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Unformatted text preview: Hurricane Elena in the Gulf of Mexico. Unlike most smallscale fluids engineering applications,
hurricanes are strongly affected by the Coriolis acceleration due to the rotation of the earth, which
causes them to swirl counterclockwise in the Northern Hemisphere. The physical properties and
boundary conditions which govern such flows are discussed in the present chapter. (Courtesy of
NASA/ColorPic Inc./E.R. Degginger/ColorPic Inc.)  v v 2  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 1
Introduction 1.1 Preliminary Remarks Fluid mechanics is the study of fluids either in motion (fluid dynamics) or at rest (fluid
statics) and the subsequent effects of the fluid upon the boundaries, which may be either solid surfaces or interfaces with other fluids. Both gases and liquids are classified
as fluids, and the number of fluids engineering applications is enormous: breathing,
blood flow, swimming, pumps, fans, turbines, airplanes, ships, rivers, windmills, pipes,
missiles, icebergs, engines, filters, jets, and sprinklers, to name a few. When you think
about it, almost everything on this planet either is a fluid or moves within or near a
fluid.
The essence of the subject of fluid flow is a judicious compromise between theory
and experiment. Since fluid flow is a branch of mechanics, it satisfies a set of welldocumented basic laws, and thus a great deal of theoretical treatment is available. However, the theory is often frustrating, because it applies mainly to idealized situations
which may be invalid in practical problems. The two chief obstacles to a workable theory are geometry and viscosity. The basic equations of fluid motion (Chap. 4) are too
difficult to enable the analyst to attack arbitrary geometric configurations. Thus most
textbooks concentrate on flat plates, circular pipes, and other easy geometries. It is possible to apply numerical computer techniques to complex geometries, and specialized
textbooks are now available to explain the new computational fluid dynamics (CFD)
approximations and methods [1, 2, 29].1 This book will present many theoretical results while keeping their limitations in mind.
The second obstacle to a workable theory is the action of viscosity, which can be
neglected only in certain idealized flows (Chap. 8). First, viscosity increases the difficulty of the basic equations, although the boundarylayer approximation found by Ludwig Prandtl in 1904 (Chap. 7) has greatly simplified viscousflow analyses. Second,
viscosity has a destabilizing effect on all fluids, giving rise, at frustratingly small velocities, to a disorderly, random phenomenon called turbulence. The theory of turbulent flow is crude and heavily backed up by experiment (Chap. 6), yet it can be quite
serviceable as an engineering estimate. Textbooks now present digitalcomputer techniques for turbulentflow analysis [32], but they are based strictly upon empirical assumptions regarding the time mean of the turbulent stress field.
1 Numbered references appear at the end of each chapter.  v v 3  eText Main Menu  Textbook Table of Contents  Study Guide 4 Chapter 1 Introduction Thus there is theory available for fluidflow problems, but in all cases it should be
backed up by experiment. Often the experimental data provide the main source of information about specific flows, such as the drag and lift of immersed bodies (Chap. 7).
Fortunately, fluid mechanics is a highly visual subject, with good instrumentation [4,
5, 35], and the use of dimensional analysis and modeling concepts (Chap. 5) is widespread. Thus experimentation provides a natural and easy complement to the theory.
You should keep in mind that theory and experiment should go hand in hand in all
studies of fluid mechanics.  v v 1.2 The Concept of a Fluid From the point of view of fluid mechanics, all matter consists of only two states, fluid
and solid. The difference between the two is perfectly obvious to the layperson, and it
is an interesting exercise to ask a layperson to put this difference into words. The technical distinction lies with the reaction of the two to an applied shear or tangential stress.
A solid can resist a shear stress by a static deformation; a fluid cannot. Any shear
stress applied to a fluid, no matter how small, will result in motion of that fluid. The
fluid moves and deforms continuously as long as the shear stress is applied. As a corollary, we can say that a fluid at rest must be in a state of zero shear stress, a state often called the hydrostatic stress condition in structural analysis. In this condition, Mohr’s
circle for stress reduces to a point, and there is no shear stress on any plane cut through
the element under stress.
Given the definition of a fluid above, every layperson also knows that there are two
classes of fluids, liquids and gases. Again the distinction is a technical one concerning
the effect of cohesive forces. A liquid, being composed of relatively closepacked molecules with strong cohesive forces, tends to retain its volume and will form a free surface in a gravitational field if unconfined from above. Freesurface flows are dominated by gravitational effects and are studied in Chaps. 5 and 10. Since gas molecules
are widely spaced with negligible cohesive forces, a gas is free to expand until it encounters confining walls. A gas has no definite volume, and when left to itself without confinement, a gas forms an atmosphere which is essentially hydrostatic. The hydrostatic behavior of liquids and gases is taken up in Chap. 2. Gases cannot form a
free surface, and thus gas flows are rarely concerned with gravitational effects other
than buoyancy.
Figure 1.1 illustrates a solid block resting on a rigid plane and stressed by its own
weight. The solid sags into a static deflection, shown as a highly exaggerated dashed
line, resisting shear without flow. A freebody diagram of element A on the side of the
block shows that there is shear in the block along a plane cut at an angle through A.
Since the block sides are unsupported, element A has zero stress on the left and right
sides and compression stress
p on the top and bottom. Mohr’s circle does not
reduce to a point, and there is nonzero shear stress in the block.
By contrast, the liquid and gas at rest in Fig. 1.1 require the supporting walls in order to eliminate shear stress. The walls exert a compression stress of p and reduce
Mohr’s circle to a point with zero shear everywhere, i.e., the hydrostatic condition. The
liquid retains its volume and forms a free surface in the container. If the walls are removed, shear develops in the liquid and a big splash results. If the container is tilted,
shear again develops, waves form, and the free surface seeks a horizontal configura  eText Main Menu  Textbook Table of Contents  Study Guide 1.2 The Concept of a Fluid
Free
surface Static
deflection Fig. 1.1 A solid at rest can resist
shear. (a) Static deflection of the
solid; (b) equilibrium and Mohr’s
circle for solid element A. A fluid
cannot resist shear. (c) Containing
walls are needed; (d ) equilibrium
and Mohr’s circle for fluid
element A. A A
Solid A
Liquid Gas (a) (c)
p σ1
θ 5 θ τ1 τ=0 p 0
0 A p A –σ = p –σ = p τ τ (1)
2θ Hydrostatic
condition σ –p σ –p (b) (d )  v v tion, pouring out over the lip if necessary. Meanwhile, the gas is unrestrained and expands out of the container, filling all available space. Element A in the gas is also hydrostatic and exerts a compression stress p on the walls.
