PROBLEMSFORCHAPTER44-1Velocityprofileisgivenasfollows:

𝑢𝑈=32𝑦𝛿−12𝑦𝛿

L

Forabovevelocityprofile,compute:a) R

S𝑅𝑒

b) V∗

S𝑅𝑒

c) VS𝑅𝑒

d) 𝐶Z 𝑅𝑒

e) 𝐶[ 𝑅𝑒

𝛿∗ = 1 −𝑢𝑈

𝑑𝑦V

]= 1 − 𝑓 𝜂 𝛿𝑑𝜂

`

]

𝜃 =𝑢𝑈

1 −𝑢𝑈

𝑑𝑦V

]= 𝑓 𝜂 1 − 𝑓 𝜂 𝛿𝑑𝜂

`

]

𝑢𝑈 =

32𝑦𝛿−12𝑦𝛿

L

𝑓 𝜂 =32𝜂 −

12𝜂L

𝑓 𝜂 1 − 𝑓 𝜂 =32𝜂 −

12𝜂L 1 −

32𝜂 +

12𝜂L

𝜃 = 𝛿 𝑓 𝜂 1 − 𝑓 𝜂 𝑑𝜂`

] = 𝛿

32𝜂c

2−94𝜂L

3−12𝜂e

4+64𝜂g

5−14𝜂i

7 ]

`

𝜃 =39280

𝛿

𝑢𝑈 =

32𝑦𝛿−12𝑦𝛿

L

𝑢 =32𝑈

𝑦𝛿−12𝑈

𝑦𝛿

L

𝑑𝑢𝑑𝑦 =

32𝑈𝛿−𝑈2𝛿3𝑦c

𝑑𝑢𝑑𝑦 mn]

=32𝑈𝛿

𝜏 = 𝜇𝑑𝑢𝑑𝑦 mn]

= 𝜇32𝑈𝛿 [1]

Von-Karmanequation:

𝜏 = 𝜌𝑈c𝑑𝜃𝑑𝑥

= 𝜌𝑈c𝑑𝑑𝑥

39280

𝛿 =39280

𝜌𝑈c𝑑𝛿𝑑𝑥

[2]

From[1]and[2]:

𝜇32𝑈𝛿 =

39280

𝜌𝑈c𝑑𝛿𝑑𝑥

𝛿𝑑𝛿 = 10.7692𝜇𝜌𝑈

1𝑑𝑥

𝛿c

2 = 10.7692

𝜇𝑥𝜌𝑈

+ 𝐶

at𝑥 = 0, 𝛿 = 0 → 𝐶 = 0

𝛿c = 21.5384𝜇𝑥𝜌𝑈

𝑥𝑥= 21.5384𝑥c

𝜇𝜌𝑈𝑥

= 21.5384𝑥c1𝑅𝑒

𝛿 =4.64𝑥𝑅𝑒

𝜃 =39280

𝛿 =39280

4.64𝑥𝑅𝑒

𝜃 =0.646𝑥𝑅𝑒

𝛿∗ = 1 −𝑢𝑈

𝑑𝑦V

]

= 1 − 𝑓 𝜂 𝛿𝑑𝜂`

]

= 1 −32𝜂 +

12𝜂L 𝛿𝑑𝜂

`

]

=38𝛿

𝛿∗ =1.740𝑥𝑅𝑒

Knownthat:

𝜏 = 𝜇𝑑𝑢𝑑𝑦 mn]

= 𝜇32𝑈𝛿

𝛿 = 4.64𝑥𝜇𝜌𝑈𝑥

=4.64𝑥𝑅𝑒

Knownthatshearstress𝜏:

𝜏 = 𝐶Z12𝜌𝑈c = 𝜇

3𝑈2𝛿

= 2𝜇3𝑈2

1𝜌𝑈c

𝜌𝑈𝑥4.64𝑥 𝜇

= 0.646𝜇𝜌𝑈𝑥

𝐶Z =0.646𝑅𝑒

𝐶[ =𝐹[

12𝜌𝐴𝑈

c

𝐹[ = 𝜏𝐵𝑑𝑥}

]= 𝜇

3𝑈2𝛿

𝐵𝑑𝑥}

]= 0.323 𝑈 𝜌𝑈𝜇 𝑥~

`c 𝐵𝑑𝑥

}

]

