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 10o C. (1)Estimate the distance from the bow on the full-scaleship 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 10o C.Estimate the distance from the bow where transition occurs.Where should transition be stimulated on model?

Things we know: Re = UL/ ReL transitions ~ 5 x 105 (experimentally determined for flat plate and dP/dx = 0)

Want same fraction of L to be turbulent for both cases: [xtr/L]model = [xtr/L]boat so ration of xtr = ratio of boat lengths

= 13.5

?

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

(*/ = 0.344)

0.344

y /

u / U

Sinusoidal, parabolic, cubic look similar to Blasius solution.

u/Ue = 2(y/) – (y/)2

u/Umax = (y/)1/7

R

R

(interesting)

u = (-R2/4)(dp/dx)(1-[r/R])2 eq. 8.12

wall = (R/2)(dp/dx) Umax = (-R2/4)(dp/dx)

du/dr = -2Umax/Rdu/dy = 2Umax/R

*

*

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

FDrag

cd

d c

U1 = Uo = 80 ft/sec1= 0.8 in.

U2 = ?P2 – P1 = ?2= 1.2 in.

Continuity U1A1eff = U2A2eff

A1eff = (h -2 *1 ) (h -2 *1 ) A2eff = (h -2 *2 ) (h -2 *2 )

*1

*1

* = 0.8 in *1

u/U = y/ =

Continuity U1A1eff = U2A2eff

A1eff = (h -2 *1 ) (h -2 *1 ) A2eff = (h -2 *2 ) (h -2 *2 )

*2

*2

*2 = 1.2 in *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/

*/ = 1o (1-u/U) d

/ = 1o u/U(1-u/U) d

w = U2 d/dx = u/y0

w = u/y0 = (U/)d(u/U)/d(y/)0

w = (U/)dsin(/2)/d0

w = (U/)(/2)cos(/2)0

w = (U/)(/2) = U/(2)

u/U = sin(/2); = y/

*/ = 1o (1-u/U) d

u/U = sin(/2); = y/

*/ = 1o (1 – sin(/2) )d

*/ = + (2/)cos (/2)from 1 to 0

*/ = 1 - 2/ = 0.363* = 0.363

/ = 1o u/U(1-u/U) d

= 0.137

Finding */ = 1o (1-u/U) d

/ = 1o u/U(1-u/U) d

w = U2 d/dx

w = U2 d/dxw = u/y0

…..

u/U = sin(/2); = y/

u/U = sin(/2); = y/

Fdrag = 0L wbdx = 0

L U2(d/dx)bdx = 0

L U2bd = U2b0L

Fdrag = U2bL

(whew!)

Calculating drag on a flat plate, zero pressure gradient – turbulent flow

*

*u/y blows up at y = 0

Can’t usewall = du/dy y=0*

Turbulent Flow

Tripped at leading edge so turbulent flow everywhere on plate

= 0.424 Rex-1/5x

w = 0.0243 U2 (/(U))1/4

w = 0.0243 U2 (/(U 0.424 Rex-1/5x ))1/4

= 0.424 Rex-1/5x

w = 0.0301 U2 Rex-1/5

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

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 + c2 + d 3

What are boundary conditions?

Given: u/U = a + b + c2 + d 3

= y/

= 0; u = 0; a = 0= 0; u/y = 0; b = 0= ; u =U; 1 = c + d= ; u/y = 0; 2c + 3d = 02(1-d) + 3d = 2 + d = 0 so d = -2 and c = 3

Separatingu/y = 0

u/U = 3 2 -2 3

dp/dx = 0

Separating Flowu/U = 3 2 -2 3

dp/dx > 0

U

If Re transition occurs at 500,000 evaluate ReL, xtr, and power to overcome skin friction drag on fin

ReL = UL/ = 1.54 x 107

Calculating transition location, x:Assume xtr occurs at 500,000

Rex = Ux/ = 500,000Rex/ReL = xtr/L = 500,000 / 1.54 x 107

xtr = 0.0325 L = 0.0325 x 1.65 = 0.0536 m

FLAT PLATE dp/dx = 0

LAMINAR:From Theory

CD = 1.33 / ReL1/2

Eq. 9.33

TURBULENT:From Experiment

5x105< ReL<107

xtransition = 0CD = 0.0742/ReL

1/5

ReL<109

xtransition = 0CD = 0.455/(logReL)2.58

For Re transition of 500,000 and 5x105 < ReL < 109

CD = 0.455/(log ReL)2.58 – 1610/ReL (Eq. 9.37b) CD = 0.00281 – 0.000105 = 0.0027

~ 4% less

FD = CD2LH(1/2)U2 = 91.6 N2 sides so 2 x Area

Smooth flat plate, dp/dx = 0; flow parallel to plate

CD ~ 0.0028

Reh = Uh/, don’t know UAssume: turbulent flow, transition at leading edge,

doesn’t flutter, 5 x 105 < ReL < 107

Then CD = 0.0742/ReL1/5

Plastic plate falling in water, find terminal velocity -

Sum of forces = 0Drag Force = f (U)

DRAGBOUYANCY

WEIGHT

+ z

- z

Net Force = 0 = FB + FD - W

FB – W = (plastic - water) g (hLt)FD = CD ½ U2A = [0.0742/ReL

1/5] ½ water U2AReh = Uh/

Solving for U get 11.0 ft /sec

Check Re number: 1.86 x 106

5 x 105 < ReL < 107

mg = 120 kgU = 6 m/s

FD = CD ½ U2A

Sum of forces = 0,pick A so U = 6 m/s

mg = 120 kgU = 6 m/s

FD = CD ½ U2A

Sum of forces = 0,pick A so U = 6 m/s

CD = 1.42 for open hemisphereFD = CD ½ U2A = W = mg

A = d2/4d2 = [mg]/[CD U2 /8]

d = [(8/)(120kg)(9.8 m/s2) (1/1.42)(1/1.23 kg/m3)(1/6m/s)2

d = 6.9 m

FD = CD ½ U2A

FD = ma = m(dU/dt) = m(dU/dx)(dx/dt)

CD ½ U2A = m (dU/dx) U

CD ½ U2A = m (dU/dx) UdU/U = CD A dx /(2m)

Integrating from x=0

to x=x gives:ln Uf – lnUi = CD Ax /(2m)CD = 2m ln {Uf/Ui} /[ Ax]

CD = 0.299

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

air

8.33 m/s

2.78 m/s

FD = CD ½ U2AA = 23.4 ft2

CD = 0.5FR = 0.015 x 4500 lbf

FD = FR to find U where aerodynamic force = frictional force

Total Power = FDU + FRU

FD [CD ½ U2A] = FR[.015xW](0.5)½ 0.00238slug/ft3U2(23.4ft2)

= 0.015 x 4500 lbf

U2 = 67.5/0.0139

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)