In the above discussion, clear decisions could be made about solids, liquids, and
gases. Most engineering fluidmechanics problems deal with these clear cases, i.e., the
common liquids, such as water, oil, mercury, gasoline, and alcohol, and the common
gases, such as air, helium, hydrogen, and steam, in their common temperature and pressure ranges. There are many borderline cases, however, of which you should be aware.
Some apparently “solid” substances such as asphalt and lead resist shear stress for short
periods but actually deform slowly and exhibit definite fluid behavior over long periods. Other substances, notably colloid and slurry mixtures, resist small shear stresses
but “yield” at large stress and begin to flow as fluids do. Specialized textbooks are devoted to this study of more general deformation and flow, a field called rheology [6].
Also, liquids and gases can coexist in twophase mixtures, such as steamwater mixtures or water with entrapped air bubbles. Specialized textbooks present the analysis  eText Main Menu  Textbook Table of Contents  Study Guide 6 Chapter 1 Introduction of such twophase flows [7]. Finally, there are situations where the distinction between
a liquid and a gas blurs. This is the case at temperatures and pressures above the socalled critical point of a substance, where only a single phase exists, primarily resembling a gas. As pressure increases far above the critical point, the gaslike substance becomes so dense that there is some resemblance to a liquid and the usual thermodynamic
approximations like the perfectgas law become inaccurate. The critical temperature
and pressure of water are Tc 647 K and pc 219 atm,2 so that typical problems involving water and steam are below the critical point. Air, being a mixture of gases, has
no distinct critical point, but its principal component, nitrogen, has Tc 126 K and
pc 34 atm. Thus typical problems involving air are in the range of high temperature
and low pressure where air is distinctly and definitely a gas. This text will be concerned
solely with clearly identifiable liquids and gases, and the borderline cases discussed
above will be beyond our scope. 1.3 The Fluid as a Continuum We have already used technical terms such as fluid pressure and density without a rigorous discussion of their definition. As far as we know, fluids are aggregations of molecules, widely spaced for a gas, closely spaced for a liquid. The distance between molecules is very large compared with the molecular diameter. The molecules are not fixed
in a lattice but move about freely relative to each other. Thus fluid density, or mass per
unit volume, has no precise meaning because the number of molecules occupying a
given volume continually changes. This effect becomes unimportant if the unit volume
is large compared with, say, the cube of the molecular spacing, when the number of
molecules within the volume will remain nearly constant in spite of the enormous interchange of particles across the boundaries. If, however, the chosen unit volume is too
large, there could be a noticeable variation in the bulk aggregation of the particles. This
situation is illustrated in Fig. 1.2, where the “density” as calculated from molecular
is plotted versus the size of the unit volume. There
mass m within a given volume
is a limiting volume
* below which molecular variations may be important and Microscopic
uncertainty ρ
Elemental
volume ρ = 1000 kg/m3
δ ρ = 1200 Fig. 1.2 The limit definition of continuum fluid density: (a) an elemental volume in a fluid region of
variable continuum density; (b) calculated density versus size of the
elemental volume. 0 δ * ≈ 109 mm3 Region containing fluid
(a) (b) One atmosphere equals 2116 lbf/ft2 v v 1200 ρ = 1300 2  Macroscopic
uncertainty ρ = 1100  eText Main Menu  101,300 Pa. Textbook Table of Contents  Study Guide δ 1.4 Dimensions and Units above which aggregate variations may be important. The density
defined as
lim
→ m
* 7 of a fluid is best (1.1) The limiting volume
* is about 10 9 mm3 for all liquids and for gases at atmospheric
pressure. For example, 10 9 mm3 of air at standard conditions contains approximately
3 107 molecules, which is sufficient to define a nearly constant density according to
Eq. (1.1). Most engineering problems are concerned with physical dimensions much larger
than this limiting volume, so that density is essentially a point function and fluid properties can be thought of as varying continually in space, as sketched in Fig. 1.2a. Such a
fluid is called a continuum, which simply means that its variation in properties is so smooth
that the differential calculus can be used to analyze the substance. We shall assume that
continuum calculus is valid for all the analyses in this book. Again there are borderline
cases for gases at such low pressures that molecular spacing and mean free path3 are comparable to, or larger than, the physical size of the system. This requires that the continuum approximation be dropped in favor of a molecular theory of rarefiedgas flow [8]. In
principle, all fluidmechanics problems can be attacked from the molecular viewpoint, but
no such attempt will be made here. Note that the use of continuum calculus does not preclude the possibility of discontinuous jumps in fluid properties across a free surface or
fluid interface or across a shock wave in a compressible fluid (Chap. 9). Our calculus in
Chap. 4 must be flexible enough to handle discontinuous boundary conditions. 1.4 Dimensions and Units A dimension is the measure by which a physical variable is expressed quantitatively.
A unit is a particular way of attaching a number to the quantitative dimension. Thus
length is a dimension associated with such variables as distance, displacement, width,
deflection, and height, while centimeters and inches are both numerical units for expressing length. Dimension is a powerful concept about which a splendid tool called
dimensional analysis has been developed (Chap. 5), while units are the nittygritty, the
number which the customer wants as the final answer.
Systems of units have always varied widely from country to country, even after international agreements have been reached. Engineers need numbers and therefore unit
systems, and the numbers must be accurate because the safety of the public is at stake.
You cannot design and build a piping system whose diameter is D and whose length
is L. And U.S. engineers have persisted too long in clinging to British systems of units.
There is too much margin for error in most British systems, and many an engineering
student has flunked a test because of a missing or improper conversion factor of 12 or
144 or 32.2 or 60 or 1.8. Practicing engineers can make the same errors. The writer is
aware from personal experience of a serious preliminary error in the design of an aircraft due to a missing factor of 32.2 to convert pounds of mass to slugs.
In 1872 an international meeting in France proposed a treaty called the Metric Convention, which was signed in 1875 by 17 countries including the United States. It was
an improvement over British systems because its use of base 10 is the foundation of
our number system, learned from childhood by all. Problems still remained because
3  v v The mean distance traveled by molecules between collisions.  eText Main Menu  Textbook Table of Contents  Study Guide 8 Chapter 1 Introduction even the metric countries differed in their use of kiloponds instead of dynes or newtons, kilograms instead of grams, or calories instead of joules. To standardize the metric system, a General Conference of Weights and Measures attended in 1960 by 40
countries proposed the International System of Units (SI). We are now undergoing a
painful period of transition to SI, an adjustment which may take many more years to
complete. The professional societies have led the way. Since July 1, 1974, SI units have
been required by all papers published by the American Society of Mechanical Engineers, which prepared a useful booklet explaining the SI [9]. The present text will use
SI units together with British gravitational (BG) units. Primary Dimensions In fluid mechanics there are only four primary dimensions from which all other dimensions can be derived: mass, length, time, and temperature.4 These dimensions and their units
in both systems are given in Table 1.1. Note that the kelvin unit uses no degree symbol.
