Solution Manual for Introduction to Fluid Mechanics, 6th Edition by William S. Janna, 9780367341275, Covering Chapters 1-13 Includes Rationales

Chapter 1 • Introduction
k k
k




1.1 Ak gask atk 20Ck mayk bek rarefiedk ifk itk containsk lessk thank1012k moleculeskperkmm3.kIfkA
vogadro’sknumberkisk6.023E23kmoleculeskperkmole,kwhatkairkpressurekdoeskthiskrepresent?

Solution: Thekmasskofkonekmoleculekofkairkmaykbekcomputedkas
Molecularkweight 28.97kmol−1
mk= =
=k4.81E−23kgkAvogadro’sknumber
6.023E23kmolecules/gkmol

Thenkthekdensitykofkairkcontainingk1012kmoleculeskperkmm3kis,kinkSIkunits,

k=k⎛1012k
moleculesk⎞⎛k g ⎞
4.81E−23
|
⎝ mm3 ⎠⎝
|| moleculek|⎠
gk =k4.81E−5k kg
=k4.81E−11
mm3 m3
Finally,kfromkthekperfectkgasklaw,kEq.k(1.13),katk20Ck=k293kK,kwekobtainkthekpressure:
⎛ kgk ⎞ ⎛ m2k k ⎞
pk=kkRTk=k|k4.81E−5
3|
| k287 s2Kk|k(293kK)k=k4.0kPa ns.
⎝ m ⎠⎝ ⎠


1.2 Thekearth’skatmospherekcankbekmodeledkaskakuniformklayerkofkairkofkthicknessk20kkmka
ndkaveragekdensityk0.6kkg/m3k(seekTablekA-
6).kUsekthesekvaluesktokestimatekthektotalkmasskandktotalknumberkofkmoleculeskofkairkinktheken
tirekatmospherekofkthekearth.

Solution:
Letk Rek bek thek earth’sk radiusk k 6377k km.k Thenk thek totalk massk ofk airk ink theka
tmospherekis

mtk =k∫kkdVolk=kavgk(AirkVol)kkavgk4kR2ek(Airkthickness)
=k(0.6kkg/m3k)4k(6.377E6km)2k(20E3km)kk6.1E18kkg Ans.
Dividingkbykthekmasskofkonekmoleculekk4.8E−23kgk(seekProb.k1.1kabove),kwekobtainkthektot
alknumberkofkmoleculeskinkthekearth’skatmosphere:

N molecules =k
m(atmosphere)k
=
6.1E21kgrams k1.3E44kmolecules Ans.
m(onekmolecule) 4.8E−23kgm/molecule

,2 SolutionskManualk •k FluidkMechanics,kFifthkEdition


1.3 ForkthektriangularkelementkinkFig.kP1.3,k
showkthatkaktiltedkfreekliquidksurface,kinkcont
actkwithkankatmospherekatkpressurekpa,kmustk
undergokshearkstresskandkhencekbeginktokflo
w.

Solution:kAssumekzerokshear.kDuektokeleme Fig.k P1.3
ntkweight,kthekpressurekalongktheklowerkandkr
ightksideskmustkvaryklinearlykaskshown,ktokakh
igherkvaluekatkpointkC.kVerticalkforceskarekpre
sumablykinkbalancekwithkele-
kmentkweightkincluded.kButkhorizontalkforces

karekoutkofkbalance,kwithkthekunbalancedkforc

ekbeingktokthekleft,kduektokthekshadedkexcess-
pressurektrianglekonkthekrightksidekBC.k Thusk
hydrostatickpressureskcannotkkeepkthekeleme
ntkinkbalance,kandkshearkandkflowkresult.


1.4 Thekquantitieskviscosityk,kvelocitykV,kandksurfacektensionkYkmaykbekcombinedkintok ak
dimensionlesskgroup.kFindkthekcombinationkwhichkiskproportionalktok.kThiskgroupkhaskakcu
stomarykname,kwhichkbeginskwithkC.kCankyoukguesskitskname?

Solution: Thekdimensionskofkthesekvariableskarek{}k=k{M/LT},k{V}k=k{L/T},kandk{Y}k=
{M/T2}.k Wek mustk dividek k byk Yk tok cancelk massk {M},k thenk workk thek velocityk intok thekg
roup:
⎧ k ⎧kMk/LTk ⎧Tk ⎧kLk
=k =k , hencekmultiplykbyk{V}k=k ;
{k }k {k 2k }k {k } { }
⎩kYJkk ⎩ Mk/Tk JkL⎩kk J ⎩kTJk
Vk
finallykobtaink =k dimensionless. Ans.
Y
ThiskdimensionlesskparameterkiskcommonlykcalledkthekCapillarykNumber.


1.5 Akformulakforkestimatingkthekmeankfreekpathkofkakperfectkgaskis:
 k
lk=k1.26k =k1.26k k(RT)
k
(1)
kk(RT) p

, Chapterk1k •k Introduction 3

wherektheklatterkformkfollowskfromkthekideal-
gasklaw,kk=kpRT.kWhatkarekthekdimensionskofkthekconstantk“1.26”?kEstimatekthekmeankfree
kpathkofkairkatk20Ckandk7kkPa.kIskairkrarefiedkatkthiskcondition?




Solution:k k Wekknowkthekdimensionskofkeveryktermkexceptk“1.26”:
⎧kM ⎧kMk {R}k=k⎧{kk L2 2 } {T}k=k{}
{l}k=k{L}k {}k=k{ LT } {}k=k{k 3k} T k
⎩ J ⎩L J ⎩ J
Thereforekthekabovekformulak(firstkform)kmaykbekwrittenkdimensionallykas
{M/LT}
{L}k=k{1.26?} =k{1.26?}{L}
k



{M/L }[{L k/T kk}{}]
3 2 2


Sincekwekhavek{L}konkbothksides,k{1.26}k=k{unity},kthatkis,kthekconstantkiskdimensionless.kT
hekformulakiskthereforekdimensionallykhomogeneouskandkshouldkholdkforkanykunitksystem.

