Fluid Forces: Chapters 9 and 11: Section 1--Overview A cylinder is falling through a fluid at Velocity V Buoyancy FB Vol What are the Forces Acting? Would you expect the force Acting to be a function of Velocity? Weight FW mg V There is a pressure Difference between back and front leads to an additional force called DRAG Net Force: F FB FW FD

V2 FD C D AP 2 Drag Forcefunction of Velocity and projected Area CD A dimensionless Coef, V AP Area projected in flow direction (a circle here) A Common Form free stream velocity vel. relative to object In same direction as Drag

So Drag is a result of fluid moving over a bodyAre there other forces that can arise in this way Flow over a thin plateDrag not important Buoyancy FB Vol Small projected area Weight FW mg V But fluid will impose a shear stress recall pipe friction on Surface of plate Resulting in a Surface Resistance Force in direction of free stream velocity Net Force: F FB FW FS V2

FS C f AS 2 C f Dimensionless coef. As Surface area (parallel to flow) What Area here? Also Lift: Consider rotating Cylinder in Free stream Flow V V The Magnus Effect High Vel, Low p Why? Low Vel, Hight p Pressure Difference will lead to a Force The Lift Force NORMAL to free stream DirectionEngineers calculate this as

V2 FL C L APlan 2 C L Dimensionless coef. Aplan plan area (looking down on flow) Sothere are three forces that can occur due to flow over a body In flow direction Drag Surface Resistance V2 FD C D AP

2 V2 FS C f AS 2 Trick is finding C For given Bodies Look up in tables Correlated to Account for Drag and shear resistance see next slide and later notes ------- normal to flow Lift V2

FL C L APlan 2 Also in tables Can be calculated for plate See section 2 Relation between Stress Distribution and Flow Forces (See Fig 11.2) Free-Stream Velocity V will give rise to pressure acting normal to body and shear acting along body Lift force normal to flow Drag force in line with flow FD

p cos dA surface sin dA surface form drag friction drag V2 FD C D AP 2 One value for body and flow condition FL

p sin dA surface form drag cos dA surface friction drag V2 FL C L APlan 2 Section 2: Boundary LayersCalculating Surface Resistance Drag on a surface 2 types Pressure stress

Shear stress / skin friction drag Chapter 9 Boundary layer velocity profile Far from the surface, the fluid velocity is unaffected. In a thin region near the surface, the velocity is reduced Which is the most correct velocity profile? this is a good approximation near the front of the plate Boundary layer growth The free stream velocity is u0, but next to the plate, the flow is reduced by drag Farther along the plate, the affect of the drag is felt by more of the stream, and because of this the boundary layer grows

Fluid friction on the surface is associated with velocity reduction throughout the boundary layer du o dy y 0 Local stress & total force, skin friction du o dy y 0 Not immediately straightforward (unlike approximations we made with thin films): du/dy decreases with x & y We need to find o y 0 ( x )then

Breadth L Fs Bdx x 0 And there is more trouble Boundary layer transition to turbulence At a certain distance along a plate, viscous forces become to small relative to inertial forces to damp fluctuations Picture of boundary layer from text figure_09_04 Boundary layer transition How can we solve problems for such a complex system? (x)x))

y0 We can think about key parameters and possible dimensionless numbers Important parameters: Viscosity , density Distance, x Velocity uO uO x uO x Re x Reynolds number combines these into one number B L thickness in laminar region Self-similar shape 5x

x 5 uO Re x Re x UO x UO x Boundary layer questions How can we solve problems for such a complex system? (x)x)) y0 We can think about key parameters and possible dimensionless

numbers UO x UO x Re x What about stress? We talk about (local) stress and (total) force on a boundary in terms of local cf and average CF stress coefficients: o cf 2 Uo 2 CF FS 2

Area * UO 2 Average shear-stress coefficient On Plate of Length L Ignore this part just for a moment figure_09_12 Note New Reynolds No Example 9.6 from text A plate is 3 m long x 1 m wide Air at 20C and atmospheric pressure flows past this plate with a velocity of 30 m/s A boundary layer over a smooth, flat plate is laminar at first and then becomes turbulent. The turbulent forms of drag, etc., are reasonable above Re = 5 x 105. What is the average resistance coefficient Cf for the plate? Also, what will be the total shearing resistance force of one side of the plate?

What will be the resistance due to the turbulent part and the laminar part of the boundary layer? air 1.2kg / m 2 1.5110 5 m 2 / s A 13 3 V 30 m / s Find C f , shearing resistance on one side of plate, and resistance due to laminar flow Here it is ! V2 FS C f AS 2 air 1.2kg / m 2 1.5110 5 m 2 / s

A 13 3 V 30 m / s Find C f , shearing resistance on one side of plate, and resistance due to laminar flow 1st calculate plat Reynolds number VL 30 3 ReL 5.96 106 5 1.5110 Mixed laminar-Turbulent Cf 0.523 1520 0.00294

2 ln (0.06 ReL ) Rel air 1.2kg / m 2 1.5110 5 m 2 / s A 13 3 V 30 m / s VL 30 3 ReL 5.96 106 5 1.5110 Find C f , shearing resistance on one side of plate, and resistance due to laminar flow Cf