The braces around a symbol like {M} mean “the dimension” of mass. All other variables
in fluid mechanics can be expressed in terms of {M}, {L}, {T}, and { }. For example, acceleration has the dimensions {LT 2}. The most crucial of these secondary dimensions is
force, which is directly related to mass, length, and time by Newton’s second law
F ma (1.2)
2 From this we see that, dimensionally, {F} {MLT }. A constant of proportionality
is avoided by defining the force unit exactly in terms of the primary units. Thus we
define the newton and the pound of force
1 newton of force
1 pound of force 1 lbf 1N 1 kg m/s2 1 slug ft/s2 (1.3) 4.4482 N In this book the abbreviation lbf is used for poundforce and lb for poundmass. If instead one adopts other force units such as the dyne or the poundal or kilopond or adopts
other mass units such as the gram or poundmass, a constant of proportionality called
gc must be included in Eq. (1.2). We shall not use gc in this book since it is not necessary in the SI and BG systems.
A list of some important secondary variables in fluid mechanics, with dimensions
derived as combinations of the four primary dimensions, is given in Table 1.2. A more
complete list of conversion factors is given in App. C.
Table 1.1 Primary Dimensions in
SI and BG Systems Primary dimension SI unit
Kilogram (kg)
Meter (m)
Second (s)
Kelvin (K) Mass {M}
Length {L}
Time {T}
Temperature { } BG unit Conversion factor Slug
Foot (ft)
Second (s)
Rankine (°R) 1
1
1
1 slug 14.5939 kg
ft 0.3048 m
s 1s
K 1.8°R  v v 4
If electromagnetic effects are important, a fifth primary dimension must be included, electric current
{I}, whose SI unit is the ampere (A).  eText Main Menu  Textbook Table of Contents  Study Guide 1.4 Dimensions and Units
Table 1.2 Secondary Dimensions in
Fluid Mechanics Secondary dimension SI unit BG unit 2 2 2 Conversion factor m
m3
m/s
m/s2 ft
ft3
ft/s
ft/s2 1m
10.764 ft2
1 m3 35.315 ft3
1 ft/s 0.3048 m/s
1 ft/s2 0.3048 m/s2 Pa N/m2
s1 lbf/ft2
s1 1 lbf/ft2 47.88 Pa
1s 1 1s 1 J Nm
W J/s
kg/m3
kg/(m s)
m2/(s2 K) Area {L }
Volume {L3}
Velocity {LT 1}
Acceleration {LT 2}
Pressure or stress
{ML 1T 2}
Angular velocity {T 1}
Energy, heat, work
{ML2T 2}
Power {ML2T 3}
Density {ML 3}
Viscosity {ML 1T 1}
Specific heat {L2T 2 1} 9 2 ft lbf
ft lbf/s
slugs/ft3
slugs/(ft s)
ft2/(s2 °R) 1
1
1
1
1 ft lbf 1.3558 J
ft lbf/s 1.3558 W
slug/ft3 515.4 kg/m3
slug/(ft s) 47.88 kg/(m s)
m2/(s2 K) 5.980 ft2/(s2 °R) EXAMPLE 1.1
A body weighs 1000 lbf when exposed to a standard earth gravity g 32.174 ft/s2. (a) What is
its mass in kg? (b) What will the weight of this body be in N if it is exposed to the moon’s standard acceleration gmoon 1.62 m/s2? (c) How fast will the body accelerate if a net force of 400
lbf is applied to it on the moon or on the earth? Solution
Part (a) Equation (1.2) holds with F
F weight and a W mg gearth:
(m slugs)(32.174 ft/s2) 1000 lbf or 1000
(31.08 slugs)(14.5939 kg/slug) 453.6 kg
Ans. (a)
32.174
The change from 31.08 slugs to 453.6 kg illustrates the proper use of the conversion factor
14.5939 kg/slug.
m Part (b) The mass of the body remains 453.6 kg regardless of its location. Equation (1.2) applies with a
new value of a and hence a new force
F Part (c) Wmoon mgmoon (453.6 kg)(1.62 m/s2) 735 N Ans. (b) This problem does not involve weight or gravity or position and is simply a direct application
of Newton’s law with an unbalanced force:
F 400 lbf ma (31.08 slugs)(a ft/s2) or
a 400
31.08 12.43 ft/s2 3.79 m/s2  v v This acceleration would be the same on the moon or earth or anywhere.  eText Main Menu  Textbook Table of Contents  Study Guide Ans. (c) Chapter 1 Introduction Many data in the literature are reported in inconvenient or arcane units suitable only
to some industry or specialty or country. The engineer should convert these data to the
SI or BG system before using them. This requires the systematic application of conversion factors, as in the following example. EXAMPLE 1.2
An early viscosity unit in the cgs system is the poise (abbreviated P), or g/(cm s), named after
J. L. M. Poiseuille, a French physician who performed pioneering experiments in 1840 on water flow in pipes. The viscosity of water (fresh or salt) at 293.16 K 20°C is approximately
0.01 P. Express this value in (a) SI and (b) BG units. Solution
Part (a) [0.01 g/(cm s)] Part (b) 1 kg
(100 cm/m)
100 0 g [0.001 kg/(m s)]
2.09 10 5 0.001 kg/(m s) Ans. (a) 1 slug
(0.3048 m/ft)
14.59 kg slug/(ft s) Ans. (b) Note: Result (b) could have been found directly from (a) by dividing (a) by the viscosity conversion factor 47.88 listed in Table 1.2. We repeat our advice: Faced with data in unusual units, convert them immediately
to either SI or BG units because (1) it is more professional and (2) theoretical equations in fluid mechanics are dimensionally consistent and require no further conversion
factors when these two fundamental unit systems are used, as the following example
shows. EXAMPLE 1.3
A useful theoretical equation for computing the relation between pressure, velocity, and altitude
in a steady flow of a nearly inviscid, nearly incompressible fluid with negligible heat transfer