𝐹[ = 0.646𝑈𝐵 𝜌𝑈𝜇𝑥

𝐶[ =0.646𝑈𝐵 𝜌𝑈𝜇𝑥

12𝜌𝐴𝑈

c= 1.292

𝜇𝜌𝑈𝑥

𝐶[ =1.292𝑅𝑒

4-3Schlichting(1979,pg.206)pointsoutthatthesimpleflat-platevelocityprofileapproximation:𝑢 ≈ 𝑈 sin �m

cVor�

�= sin �m

cV

givesmuchbetteraccuracyfor𝐶Z, 𝜃and𝛿∗(±2%)thantheparabolicprofileofEq.4-11.Verifythisbycomputing𝐶Z.Doesthissine-waveshapesatisfyanyadditionalboundaryconditionscomparedtoEq.4-11.

𝛿∗ =411𝛿

𝜃 =322𝛿

𝛿 =4.795𝑥𝑅𝑒

(i) Boundarylayerthickness,𝛿:

𝜏]𝜌𝑈c

=𝑑𝑑𝑥

𝑢𝑈

1 −𝑢𝑈

𝑑𝑦V

]=𝑑𝑑𝑥

sin𝜋𝑦2𝛿

1 − sin𝜋𝑦2𝛿

𝑑𝑦V

]

=𝑑𝑑𝑥

sin𝜋𝑦2𝛿

− sinc𝜋𝑦2𝛿

𝑑𝑦V

]=𝑑𝑑𝑥

sin𝜋𝑦2𝛿

−1 − cos 𝜋𝑦

𝛿2

𝑑𝑦V

]

=𝑑𝑑𝑥

2𝛿𝜋−𝛿2

𝜏] =4 − 𝜋2𝜋

𝜌𝑈c𝑑𝛿𝑑𝑥 [1]

𝜏] = 𝜇𝑑𝑢𝑑𝑦 mn]

𝑢 = 𝑈 sin𝜋𝑦2𝛿

𝑑𝑢𝑑𝑦 = 𝑈 cos

𝜋𝑦2𝛿

𝜋2𝛿

𝜏] = 𝜇𝑈 cos𝜋𝑦2𝛿

𝜋2𝛿

=𝜇𝑈𝜋2𝛿

[2]

Fromequation[1]and[2]:

4 − 𝜋2𝜋

𝜌𝑈c𝑑𝛿𝑑𝑥

=𝜇𝑈𝜋2𝛿

𝛿𝑑𝛿 = 11.4975𝜇𝜌𝑈

𝑑𝑥

𝛿c

2 = 11.4975

𝜇𝜌𝑈

𝑥 + 𝐶 at𝑥 = 0, 𝛿 = 0, ∴ 𝐶 = 0

𝛿 =4.795𝑥𝑅𝑒S

(ii) Shearstress,𝜏]:

𝜏] =𝜇𝑈𝜋2𝛿

= 0.327𝜇𝑈𝑥

𝑅𝑒S

(iii) Localcoefficientofdrag,𝐶�:

𝜏] = 0.327𝜇𝑈𝑥

𝑅𝑒S

𝜏] = 𝐶[12𝜌𝑈c

𝐶� =0.654𝑅𝑒S

(iv) Coefficientofdrag,𝐶[:

𝐶[ =𝐹[

12 𝜌𝐴𝑈

c

𝐹[ = 𝜏]𝐵𝑑𝑥}

]= 0.327

𝜇𝑈𝐵𝑥

𝑅𝑒S𝑑𝑥}

]

= 0.327𝜇𝑈𝐵𝑥

𝜌𝑈𝑥𝜇

𝑑𝑥}

]= 0.654𝜇𝑈𝐵

𝜌𝑈𝐿𝜇

𝐶[ =0.654𝜇𝑈𝐵 𝜌𝑈𝐿

𝜇12 𝜌𝐿𝐵𝑈

c=1.31𝑅𝑒}

4-2Repeat the integral heat-transfer analysis of section 4-1.7 by replacing Eq.4-23 by the quartictemperatureprofileapproximation

