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 g
4.81E
3 23
mm molecule
g kg
4.81E 11 4.81E 5
mm m3
3
Finally, from the perfect gas law, Eq. (1.13), at 20 C 293 K, we obtain the pressure:
kg m2
p RT 4.81E 5 (293 K) 4.0 Pa ns.
3 287 2
m s 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:
bz
matmosphere d (vol) [ e ](4 R2 rth dz)
o ea
0
2 (1.2255 kg / m )4 (6.377E6 m)2
3
o4 Rea
rth
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:
5.7E21 grams 1.2Ε4 molecules Ans.
N 4
m(atmospher
e)
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.
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McGraw-Hill Education.
, Chapter 1 Introduction 1-3
P1.3 For the triangular element in Fig.
P1.3, show that a tilted free liquid
surface, in contact with an atmosphere
at pressure pa, must undergo shear
stress and hence begin to flow.
Fig. P1.3
Solution: Assume zero shear. Due to
element weight, the pressure along the
lower and right sides must vary linearly as
shown, to a higher value at point C.
Vertical forces are presumably in
balance with ele-ment weight included.
But horizontal forces are out of
balance, with the unbalanced force
being to the left, due to the shaded
excess- pressure triangle on the right side
BC. Thus hydrostatic pressures cannot
keep the element in balance, and shear
and flow result.
P1.4 Sand, and other granular materials, definitely flow, that is, you can pour them
from a container or a hopper. There are whole textbooks on the “transport” of
granular materials [54]. Therefore, is sand a fluid? Explain.
Solution: Granular materials do indeed flow, at a rate that can be measured by
“flowmeters”. But they are not true fluids, because they can support a small shear
stress without flowing. They may rest at a finite angle without flowing, which is not
possible for liquids (see Prob. P1.3). The maximum such angle, above which sand
begins to flow, is called the angle of repose. A familiar example is sugar, which pours
easily but forms a significant angle of repose on a heaping spoonful. The physics of
granular materials are complicated by effects such as particle cohesion, clumping,
vibration, and size segregation. See Ref. 48 to learn more.
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, Chapter 1 Introduction 1-4
P1.5 A formula for estimating the mean free path of a perfect gas is:
ℓ 1.26 1.26 RT (1)
RT p
where the latter form follows from the ideal-gas law, p RT. What are the
dimensions of the constant “1.26”? Estimate the mean free path of air at 20 C and 7
kPa. Is air rarefied at this condition?
Solution: We know the dimensions of every term except “1.26”:
M M L2
{ℓ} {L} { } { } {R} {T} { }
3 2
LT L T
Therefore the above formula (first form) may be written dimensionally as
{M/L T}
{L} {1.26?} {1.26?}{L}
{M/L } [{L2 /T 2
3
}{ }]
Since we have {L} on both sides, {1.26} {unity}, that is, the constant is dimensionless.
The formula is therefore dimensionally homogeneous and should hold for any unit
system.
For air at 20 C 293 K and 7000 Pa, the density is p RT (7000)/[(287)(293)]
0.0832 kg m . From Table A-2, its viscosity is 1.80E 5 N s/m2. Then the
3
formula predicts a mean free path of
1.80E 5
ℓ 1.26 9.4E 7 m Ans.
(0.0832)[(287)(293)]1/2
This is quite small. We would judge this gas to approximate a continuum if the
physical scales in the flow are greater than about 100 ℓ, that is, greater than about 94
m.
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McGraw-Hill Education.
, Chapter 1 Introduction 1-5
P1.6 Henri Darcy, a French engineer, proposed that the pressure drop Δp for
flow at velocity V through a tube of length L could be correlated in the form
p
LV 2
If Darcy’s formulation is consistent, what are the dimensions of the coefficient α?
Solution: From Table 1.2, introduce the dimensions of each variable:
p ML 1T 2 L2 2 L2
L
ML 3
L T2
V
T2
Solve for {α} = {L-1} Ans.