and shaft work5 is the Bernoulli relation, named after Daniel Bernoulli, who published a hydrodynamics textbook in 1738:
p0
where p0
p
V
Z
g p 1
2 V2 gZ (1) stagnation pressure
pressure in moving fluid
velocity
density
altitude
gravitational acceleration 5 That’s an awful lot of assumptions, which need further study in Chap. 3.  v v 10  eText Main Menu  Textbook Table of Contents  Study Guide 1.4 Dimensions and Units 11 (a) Show that Eq. (1) satisfies the principle of dimensional homogeneity, which states that all
additive terms in a physical equation must have the same dimensions. (b) Show that consistent
units result without additional conversion factors in SI units. (c) Repeat (b) for BG units. Solution
Part (a) We can express Eq. (1) dimensionally, using braces by entering the dimensions of each term
from Table 1.2:
{ML 1T {ML 1T } 2 {ML 1T Part (b) 2 2 } {ML 3}{L2T 2 } {ML 3}{LT 2}{L} } for all terms Ans. (a) Enter the SI units for each quantity from Table 1.2:
{N/m2} {N/m2} {kg/m3}{m2/s2} {N/m2} {kg/m3}{m/s2}{m} {kg/(m s2)} The righthand side looks bad until we remember from Eq. (1.3) that 1 kg
{kg/(m s2)} {N s2/m }
{m s2} 1 N s2/m. {N/m2} Ans. (b) Thus all terms in Bernoulli’s equation will have units of pascals, or newtons per square meter,
when SI units are used. No conversion factors are needed, which is true of all theoretical equations in fluid mechanics. Part (c) Introducing BG units for each term, we have
{lbf/ft2} {lbf/ft2} {slugs/ft3}{ft2/s2} {lbf/ft2} {slugs/(ft s2)} But, from Eq. (1.3), 1 slug {slugs/ft3}{ft/s2}{ft} 1 lbf s2/ft, so that {slugs/(ft s2)} {lbf s2/ft}
{ft s2} {lbf/ft2} Ans. (c) All terms have the unit of poundsforce per square foot. No conversion factors are needed in the
BG system either. There is still a tendency in Englishspeaking countries to use poundforce per square
inch as a pressure unit because the numbers are more manageable. For example, standard atmospheric pressure is 14.7 lbf/in2 2116 lbf/ft2 101,300 Pa. The pascal is a
small unit because the newton is less than 1 lbf and a square meter is a very large area.
4
It is felt nevertheless that the pascal will gradually gain universal acceptance; e.g., repair manuals for U.S. automobiles now specify pressure measurements in pascals.  Note that not only must all (fluid) mechanics equations be dimensionally homogeneous,
one must also use consistent units; that is, each additive term must have the same units.
There is no trouble doing this with the SI and BG systems, as in Ex. 1.3, but woe unto v v Consistent Units  eText Main Menu  Textbook Table of Contents  Study Guide 12 Chapter 1 Introduction those who try to mix colloquial English units. For example, in Chap. 9, we often use
the assumption of steady adiabatic compressible gas flow:
12
V
2 h constant where h is the fluid enthalpy and V2/2 is its kinetic energy. Colloquial thermodynamic
tables might list h in units of British thermal units per pound (Btu/lb), whereas V is
likely used in ft/s. It is completely erroneous to add Btu/lb to ft2/s2. The proper unit
for h in this case is ft lbf/slug, which is identical to ft2/s2. The conversion factor is
1 Btu/lb 25,040 ft2/s2 25,040 ft lbf/slug. Homogeneous versus
Dimensionally Inconsistent
Equations All theoretical equations in mechanics (and in other physical sciences) are dimensionally homogeneous; i.e., each additive term in the equation has the same dimensions.
For example, Bernoulli’s equation (1) in Example 1.3 is dimensionally homogeneous:
Each term has the dimensions of pressure or stress of {F/L2}. Another example is the
equation from physics for a body falling with negligible air resistance:
S S0 V0t 1
2 gt2 where S0 is initial position, V0 is initial velocity, and g is the acceleration of gravity. Each
1
term in this relation has dimensions of length {L}. The factor 2 , which arises from integration, is a pure (dimensionless) number, {1}. The exponent 2 is also dimensionless.
However, the reader should be warned that many empirical formulas in the engineering literature, arising primarily from correlations of data, are dimensionally inconsistent. Their units cannot be reconciled simply, and some terms may contain hidden variables. An example is the formula which pipe valve manufacturers cite for liquid
volume flow rate Q (m3/s) through a partially open valve:
Q CV p
SG 1/2 where p is the pressure drop across the valve and SG is the specific gravity of the
liquid (the ratio of its density to that of water). The quantity CV is the valve flow coefficient, which manufacturers tabulate in their valve brochures. Since SG is dimensionless {1}, we see that this formula is totally inconsistent, with one side being a flow
rate {L3/T} and the other being the square root of a pressure drop {M1/2/L1/2T}. It follows that CV must have dimensions, and rather odd ones at that: {L7/2/M1/2}. Nor is
the resolution of this discrepancy clear, although one hint is that the values of CV in
the literature increase nearly as the square of the size of the valve. The presentation of
experimental data in homogeneous form is the subject of dimensional analysis (Chap.