𝑇 − 𝑇� ≈ 𝑇� − 𝑇� 1 − 2𝜂 + 2𝜂L − 𝜂e

𝑢𝑈 = 2

𝑦𝛿−

𝑦𝛿

c

𝑇 − 𝑇� = 𝑇� − 𝑇� 1 − 2𝑦𝛿�

+ 2𝑦𝛿�

L−

𝑦𝛿�

e

𝑞� =𝑑𝑑𝑥

𝜌𝑐�𝑢 𝑇 − 𝑇��

]𝑑𝑦

𝑞� =𝑑𝑑𝑥

𝜌𝑐�𝑈 𝑇� − 𝑇�2𝑦𝛿−

𝑦𝛿

c1 − 2

𝑦𝛿�

+ 2𝑦𝛿�

L−

𝑦𝛿�

e𝑑𝑦

V�

]

𝑞� =𝑑𝑑𝑥

𝜌𝑐�𝑈 𝑇� − 𝑇� 𝛿2𝐵c

15−𝐵L

42

Where: 𝐵 =𝛿�𝛿

4-4Airat20°Cand1atmflowspastasmoothflatplateasinFig.P4-4.Apitotstagnationtube,placed2mmfromthewall,developsawatermanometerhead,h=21mm.Usethis informationwiththeBlasiussolution,Table4-1,toestimatetheposition𝑥ofthepitottube.Checktoseeiftheflowislaminar.

Forair20°,1atm:

𝜌 = 1.205kgmL ,𝜇 = 1.81×10~gkg/ms

Estimatethevelocity,𝑢at𝑦 = 2mm

BernoulliEquationatpoint(1) = BernoulliEquationatpoint(2)

𝑃𝜌𝑔

+𝑉 c

2𝑔+ 𝑧` =

𝑃c𝜌𝑔

+𝑉cc

2𝑔+ 𝑧c

𝑃c = 𝑃� = 𝑃  = 𝜌�𝑔ℎ + 𝑃¢£¤

𝑃� = 998 9.81 0.021

𝑃c = 205Pa

𝑃 = 𝑃¢£¤ = 0

𝑃c = 205 = 𝜌𝑔ℎ 𝑉c = 2𝑔ℎ

ℎ = 17.34195𝑚 𝑉c = 𝑢 = 18.45m/s

𝑈 = 20m/s

𝑢𝑈= 0.9225

UseTable4-1(Pg233) InBlasiussolution,�

�= 𝑓§ 𝜂 = 0.9225

FromTable4-1,incolumn𝑓§ 𝜂 weneedtofindthevalueof0.9225Fromtable: 𝜂 = 2.4 → 𝑓§ 𝜂 = 0.90107 𝜂 = 2.6 → 𝑓§ 𝜂 = 0.93060Usinglinearinterpolationfor𝑓§ 𝜂 = 0.9225

𝜂 = 2.5451DariBlasiusexactsolution,

𝜂 = 𝑦𝜌𝑈2𝜇𝑥

𝑥 = 41cm

ChecktheReynoldsnumber,

𝑅𝑒 =𝜌𝑉𝐷𝜇

= 5.46×10g

Forsmoothplate,transitiontoturbulenceoccuratReynoldnumber

𝑅𝑒 > 10ªSo,wecouldsaythatflowat𝑥 = 41cmislaminar.Ingeneral,thetransitionfromlaminartoturbulentboundarylayerisoccuratReynoldsnumber5×10g.

4-11Athinequilateraltriangleplateisimmersedparalleltoa12m/sstreamofairat20°Cand1atm,asinFig.P4-12.Assuminglaminarflow,estimatethedragofthisplateinNewton?