[The complete Darcy correlation is α = f /(2D), where D is the tube diameter, and f is a
dimensionless friction factor (Chap. 6).]
P1.7 Convert the following inappropriate quantities into SI units: (a) 2.283E7
U.S. gallons per day; (b) 4.48 furlongs per minute (racehorse speed); and (c)
72,800 avoirdupois ounces per acre.
Solution: (a) (2.283E7 gal/day) x (0.0037854 m3/gal) ÷ (86,400 s/day) =
1.0 m3/s Ans.(a)
(b) 1 furlong = (⅛)mile = 660 ft.
Then (4.48 furlongs/min)x(660 ft/furlong)x(0.3048 m/ft)÷(60 s/min) =
15 m/s Ans.(b)
(c) (72,800 oz/acre)÷(16 oz/lbf)x(4.4482 N/lbf)÷(4046.9 acre/m2) =
5.0 N/m2 = 5.0 Pa Ans.(c)
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, Chapter 1 Introduction 1-6
P1.8 Suppose that bending stress in a beam depends upon bending moment M
and beam area moment of inertia I and is proportional to the beam half-thickness y.
Suppose also that, for the particular case M 2900 in lbf, y 1.5 in, and I 0.4 in4, the
predicted
stress is 75 MPa. Find the only possible dimensionally homogeneous formula for .
Solution: We are given that y fcn(M,I) and we are not to study up on strength
of materials but only to use dimensional reasoning. For homogeneity, the right
hand side must have dimensions of stress, that is,
M
{ } {y}{fcn(M,I)}, or: {L}{fcn(M,I)}
LT 2
M
or: the function must have dimensions {fcn(M,I)}
L2 T 2
Therefore, to achieve dimensional homogeneity, we somehow must combine
bending moment, whose dimensions are {ML2T–2}, with area moment of inertia, {I}
{L4}, and end up with {ML–2T–2}. Well, it is clear that {I} contains neither mass {M}
nor time {T} dimensions, but the bending moment contains both mass and time and in
exactly the com-
bination we need, {MT–2}. Thus it must be that is proportional to M also. Now we
have reduced the problem to:
M ML2 4
yM fcn(I), {L} {fcn(I)}, or: {fcn(I)} {L }
o LT 2 T2
r
We need just enough I’s to give dimensions of {L–4}: we need the formula to be
exactly
inverse in I. The correct dimensionally homogeneous beam bending formula is thus:
My
C , where {C} {unity} Ans.
I
The formula admits to an arbitrary dimensionless constant C whose value can only
be obtained from known data. Convert stress into English units: (75
MPa) (6894.8) 10880 lbf in2. Substitute the given data into the proposed formula:
1088 lbf C
My
C
(2900 lbf in)(1.5 in)
, C 1.00 Ans.
0
or
:
in2 I 0.4 in4
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McGraw-Hill Education.
, Chapter 1 Introduction 1-7
The data show that C 1, or My/I, our old friend from strength of materials.
P1.9 A hemispherical container, 26 inches in diameter, is filled with a liquid at 20 C
and weighed. The liquid weight is found to be 1617 ounces. (a) What is the density
of the fluid, in kg/m3? (b) What fluid might this be? Assume standard gravity, g =
9.807 m/s2.
Solution: First find the volume of the liquid in m3:
1 1 3
Hemisphere
in volume D3 26 in 4601 in 3 3
0.0754m3
61024
2 6 2 6 m3
kg
Liquid mass 1617 oz 16 101 lbm 0.45359 45.84 kg
lbm
45.84 kg kg
Then the liquid density 607 Ans.(a)
0.0754 m3 m3
From Appendix Table A.3, this could very well be ammonia. Ans.(b)
P1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is:
9
F 3 DV V 2D2
16
where D sphere diameter, viscosity, and density. Is the formula
homogeneous?
Solution: Write this formula in dimensional form, using Table 1-2:
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McGraw-Hill Education.
