XAMPLE 9.I3GIVEN A 100 km/hr (i.e., 28 m/s) wind blows past the watertower shown in Fig. E9.l3a.

FIND Estimate the moment (torque), M, needed at the base tokeep the tower from tipping over.

U = 100 km/hr= 28 m/s----)

D,= L2 mD.= 5 mb=15m

b)IFIGURE E9.13

9.67 The structure ihown in Fig. W.67 consists of threecylindrical support posts to which an elliptical flat-plate sign isattached. Estimate the drag on the structure when a 80 km/hr windblows against it.

IFIGURE P9.67

Ql. (a)

-L' sME2323D€fin€ the bomdrylayer thickness, 6

Describe 6e formaim ofthe bondry layer.

(5nrk+)

O) Ddermine tbe valuc of Cr md Cz in fte Table Q l. Show clerly yoE oalcotdioa

Tabte Ql: Ft*plde momentum inqgral rcsults forvcious assmed ltminarflow vclocity Profrcs

Piofil€ Charactert{RE, cr@ cD@

(r)

(D

I

Blasius solution

Lins

(v=z\\u 6)Prabolic

t;2!=zL-[ Zlu 6 \d/Sinwerre

g=*iokz-)u \2dl

5,00

3-46

5.48

0.664

0.578

0.7

o.655

1.328

Cr

r.ffi

t.3loe4

(lSnats)

( c ) Determine thc total pm'€r required to p'ull ec timb6 (diet€r 0.4 m md leogth f5m) witt rdocity O8 ds. Only half of the timbcr on ftewdEr sufaca Ncgld ttedr€ force due to fimt srrfrce.

Tatre cD..- =#, co*=H, p*,=lwfulm!,u--lxloam2 ls

(5 malcs)

Question:a) Determinetheequationfordragcoefficient67fora“one-seventhpower”turbulentvelocityprofile.

TheBlasiusresistanceequationisgivenas:

DE = 0.0225JKLMJKN

OP

b) Atimberispulledbyaboatalongtheriverwithforce200NasshowninFigureQ2.Onlyhalfofthetimberabovethewater.Determinethedragforceduetofrontsurfaceofthetimber.Takethedragcoefficientforturbulentderivedinquestion(a).Take,67(XYZ[\Y]) =

O.^L_`a,JEYba] =

OcccdeZf MEYba] = 1×10i^jk/mL

Answer:a)

nK=

oN

Op DE = 0.0225JKL

MJnN

Op

Diketahui;

r =nK

1 −nK

t

cuo =

772N

D = JKLuruw

= JKLuuw

7N72

Buatpersamaanserentak;

0.0225JKLMJKN

Op=

772

JKLuNuw

Selesaikan

N =0.3721w

z{O|

Diketahui

r =772N =

772

0.3721w

z{O|

=0.036w

z{O|

67 =D~uw

12 JK

L~�=

JKL uruw ~uw12 JK

L~�=2r�=2�

0.036w

z{O|

=0.072

z{O|

b)Reynoldnumberattheendoftimber

z{ =JKwM

=(1000)(2 + 1)(10)

1×10i^= 3×10p(turbulent)

PanjangaliranBLlaminar

z{Ö = 5×10| =JKwM

w =5×10| 1×10i^

(1000)(2 + 1)= 0.17 m

Lessthan10%fromtotallengthoftimberLaminarBLcanbeignored

Dragforcecausebysidesurface

á =12×2àâ×� = 6.286mL

ã7 = 6712JKLá =

0.072

z{O|×12JKL 6.286 = 65.1 j

Totalpullingforce= Dragforceatside+Dragforceatfrontsurface200 = 65.1 + ww = 134.9j

Question a) What is drag? And why do we usually try to minimize it? b) Assuming the velocity profile in the turbulent boundary layer as:

!"=

$%

&'

Determine the equation of drag coefficient () in term of Reynold number using momentum integral equation

*+ = ,"-./.0

And the Blasius resistance equation

*+ = 0.0225,"-5"%

&6

c) Figure Q2 shows a flat plate 3 m x 1.5 m is held in water moving at 5 m/s parallel to its length. Determine the power required to overcome the drag. Take , = 10009:/<= and > = 0.001?@/<-

()ABCDEBF =1.328

IJAKLMIJ ≤ 5×10PQR.