FromtheBlasiusexactsolution:

𝐶[ =1.328𝑅𝑒

=1.328𝜌𝑈𝐿𝜇

`/c =1.328𝜌𝑈𝜇

`/c 𝐿~`/c

𝐹[ = 𝐶[ ∙12𝜌𝐴𝑈c

Dragforceforasmallhorizontalarea:

𝑑𝐹[ = 𝐶[ ∙12𝜌𝑈c ∙ 𝑑𝐴 = 𝐶[ ∙

12𝜌𝑈c ∙ 𝐿 ∙ 𝑑𝑦

=1.328𝜌𝑈𝜇

`/c 𝐿~`/c ∙

12𝜌𝑈c ∙ 𝐿 ∙ 𝑑𝑦 =

1.328𝜌𝑈𝜇

`/c ∙12𝜌𝑈c ∙ 𝐿`/c ∙ 𝑑𝑦

=1.328𝜌𝑈𝜇

`/c ∙12𝜌𝑈c ∙ 𝐿

`c ∙

11.73205

∙ 𝑑𝐿 =1.328𝜌𝑈𝜇

`/c ∙12𝜌𝑈c ∙

11.73205

∙ 𝐿`/c ∙ 𝑑𝐿

𝐹[ =1.328𝜌𝑈𝜇

`/c ∙12𝜌𝑈c ∙

11.73205

∙ 𝐿`/c`.iLc]g

]∙ 𝑑𝐿

}¯°±m¯°±

= `.iLc]g`

= }m,𝑑𝑦 = `

`.iLc]g∙ 𝑑𝐿

At𝑦 = 0,𝐿 = 0At𝑦 = 1,𝐿 = 1.73205

𝐹[ =1.328𝜌𝑈𝜇

`/c ∙12𝜌𝑈c ∙

11.73205

∙ 𝐿`/c`.iLc]g

]∙ 𝑑𝐿

=1.328𝜌𝑈𝜇

`/c ∙12𝜌𝑈c ∙

11.73205

∙23𝐿L/c

]

`.iLc]g

Substituteallvalues:𝜌 = 1.205kg/mL

𝜇 = 1.81×10~gkg/ms𝑈 = 12m/s

𝐹[ = 195.44𝑁

Forbothsideofplate:𝐹�³£¢´ = 390.88𝑁

4-13Flowstraightenersconsistofarraysofnarrowductsplaced ina flowtoremoveswirlandothertransverse(secondary)velocities.OneelementcanbeidealizedasasquareboxwiththinsidesasinFig.P4-13.Usingflatplatetheory,deriveaformulaforthepressuredrop∆𝑝acrossan𝑁×𝑁bundleofsuchboxes.

Considertheequationforlaminarflowoverflatplate.

𝐶[ =

1.328

𝜌𝑈𝐿𝜇

`c

𝐹[ = 𝐶[12𝜌𝐴𝑈c =

1.328

𝜌𝑈𝐿𝜇

`c

12𝜌𝑎𝐿𝑈c = 0.664 𝜌𝜇𝐿

`c𝑈

Lc𝑎

Calculatethedragover4wallsand 𝑁c boxes.

𝐹£³£¢´ = 0.664 𝜌𝜇𝐿`c𝑈

Lc𝑎 4 𝑁c

= 2.656𝑁c 𝜌𝜇𝐿`c𝑈

Lc𝑎

(b)Calculatethepressuredropacrossthearray

∆𝑝¢¸¸¢m =𝐹£³£¢´

totalarea

=2.656𝑁c 𝜌𝜇𝐿

`c𝑈

Lc𝑎

𝑁𝑎 c

=2.656 𝜌𝜇𝐿

`c𝑈

Lc

𝑎

4-17Airat20°Cand1atmissuesfromanarrowslotandformsatwo-dimensionallaminarjet.At50cmdownstreamoftheslot,themaximumvelocityis20cm/s.Estimateatthisposition;(a) Thejetwidth(b) Thejetmassflowperunitdepth(c) AnappropriateReynoldsnumberforthejet

(a) ThejetwidthAt20°C: 𝜌 = 1.205kg/mL , 𝜇 = 1.81×10~gkg/m ∙ s

𝑢¤¢S = 0.4543𝐽c

𝜌𝜇𝑥

`L (Eq.4-104)

𝐽 = 0.000965kg/sc

width = 2 𝑦 `%~�¯°± = 𝑏 = 21.8𝑥c𝜇c

𝐽𝜌

`L (Eq.4-106)

𝑏 = 0.09m

(b) Jetmassflow

𝑚 = 36𝐽𝜌𝜇𝑥`L (Eq.4-107)

𝑚 = 0.007236kg/s

(c) Reynoldsnumber

𝑅𝑒 =𝑚𝜇= 400 TheappropriateReynoldsnumberinthisproblem.

