer – air
sticks to ball
high pressure region (viscosity) – air
low pressure region dragged around with
ball
lift
Drag Coefficient
For slow flow around a sphere and Re <10
24 24
Cd
Re Du0
2
Recall:
C d A u 0
FD
2
Stokes’ Law for Creeping Flow Around Sphere
FD 3 Du0
Flow past an object
Character of the
steady, viscous flow
past a circular
cylinder: (a) low
Reynolds number
flow, (b) moderate
Reynolds number
flow, (c) large
Reynolds number
flow.
Effect of pressure gradient
inviscid flow viscous flow
Examples
• b/h =1 square, CD = 1.18; (disk; CD = 1.17)
• CD independent of Re for Re > 1000
Question: CD = FD/(1/2 U2A)
What happens to CD if double area (b/h 2b/2h)?
What happens to FD if double area (b/h 2b/2h)?
Drag dependence
Drag Coefficient
Re 10 Cd 24 Re Re 1000 Cd 0.44
for external flow: Re > 100 dominated by inertia, Re < 1 – by viscosity
Why Different Regions?
As the flow rate increases wake drag becomes an important factor.
The streamline pattern becomes mixed at the rear of the particle thus
causing a greater pressure at the front of the particle and thus an extra
force term due to pressure difference. At very high Reynolds numbers
completely separate in the wake.
Streamline separation
Example
A cylindrical bridge pier 1 meter in diameter is submerged to a depth
of 10m in a river at 20°C. Water is flowing past at a velocity of 1.2
m/s. Calculate the force in Newtons on the pier.
water 998.2kg m 3
water 1.005 x10 3 kg m s u0 1.2m s
Cd A u02
Fk
2
3
u0 D 998.2 kg m 1.2 m s 1m 6
Re 1.192 10
1.005 10 3 kg m s
From figure Cd ≈ 0.35
Projected Area = DL = 10 m2
2
0.35 kg 2 m
Fk 10m 2 998.2 3 1.2 2 2,515 N
2 m s
28
29
STREAMLINING
Streamlining is the attempt to reduce the drag on a body
CD ~ 2 for flat plate CD ~ 0.06
Streamlining
The less drag you have…
◦ Flying a glider: the further you
can fly
◦ Flying an airplane: the less fuel
you use
Therefore streamlining is important
◦ A design device by which a body
is shaped to minimize drag
In general, the importance of
streamlining to reduce drag.
2-D rectangular cylinder
STREAMLINING
STREAMLINING
2 cm
~ same drag AND wake
Non-circular Channels
Equivalent diameter defined as 4 times the hydraulic radius
(rH).
A
r
H
L p
Where, A = cross-sectional area of channel
Lp = perimeter of channel in contact with fluid
Hydraulic radius of circular tube,
D 4 D
2
r
H
D 4
The equivalent diameter is 4 rH.
For a rectangular duct with width W and height H, the
hydraulic diameter is
4A 4WH 2WH H
Dh
P 2(W H ) W H
W
Annulus between two circular pipes
D D 2 2
0
D D
i
r 4 4 0 i
D D 4 i 0
Sphericity
Surface area of sphere, Sp = 4 r 2 = Dp2
Volume of sphere, Vp = (4/3) r 3 = (1/6) Dp3
Sphericity (s) : The surface-volume ratio for a sphere of
diameter Dp divided by the surface-volume ratio for the
particle whose Nominal size is Dp.
6
D S 6
S
p
p
V D
s
v p
p s p
37
38
39
40
41
42
Response to Superficial Velocities
44
45
46
Problem
Air ( = 1.22 Kg/m3, = 1.9 X 10 –-5 pa.s) is flowing
in a fixed bed of a diameter 0.5 m and height 2.5 m. The
bed is packed with spherical particles of diameter 10 mm.
The void fraction is 0.38. The air mass flow rate is 0.5
kg/s. Calculate the pressure drop across the bed of
particles.
47 1
Solution
Q = volumetric flow rate = = 0.41 m3/s
A = D2 = (0.5)2 = 0.1963 m2
u = = = 2.1 m/s
Rep = =
Rep = 2174
fp = + 1.75 = 1.819 =
0.276 x 105 pa.
48 1
