u/U = sin(  /2);  = y/  Given U and viscosity table 9.2 Sketch  (x),  *(x),  w (x) LAMINARLAMINAR T A B L E 9.2

Sketch  (x),  *(x),  w (x) u/U = sin(  /2);  = y/ 

….. u/U = sin(  /2);  = y/  

WHICHHASMOREFD?WHICHHASMOREFD? A VS B

A B U Which plate experiences the most drag? LAMINAR / dp/dx = 0 Incompressible, steady, 2-D…

A B U Which plate experiences the most drag? F D =  w dA A = bL C f =  w /( ½  U 2 ) = 0.730/Re x F D = 0.73b(  L  U 3 ) 1/2 F D (A) = 0.73(4L)(  L  U 3 ) 1/2 F D (B) = 0.73(L)(  4L  U 3 ) 1/2 F D (A)/F D (B) = 2 b=4L b=L L=4L L=L

A U How Does Drag Change if 2U? b L

A U How Does Drag Change if 2U? F D =  w dA A = bL C f =  w /( ½  U 2 ) = 0.730/Re x F D = 0.73b(  L  U 3 ) 1/2 If U goes to 2U, F D goes to (8) 1/2 b L

(  */  = 0.344) Table 9.2

(  */  = 0.344)

0.344

y /  u / U Sinusoidal, parabolic, cubic look similar to Blasius solution.

u/U e = 2(y/  ) – (y/  ) 2 u/U max = (y/  ) 1/7

I give you Re x and x, how do you get  w (x)? D L ?

Laminar boundary-layer over flat plate, zero pressure gradient 1.8 m 0.9 m U = 3.2 m/s u/U = sin[(  /2)(  /y)] Drag = ? using Table 9.2

F D =  w dA =   U 2 d  /dx bdx =  U 2 b  L

x=L

 u/  y| Wall =  /  y U[(2y/  ) – (y/  ) 2 ] y=0  u/  y| Wall = U2/  PIPE:  u/  y| Wall = U cl 2/R Fully developed laminar flow

 u/  y =  /  y {y/  } 1/7 = 1/7 {y/  } -6/7  u/  y| y=0 = 

So how get  wall for turbulent velocity profile u/U 0 = (y/  ) 1/7 ?

u/U o = (y/  ) 1/7  w =  (V) 2 [ /(RV)] 1/4 Eq(8.39) V/U c/l = (2n 2 )/[(n+1)(2n+1)] Eq(8.24)  w =  (U 0 ) 2 [ /(  U o )] 1/4

If Re transition occurs at 500,000 what is the skin friction drag on fin?

F D = C D (½  U o 2 A) Laminar Flow – starting at x=0 C D = 1.33/Re L 1/2 Turbulent Flow – starting at x=0 C D = /Re 1/5 {u/U = (y/  ) 1/7 } for 5x10 5 <Re L <10 7 C D = /(log Re L ) 2.58 For Re L < 10 9

Calculating transition location, x: Assume x tr occurs at 500,000 Re x = Ux/ = 500,000 Re L = UL/ = 1.54 x 10 7 Re x /Re L = x tr /L = 500,000 / 1.54 x 10 7 x tr = L = x 1.65 = m

Re L = UL/ = 1.54 x 10 7 Turbulent Flow – starting at x=0 C D = /(log Re L ) 2.58 L = 1.65m x tr = 0.054m

FLAT PLATE dp/dx = 0 LAMINAR: From Theory C D = 1.33 / Re L 1/2 Eq TURBULENT: From Experiment 5x10 5 < Re L <10 7 x transition = 0 C D = /Re L 1/5 Re L <10 9 x transition = 0 C D = 0.455/(logRe L ) 2.58

For Re transition of 500,000 and 5x10 5 < Re L < 10 9 C D = 0.455/(log Re L ) 2.58 – 1610/Re L (Eq. 9.37b) C D = – = ~ 4% less F D = C D 2LH(1/2)  U 2 = 91.6 N 2 sides so 2 x Area

Smooth flat plate, dp/dx = 0; flow parallel to plate C D ~ C D = F D /( ½  U o 2 ); C f (x) =  w (x) /( 1 / 2  U o 2 )

Where does transition occur for these data? These data include rough plates?

C D = F D /( ½  U o 2 ); C f (x) =  w (x) /( 1 / 2  U o 2 ) Re L

breath

Plastic plate falling in water, find terminal velocity -

Sum of forces = 0 DRAG BOUYANCY WEIGHT + z - z F D = C D ½  U 2 A Given  of water and plastic

Assume: turbulent flow, transition at leading edge, doesn’t flutter, 5 x 10 5 < Re L < 10 7 Then C D = /Re h 1/5 Plastic plate falling in water, find terminal velocity - Sum of forces = 0 Drag Force = f (U) DRAG BOUYANCY WEIGHT + z - z

Net Force = 0 = F B + F D - W F B – W = (  plastic -  water ) g (hLt) F D = C D ½  U 2 A = [0.0742/Re h 1/5 ] ½  water U 2 A Re h = Uh/ Solving for U get 11.0 ft /sec Check Re number: 1.86 x x 10 5 < Re L < 10 7

THE END

THETHE ENDEND

Extra Examples

U 1 = U o = 80 ft/sec  1 = 0.8 in. U 2 = ? P 2 – P 1 = ?  2 = 1.2 in.