𝑅𝑒 =𝐽𝜌𝑥𝜇c

`/L= 121 Reynoldsnumberistoosmall.

𝑅𝑒 =𝜌𝑈𝑏𝜇

= 1200 Reynoldsnumberistoobig.

4-18Airat20°Cand1atmflowsat1m/spastaslendertwo-dimensionalbody,oflengthL=30cm,whosedragcoefficientis0.05basedon“plan”area𝑏𝐿.Assuminglaminarflowatapoint3mdownstreamofthetrailingedge,estimate:(a) Themaximumwakevelocitydefect(b) The“one-percent”wakethickness(c) ThewakethicknessReynoldsnumber

At20°C: 𝜌 = 1.205kg/mL , 𝜇 = 1.81×10~gkg/m ∙ s(a) BodylengthRenumber,𝑅𝑒}

𝑅𝑒} =𝜌𝑢𝐿𝜇

= 20,000

For𝑥 = 3m,thecenterlinewakedefectvelocityiscomputedbyapplyingEq(4-112),at𝑦 = 0

𝑢`𝑈]

= 𝐶[𝑅𝑒}16𝜋

`c 𝐿𝑥

`cexp −

𝑈]𝑦c

4𝑥𝜐= 0.315

𝑢` = 1 0.315

𝑢` 𝑦 = 0 = 0.315m/s = ∆𝑢¤¢S

(b) Wakehalf-thicknesscouldbedefinedasthepointwherethedefectvelocitydropsto1%ofitsmaximumvelocity.

FromGaussianprofile:

exp −𝑈𝑦c

4𝜐𝑥 = 0.01

− 𝑦 c`% =

ln(0.01) 4𝜐𝑥𝑈

=ln(0.01) 4𝑥𝜇

𝑈𝜌

𝑦 `% = 0.0288 m

wake"half − thickness" =𝑦 `%

2

= 0.0144 m

width, 𝑏 = 2 𝑦 `% = 0.0288 m (c) WakeReynoldsnumber

𝑅𝑒 =𝜌∆𝑢¤¢S𝑏

𝜇= 604

4-34InEq.4-144for𝑅𝑒¢ = 9500, �

�Ã= 1.814 S

¢nearthefrontofthecylinder.Forairat20°Cand1

atm,thiscorrespondsapproximatelyto𝑎 = 5𝑐𝑚,𝑈� = 2.85𝑚/𝑠.UsingtheFalkner-Skantheory,Table4-2and4-3,andatemperaturedifference𝑇� − 𝑇� = 12°𝐶,Estimate:

(a) Themomentumthicknessinmm. (b) TheheattransferrateinW/m2atthefrontofthiscylinder.

𝜌 = 1.205kg/mL𝜇 = 1.81×10~gkg/m ∙ s

𝑃 = 0.71𝑘 = 0.026W/m ∙ K

𝜐 =𝜇𝜌= 1.5×10~gmc/s

UseTable4-2(pg243) When𝑚 = 𝛽 = 1 stagnationflow ,𝜃∗ = 0.29235

𝜂 = 𝑦𝑚 + 1 𝑈2𝜐𝑥

𝜃∗ = 0.29235 = 𝜃𝑚 + 1 𝑈2𝜐𝑥

= 𝜃1 + 1 1.814𝑈�

𝑥𝑎

2𝜐𝑥

`c

𝜃 = 0.111 mm

UseTable4-3(pg246)

for𝑚 = 𝛽 = 1 stagnationflow Eq4-76

for𝛽 = 1

𝑁𝑢S

𝑅𝑒S`c = 0.57𝑃𝑟].e

𝑁𝑢 =ℎ𝑥𝑘= 0.57𝑃𝑟].e 𝑅𝑒

`c

ℎ𝑥𝑘 = 0.57𝑃𝑟].e 1.814𝑈�

𝑥𝑎

𝑥𝜐

`c

ℎ = 34W/mcK

𝑞� = ℎ 𝑇� − 𝑇�

= 408 W/mc

4-35

Foraflatplate,𝑈 = 𝑈],andawalltemperaturedistribution𝑇� − 𝑇� = ∆𝑇] 1 − S}

L.

