Drag force lift force

Submerged bodies

1. DRAG FORCE & LIFT

2. BODI E S

3. You all must have come across soothing breeze flow past around your face. It gives a delightful pleasure to one’s heart. But have you ever wondered about science behind it ????

4. When a body moves through a moving a fluid, it is acted upon by two forces exerted by the flow upon it:- Shear Force. Pressure Force. Resolving these forces into horizontal and vertical components:- The resultant of shear & pressure forces acting in the direction of flow, is called Drag Force . The resultant of shear & pressure forces acting normally to the direction of fluid flow, is called Lift force .

5. D R A G 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. What does drag look like? C d = F d 1 ρv 2 C d Drag Coefficient A F d Drag force 2

6. LIFT Represents a force that acts perpendicular to the direction of relative motion between fluid & body. Created by different pressures on opposite sides of a blunt body due to boundary layer separation. Bernoulli’s Principle :- Pressure is inversely propor tional to velocity. Therefore, Fast relative velocity, lower pressure. Slow relative velocity, higher pressure. What does Lift look like ? C L Lift Coefficient F L Lift Force • .

7. Wind Mill RBC Transportation in Body

8. Parachuting is made possible only due to Drag . Motion of sub-marine is due to drag force acting on body immersed in water bodies . Generation of power using turbine mills .

9. Streamlined Bodies- A streamlined body is a shape that lowers the friction between fluid like water and air, and object moving through that fluid. Blunt Bodies- A blunt body is the one that as a result of its shape has separated flow over a substantial part of its surface.

10. 2 -Dimensional fluid flow :- The flow over a body is said to be 2 dimensional when the body is very long & of constant cross-section area & the flow is normal to the body. A xis – S ymmetric fluid flow:- The flow in which the body possess rotational symmetry about an axis in flow direction. 3- Dimensional fluid flow:- Flow over a body that cannot be modeled as two- dimensional or axis-symmetric, is called as 3- dimensional fluid flow.

11. Shear / Skin Friction Drag :- The part of the drag that is due directly to wall shear stress( τ w ) is called skin friction drag. C D,friction = F D,friction 1 ρv 2 A 2 Pressure / Form Drag :- The part of drag that is directly due to pressure P, is called the Pressure / Form Drag. C D,Pressure = F D,Pressure 1ρv 2 A 2

12. r w e r On orientation of body. Magnitude of wall shea stress. Viscosity of fluid Reynolds number of flo Planform surface area Surface roughness und turbulent conditions.

13. Becomes significant only when fluid velocity is very high. Layer separation. Orientation of body.

14. The wings of air planes are streamlined so as to minimize the drag force experienced by it & maximize the lift force that helps it in initial take off. The wings are inclined at an angle of 5 (ranging between 5 to 15 ), called angle of attack . In case of turbulent flows, pressure drag dominates over shear drag, which is reduced simply by reducing the frontal surface area, which in turn reduces boundary layer separation, thus minimizing pressure drag.

15. Methods of Preventing the Separation of Boundary Layer: When BL Separates certain portion adjacent to the surface has back flow and eddies are continuously formed in this region and hence continuous loss of energy takes place. Thus separation of BL is undesirable and attempts should be made to avoid separation by various methods. Methods for prevent Separation of BL Suction of the slow moving fluid by a Suction slot Supplying additional energy from a blower Providing a by pass in the slotted wing Rotating boundary in the direction of flow Providing small divergence in a diffuser Providing guide blades in a bend Providing a trip-wire in the laminar region for the flow over a sphere

16. When a blunt body obstructs the path of a fluid flow, the flow layer unable to remain in contact in surface gets separated, resulting in formation of wake region . The fluid re-circulate in low pressure wake region, the phenomena known as Vortexing . The point where the flow layer separates is called separation point . The whole phenomena results in formation of high & low pressure regions at opposite ends of body, thus increasing the pressure drag acting on body.

17. A flat plate 1.5m x 1.5m moves at 50km/h in stationary air of specific weight 1.15kgf/m 3 . If the coefficient of drag and lift are 0.15 and 0.75 respectively, Determine : The lift force, The drag force, The resultant force.