5). There we shall learn that a homogeneous form for the valve flow relation is
Q Cd Aopening p 1/2  v v where is the liquid density and A the area of the valve opening. The discharge coefficient Cd is dimensionless and changes only slightly with valve size. Please believe —until we establish the fact in Chap. 5—that this latter is a much better formulation of the data.  eText Main Menu  Textbook Table of Contents  Study Guide 1.4 Dimensions and Units 13 Meanwhile, we conclude that dimensionally inconsistent equations, though they
abound in engineering practice, are misleading and vague and even dangerous, in the
sense that they are often misused outside their range of applicability. Convenient Prefixes in
Powers of 10 Engineering results often are too small or too large for the common units, with too
many zeros one way or the other. For example, to write p 114,000,000 Pa is long
and awkward. Using the prefix “M” to mean 106, we convert this to a concise p
114 MPa (megapascals). Similarly, t 0.000000003 s is a proofreader’s nightmare
compared to the equivalent t 3 ns (nanoseconds). Such prefixes are common and
convenient, in both the SI and BG systems. A complete list is given in Table 1.3. Table 1.3 Convenient Prefixes
for Engineering Units
Multiplicative
factor Prefix Symbol 1012
109
106
103
102
10
10 1
10 2
10 3
10 6
10 9
10 12
10 15
10 18 tera
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
nano
pico
femto
atto T
G
M
k
h
da
d
c
m
n
p
f
a EXAMPLE 1.4
In 1890 Robert Manning, an Irish engineer, proposed the following empirical formula for the
average velocity V in uniform flow due to gravity down an open channel (BG units):
V
where R
S
n 1.49 2/3 1/2
RS
n (1) hydraulic radius of channel (Chaps. 6 and 10)
channel slope (tangent of angle that bottom makes with horizontal)
Manning’s roughness factor (Chap. 10) and n is a constant for a given surface condition for the walls and bottom of the channel. (a) Is
Manning’s formula dimensionally consistent? (b) Equation (1) is commonly taken to be valid in
BG units with n taken as dimensionless. Rewrite it in SI form. Solution
Part (a) Introduce dimensions for each term. The slope S, being a tangent or ratio, is dimensionless, denoted by {unity} or {1}. Equation (1) in dimensional form is
1.49
{L2/3}{1}
n L
T This formula cannot be consistent unless {1.49/n} {L1/3/T}. If n is dimensionless (and it is
never listed with units in textbooks), then the numerical value 1.49 must have units. This can be
tragic to an engineer working in a different unit system unless the discrepancy is properly documented. In fact, Manning’s formula, though popular, is inconsistent both dimensionally and
physically and does not properly account for channelroughness effects except in a narrow range
of parameters, for water only. Part (b) From part (a), the number 1.49 must have dimensions {L1/3/T} and thus in BG units equals
1.49 ft1/3/s. By using the SI conversion factor for length we have
(1.49 ft1/3/s)(0.3048 m/ft)1/3 1.00 m1/3/s Therefore Manning’s formula in SI becomes  v v V  eText Main Menu  1.0 2/3 1/2
RS
n Textbook Table of Contents  Ans. (b) (2) Study Guide 14 Chapter 1 Introduction
with R in m and V in m/s. Actually, we misled you: This is the way Manning, a metric user, first
proposed the formula. It was later converted to BG units. Such dimensionally inconsistent formulas are dangerous and should either be reanalyzed or treated as having very limited application. 1.5 Properties of the
Velocity Field In a given flow situation, the determination, by experiment or theory, of the properties
of the fluid as a function of position and time is considered to be the solution to the
problem. In almost all cases, the emphasis is on the spacetime distribution of the fluid
properties. One rarely keeps track of the actual fate of the specific fluid particles.6 This
treatment of properties as continuumfield functions distinguishes fluid mechanics from
solid mechanics, where we are more likely to be interested in the trajectories of individual particles or systems. Eulerian and Lagrangian
Desciptions There are two different points of view in analyzing problems in mechanics. The first
view, appropriate to fluid mechanics, is concerned with the field of flow and is called
the eulerian method of description. In the eulerian method we compute the pressure
field p(x, y, z, t) of the flow pattern, not the pressure changes p(t) which a particle experiences as it moves through the field.
The second method, which follows an individual particle moving through the flow,
is called the lagrangian description. The lagrangian approach, which is more appropriate to solid mechanics, will not be treated in this book. However, certain numerical
analyses of sharply bounded fluid flows, such as the motion of isolated fluid droplets,
are very conveniently computed in lagrangian coordinates [1].
Fluiddynamic measurements are also suited to the eulerian system. For example,
when a pressure probe is introduced into a laboratory flow, it is fixed at a specific position (x, y, z). Its output thus contributes to the description of the eulerian pressure
field p(x, y, z, t). To simulate a lagrangian measurement, the probe would have to move
downstream at the fluid particle speeds; this is sometimes done in oceanographic measurements, where flowmeters drift along with the prevailing currents.
The two different descriptions can be contrasted in the analysis of traffic flow along
a freeway. A certain length of freeway may be selected for study and called the field
of flow. Obviously, as time passes, various cars will enter and leave the field, and the
identity of the specific cars within the field will constantly be changing. The traffic engineer ignores specific cars and concentrates on their average velocity as a function of
time and position within the field, plus the flow rate or number of cars per hour passing a given section of the freeway. This engineer is using an eulerian description of the
traffic flow. Other investigators, such as the police or social scientists, may be interested in the path or speed or destination of specific cars in the field. By following a
specific car as a function of time, they are using a lagrangian description of the flow. The Velocity Field Foremost among the properties of a flow is the velocity field V(x, y, z, t). In fact, determining the velocity is often tantamount to solving a flow problem, since other prop  v v 6
One example where fluidparticle paths are important is in waterquality analysis of the fate of
contaminant discharges.  eText Main Menu  Textbook Table of Contents  Study Guide 1.5 Properties of the Velocity Field 15 erties follow directly from the velocity field. Chapter 2 is devoted to the calculation of
the pressure field once the velocity field is known. Books on heat transfer (for example, Ref. 10) are essentially devoted to finding the temperature field from known velocity fields.
In general, velocity is a vector function of position and time and thus has three components u, v, and w, each a scalar field in itself:
V(x, y, z, t) iu(x, y, z, t) jv(x, y, z, t) kw(x, y, z, t) (1.4) The use of u, v, and w instead of the more logical component notation Vx, Vy, and Vz
is the result of an almost unbreakable custom in fluid mechanics.
Several other quantities, called kinematic properties, can be derived by mathematically manipulating the velocity field. We list some kinematic properties here and give
more details about their use and derivation in later chapters:
1. Displacement vector: r 2. Acceleration: a 3. Volume rate of flow: Q 4. Volume expansion rate: 5. Local angular velocity: V dt (Sec. 1.9) dV
dt (Sec. 4.1) (V n) dA 1d
dt V
1
2 (Sec. 3.2)
(Sec. 4.2) V (Sec. 4.8) We will not illustrate any problems regarding these kinematic properties at present. The
point of the list is to illustrate the type of vector operations used in fluid mechanics and
to make clear the dominance of the velocity field in determining other flow properties.
Note: The fluid acceleration, item 2 above, is not as simple as it looks and actually involves four different terms due to the use of the chain rule in calculus (see Sec. 4.1). EXAMPLE 1.5
Fluid flows through a contracting section of a duct, as in Fig. E1.5. A velocity probe inserted at
section (1) measures a steady value u1 1 m/s, while a similar probe at section (2) records a
steady u2 3 m/s. Estimate the fluid acceleration, if any, if x 10 cm.