UsethesuperpositionmethodofSec.4-8.4tocomputethevalueof𝑥atwhichthelocalheattransfer𝑞�changessign.RefertoEq(4-172)andEq(4-173)FromEq(4-173)

𝑎] + 1.6123𝑎`𝑥 + 1.9556𝑎c𝑥c + 2.2091𝑎L𝑥L + 2.4151𝑎e𝑥e + ⋯Inthiscase,allpolynomialcoefficientarezeroexcept𝑎]and𝑎LThus,Eq4-173becomes:

𝑎] + 2.2091𝑎L𝑥L

Eq4-172:

𝑞� =0.332𝑘𝑥

𝑃𝑟`L 𝑅𝑒

`c 𝑎] +

43𝑗𝑎; 𝑥Ê

𝑇§ 4𝑗3 𝑇§ 2

3𝑇§ 4𝑗

3 +23

Ë

Ìn`

=0.332𝑘𝑥

𝑃𝑟`L 𝑅𝑒

`c 1 − 2.2091

𝑥L

𝐿L

𝑞� = 0when2.2091𝑥L

𝐿L= 0

𝑥𝐿=

12.2091

`L= 0.7678

4-40Airat20°Cand1atmissuesfromacircularholeandformsaroundlaminarjet.At20cmdownstreamofthehole,themaximumvelocityis35cm/s.Estimate,atthisposition:(a) The“one-percent”jetthickness(b) Thejetmassflow(c) AnappropriateReynoldsnumberforthejet

𝜌 = 1.205kg/mL

𝜇 = 1.81×10~gkg/ms𝜐 = 1.5×10~gmc/s

𝑢¤¢S =3𝐽8𝜋𝜇𝑥

Eq(4-207)

𝐽 = 1.06×10~g kgm/sc

𝐶 =3𝐽

16𝜋𝜌𝜐c

`c

𝐶 = 48.3

(a) The1%jetthicknessoccurswhentheprofileshapefunctioninEq4-207equals0.01

1 +𝐶c𝜂c

4

~c

= 0.01

1 +𝐶c𝜂c

4 = 0.01 ~`c

𝜂 = 0.1242

𝜂 =𝑟(`%)𝑥

= 0.1242

𝑟(`%) = 0.1242𝑥

Jetthickness

𝑏 = 2𝑟(`%) = 0.1242(m)

Jetspreadsoutwardatahalf-angleoftan~` 0.1242 = 7.08°

(b) Jetmassflow

𝑚 = 8𝜋𝜇𝑥

= 9.1×10~g kg/s (c) LocaljetReynoldsnumber

𝑅𝑒 =𝐽𝜌𝜐c

= 39096

𝑅𝑒 =𝑢¤¢S𝑏𝜐

= 1157

4-41Airat20°Cand1atmflowsat1m/spastaslenderbodyofrevolution,oflengthL=15cm,whosedragcoefficientis0.008basedonarea(L2).Assuminglaminarflowatpoint3mdownstreamofthetrailingedge,estimate:(a) Themaximumwakevelocitydefect(b) Theonepercentwakethickness(c) ThewakethicknessReynoldsnumber

𝜌 = 1.205kg/mL

𝜇 = 1.81×10~gkg/ms𝜐 = 1.5×10~gmc/s

(a) Eq(4-211)

𝑢`𝑈] = 𝐶[

𝑈]𝐿8𝜋𝜐

𝐿𝑥exp −

𝑈]𝑟c

4𝑥𝜐

𝑢` = 𝑚𝑎𝑥apabila𝑒𝑥𝑝 = 1

𝑢¤¢S =𝐶[𝑈]c𝐿c

8𝜋𝜐𝑥= 15.9cm/s

(b) 1%thicknessoccurwhentheGaussianprofileinEq4-210equal0.01

exp −𝑈]𝑟c

4𝑥𝜐 = 0.01

𝑟 = 𝑟 `% = 0.0288 m

Wakethickness

𝑏 = 2 𝑟 `% = 5.76 cm (c)

