SOLUTION MANUAL FOR FLUID MECHANICS, 9TH EDITION BY FRANK M. WHITE | COMPLETE STEP-BY-STEP SOLUTIONS FOR ALL CHAPTERS 1-11

ALL 11 CHAPTERS COVERED




SOLUTIONS MANUAL

,TABLE OF CONTENTS

Chapter 1 Introduction
Chapter 2 Pressure Distribution in a Fluid
Chapter 3 Integral Relations for a Control Volume
Chapter 4 Differential Relations for Fluid Flow
Chapter 5 Dimensional Analysis and Similarity
Chapter 6 Viscous Flow in Ducts
Chapter 7 Flow Past Immersed Bodies
Chapter 8 Potential Flow and Computational Fluid Dynamics
Chapter 9 Compressible Flow
Chapter 10 Open-Channel Flow
Chapter 11 Turbomachinery

APPENDIX F

, Chapter 1 • Introduction 1-1




Chapter 1 • Introduction

P1.1 A gas at 20C may be rarefied if it contains less than 1012 molecules per mm3. If
Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent?

Solution: The mass of one molecule of air may be computed as
Molecular weight 28.97 mol−1
m= = = 4.81E−23 g
Avogadro’s number 6.023E23 molecules/gmol

Then the density of air containing 1012 molecules per mm3 is, in SI units,

 = 1012 molecules  4.81E−23 
molecule 
g
 3 
mm
  
g = 4.81E−5 kg
= 4.81E−11
mm3 m3
Finally, from the perfect gas law, Eq. (1.13), at 20C = 293 K, we obtain the pressure:
 kg  2 
p =  RT =  4.81E−5 3   287 m (293 K) = 4.0 Pa ns.
 m  s 2
K





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, Chapter 1 • Introduction 1-2



P1.2 Table A.6 lists the density of the standard atmosphere as a function of altitude. Use
these values to estimate, crudely, say, within a factor of 2, the number of molecules of air in
the entire atmosphere of the earth.

Solution: Make a plot of density  versus altitude z in the atmosphere, from Table A.6:


1.2255 kg/m3
Density in the Atmosphere




0 z 30,000 m

This writer’s approximation: The curve is approximately an exponential,   o exp(-bz), with
b approximately equal to 0.00011 per meter. Integrate this over the entire atmosphere, with the
radius of the earth equal to 6377 km:
m =  d (vol)  
[ e−b z ](4 R2 dz) =
atmosphere  0 o earth

o 4 Rearth
2
(1.2255 kg / m3 )4 (6.377E6 m)2
= =  5.7E18 kg
b 0.00011 / m

Dividing by the mass of one molecule  4.8E−23 g (see Prob. 1.1 above), we obtain
the total number of molecules in the earth’s atmosphere:

N molecules =
m(atmosphere)
=
5.7E21 grams  1.2Ε44 molecules Ans.
m(one molecule) 4.8E−23 gm/molecule

This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere.




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McGraw-Hill Education.