0.523 1520 0.00294 ln 2 (0.06 ReL ) Rel V2 (30) 2 FS C f AS 0.00294 1.2 3 4.76 N 2 2 air 1.2kg / m 2 1.5110 5 m 2 / s A 13 3 V 30 m / s

VL 30 3 ReL 5.96 106 5 1.5110 Find C f , shearing resistance on one side of plate, and resistance due to laminar flow Cf 0.523 1520 0.00294 ln 2 (0.06 ReL ) Rel V2 (30) 2 FS C f AS 0.00294 1.2 3 4.76 N

2 2 Now Calculate Transition point Vxt 500,000 xt 500,000 / V 0.252m air 1.2kg / m 2 1.5110 5 m 2 / s A 13 3 V 30 m / s VL 30 3 ReL 5.96 106 5 1.5110

Find C f , shearing resistance on one side of plate, and resistance due to laminar flow Cf 0.523 1520 0.00294 ln 2 (0.06 ReL ) Rel V2 (30) 2 FS C f AS 0.00294 1.2 3 4.76 N 2 2 Now Calculate Transition point Vxt 500,000 xt 500,000 / V 0.252m

So laminar layer Coefficient is C f 1.33 / Retr 0.00188 air 1.2kg / m 2 1.5110 5 m 2 / s A 13 3 V 30 m / s VL 30 3 ReL 5.96 106 5 1.5110 Find C f , shearing resistance on one side of plate, and resistance due to laminar flow Cf

0.523 1520 0.00294 ln 2 (0.06 ReL ) Rel V2 (30) 2 FS C f AS 0.00294 1.2 3 4.76 N 2 2 Now Calculate Transition point Vxt 500,000 xt 500,000 / V 0.252m So laminar layer Coefficient is C f 1.33 / Retr 0.00188 And laminar force is

V2 (30)2 Flam C f Atr 0.00188 1.2 10.252 0.256N 2 2 Average shear-stress coefficient figure_09_12 Laminar, Turbulence, Induced Turbulence Laminar 5x (x) Turbulent 0.16 x 1/ 7

Re X Re X Induced 0.16 x Re X 1/ 7 2 UO cf 2 0 x O cf 2 U O 2

L FS Bdx 0.664 0.455 ln 2 0.06 Re X Re X 0.027 Re X 1/ 7 2 Uo Area * Cf 2

x 0 Cf FS 1.33 2 BLUO 2 UO x UO x Re x ReL

0.523 1520 ln 2 0.06 ReL ReL 0.032 ReL 1/ 7 UOL ReL BACK TO DRAG ON SUBMERGED OBJECTS Drag on a surface 2 types Pressure stress / form drag Shear stress / skin friction drag

A boundary layer forms due to skin friction For shapes more complex than a plane, these result in total drag forces which are usually hard to solve analytically Shortcuts for total drag For less precise design and/or well-known / well-studied (simple) objects, we rely on charts for an average coefficient of drag FD CD V 2 A 2 Frontal Area + shear Drag coefficients for 2d or infinitely long objects figure_11_07 figure_11_04

Vd Re for 3d bodies or Vb P 11.18, 9 edition th Compute the overturning moment exerted by a 35 m/s wind on a smokestack that has a diameter of 2.5 m and a height of 75 m. Assume that the air temperature is 20 C and that the atmospheric pressure is 99

kPa absolute. 2.5 m 1.17kg / m3 , 1.5110 5 m 2 / s V= 35m/s Object is ~an infinite cylinder 75m Vd Re 5.79 10 6 d=2.5 m 1.17kg / m3 , 1.5110 5 m 2 / s V= 35m/s Object is ~an infinite cylinder 75m Vd

Re 5.79 10 6 C D 0.62 d=2.5 m 1.17kg / m3 , 1.5110 5 m 2 / s V= 35m/s Object is ~an infinite cylinder 75m Vd Re 5.79 10 6 C D 0.62 2 ( 35

) FD C D V 2 A 2 0.62 1.17 (2.5 75) 83.31kN 2 Then turning moment h M FD 37.5 83.31 3.12MN .m 2 Lift Total lift Similar to our calculations of total drag, we rely on charts for an average coefficient of lift FL C L V 2 A 2 A is a reference area, usually planform area Example 11.6 Lift on a Rotating Sphere A ping-pong ball is moving at 10 m/s in air and is spinning CW at 6000 rpm

as shown. The ball diameter = 3 cm. Calculate the lift and drag forces and indicate the direction of each. Assume standard atmospheric pressure and a temperature of 20 C. How does the answer change if the ball is spinning CCW? 6000 1.2kg / m 3 0.03 Rotation parameter r V Find Lift and Drag Forces on Ping-Pong 6000 1.2kg / m 3 0.03 Rotation rate

100 (rev / s ) 2 (rad / rev ) 628 rad / s Rotation parameter r 628 0.015 .942 V 10 C L 0.26, C D 0.64 r Rotation parameter V FL C L V 2 A 2 1.110 2 N FD C D V 2 A 2 2.7110 2 N Sothere are three forces that can occur due to flow over a body

In flow direction Drag Surface Resistance V2 FD C D AP 2 V2 FS C f AS 2 Trick is finding C For given Bodies Look up in tables ch 11 Correlated to Account for Drag and shear resistance

see next slide and later notes ------- normal to flow Lift V2 FL C L APlan 2 Also in tables ch 11 Can be calculated for plate See chapter 9