𝑅𝑒 =𝑢¤¢S𝑏𝜐

= 610

4-47Averticalisothermalplate40cmhighand30cmwideisimmersedinairat20°Cand1atm.Eachsideoftheplateistodissipate100Wofheattotheair.Whatisthewalltemperature?Inthiscase

𝑘 = 0.0276W/mK𝜐 = 1.8×10~gmc/s

Averagefluid=50°C

Forair,Prnumber=0.71

𝑁𝑢} =𝑞�𝐿

𝐴�𝑘∆𝑇=12077∆𝑇

𝑁𝑢 = 0.472𝑔𝛽𝜐c𝐿L∆𝑇

`e= 23.4∆𝑇` e

𝑁𝑢 =12077∆𝑇

= 23.4∆𝑇` e

∆𝑇 = 148℃ = 𝑇� − 𝑇�

𝑇� = 168℃ CheckGrnumber

𝐺𝑟 =𝑔𝛽 ∆𝑇 𝐿L

𝜐c

=9.81 1

323 148 0.4 L

1.8×10~g c = 887,898

4-48Ahorizontalpipeofouterdiameter5cmisimmersedinairat20°Cand1atm.Ifthecylindersurfaceisat300°C,howmuchheat(inW)islosttotheairpermeterofpipelength?Estimatethefluidpropertiesataveragetemperature:

𝑇¢ÏÐ =20 + 300

2= 160℃ = 433K

FromTableA-2,pg575

𝑘 = 0.035W/mK𝜐 = 2.96×10~gmc/s

𝛽 =1𝑇=

1433

Rayleighnumberofthecylinder:

𝑅𝑎 =𝑔𝛽𝜐c𝐷L∆𝑇𝑃𝑟

= 642575

Therelevantheat-transfercorrelationforfreeconvectiontohorizontalcylinder.Eq(4-265)

𝑁𝑢

`c = 0.60 +

0.387 𝑅𝑎Ñ`ª

1 + 0.559𝑃𝑟

Ñ`ª

Òci

𝑁𝑢 = 12.85 =ℎ𝐷𝑘

ℎ = 8.995W/mck Thetotalheatlossfromcylindertoairisthusestimatedas:

𝑞� = ℎ 𝜋𝐷𝐿 ∆𝑇

= 395.7W/m

4-50Airatabout1atmand20°Cflowsthrougha12cm“square-duct”at0.4m3/s,asinFig.P4-50.Twohundreds(200)thinplatesof1cmchordlengtharestretchedacrosstheductatrandompositions.They do not interfere with each other. How much additional pressure drop do these platescontributetotheductflowloss.

𝑄 = 𝐴𝑉 → 𝑉 =𝑄𝐴= 27.8m/s

ChecktheReynoldsnumber

𝑅𝑒 =𝜌𝑉𝐷𝜇

= 222000

Re>2000,flow→turbulent

Lengthofplateis0.01(m)Reynoldsnumberontheplateis:

𝑅𝑒 =𝜌𝑉𝐿𝜇

= 18507

18507 < 5×10g → Laminar

CalculatethedragforceuseBlasiuseq.:𝐶[ =1.328𝑅𝑒

= 0.00976

Dragforcefor1plate

𝐹[ = 𝐶[12𝜌𝐴𝑉c = 0.0109 N

Dragforcefor200plates

𝐹£³£¢´ = 2.18 N

Afreebodyoftheplatesshowsthatthisforcecanonlybebalancedbyanoverallpressuredragacrossthearrayofplates.

∆𝑃 =𝐹£³£¢´𝐴�� £

= 151.4 Pa

4-52AconicaldiffuserofinitialradiusRexpandsatauniformangle𝜃,asinFig.P4-52.Theflowentersatuniformvelocity,U0.Assumingaone-dimensionalfreestream,useanylaminarboundarylayermethodofyourchoosingtoestimatetheangle𝜃forwhichflowseparationoccursat𝑥 = 2𝑅.

