THEPROWESS STUVIA
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 20C 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/gmol
Then the density of air containing 1012 molecules per mm3 is, in SI units,
= 1012 molecules 4.81E−23 g
3 molecule
mm
g = 4.81E−5 kg
= 4.81E−11
mm3 m3
Finally, from the perfect gas law, Eq. (1.13), at 20C = 293 K, we obtain the pressure:
kg 2
p = RT = 4.81E−5 3 287 m 2 (293 K) = 4.0 Pa ns.
m s K
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, 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
2
o 4 Rearth (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:
m(atmosphere) 5.7E21 grams 1.2Ε44 molecules Ans.
N molecules = =
m(one molecule) 4.8E−23 gm/molecule
This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
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 20C 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/gmol
Then the density of air containing 1012 molecules per mm3 is, in SI units,
= 1012 molecules 4.81E−23 g
3 molecule
mm
g = 4.81E−5 kg
= 4.81E−11
mm3 m3
Finally, from the perfect gas law, Eq. (1.13), at 20C = 293 K, we obtain the pressure:
kg 2
p = RT = 4.81E−5 3 287 m 2 (293 K) = 4.0 Pa ns.
m s K
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, 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
2
o 4 Rearth (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:
m(atmosphere) 5.7E21 grams 1.2Ε44 molecules Ans.
N molecules = =
m(one molecule) 4.8E−23 gm/molecule
This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