18. Sol: Area of plate(A) =1.5 x 1.5= 2.25m 2 1 3 6 m/sec velocity of plate(U) =50km/h = 50 x specific weight of air, w= 1.15kgf/m 3 𝑤 1.15 Density of air , ρ = = =0.1172 kg/m 3 𝑈 2 13.89 2 2 𝑔 9.81 coefficient of drag ,C D =0.15 Coefficient of lift, C L = 0.75 Using equation F L = C L Aρ 2 = 0.75 x 2.25 x0.1172 x =19.078 kgf 𝑈 2 Using equation, F D = C D Aρ 2 13.89 2 2 =0.15 x 2.25 x0.1172 x = 3.815 kgf 2 2 Resultant force, F R = F L + F D = 19.078 2 + 3.815 2 = 363.47 + 14.554 kgf =19.455 kgf

19. Boundary Layer Concept N m a g n et S Free stream velocity Flow over a plate Laminar boundary layer Leading edge Laminar sub layer Turbulent boundary layer Highly viscous region

20. Laminar boundary layer Re = 5×10 5 Inertial force U×X×ρ viscous force = μ = where ,U= Free stream velocity X= Distance from leading edge μ = dynamic viscosity of fluid ρ=density of fluid Turbulent boundary layer R e > 5×10 5 Laminar sub layer τ = μ ∂u ∂y Boundary layer thickness

21. = ρ × u × b × dy. Distance BC= Let y = distance of elemental strip from the plate dy = thickness of elemental strip u= velocity of fluid at elemental strip D=width of plate Area of strip dA =D × dy Mass of fluid per second flowing through the element strip =ρ × velocity × area of s C t o ri n p sider no plate is present, so mass of fluid flowing through elemental strip= ρ × velocity × area = ρ× U × bdy The reduction in mass per second flowing through elemental strip = ρ Ubdy- ρ ubdy = ρ b( U - u) d y Total reduction in mass per second flowing through BC due to plate = ρ b ∫ (U-u) dy Let the plate is displaced by *, then the velocity of flow per the distance * is equal to the free stream velocity So, the loss of mass of fluid per second will be at *= ρ × velocity × area = ρ× U × * × b

22. Equating equation 3 and 4 ,we get; = ρ b ∫ (U - u ) dy = ρ U *b * = ∫ (1- u ) dy. U

23. Momentum thickness The distance by which a boundary layer should be displaced for the compensate for the reduction in momentum of the flowing fluid on account of boundary layer formation. The area of element strip = b dy Mass of fluid on elemental area= ρ × velocity × area = ρubdy Momentum of fluid = mv =ρubdy u = ρu 2 bdy If plate is not considered; Momentum of fluid= ρuUbdy Lass of momentum= ρb(uU-u 2) dy T ot a l l o ss in mome n tum= ρbU ∫ u(1-u ) dy U

24. let θ = distance by which plate is displaced when the fluid is flowing with constant velocity U , Therefore loss of momentum per second of fluid flowing through distance θ with velocity U =mass of fluid through θ × velocity =ρ × velocity × area =(ρθ×b ×U) ×U = ρθbU 2 Equating eq.1 and 2; θ= ∫ u (1- u ) dy U U

25. The shear stress is given by =(μ 𝜕𝑢 ) 𝜕𝑦 y=0 Drag force or shear force on the small distance is given by ∆F d = ∆xb Drag force ∆F d must be equal to the rate of change of momentum over the distance ∆x The mass rate of flow entering through AD = ∫ ( ρ × velo c ity × area of thic k ne s s dy) = ∫ ρ ubdy The mass rate of flow leaving side BC 𝜕𝑥 mass through AD+ 𝜕 (mass through AD) × ∆𝑥 = 𝛿 𝜌𝑢𝑏𝑑𝑦 + 𝜕 [ 𝜕𝑥 𝛿 𝜌𝑢𝑏𝑑𝑦 ] ∆𝑥 From continuity equation for a steady incompressible flow ,we have M AD +M DC =M BC

26. M DC = M BC – M AD 𝜕 𝛿 = 𝜕𝑥 [𝜌𝑢𝑏𝑑𝑦 ]∆x 𝛿 momentum flux entering through AD= [𝜌𝑢𝑏𝑑𝑦] x [u] 𝛿 = 𝜌 u 2 bdy 𝜕𝑥 0 0 𝛿 𝜕 𝛿 Momentum flux entering through side BC = 𝜌 u 2 bdy = [ 𝜌 u 2 bdy] ∆𝑥 𝜕 𝑥 𝜕 𝛿 So, momentum flux through DC = [ 𝜌 u bdy] ∆𝑥 x U Rate of change of momentum of control volume = MF BC - MF AD –MF DC 𝜕 𝑥 𝛿 𝜕 𝛿 { 𝜌 u 2 bdy + [ 𝜌 u 2 bdy] ∆𝑥 }-{ 𝛿 2 𝜌 u bdy }- { 𝜕 𝜕 𝑥 𝛿 [ 𝜌 u bdy] ∆𝑥 x u} 𝜕 𝑥 = 𝜌𝑏 𝜕 [ 𝛿 (𝑢 2 - 𝑢𝑈)𝑑𝑦] × ∆𝑥

