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

1

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> ., T0 (( te r. ;

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•••■ • J'.., -....Z.:02•- .0

■• • . _••.., . •••• „.....„ .. ••......

y

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

- 1010 2

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

T1

4 6110 '! 4 MIMO 1 2 4 210 4 210 2 4 MO. 2 4 Me 2 4 MOS I0

' ' •

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

36 Ht 2.0

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100 80 80

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Variation of CD with R e for a circular cylinder (SouRc.C. SChike-h+L1'3,1768)