(1) Solution (2) u1 u2 The flow is steady (not timevarying), but fluid particles clearly increase in velocity as they pass
from (1) to (2). This is the concept of convective acceleration (Sec. 4.1). We may estimate the
acceleration as a velocity change u divided by a time change t
x/uavg:
ax x velocity change
time change u2 u1
x/[ 1 (u1 u2)]
2 (3.0 1.0 m/s)(1.0
2(0.1 m) 3.0 m/s) 40 m/s2 Ans. A simple estimate thus indicates that this seemingly innocuous flow is accelerating at 4 times  v v E1.5  eText Main Menu  Textbook Table of Contents  Study Guide 16 Chapter 1 Introduction
the acceleration of gravity. In the limit as x and t become very small, the above estimate reduces to a partialderivative expression for convective xacceleration:
ax,convective lim t→0 u
t u u
x In threedimensional flow (Sec. 4.1) there are nine of these convective terms. 1.6 Thermodynamic Properties
of a Fluid While the velocity field V is the most important fluid property, it interacts closely with
the thermodynamic properties of the fluid. We have already introduced into the discussion the three most common such properties
1. Pressure p
2. Density
3. Temperature T
These three are constant companions of the velocity vector in flow analyses. Four other
thermodynamic properties become important when work, heat, and energy balances are
treated (Chaps. 3 and 4):
4.
5.
6.
7. Internal energy e
Enthalpy h û p/
Entropy s
Specific heats cp and cv In addition, friction and heat conduction effects are governed by the two socalled transport properties:
8. Coefficient of viscosity
9. Thermal conductivity k
All nine of these quantities are true thermodynamic properties which are determined
by the thermodynamic condition or state of the fluid. For example, for a singlephase
substance such as water or oxygen, two basic properties such as pressure and temperature are sufficient to fix the value of all the others:
( p, T ) h h( p, T ) ( p, T ) (1.5)  v v and so on for every quantity in the list. Note that the specific volume, so important in
thermodynamic analyses, is omitted here in favor of its inverse, the density .
Recall that thermodynamic properties describe the state of a system, i.e., a collection of matter of fixed identity which interacts with its surroundings. In most cases
here the system will be a small fluid element, and all properties will be assumed to be
continuum properties of the flow field:
(x, y, z, t), etc.
Recall also that thermodynamics is normally concerned with static systems, whereas
fluids are usually in variable motion with constantly changing properties. Do the properties retain their meaning in a fluid flow which is technically not in equilibrium? The
answer is yes, from a statistical argument. In gases at normal pressure (and even more
so for liquids), an enormous number of molecular collisions occur over a very short
distance of the order of 1 m, so that a fluid subjected to sudden changes rapidly ad  eText Main Menu  Textbook Table of Contents  Study Guide 1.6 Thermodynamic Properties of a Fluid 17 justs itself toward equilibrium. We therefore assume that all the thermodynamic properties listed above exist as point functions in a flowing fluid and follow all the laws
and state relations of ordinary equilibrium thermodynamics. There are, of course, important nonequilibrium effects such as chemical and nuclear reactions in flowing fluids which are not treated in this text. Pressure Pressure is the (compression) stress at a point in a static fluid (Fig. 1.1). Next to velocity, the pressure p is the most dynamic variable in fluid mechanics. Differences or
gradients in pressure often drive a fluid flow, especially in ducts. In lowspeed flows,
the actual magnitude of the pressure is often not important, unless it drops so low as to
cause vapor bubbles to form in a liquid. For convenience, we set many such problem
assignments at the level of 1 atm 2116 lbf/ft2 101,300 Pa. Highspeed (compressible)
gas flows (Chap. 9), however, are indeed sensitive to the magnitude of pressure. Temperature Temperature T is a measure of the internal energy level of a fluid. It may vary considerably during highspeed flow of a gas (Chap. 9). Although engineers often use Celsius or Fahrenheit scales for convenience, many applications in this text require absolute (Kelvin or Rankine) temperature scales:
°R
K °F
°C 459.69
273.16 If temperature differences are strong, heat transfer may be important [10], but our concern here is mainly with dynamic effects. We examine heattransfer principles briefly
in Secs. 4.5 and 9.8. Density The density of a fluid, denoted by (lowercase Greek rho), is its mass per unit volume. Density is highly variable in gases and increases nearly proportionally to the pressure level. Density in liquids is nearly constant; the density of water (about 1000 kg/m3)
increases only 1 percent if the pressure is increased by a factor of 220. Thus most liquid flows are treated analytically as nearly “incompressible.”
In general, liquids are about three orders of magnitude more dense than gases at atmospheric pressure. The heaviest common liquid is mercury, and the lightest gas is hydrogen. Compare their densities at 20°C and 1 atm:
Mercury: 13,580 kg/m3 Hydrogen: 0.0838 kg/m3 They differ by a factor of 162,000! Thus the physical parameters in various liquid and
gas flows might vary considerably. The differences are often resolved by the use of dimensional analysis (Chap. 5). Other fluid densities are listed in Tables A.3 and A.4 (in
App. A). Specific Weight The specific weight of a fluid, denoted by (lowercase Greek gamma), is its weight
per unit volume. Just as a mass has a weight W mg, density and specific weight are
simply related by gravity:  v v g  eText Main Menu  Textbook Table of Contents (1.6)  Study Guide 18 Chapter 1 Introduction The units of are weight per unit volume, in lbf/ft3 or N/m3. In standard earth gravity, g 32.174 ft/s2 9.807 m/s2. Thus, e.g., the specific weights of air and water at
20°C and 1 atm are approximately
air
water (1.205 kg/m3)(9.807 m/s2)
(998 kg/m3)(9.807 m/s2) 11.8 N/m3 0.0752 lbf/ft3 9790 N/m3 62.4 lbf/ft3 Specific weight is very useful in the hydrostaticpressure applications of Chap. 2. Specific weights of other fluids are given in Tables A.3 and A.4. Specific Gravity Specific gravity, denoted by SG, is the ratio of a fluid density to a standard reference
fluid, water (for liquids), and air (for gases):
gas SGgas gas air
liquid SGliquid (1.7) 1.205 kg/m3
liquid 998 kg/m3 water For example, the specific gravity of mercury (Hg) is SGHg 13,580/998 13.6. Engineers find these dimensionless ratios easier to remember than the actual numerical
values of density of a variety of fluids. Potential and Kinetic Energies In thermostatics the only energy in a substance is that stored in a system by molecular activity and molecular bonding forces. This is commonly denoted as internal energy û. A commonly accepted adjustment to this static situation for fluid flow is to add
two more energy terms which arise from newtonian mechanics: the potential energy
and kinetic energy.
The potential energy equals the work required to move the system of mass m from
the origin to a position vector r i x j y k z against a gravity field g. Its value is
mg r, or g r per unit mass. The kinetic energy equals the work required to change
the speed of the mass from zero to velocity V. Its value is 1 mV2 or 1 V2 per unit mass.