27. Total external force is in the direction of rate of change of momentum; 𝜕𝑥 - 𝜏 ∆ 𝑥 b = 𝜌𝑏 𝜕 [ 𝛿 (𝑢 2 - 𝑢𝑈)𝑑𝑦] × ∆𝑥 𝜏 𝜌𝑢 2 𝜕𝑥 𝑈 𝑈 = [ 𝜕 𝛿 𝑢 (1- 𝑢 ) 𝑑𝑦 ] 𝜏 𝜌𝑢 2 = 𝜕 𝜃 𝜕𝑥 Where ; θ= Momentum thickness This is Von karman momentum integral equation This equation is applicable for all type of boundary layers.

28. Local coefficient of Drag [C D *] D C * = 𝜏 𝑈 2 𝜌 2 Average drag coefficient [C D ] D C = 𝐹 𝐷 𝑈 2 𝐴𝜌 2 Wh e re; A = Area of plate U= free stream velocity ρ =mass density of fluid

29. Friction coefficient The friction coefficient for laminar flow can be determined experimentally by using conservation of mass and momentum together. Average friction coefficient for entire plate can be determined as: For laminar flow; Boundary layer thickness( 𝜹 )= for C f,x = 0.664 Re< 5×10 5 4. 91x R e 1/2 R e 1/2 The corresponding relation for turbulent flow are 𝜹 = 0.38 x R e 1/5 for C f,x = 0.05 9 5×10 5 <Re<10 7 R e 1/5 C f = 1 ∫ C f,x dx L

30. In some cases a flat plate is sufficiently long enough to become turbulent but it is not long enough to disregard the laminar flow regimes So , Average friction coefficient for determining the friction coefficient for the region = ∫ X C f,x laminar dx + ∫ L C f,x turbulent dx C f = 0.074 - 1742 5×10 5 <Re<10 7 R e 1/5 R e For laminar flow, the friction coefficient depends on only the Reynolds number ,and the surface roughness has no effect. further flow, however, surface roughness causes the friction coefficient.

31. According to Navier stokes equation, drag force on sphere: F D = 3 𝜋𝜇𝐷𝑈 ,

32. Expression for C D for Reynolds number less than 0.2 F D = C D 2 A 𝜌𝑈 2 F D =3 𝜋𝜇𝐷𝑈 Equating both we get; 3 𝜋𝜇𝐷𝑈 = C D 2 A 𝜌𝑈 2 4 2 Take A= 𝜋𝐷 ; C D = 24 𝑅 𝑒 Expression for C D when 0.2< R e <5 D C = 24 [1+ 3 𝑅 𝑒 16𝑅 𝑒 ] Note: 1:) for 5<R e <1000; C D =0.4 2:) For 10 3 <R e <10 5 ; C D = 0.5 3:)For R e >10 5 ; C D =0.2 

33. Calculate the weight of ball of diameter 8cm which is just supported in a vertical airstream which is flowing at a velocity of 7m/sec. The specific weight of air is given as 1.25kg/m 3 .The kinematic viscosity of air =1.5 stokes.

34. Diameter of ball D = 8cm =0.08m Velocity of air U = 7m/sec Specific weight of air, w =1.25kgf/m 3 so, density of air , ρ= 𝑤 = 1.25 =0.1274 slug/m 3 𝑔 9.81 Kinematic viscosity 𝜗 = 1.5stokes =1.5 x 10 -4 m 2 /sec Reynolds number, Re= 𝑈 𝑋 𝐷 = 7 𝑋 0.08 x 10 4 =3730 𝜗 1.5 Thus the value of Re lies between 1000 to 100000, And hence C D =0.5 F D = C D Aρ 𝑈 2 2 =0.2 x 0.005026 x .01274 ×7 ×7 2 = 0.007843 kgf