()STFUTAVES =0.074

IJA&P

KLMIJ ≥ 5×10P

Answer:

/ =9110

%

* = * = ,"-./.0

= ,"-..0

.110

% = ,"-9110

.%

.0

Diberi dari penyelesaian Blasius ;

* = 0.0225,"-5"%

&6

Oleh itu ;

% = 0.42565"

&P06P =

0.42560

(IJ)&P

Shear stress ;

* = 0.0225,"-5"%

&P

Masukkan nilai % ;

* =0.02795

&P,"-

"&P0

&P

Drag force ;

)̂ = *_.0 =0.02795

&P,"-

"&P0

&P

_.0`

a= 0.0349,"-_

5&P

"&P

b6P

Drag coefficient ;

() =

)̂12 ,"

-c=

0.0349,"-_ 5&P

"&Pb6P

12 ,"

-_b= 0.0698

5&P

"&Pb

&P

=0.0698

IJ&P

, = 10009:/<=, > = 0.001?@/<-

Nilai nombor Re,

IJ =,"0>

=(1000)(5)(3)

0.001= 1.5×10d = Turbulent

Re number at transition ;

5×10P =,"0`>

0` =5×10P>,"

=5×10P (0.001)(1000)(5)

= 0.1<

Hanya 3.3% sahaja kawasan akan mengalami laminar boundary layer. Oleh itu laminar boundary layer boleh diabaikan

Diberi dari soalan, nilai () untuk turbulent,

() =0.074

(IJ)&P

=0.074

(1.5×10d)&P

= 2.71×10m=

Drag force ;

)̂ = ()12,"-c = 2.71×10m=

12

1000 5 - 3 1.5 = 152.4 ?

Tetapi BL akan berlaku dikedua-dua belah permukaan plat, oleh itu jumlah drag force ;

)̂ = 152.4×2 = 304.8 ? Power = Drag×Velocity = 304.8×5 = 1524 Watt

Question1a) Transitionfromlaminartoturbulentboundarylayerflowactuallyoccursoverafinitelengthofthe

surface.Duringwhich the velocityprofile andwall shear stress adjust from laminar to turbulentforms.Ausefulapproximationduringtransition is that themomentumthicknessof theboundary

layer remains constant. Assuming constantmomentum thickness, find the ratio (CDEFGEHIJDCHKLMJKF

) fortransitionfromaparaboliclaminarvelocityprofiletoa“1/7=power”turbulentvelocityprofile.

b) Thevelocityprofile ina turbulentboundary layeroften isapproximatedby the ‘”1/7power law”equation.EvaluateV = W∗/Yforthepowerlowprofile

Answer:(a)Paraboliclaminarvelocityprofile,

\] = 2_ − _a

Momentumthickness,

Ycdefgdh = W i(_) 1 − i(_)j

kl_ =

215 Wcdefgdh

1/7-powerturbulentvelocityprofile,

\] = _

jo

Momentumthickness,

Ypqhrqcsgp = W i(_) 1 − i(_)j

kl_ =

772 Wpqhrqcsgp

WpqhWcde

=215×

727 = 1.37

(b)1/7-powerlaw,

\] = _

jo

Where,

Y = W772

W∗ = W18

V =W∗

Y =W 18

W 772

= 1.286

Question1

a) Explainclearlywithsketches,theboundarylayerthicknessforaflowpastaflatplate,andwriteamathematicexpressionoftheboundarylayerthickness

b) Fluidhavingadensityof1.24BC/EFandkinematicviscosityof1.5×10JKEL/Mflowingwithvelocity4.0E/Moveraplateof0.5mlongand0.6mwide.ThevelocityprofileoftheflowisintheformofP = R + TU + VULDetermine:

i. Thevelocityprofileii. Theboundarylayerthicknessatthetrailingedgeoftheplateiii. Theshearstressatthemiddleoftheplateiv. Thedragforceononesideoftheplate

(20marks)

Answer:(b) P = R + TU + VUL

i. Denganmenggunakanboundarycondition:U = 0, P = 0

U = [, P = \

U = [,]P]U

= 0

P\= 2

U[−

U[

L

ii.