Continuity U 1 A 1eff = U 2 A 2eff A 1eff = (h -2  * 1 ) (h -2  * 1 ) A 2eff = (h -2  * 2 ) (h -2  * 2 ) *1*1 *1*1  * = 0.8 in *1*1 u/U = y/  = 

Continuity U 1 A 1eff = U 2 A 2eff A 1eff = (h -2  * 1 ) (h -2  * 1 ) A 2eff = (h -2  * 2 ) (h -2  * 2 ) *2*2 *2*2  * 2 = 1.2 in *2*2

First a bit of mathematic legerdemain before attempting Ex. 9.3 ~ = ?

L = 1.8 m U = 3.2 m/s  = ?;  * = ?;  w = ? b = 0.9m Laminar and assume u/U = sin(  /2);  = y/   */  =  1 o (1-u/U) d   /  =  1 o u/U(1-u/U) d   w =  U 2 d  /dx =  u/  y  0

 w =  u/  y  0 =  (U/  ) d(u/U)/d(y/  )  0  w =  (U/  ) dsin(  /2) /d   0  w =  (U/  )(  /2) cos(  /2)  0  w =  (U/  )(  /2) =   U/(2  ) u/U = sin(  /2);  = y/ 

 */  =  1 o (1-u/U) d  u/U = sin(  /2);  = y/   */  =  1 o (1 – sin(  /2) )d   */  =  + (2/  )cos (  /2)  from 1 to 0  */  = 1 - 2/  =  * = 

 /  =  1 o u/U(1-u/U) d   = 

Finding   */  =  1 o (1-u/U) d   /  =  1 o u/U(1-u/U) d   w =  U 2 d  /dx

 w =  u/  y  0

….. u/U = sin(  /2);  = y/ 

F drag =  0 L  w bdx =  0 L  U 2 (d  /dx)bdx =  0 L  U 2 bd  =  U 2 b  0 L F drag =  U 2 b  L (whew!)

Calculating drag on a flat plate, zero pressure gradient – turbulent flow * *  u/  y blows up at y = 0 Can’t use  wall =  du/dy  y=0 *

Turbulent Flow Tripped at leading edge so turbulent flow everywhere on plate  = Re x -1/5 x

 w =  U 2 ( /(U  )) 1/4  w =  U 2 ( /(U Re x -1/5 x )) 1/4  = Re x -1/5 x  w =  U 2 Re x -1/5

u/U = (y /  ) 1/6 u/U = (y /  ) 1/7 u/U = (y /  ) 1/8 u/U y/  Re increases, n increases, wall shear stress increases, boundary layer increases, viscous sublayer decrease

LAMINAR BOUNDARY LAYER AT SEPARATION Given: u/U = a + b  + c  2 + d  3 What are boundary conditions?

Given: u/U = a + b  + c  2 + d  3  = y/   = 0; u = 0; a = 0  = 0;  u/  y = 0; b = 0  =  ; u =U; 1 = c + d  =  ;  u/  y = 0; 2c + 3d = 0 2(1-d) + 3d = 2 + d = 0 so d = -2 and c = 3 Separating  u/  y = 0 u/U = 3  2 -2  3

dp/dx = 0 Separating Flow u/U = 3  2 -2  3 dp/dx > 0

U

mg = 120 kg U = 6 m/s F D = C D ½  U 2 A Sum of forces = 0, pick A so U = 6 m/s

mg = 120 kg U = 6 m/s F D = C D ½  U 2 A Sum of forces = 0, pick A so U = 6 m/s C D = 1.42 for open hemisphere F D = C D ½  U 2 A = W = mg A =  d 2 /4 d 2 = [mg]/[C D  U 2  /8] d = [(8/  )(120kg)(9.8 m/s 2 ) (1/1.42)(1/1.23 kg/m 3 )(1/6m/s) 2 d = 6.9 m

F D = C D ½  U 2 A F D = ma = m(dU/dt) = m(dU/dx)(dx/dt) C D ½  U 2 A = m (dU/dx) U

dU/U = C D  A dx /(2m) Integrating from x=0 to x=  x gives: ln U f – lnU i = C D  A  x /(2m) C D = 2m ln {U f/ U i } /[  A  x] C D = 0.299

Find POWER = F U for this condition And then see if that is enough to win bet.

air

8.33 m/s 2.78 m/s

F D = C D ½  U 2 A A = 23.4 ft 2 C D = 0.5 F R = x 4500 lbf F D = F R to find U where aerodynamic force = frictional force Total Power = F D U + F R U

F D [ C D ½  U 2 A] = F R [.015xW] (0.5)½ slug/ft 3 U 2 (23.4ft 2 ) = x 4500 lbf U 2 = 67.5/ U = 69.7 ft/s = 47.5 mph Where does aerodynamic force = frictional force ?

Average of 44 sport cars is 0 43, not much better than the 0.47 of ’37 the Lincoln Zephyr

(245 km/hr)

A model of a river boat is to be tested at 1:13.5 scale. The boat is designed to travel at 9mph in fresh water at 10 o C. (1)Estimate the distance from the bow on the full-scale ship where transition occurs. (2) Where should transition be stimulated on model? What equations are relevant to this problem? What approach should we take?

A model of a river boat is to be tested at 1:13.5 scale. The boat is designed to travel at 9mph in fresh water at 10 o C. Estimate the distance from the bow where transition occurs. Where should transition be stimulated on model? Things we know: Re = UL/ Re L transitions ~ 5 x 10 5 (experimentally determined for flat plate and dP/dx = 0) Want same fraction of L to be turbulent for both cases: [x tr /L]model = [x tr /L]boat so ration of x tr = ratio of boat lengths = 13.5

?

Three unknowns, A, B, and C- will need three boundary conditions. What are they?

1 m U = 2.7 m/s cd

(5) Two-Dimensional (6) Incompressible assumed assumptions

cd u(x,y)/U = y/  =  (y) 1 0

cd

F Drag cd 

d c