2
2
Then by common convention the total stored energy e per unit mass in fluid mechanics is the sum of three terms:
e 1
2 û V2 ( g r) (1.8) Also, throughout this book we shall define z as upward, so that g
gz. Then Eq. (1.8) becomes
e û 1
2 V2 gk and g r gz (1.9) The molecular internal energy û is a function of T and p for the singlephase pure substance, whereas the potential and kinetic energies are kinematic properties. Thermodynamic properties are found both theoretically and experimentally to be related to each other by state relations which differ for each substance. As mentioned,  v v State Relations for Gases  eText Main Menu  Textbook Table of Contents  Study Guide 1.6 Thermodynamic Properties of a Fluid 19 we shall confine ourselves here to singlephase pure substances, e.g., water in its liquid phase. The second most common fluid, air, is a mixture of gases, but since the mixture ratios remain nearly constant between 160 and 2200 K, in this temperature range
air can be considered to be a pure substance.
All gases at high temperatures and low pressures (relative to their critical point) are
in good agreement with the perfectgas law
p RT R cp cv gas constant (1.10) Since Eq. (1.10) is dimensionally consistent, R has the same dimensions as specific
heat, {L2T 2 1}, or velocity squared per temperature unit (kelvin or degree Rankine). Each gas has its own constant R, equal to a universal constant divided by the
molecular weight
Rgas
where
49,700 ft2/(s2 °R)
for air, with M 28.97:
Rair (1.11) Mgas 8314 m2/(s2 K). Most applications in this book are 1717 ft2/(s2 °R) 287 m2/(s2 K) (1.12) Standard atmospheric pressure is 2116 lbf/ft2, and standard temperature is 60°F
520°R. Thus standard air density is
air 2116
(1717)(520) 0.00237 slug/ft3 1.22 kg/m3 (1.13) This is a nominal value suitable for problems.
One proves in thermodynamics that Eq. (1.10) requires that the internal molecular
energy û of a perfect gas vary only with temperature: û û(T). Therefore the specific
heat cv also varies only with temperature:
û
T cv
or dû
dT dû cv(T) cv(T) dT (1.14) In like manner h and cp of a perfect gas also vary only with temperature:
h û p
h
T cp dh û p RT
dh
dT h(T)
cp(T) (1.15) cp(T) dT The ratio of specific heats of a perfect gas is an important dimensionless parameter in
compressibleflow analysis (Chap. 9)  v v k  eText Main Menu  cp
cv k(T) Textbook Table of Contents  1 Study Guide (1.16) Chapter 1 Introduction As a first approximation in airflow analysis we commonly take cp, cv, and k to be constant
kair
R cv k
k 1.4 4293 ft2/(s2 °R) 718 m2/(s2 K) 6010 ft2/(s2 °R) 1 1005 m2/(s2 K) kR cp 1 (1.17) Actually, for all gases, cp and cv increase gradually with temperature, and k decreases
gradually. Experimental values of the specificheat ratio for eight common gases are
shown in Fig. 1.3.
Many flow problems involve steam. Typical steam operating conditions are relatively close to the critical point, so that the perfectgas approximation is inaccurate.
The properties of steam are therefore available in tabular form [13], but the error of
using the perfectgas law is sometimes not great, as the following example shows. EXAMPLE 1.6
Estimate and cp of steam at 100 lbf/in2 and 400°F (a) by a perfectgas approximation and
(b) from the ASME steam tables [13]. Solution
First convert to BG units: p 100 lbf/in2 14,400 lb/ft2, T 400°F 860°R. From Table A.4
the molecular weight of H2O is 2MH MO 2(1.008) 16.0 18.016. Then from Eq. (1.11)
the gas constant of steam is approximately Part (a) 49,700
18.016 R 2759 ft2/(s2 °R) whence, from the perfectgas law,
p
RT 14,400
2759(860) 0.00607 slug/ft3 Ans. (a) From Fig. 1.3, k for steam at 860°R is approximately 1.30. Then from Eq. (1.17),
cp kR
k 1 1.30(2759)
1.30 1 12,000 ft2/(s2 °R) Ans. (a) From Ref. 13, the specific volume v of steam at 100 lbf/in2 and 400°F is 4.935 ft3/lbm. Then
the density is the inverse of this, converted to slugs: Part (b) 1
(4.935 ft2/lbm)(32.174 lbm/slug) 1
v 0.00630 slug/ft3 Ans. (b) This is about 4 percent higher than our idealgas estimate in part (a).
Reference 13 lists the value of cp of steam at 100 lbf/in2 and 400°F as 0.535 Btu/(lbm °F).
Convert this to BG units:
cp [0.535 Btu/(lbm °R)](778.2 ft lbf/Btu)(32.174 lbm/slug)
13,400 ft lbf/(slug °R)  v v 20  eText Main Menu  13,400 ft2/(s2 °R) Textbook Table of Contents  Study Guide Ans. (b) 1.6 Thermodynamic Properties of a Fluid 21 This is about 11 percent higher than our idealgas estimate in part (a). The chief reason for the
discrepancy is that this temperature and this pressure are quite close to the critical point and saturation line of steam. At higher temperatures and lower pressures, say, 800°F and 50 lbf/in2, the
perfectgas law gives and cp of steam within an accuracy of 1 percent.
Note that the use of poundmass and British thermal units in the traditional steam tables requires continual awkward conversions to BG units. Newer tables and disks are in SI units. State Relations for Liquids The writer knows of no “perfectliquid law” comparable to that for gases. Liquids are
nearly incompressible and have a single reasonably constant specific heat. Thus an idealized state relation for a liquid is
const cp cv const dh cp dT (1.18) Most of the flow problems in this book can be attacked with these simple assumptions. Water is normally taken to have a density of 1.94 slugs/ft3 and a specific heat cp 25,200 ft2/(s2 °R). The steam tables may be used if more accuracy
is required. 1.7
Ar
1.6 Atmospheric pressure
1.5 H2 1.4
cp
k=c
υ CO 1.3 O2 Air and
N2 Steam
1.2
CO2
1.1  v v Fig. 1.3 Specificheat ratio of eight
common gases as a function of temperature. (Data from Ref. 12.)  1.0 0 eText Main Menu 1000 2000 3000 4000 5000 Temperature, ° R  Textbook Table of Contents  Study Guide 22 Chapter 1 Introduction The density of a liquid usually decreases slightly with temperature and increases
moderately with pressure. If we neglect the temperature effect, an empirical pressuredensity relation for a liquid is
n p
pa (B 1)
B (1.19) a where B and n are dimensionless parameters which vary slightly with temperature and
pa and a are standard atmospheric values. Water can be fitted approximately to the
values B 3000 and n 7.
