Friction Drag + Pressure Drag

Home , Inviscid flow

Objective: To determine the aerodynamic lift and drag forces, FL and FD, respectively, experienced by a circular cylinder placed in a uniform free-stream velocity, U∞

FD Flow separates U∞

Whenever there is relative motion between a solid body and the fluid in which It is immersed, the body experience a net force F due to the action of the fluid

Drag is the component of force on a body acting parallel to direction of motion

Lift is the component of resultant force perpendicular to the fluid motion F = ∫ dF shear + ∫ dF pressure body surface body surface

dF shear = τ wdA

i d F pressure = − pd A j F = F i + F j L D Flow direction

The resultant force F can be resolved into components, parallel and perpendicular to the direction of motion, FD (Drag) and FL (Lift) Drag

Drag is the component of force on a body acting parallel to the direction of motion

Total drag D = Friction Drag + Pressure Drag

Caused by shear stress Caused by pressure gradient across the body due to flow separation Consider uniform flow over a flat plate : Friction Drag U Boundary Layer profile U U

Laminar Transition Turbulent Friction Drag U Boundary Layer profile U U u

Laminar Transition Turbulent

for uniform form flow, pressure gradient is zero

Total drag D = Friction Drag + Pressure Drag

du F = ∫ τ dA τ = µ D plate surface w w dy y=0

In this case CD is strong function of Re Pressure Drag Consider flow over a flat plate normal to flow :

F D Wake

Low pressure

Wall shear stress does not contribute to drag force in this case

Flow separation causes pressure drag

Total drag D = Friction Drag + Pressure Drag

F = ∫ pdA D plate surface Drag coefficient For sphere it is πr2 FD C = A is projected area D 1 For cylinder it is DL ρV 2 A 2 D Dynamic pressure x A L

Source: NASA Drag Coefficients of smooth 2D bodies at low Mach numbers

Source: F M White Flow over cylinder and sphere

In case of flow over sphere or cylinder both friction drag and pressure drag contribute to total drag

Total drag D = Friction Drag + Pressure Drag

Outer stream grossly perturbed by broad Thin front BL flow separation and wake

For Inviscid flow how stream pattern look like? Source : FM White Flow separation in boundary layer

Ue Boundary layer

xy ∂u u = 0 ∂y y y=0 x

∂u Flow will separates from the surface If = 0 ∂ y y=0

Such a case, the fluid layer NEAR the solid surface will also brought to zero VELOCITY. Flow from behind will deflected off from the surface and the flow is called ‘separated’ Favorable and Adverse pressure gradients Consider flow over a curved surface

u(x) – increases u(x) – decreases p – decreases p – increases dp dp Favorable pressure gradient < 0 > 0 adverse pressure gradient dx dx

Flow converges Flow diverges

V p

dp Upstream highest point of the surface flow converges and velocity < 0 increases and pressure decreases. (Favorable pressure gradient) dx dp Downstream of the highest point streamlines diverges resulting in > 0 decrease in u(x) and a rise in pressure (Adverse pressure gradient) dx Velocity profiles across BL with favorable and adverse pressure gradients dp < 0 is favorable pressure gradient (flow from high pressure to low pressure) dx dp > 0 is adverse pressure gradient (flow from low pressure to high pressure) dx

Source : Kundu & Cohen Flow separation

Boundary layer in a decelerating stream has a point of inflection I, (PI) and grows rapidly.

The point of inflection implies slowing down of fluid layer next to wall, a consequence of adverse pressure gradient.

Under a strong adverse pressure gradient, the flow next to the wall reverses direction resulting in a region of backward flow

The reversed flow meets the forward flow at some point S, at which the fluid near the surface is transported out into the mainstream or the ‘ flow is separated’

The separation point S is defined as the boundary between the ∂u forward flow and backward flow of the fluid near the wall where = 0 the shear stress vanish ∂ y y=0 At low Re number, the reversed flow down stream of the point of separation forms part of a large steady vortex behind the surface

Flow separation over a cylinder at different Re numbers

Source : Kundu and Cohen Flow separation over cylinder in laminar and turbulent flow

Why CD reduced?

Measured drag coefficient of a cylinder Source : FM White Stream lining to reduce the drag

•The ability of Boundary layer to withstand the adverse pressure gradient without undergoing separation depends mainly on • -the geometry of the flow and -whether BL is laminar or turbulent

•A steep pressure gradient, such as that behind a blunt body, in variably leads to a quick separation

•Boundary layer on the trailing surface of a thin body can overcome the weak adverse pressure gradients involved

•Therefore, to avoid separation and large drag, the trailing section of a submerged body should be gradually reduced in size, giving it a so-called ‘stream line shape.

