DRAG AND LIFT FORCES ‘C wp - IIT Bombay
Transcript of DRAG AND LIFT FORCES ‘C wp - IIT Bombay
DRAG AND LIFT FORCES ‘C ""wp CDEEP
IIT Bombay
36 Slide
fluid causes a net force to act on the body
• Force acting on an elemental area dA of the body will be
neither normal nor parallel to the surface
• The resultant force can be resolved into components parallel
to the direction of flow i.e. the free stream, and
perpendicular to the flow
• The force parallel to the direction of motion is known as
drag
• The component normal to the flow direction is called the lift
force
• The relative motion between a solid body and the
DRAG FORCES CDEEP
IIT Bombay
CE 223 I 3L 2.. • In the case of real fluids, both the shear forces and
the pressure forces act simultaneously on the surface
• The part of the drag force arising out of the viscous action is
known as the viscous drag or the skin- friction drag
• The other part of the drag force is caused due to pressure
force acting on the surface and is known as the pressure
drag, or the form drag
• One can write
FD (total drag force) = Fviscous + Fform
DRAG FORCES (Contd.) CDEEP
IIT Bombay
CE 223 1,31:1 /Slicle3
• The relative importance of the viscous and form
drags depends on the shape and orientation of the body
• When a thin flat plate or a disk is held parallel to the free
stream, viscous force predominates and pressure drag force
is zero
• If the plate is held normal to the direction of flow, the drag
force experienced by the body is due to the pressure
differentials and viscous drag is zero
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DRAG FORCES (Contd.)
CDEEP IIT Bombay
CE 223 L3k)
(a) Viscous drag (b) Pressure drag
COEFFICIENT OF DRAG Co
Body shape CD Range of Re
2-D Circular cylinder 1.2 104 to 1.5x10 5
Elliptical cylinder (4:1) 0.32 2.5x104 to 105
2-D Square cylinder 2.0 3.5x104
Circular disc 1.17 > 103
Open hemisphere (concave
side facing flow) 1.4 > 104
Open hemisphere (convex
side facing flow) 0.4 104
CDEEP IIT Bombay
CE 223 L3L Slide
CD D rveL
1.328
DRAG COEFFICIENT FOR FLAT PLATES CDEEP
IIT Bombay
CE 223 L34 /Slide
Laminar boundary layer
CD Turbulent boundary layer beginning from R
the leading edge, 5x10 5 < Rey < 10 7
• For ReL <109 , the empirical equation of Schlichting may be
used
CD= 0 . 455/(log Re L) 2.58
0.074
5 eL
DRAG COEFFICIENT FOR FLAT PLATES (Cont.) CDEEP
IIT Bombay
CE 223 1_315/Slide/
The revised values of the coefficient of drag for a flat
plate, accounting for laminar boundary layer, can be written as
CD 0.074 1740
1 ReL
R eL
(5x105 <Ret_<107 )
CD 9-940 0,45 1610
(Log R e L) 2 . 58
ReL (5x105 <R,L <109 )
Where Rey , is the plate Reynolds number (= VL/v)
c
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tO 10 0
- IQ 10 •
- -t0101
"Z.:6•■•44"... ••••■■ "•-■
"•• ■■•.."1„, 6•••■
DRAG COEFFICIENT OF SMOOTH PLATES CDEEP
IIT Bombay
CE 223 L2, /Slide 8/
10 1D6
10'
RF =U * L/V
Variation of the coefficient of drag with the Reynolds number
DRAG COEFFICIENT OF SMOOTH PLATES (Contd.) CDEEP
IIT Bombay
22:-) [3i2 .'Slide
• The lower straight line is meant for a laminar
boundary layer and the upper curves are for turbulent
boundary layers
• The curve in the middle, connecting the two, represents the
variation of CD in the transition flow regime.
• Contribution of the laminar drag is insignificant for ReL>107.
TUTORIAL CDEEP
IIT Bombay
CE 223 1,36 -;11(1> to
A super tanker is 360 m long and has a beam width of
70 m and a draft of 25 m. Estimate the force and power
required to overcome skin friction drag at a cruising speed of
13 knots in seawater at 10° C.
