Solution Manual for Fluid Mechanics (9th Edition) by Frank White | Complete Solutions for All 11 Chapters + Appendices

ALL 11 CHAPTERS COVERED
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SOLUTIONS MANUAL
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,TABLE OF CONTENTS
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Chapter 1 Introduction o o




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 Di
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mensional Analysis and Similarity Chapter 6 Visco
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us 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 Ch
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apter 9 Compressible Flow
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Chapter 10 Open- o o




Channel Flow Chapter 11 Turbomachi
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nery

APPENDIX F o

, Chaptero1o o Introduction 1-1




Chapter 1  Introductiono o
o




P1.1
Aogasoato20Comayobeorarefiedoifoitocontainsolessothano1012omoleculesoperomm3.oIfoA
vogadro’sonumberoiso6.023E23omoleculesoperomole,owhatoairopressureodoesothisorepresent?

Solution: Theomassoofooneomoleculeoofoairomayobeocomputedoas
Molecularoweight 28.97omol1
mo 
o4.81E23ogoAvogadro’sonumber
6.023E23omolecules/gmol

Thenotheodensityoofoairocontainingo1012omoleculesoperomm3ois,oinoSIounits,

oo 1012 moleculesoo4.81E23 
moleculeo
g
 3 
mm
  
go o4.81E5o kg
o4.81E11
mm3 m3
Finally,ofromotheoperfectogasolaw,oEq.o(1.13),oato20Coo293oK,oweoobtainotheopressure:
 kgo  2o o 
poooRTooo4.81E5o 3o   o287o 2m (293oK)oo4.0oPa ns.
 m  s K

o


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





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, Chaptero1o o Introduction 1-2


P1.2o TableoA.6olistsotheodensityoofotheostandardoatmosphereoasoaofunctionoofoaltitude.o Useoth
eseovaluesotooestimate,ocrudely,osay,owithinoaofactoroofo2,otheonumberoofomoleculesoofoairoinotheo
entireoatmosphereoofotheoearth.

Solution:o o Makeoaoplotoofodensityooversusoaltitudeozoinotheoatmosphere,ofromoTableoA.6:


1.2255okg/m3
DensityoinotheoAtmosphere





0 z 30,000om

Thisowriter’soapproximation:o Theocurveoisoapproximatelyoanoexponential,ooooo exp(-
bz),owithoboapproximatelyoequalotoo0.00011operometer.o Integrateothisooverotheoentireoatmosphere,o
withotheoradiusoofotheoearthoequalotoo6377okm:
m  odo(vol)o  
[o ebozo](4oR2 dz)o 
atmosphere  0 o earth

oo4oR earth
2
(1.2255okgo/om3o)4o(6.377E6om)2
 o o 5.7E18o kg
b 0.00011o/om

Dividingobyotheomassoofooneomoleculeoo4.8E23ogo(seeoProb.o1.1oabove),oweoobtainoth
eototalonumberoofomoleculesoinotheoearth’soatmosphere:

N molecules o
m(atmosphere)o

5.7E21ograms o 1.2Ε44 molecules Ans.
m(oneomolecule) 4.8E23ogm/molecule

Thisoestimate,othoughocrude,oisowithino10operocentoofotheoexactomassoofotheoatmosphere.




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HilloEducation.oAllorightsoreserved.oNooreproductionoorodistributionowithoutotheopriorowrittenoconsentoof
McGraw-HilloEducation.