(a) Rectangular cylinder (b) Rounded nose

The importance of streamlining in reducing drag of a body ( CD is based on frontal area

Source : FM White Aero dynamic design reduces Drag considerably in automobiles Flow past a circular cylinder- Surface Pressure distribution

p∞ θ = 0 p − p∞ θ C p = 1 2 1 2 ρV p = pt = p∞ + ρV 2 2 C p = 1

Cp is coefficient of pressure

For Inviscid flow Cp =1-4sin2θ Lift and Drag from surface pressure measurements

F = ∫ (p − p∞ )dA = FDi + FL j p − p∞

px = p cosθ p

y W = p

p rd

sin θ θ

d θ r θ Pressure ports on the surface dA = Wrdθ

2π FD = ∫ (p − p∞ )cosθ rdθ Drag per unit length W 0 2π FL = ∫ (p − p∞ )sinθ rdθ Lift per unit length W 0 2π FD = W ∫ (p − p∞ )cosθ rdθ Total drag force 0 F C = D A = W × D D 1 ρV 2WD 2 p − p∞ C p = 2π 1 FD (p − p∞ ) r ρV 2 = ∫ cosθ dθ 2 1 2 0 1 2 D ρV WD ρV 2 2 Cp is coefficient of pressure

2π r CD = ∫ C p cosθdθ For Inviscid flow 0 D Cp =1-4sin2θ

1 2π 1 2π = ∫ θ θ CD C p cos d CL = ∫ C p sinθdθ 2 0 2 0 D= 50 ± 0.3mm Cylinder exposed to free stream

V= 20 ± 1 m/s. (Density ρ= 1.2 kg/m3)

Pressure measured at port 1=?

Calculate uncertainty in P

1 2 p = p∞ + ρV t1 2

5 Patm= 14.6 psi or 10 Pascal's Drag & Lift calculation for Invicid flow

p − p C = ∞ p 1 ρV 2 2

For Inviscid flow Cp=1- 4sin2θ

1 2π CD = ∫ C p cosθdθ 2 0

How stream flow patterns looks like? Drag prediction from wake measurements Objective: To find drag and lift For a steady flow, – Measure the velocity profile in the wake Flow separates U – Use conservation of linear momentum ∞ FD

The drag force experienced by the cylinder can be determined from this Circular cylinder in a uniform free stream

Momentum in x direction @ II Momentum in x direction @ I = mass flow * velocity = mass flow * velocity Wind tunnel U ∞ u1 d M 2 = ∫ ρWu1 *.U ∞ dy M1 = ∫ ρWu1 *.u1dy

Effective mass flow mass flow @1 II I

Change in momentum = M 2 − M1 = Drag = FD

FD = ρW ∫ u1 (U ∞ − u1 )dy p2 = p1 = p∞ This is true if the pressure at locations 1 and 2 FD = ρW ∫ u1 (U ∞ − u1 )dy (1) are same i.e. p∞

W is width of body Which is true for a cross section far downstream (section I) 100d Section II Section I (Hypothetical) p = p∞ 2 p1 p∞ d p p U ∞ t2 t1 p u (x) t∞ 2 u1 (x) II I

The actual wake profile is measured at section II close the cylinder

Assuming no pressure drop between section I and II, ( total pressure is same) and using equation of continuity a relation can be established between u1(x) and u2(x)

∫ ρ = ∫ ρ pt1 = pt2 u1dy I u2dy II (2) Q. How velocity profile looks like @ II considering the wall effects? ∫ ρu dy = ∫ ρu dy (2) FD = ρW ∫ u1 (U ∞ − u1 )dy (1) 1 I 2 II

FD = ρW ∫ u2 (U ∞ − u1 )dy (3)

1 2 (pt2 − p2 ) pt2 = p2 + ρu2 u = 2 (a) 2 2 ρ

1 2 (pt∞ − p∞ ) pt∞ = p∞ + ρU ∞ U = 2 (b) 2 ∞ ρ − 1 2 (pt1 p1 ) p = p + ρu u1 = 2 t1 1 2 1 ρ

(p − p∞ ) p = p p = p u = 2 t2 (c) 1 ∞ t1 t2 1 ρ (a) (b) and (c) in (3) FD = ρW ∫ u2 (U ∞ − u1 )dy (3)

FD = 2W ∫ (pt2 − p2 )[ (pt∞ − p∞ ) − (pt2 − p∞ )] dy (3)

Section II

II p p p2 t p∞ t∞ 2

U ∞ u2 (y)