[1 knot = 1852 m per hour; at 10° C viscosity j = 1.4 x 10 -6
M 2 Is]
SOLUTION CDEEP
IIT Bombay
CE 223 LILSlide I I
L = 360 m, U = 1852x13
= 6.69 m/s, v = 1.4 x 3600
10 -6 m 2/s
RL = 6.69x360 = 1.72 x 10 9 1.4x 10 -6
0.455 CD! = (log RL)2.58
1610 (Valid for 5 x 10 5 < RL < 109 )
R L
= 0.00147 — 0.0000016 = 0.00147
pU2 = Z x 1020 x 6.69 2 = 22825.6 N/m 2
Total area to be considered = bottom + sides
SOLUTION (Contd.) CDEEP
IIT Bombay
Cr 223 L312.Slidel2
= (360 x 70) + (360 x 25) x 2
= 25,200 + 18,000 = 43,200 m 2
Therefore, F=CDf 1x 2-xpxU2 xA= 0.00147 x
43,200 x 22,825.6
= 1.45 x 10 6 N = 1.45 M-N
Power = F. U = 1.45 x 10 6 x 6.69 = 9.7 x 10 6 N- m/s =
9.7M- W
FORCES ON BLUFF BODIES C DEEP
IIT Bombay
CL 223 33 I i '3
• Bodies which create a large wake in the flow are classified
as bluff bodies
• Circular cylinders, spheres, elliptical cross sections,
rectangular cross-sections of finite aspect ratios, a flat plate
held normal to the flow, are some examples of the bluff
bodies
• For bluff bodies the frictional drag component is very small
relative to the form drag
• The coefficient of drag CD is independent of the Reynolds
number, above a threshold value of Re .
FORCES ON BLUFF BODIES (Contd.) CDEEP
IIT Bombay
CE 223 L36 -,1;1(‘ 14
▪ Drag coefficient curve is flat in the range 10 3 <Re<3*10 5
• For Re of around 3x10 5 , the laminar boundary layer on the
font part of the sphere undergoes a change and the
boundary layer becomes turbulent
• In a laminar boundary layer the fluid particles moving close
the surface are able to overcome the resistance, due to
viscous action, in the presence of a favorable pressure
gradient (dp/dx<0) in the front half of the sphere
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' ' •
SMOOTH SPHERE CDEEP
IIT Bombay
• The Reynolds number for a smooth sphere at which
this sudden drop in Co takes place, as shown in
Figure below is known as the critical Reynolds number
• The drag coefficient with a turbulent boundary layer is
approximately 20% of that with the laminar boundary layer
Variation of Co with Re for a smooth sphere
FORCES ON BLUFF BODIES (Contd.) CDEEP
IIT Bombay
CE 223 Laix/Slidel5
• The separation of flow occurs just upstream of the
midsection, a little before the fluid particles are subject to a
'pressure-hill' on the rear half
• The pressure difference between the front and the rear is
the main cause for the drag
• Slow moving particles around the midsection acquire more
momentum and the turbulent boundary layer is able to
resist flow separation for some more distance over the
sphere
Theoretical
(Inviscid)
Turbulent
1.0
C,
0.5 Laminar
1.0
1.5 0 60 120 180
e (degrees)
FORCES ON BLUFF BODIES (Contd.) C DEEP
IIT Bombay
CE 223 L34 /Slide /6
Pressure distribution around a smooth sphere for laminar and turbulent
boundary-layer flow, compared with inviscid flow
CIRCULAR CYLINDER CDEEP
IIT Bombay
GE 223 L3L/Sliciell
• The velocity is zero at the stagnant points located at the
front and rear of the cylinder
• The maximum value of v 0 =-2U and it occurs for 0=90°
• The pressure is maximum at the upstream stagnation point,
drops to a minimum at 0=90° and recovers to attain a
maximum value at the downstream stagnation point
• The net force due to the differential pressures on the
circular cylinder is zero
CIRCULAR CYLINDER (Contd.)
Cp -
6.7 x 105
Re =1.1 x 104
20 40 60 eao 100 120 140 160 180
Angle from Forward Stagnation point (degrees)
Pressure distribution around a smooth cylinder
CIRCULAR CYLINDER (Contd.) CDEEP
IIT Bombay
CE 223 3E) !Side
• The pressure difference on the front and the rear surface of
the cylinder gives rise to a significant drag force, the
pressure drag
• The experimental data of the turbulent boundary layer
follow the potential flow results better than the laminar
boundary layer case
• The total drag experienced by the cylinder drops suddenly
at the critical Reynolds number, Rec = 3 X 10 5
CIRCULAR CYLINDER (Contd.) CDEEP
III Bombay
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Variation of CD with R e for a circular cylinder (SouRc.C. SChike-h+L1'3,1768)