𝐴`𝑉 = 𝐴c𝑉c

𝜋 𝑅 c𝑈 = 𝜋 𝑟 c𝑢

𝑢 = 𝑈𝑅c

𝑟c

in∆𝐴𝐷𝐸, tan 𝜃 =𝑟

2𝑥 + 𝐹 [1]

in∆𝐴𝐵𝐶, tan 𝜃 =𝑅𝐹

𝐹 =𝑅

tan 𝜃 [2]

From[1]and[2]:

𝑟 = 𝑅 1 +2 tan 𝜃𝑅

𝑥 = 𝑅 1 + 𝑘𝑥

Assumethat:

𝑘 =2 tan 𝜃𝑅

DariEq(4-191),forcylindricalcondition:

𝜃c

𝜐=0.45𝑟c𝑢ª

𝑟c𝑢g𝑑𝑥S

]=

0.45

𝑟c 𝑈 𝑅c𝑟c

ª 𝑟c 𝑈𝑅c

𝑟c

g

𝑑𝑥S

]

=0.45 𝑟`]

𝑈𝑅c1𝑟Ò𝑑𝑥

S

]=0.45 𝑅`] 1 + 𝑘𝑥 `]

𝑈𝑅c1

𝑅Ò 1 + 𝑘𝑥 Ò 𝑑𝑥S

]

=0.45 1 + 𝑘𝑥 `]

7𝑈𝑘1 − 1 + 𝑘𝑥 ~i

Knownthat;

𝑢 = 𝑈𝑅c

𝑟c= 𝑈

𝑅c

𝑅c 1 + 𝑘𝑥 c

𝑑𝑢𝑑𝑥

=−2𝑈𝑘1 + 𝑘𝑥 L

Then,theThwaitesparameter𝜆 𝑥 isgivenby

𝜆 =𝜃c

𝜐𝑑𝑢𝑑𝑥

=0.45 1 + 𝑘𝑥 `]

7𝑈𝑘1 − 1 + 𝑘𝑥 ~i −2𝑈𝑘

1 + 𝑘𝑥 L = −0.97

1 + 𝑘𝑥 i − 1

Separationoccurat𝜆 = −0.09

−0.09 = −0.97

1 + 𝑘𝑥 i − 1

𝑘𝑥 = 0.0788

2 tan 𝜃𝑅

2𝑅 = 0.0788

𝜃 = 1.13°

4-53Showthatthepoint-sinkboundarylayersolutionofEq.4-86and4-89maybeinterpretedthroughThwaites’methodasaconstantvalueofThwaites’parameter,𝜆 = Ñ

Ò].Howthisvalueof𝜆compare

totwo-dimensionalstagnationflow?FromFigure4-16,page250:

Assumethat,𝑢 =−𝑘𝑥

𝜃c

𝜐 =

0.45𝑟c𝑢ª

𝑟c𝑢g𝑑𝑥S

]=0.45𝑢ª

𝑢g𝑑𝑥�

S

=0.45−𝑘𝑥

ª−𝑘𝑥

g

𝑑𝑥�

S=0.45𝑥ª

𝑘ª𝑘g

4𝑥e

𝑢 =−𝑘𝑥

𝑑𝑢𝑑𝑥 =

𝑘𝑥c

𝜆 =𝜃c

𝜐𝑑𝑢𝑑𝑥

=980= 0.1125

Forstagnationflow,𝛽 = 𝑚 = 1FromTable4-2

𝑓§§ = 1.23259𝜆∗ = 0.64790𝜃∗ = 0.29235

𝑢 = 𝛽𝑥𝑑𝑢𝑑𝑥

= 𝛽

𝜃∗ = 𝜃𝑚 + 12

𝑢𝜐𝑥

= 0.29235

𝜃 = 0.29235𝜐𝛽

𝜆 =𝜃c

𝜐𝑑𝑢𝑑𝑥

=0.29235 c 𝜐

𝛽𝜐

𝛽

𝜃 = 0.0855