Fluid Mechanics 9th Edition Frank White Solution Manual 2025 I Complete Chapters 1-11 I 1360 Pages I Verified Step-by-Step Solutions I A+ Guide

ALL 11 CHAPTERS COVERED
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SOLUTIONS MANUAL
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,TABLE OF CONTENTS
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Chapter 1 Introduction
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Chapter 2 Pressure Distribution in a Fluid Chapter
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3 Integral Relations for a Control Volume Chapter
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4 Differential Relations for Fluid Flow Chapter 5 D
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imensional Analysis and Similarity Chapter 6 Visc
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ous Flow in Ducts
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Chapter 7 Flow Past Immersed Bodies
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Chapter 8 Potential Flow and Computational Fluid Dynamics C
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hapter 9 Compressible Flow
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Chapter 10 Open-k k



Channel Flow Chapter 11 Turbomachi
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nery

APPENDIX F k

, Chapterk1k •k Introductio 1-1
n




Chapter 1 • Introduction
k k k




P1.1
Akgaskatk20Ckmaykbekrarefiedkifkitkcontainsklesskthank1012kmoleculeskperkmm3.kI
fkAvogadro’sknumberkisk6.023E23kmoleculeskperkmole,kwhatkairkpressurekdoeskthiskrepresen
t?

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

Thenkthekdensitykofkairkcontainingk1012kmoleculeskperkmm3kis,kinkSIkunits,

k=k 1012 moleculeskk4.81E−23 
moleculek
g
 3 
mm
  
gk =k4.81E−5k kg
=k4.81E−11
mm3 m3
Finally,kfromkthekperfectkgasklaw,kEq.k(1.13),katk20Ck=k293kK,kwekobtainkthekpressure:
 kgk 2k k 
pk=kkRTk=kk4.81E−5k 3k   k287k m2 (293kK)k=k4.0kP ns.
a m s Kk
  




Copyrightk©kMcGraw-
HillkEducation.kAllkrightskreserved.kNokreproductionkorkdistributionkwithoutkthekpriorkwrittenkconsentkof
McGraw-HillkEducation.

, Chapterk1k •k Introductio 1-2
n


P1.2k TablekA.6klistskthekdensitykofkthekstandardkatmospherekaskakfunctionkofkaltitude.k Us
ekthesekvaluesktokestimate,kcrudely,ksay,kwithinkakfactorkofk2,ktheknumberkofkmoleculeskofkai
rkinkthekentirekatmospherekofkthekearth.

Solution:k k Makekakplotkofkdensitykkversuskaltitudekzkinkthekatmosphere,kfromkTablekA.6:


1.2255kkg/m3
DensitykinkthekAtmosphere




0 z 30,000km

Thiskwriter’skapproximation:k Thekcurvekiskapproximatelykankexponential,kkkok exp(-
bz),kwithkbkapproximatelykequalktok0.00011kperkmeter.k Integratekthiskoverkthekentirekatmospher
e,kwithkthekradiuskofkthekearthkequalktok6377kkm:
m =
kdk(vol)k  2[k e−bkzk](4kR dz)k =

atmosphere  0 o earth

ok4kRearth
2
(1.2255kkgk/km3k)4k(6.377E6km)
= =
2
k k 5.7E18k kg
b
0.00011k/km

Dividingkbykthekmasskofkonekmoleculekk4.8E−23kgk(seekProb.k1.1kabove),kwekobtain
kthektotalknumberkofkmoleculeskinkthekearth’skatmosphere:




N molecules =k
m(atmosphere)k
=
5.7E21kgrams k 1.2Ε44 molecules Ans.
m(onekmolecule) 4.8E−23kgm/molecule

Thiskestimate,kthoughkcrude,kiskwithink10kperkcentkofkthekexactkmasskofkthekatmosphere.




Copyrightk©kMcGraw-
HillkEducation.kAllkrightskreserved.kNokreproductionkorkdistributionkwithoutkthekpriorkwrittenkconsentkof
McGraw-HillkEducation.