, Chapter 1 Introduction 1-8
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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 g
4.81E
3 23
mm molecule
g kg
4.81E 11 4.81E 5
mm m3
3
Finally, from the perfect gas law, Eq. (1.13), at 20 C 293 K, we obtain the pressure:
kg m2
p RT 4.81E 5 (293 K) 4.0 Pa ns.
3 287 2
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:
bz
matmosphere d (vol) [ e ](4 R2 rth dz)
o ea
0
2 (1.2255 kg / m )4 (6.377E6 m)2
3
o4 Rea
rth
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:
5.7E21 grams 1.2Ε4 molecules Ans.
N 4
m(atmospher
e)
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.
, Chapter 1 Introduction 1-3
P1.3 For the triangular element in Fig.
P1.3, show that a tilted free liquid
surface, in contact with an atmosphere
at pressure pa, must undergo shear
stress and hence begin to flow.
Fig. P1.3
Solution: Assume zero shear. Due to
element weight, the pressure along the
lower and right sides must vary linearly as
shown, to a higher value at point C.
Vertical forces are presumably in
balance with ele-ment weight included.
But horizontal forces are out of
balance, with the unbalanced force
being to the left, due to the shaded
excess- pressure triangle on the right side
BC. Thus hydrostatic pressures cannot
keep the element in balance, and shear
and flow result.
P1.4 Sand, and other granular materials, definitely flow, that is, you can pour them
from a container or a hopper. There are whole textbooks on the “transport” of
granular materials [54]. Therefore, is sand a fluid? Explain.
Solution: Granular materials do indeed flow, at a rate that can be measured by
“flowmeters”. But they are not true fluids, because they can support a small shear
stress without flowing. They may rest at a finite angle without flowing, which is not
possible for liquids (see Prob. P1.3). The maximum such angle, above which sand
begins to flow, is called the angle of repose. A familiar example is sugar, which pours
easily but forms a significant angle of repose on a heaping spoonful. The physics of
granular materials are complicated by effects such as particle cohesion, clumping,
vibration, and size segregation. See Ref. 48 to learn more.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, Chapter 1 Introduction 1-4
P1.5 A formula for estimating the mean free path of a perfect gas is:
ℓ 1.26 1.26 RT (1)
RT p
where the latter form follows from the ideal-gas law, p RT. What are the
dimensions of the constant “1.26”? Estimate the mean free path of air at 20 C and 7
kPa. Is air rarefied at this condition?
Solution: We know the dimensions of every term except “1.26”:
M M L2
{ℓ} {L} { } { } {R} {T} { }
3 2
LT L T
Therefore the above formula (first form) may be written dimensionally as
{M/L T}
{L} {1.26?} {1.26?}{L}
{M/L } [{L2 /T 2
3
}{ }]
Since we have {L} on both sides, {1.26} {unity}, that is, the constant is dimensionless.
The formula is therefore dimensionally homogeneous and should hold for any unit
system.
For air at 20 C 293 K and 7000 Pa, the density is p RT (7000)/[(287)(293)]
0.0832 kg m . From Table A-2, its viscosity is 1.80E 5 N s/m2. Then the
3
formula predicts a mean free path of
1.80E 5
ℓ 1.26 9.4E 7 m Ans.
(0.0832)[(287)(293)]1/2
This is quite small. We would judge this gas to approximate a continuum if the
physical scales in the flow are greater than about 100 ℓ, that is, greater than about 94
m.
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, Chapter 1 Introduction 1-5
P1.6 Henri Darcy, a French engineer, proposed that the pressure drop Δp for
flow at velocity V through a tube of length L could be correlated in the form
p
LV 2
If Darcy’s formulation is consistent, what are the dimensions of the coefficient α?
Solution: From Table 1.2, introduce the dimensions of each variable:
p ML 1T 2 L2 2 L2
L
ML 3
L T2
V
T2
Solve for {α} = {L-1} Ans.