Forkairkatk20Ck=k293kKkandk7000kPa,kthekdensitykiskk=kpRTk=k(7000)/[(287)(293)]k=k0.
0832kkgm3.kFromkTablekA-
2,kitskviscositykisk1.80E−5kN ks/m2.kThenkthekformulakpredictkakmeankfreekpathkof
1.80E−5
lk=k1.26k k9.4E−7km Ans.
k


1/2
(0.0832)[(287)(293)]
Thiskiskquiteksmall.kWekwouldkjudgekthiskgasktokapproximatekakcontinuumkifkthekphysicalksc
aleskinkthekflowkarekgreaterkthankaboutk100kl,k thatkis,kgreaterkthankaboutk94km.


1.6 Ifkpkiskpressurekandkykiskakcoordinate,kstate,kinkthek{MLT}ksystem,kthekdimensionskofkth
ekquantitiesk(a)kpy;k(b)k∫kpkdy;k(c)k2py2;k(d)kp.

Solution: (a)k{ML−2T−2}; (b)k{MT−2}; (c)k{ML−3T−2}; (d)k{ML−2T−2}


1.7 Aksmallkvillagekdrawsk1.5kacre-
footkofkwaterkperkdaykfromkitskreservoir.kConvertkthiskwaterkusagekintok(a)kgallonskperkminut
e;kandk(b)kliterskperksecond.

Solution:k Onekacrek=k(1kmi2640)k=k(5280kft)2640k=k43560kft2.kThereforek1.5kacre-
ftk=k65340kft3k=k1850km3.kMeanwhile,k1kgallonk=k231kin3k=k231/1728kft3.kThenk1.5kacre-
ftkofkwaterkperkdaykiskequivalentkto
ft3 ⎛k1728k galk⎞⎛ 1 dayk⎞ gal
Qk=k65340 |k 3 || k |kk340k Ans.k(a)
day ⎝ 231 ft k ⎠⎝1440k k mink ⎠ min

, 4 SolutionskManualk •k FluidkMechanics,kFifthkEdition


Similarly,k1850km3k=k1.85E6kliters.kThenkakmetrickunitkforkthiskwaterkusagekis:
Qk=k⎛|1.85E6 dayLk ⎞||⎛ k 1 dayk⎞k L Ans.k(b)
k 86400k k seck| k21k s
⎝ ⎠⎝ ⎠


1.8 SupposekthatkbendingkstresskkinkakbeamkdependskuponkbendingkmomentkMkandkbeamk
areakmomentkofkinertiakIkandkiskproportionalktokthekbeamkhalf-
thicknessky.kSupposekalsokthat,kforkthekparticularkcasekMk=k2900kinlbf,kyk=k1.5kin,kandkIk=k0.
4kin4,kthekpredictedkstresskisk75kMPa.kFindkthekonlykpossiblekdimensionallykhomogeneouskfo
rmulakfork.

Solution:k Wekarekgivenkthatkk=kykfcn(M,I)kandkwekareknotktokstudykupkonkstrengthkofkmater
ialskbutkonlyktokusekdimensionalkreasoning.kForkhomogeneity,kthekrightkhandksidekmustkhave
kdimensionskofkstress,kthatkis,

⎧kMk
{}k=k{y}{fcn(M,I)}, or: =k{L}{fcn(M,I)}
{k 2k }
⎩ LT J ⎧k M
or: thekfunctionkmustkhavekdimensionsk{fcn(M,I)}k=k
{k }
⎩ L2T2kJ
Therefore,ktokachievekdimensionalkhomogeneity,kweksomehowkmustkcombinekbendingkmo
ment,kwhosekdimensionskarek{ML2T–
2
},kwithkareakmomentkofkinertia,k{I}k=k{L4},kandkendkupkwithk{ML–2T–
2
}.kWell,kitkiskclearkthatk{I}kcontainskneitherkmassk{M}knorktimek{T}kdimensions,kbutkthekbe
ndingkmomentkcontainskbothkmasskandktimekandkinkexactlykthekcom-
–
kbinationkwekneed,k{MT

2
}.kThuskitkmustkbekthatkkiskproportionalktokMkalso.kNowkwek havekreducedkthekproblemkto:
⎧ML2k
k = yMkfcn(I), or ⎧ k Mk k
=
{k 2k } {L}{ 2 }{fcn(I)}, or:k k {fcn(I)}
=k{L−4k}
⎩LT J ⎩
T
J
Wek needk justk enoughk I’sk tok givek dimensionsk ofk {L–4}:k wek needkthek formulak tok bek exactly
inversekinkI.kThekcorrectkdimensionallykhomogeneouskbeamkbendingkformulakiskthus:
Myk
k =kCk , wherek{C}k=k{unity} Ans.
I
ThekformulakadmitsktokankarbitrarykdimensionlesskconstantkCkwhosekvaluekcankonlykbekobtai
nedkfromkknownkdata.kConvertkstresskintokEnglishkunits:kk =k(75kMPa)(6894.8)k=k10880klb
fin2.kSubstitutekthekgivenkdatakintokthekproposedkformula:
lbfk Myk (2900klbfin)(1.5kin)k
k =k10880k =kCk =kCk , or: Ckk1.00 Ans.
2 I 4
in 0.4kin
ThekdatakshowkthatkCk=k1,korkk =kMy/I,kourkoldkfriendkfromkstrengthkofkmaterials.