_ = `\L215

][]a

=2b\[

[ =5.48a

`\ab

=5.48(0.5)

(4)(0.5)1.5×10JK

= 7.5×10JF m

iii. Shearstressatthemiddleoftheplate

_ =2b\[

[ =5.48(0.25)

(4)(0.25)1.5×10JK

= 5.306×10JF m

_ =2b\[

=2 1.24 (1.5×10JK)(4)

5.306×10JF= 0.0280 gR

iv.

hi = _j]a = _j]a = 0.73 ` k \j

\lk=

m.K

m

n

m0.011899 p

Question:

a) Explainwhatyouunderstandbyboundarylayerthicknessanddisplacementlayerthickness.

b) Assumethatinthelaminarboundarylayer,theflowobeysthelawofshearstress> = @AB

AC,where@

istheviscosity,whichleadtotheprofileof(F − H) = I(J − K)L,whereUisthefreestreamvelocity,

u is the velocity at a distance y above theplate andk is a constant.Determine thedisplacement

thickness.

c) Usingtheanswerfrom(b),findthedisplacementthicknessattheboundaryconditionsy=0andu=0

Answer:

Displacementthickness,,J∗

J∗ = 1 −H

FSK

T

U

Velocityprofile,

F − H = I(J − K)L

BahagikansemuadenganU

1 −H

F=

I

F(J − K)L

J∗ = 1 −H

FSK

T

U

=I

F(J − K)LSK

T

U

=I

F(J − K)LSKT

U

=I

FJL − 2JK + KL SK

T

U

=I

FJLC −

2JKL

2+KZ

3U

T

=I

FJZ − JZ +

JZ

3

J∗ =I

F

JZ

3

Atboundarycondition

J∗ =I

F

JZ

3 (i)

AtK = 0, H = 0

(F − H) = I(J − K)L

F = IJL (ii)

J∗ =IJZ

3

1

IJL=J

3

Question

a) Whatdoesitmeantbydisplacementthickness?

b) ProvethattheVon-KarmanMomentumIntegralequationcanbewrittenas

BC = EFGHI

HJ

Where

BC :Wallshearstress

E :Fluiddensity

F :Freestreamvelocity

I :Momentumthickness

c) FigureQ1showsauniformflowpasta flatplate.DeterminethewallshearstressBCbetween

AD=0.5m.thevelocityprofileislaminarT

U=

V

W

XG, Z[\ = 0.15]andZ[^_ = 0.16].TakeE =

1.208de/]gandh = 1.81×10jklm/]Gforair.

d) DetermineCDforT

U=

V

W

Xo= ɳ

Xo

Answer:

(c)

E = 1.208de/]g Z[\ = 0.15]

h = 1.81×10jklm/]G Z^_ = 0.16]

HZ = Z^_ − Z[\

s

F=

t

Z

XG

= ɳXG

= 0.16 − 0.15

= 0.01]

HJ = 0.05]

BC = EFGHI

HJ

=

EFG

H

HJ

s

F1 −

s

F

W

u

Ht

=

EFG

HZ

HJɳXG 1 − ɳ

XG

W

u

Hɳ

=

EFG

HZ

HJɳXG − ɳ

W

u

Hɳ

=

EFG

HZ

HJ

2

3ɳgG −

1

2ɳG

u

X

BC = EFGHZ

HJ

2

3−1

2

= EFGHZ

HJ

2

3−1

2

= (1.208)(0.5)G0.01

0.5

1

6

= 1.007×10jg l/]G

Thevelocityprofileofturbulentboundarylayer

s

F=

t

Z

Xo

= xXo

Momentumintegralequation

BC = EFGHI

HJ

Blasiusresistanceequation

BC = 0.0225EFG

z

FZ

X{

BC = EFGHI

HJ= EFG

HZ

HJ

9

10−9

11= 0.0818EFG

HZ

HJ

0.0818EFGHZ

HJ= 0.0225EFG

z

FZ

X{

ZX{HZ =

0.0225EFGzX{F

jX{

0.0818EFGHJ

ZX{HZ = 0.275z

X{F

jX{HJ

4

5Zk{ = 0.275z

X{F

jX{J + �

Z = 0.3438zX{F

jX{J

{k

�_ =

0.0698

ÄÅÇ

X

k