[The complete Darcy correlation is α = f /(2D), where D is the tube diameter, and f is a
dimensionless friction factor (Chap. 6).]
P1.7 Convert the following inappropriate quantities into SI units: (a) 2.283E7
U.S. gallons per day; (b) 4.48 furlongs per minute (racehorse speed); and (c)
72,800 avoirdupois ounces per acre.
Solution: (a) (2.283E7 gal/day) x (0.0037854 m3/gal) ÷ (86,400 s/day) =
1.0 m3/s Ans.(a)
(b) 1 furlong = (⅛)mile = 660 ft.
Then (4.48 furlongs/min)x(660 ft/furlong)x(0.3048 m/ft)÷(60 s/min) =
15 m/s Ans.(b)
(c) (72,800 oz/acre)÷(16 oz/lbf)x(4.4482 N/lbf)÷(4046.9 acre/m2) =
5.0 N/m2 = 5.0 Pa Ans.(c)
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, Chapter 1 Introduction 1-6
P1.8 Suppose that bending stress in a beam depends upon bending moment M
and beam area moment of inertia I and is proportional to the beam half-thickness y.
Suppose also that, for the particular case M 2900 in lbf, y 1.5 in, and I 0.4 in4, the
predicted
stress is 75 MPa. Find the only possible dimensionally homogeneous formula for .
Solution: We are given that y fcn(M,I) and we are not to study up on strength
of materials but only to use dimensional reasoning. For homogeneity, the right
hand side must have dimensions of stress, that is,
M
{ } {y}{fcn(M,I)}, or: {L}{fcn(M,I)}
LT 2
M
or: the function must have dimensions {fcn(M,I)}
L2 T 2
Therefore, to achieve dimensional homogeneity, we somehow must combine
bending moment, whose dimensions are {ML2T–2}, with area moment of inertia, {I}
{L4}, and end up with {ML–2T–2}. Well, it is clear that {I} contains neither mass {M}
nor time {T} dimensions, but the bending moment contains both mass and time and in
exactly the com-
bination we need, {MT–2}. Thus it must be that is proportional to M also. Now we
have reduced the problem to:
M ML2 4
yM fcn(I), {L} {fcn(I)}, or: {fcn(I)} {L }
o LT 2 T2
r
We need just enough I’s to give dimensions of {L–4}: we need the formula to be
exactly
inverse in I. The correct dimensionally homogeneous beam bending formula is thus:
My
C , where {C} {unity} Ans.
I
The formula admits to an arbitrary dimensionless constant C whose value can only
be obtained from known data. Convert stress into English units: (75
MPa) (6894.8) 10880 lbf in2. Substitute the given data into the proposed formula:
1088 lbf C
My
C
(2900 lbf in)(1.5 in)
, C 1.00 Ans.
0
or
:
in2 I 0.4 in4
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, Chapter 1 Introduction 1-7
The data show that C 1, or My/I, our old friend from strength of materials.
P1.9 A hemispherical container, 26 inches in diameter, is filled with a liquid at 20 C
and weighed. The liquid weight is found to be 1617 ounces. (a) What is the density
of the fluid, in kg/m3? (b) What fluid might this be? Assume standard gravity, g =
9.807 m/s2.
Solution: First find the volume of the liquid in m3:
1 1 3
Hemisphere
in volume D3 26 in 4601 in 3 3
0.0754m3
61024
2 6 2 6 m3
kg
Liquid mass 1617 oz 16 101 lbm 0.45359 45.84 kg
lbm
45.84 kg kg
Then the liquid density 607 Ans.(a)
0.0754 m3 m3
From Appendix Table A.3, this could very well be ammonia. Ans.(b)
P1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is:
9
F 3 DV V 2D2
16
where D sphere diameter, viscosity, and density. Is the formula
homogeneous?
Solution: Write this formula in dimensional form, using Table 1-2:
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
, Chapter 1 Introduction 1-8
THOSE WERE PREVIEW PAGES
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CLICK ON THE L.I.N.K
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