Seawater is a variable mixture of water and salt and thus requires three thermodynamic properties to define its state. These are normally taken as pressure, temperature,
ˆ
and the salinity S, defined as the weight of the dissolved salt divided by the weight of
the mixture. The average salinity of seawater is 0.035, usually written as 35 parts per
1000, or 35 ‰. The average density of seawater is 2.00 slugs/ft3. Strictly speaking,
seawater has three specific heats, all approximately equal to the value for pure water
of 25,200 ft2/(s2 °R) 4210 m2/(s2 K). EXAMPLE 1.7
The pressure at the deepest part of the ocean is approximately 1100 atm. Estimate the density
of seawater at this pressure. Solution
Equation (1.19) holds for either water or seawater. The ratio p/pa is given as 1100:
7 1100 (3001) 3000
a or
a 4100
3001 Assuming an average surface seawater density
1.046(2.00) 1/7 1.046
a 2.00 slugs/ft3, we compute
2.09 slugs/ft3 Ans. Even at these immense pressures, the density increase is less than 5 percent, which justifies the
treatment of a liquid flow as essentially incompressible. The quantities such as pressure, temperature, and density discussed in the previous section are primary thermodynamic variables characteristic of any system. There are also
certain secondary variables which characterize specific fluidmechanical behavior. The
most important of these is viscosity, which relates the local stresses in a moving fluid
to the strain rate of the fluid element. Viscosity When a fluid is sheared, it begins to move at a strain rate inversely proportional to a
property called its coefficient of viscosity . Consider a fluid element sheared in one  v v 1.7 Viscosity and Other
Secondary Properties  eText Main Menu  Textbook Table of Contents  Study Guide 1.7 Viscosity and Other Secondary Properties 23 plane by a single shear stress , as in Fig. 1.4a. The shear strain angle
will continuously grow with time as long as the stress is maintained, the upper surface moving
at speed u larger than the lower. Such common fluids as water, oil, and air show a
linear relation between applied shear and resulting strain rate
(1.20) t
From the geometry of Fig. 1.4a we see that
ut
y tan (1.21) In the limit of infinitesimal changes, this becomes a relation between shear strain rate
and velocity gradient
d
dt du
dy (1.22) From Eq. (1.20), then, the applied shear is also proportional to the velocity gradient
for the common linear fluids. The constant of proportionality is the viscosity coefficient
d
dt du
dy (1.23) Equation (1.23) is dimensionally consistent; therefore has dimensions of stresstime:
{FT/L2} or {M/(LT)}. The BG unit is slugs per footsecond, and the SI unit is kilograms per metersecond. The linear fluids which follow Eq. (1.23) are called newtonian fluids, after Sir Isaac Newton, who first postulated this resistance law in 1687.
We do not really care about the strain angle (t) in fluid mechanics, concentrating
instead on the velocity distribution u(y), as in Fig. 1.4b. We shall use Eq. (1.23) in
Chap. 4 to derive a differential equation for finding the velocity distribution u(y) — and,
more generally, V(x, y, z, t) — in a viscous fluid. Figure 1.4b illustrates a shear layer,
or boundary layer, near a solid wall. The shear stress is proportional to the slope of the y
δu δ t τ∝ u( y) δθ
δt Velocity
profile u = δu
du
δθ  v v Fig. 1.4 Shear stress causes continuous shear deformation in a fluid:
(a) a fluid element straining at a
rate / t; (b) newtonian shear distribution in a shear layer near a
wall.  δθ τ = µ du
dy dy δy eText Main Menu No slip at wall δx u=0 0 τ
(a)  Textbook Table of Contents (b)  Study Guide 24 Chapter 1 Introduction Table 1.4 Viscosity and Kinematic
Viscosity of Eight Fluids at 1 atm
and 20°C Fluid ,
kg/(m s)† Hydrogen
Air
Gasoline
Water
Ethyl alcohol
Mercury
SAE 30 oil
Glycerin 8.8 E – 6
1.8 E – 5
2.9 E – 4
1.0 E – 3
1.2 E – 3
1.5 E – 3
0.29
1.5 † 1 kg/(m s) Ratio
/ (H2) ,
kg/m3 00,0001.0
0,00002.1
00,0033
00,0114
0,00135
00,0170
033,000
170,000 0.0209 slug/(ft s); 1 m2/s m2/s† 00,000.084
00,001.20
0,0680
0,0998
0,0789
13,580
0,0891
01,264 1.05
1.51
4.22
1.01
1.52
1.16
3.25
1.18 E–4
E–5
E–7
E–6
E–6
E–7
E–4
E–3 Ratio
/ (Hg)
00,920
00,130
00,003.7
0000,8.7
000,13
0000,1.0
02,850
10,300 10.76 ft2/s. velocity profile and is greatest at the wall. Further, at the wall, the velocity u is zero
relative to the wall: This is called the noslip condition and is characteristic of all
viscousfluid flows.
The viscosity of newtonian fluids is a true thermodynamic property and varies with
temperature and pressure. At a given state ( p, T) there is a vast range of values among the
common fluids. Table 1.4 lists the viscosity of eight fluids at standard pressure and temperature. There is a variation of six orders of magnitude from hydrogen up to glycerin.
Thus there will be wide differences between fluids subjected to the same applied stresses.
Generally speaking, the viscosity of a fluid increases only weakly with pressure. For
example, increasing p from 1 to 50 atm will increase of air only 10 percent. Temperature, however, has a strong effect, with increasing with T for gases and decreasing for liquids. Figure A.1 (in App. A) shows this temperature variation for various common fluids. It is customary in most engineering work to neglect the pressure variation.
The variation ( p, T) for a typical fluid is nicely shown by Fig. 1.5, from Ref. 14,
which normalizes the data with the criticalpoint state ( c, pc, Tc). This behavior, called
the principle of corresponding states, is characteristic of all fluids, but the actual numerical values are uncertain to 20 percent for any given fluid. For example, values
of (T ) for air at 1 atm, from Table A.2, fall about 8 percent low compared to the
“lowdensity limit” in Fig. 1.5.
Note in Fig. 1.5 that changes with temperature occur very rapidly near the critical
point. In general, criticalpoint measurements are extremely difficult and uncertain. The Reynolds Number As we shall see in Chaps. 5 through 7, the primary parameter correlating the viscous
behavior of all newtonian fluids is the dimensionless Reynolds number:
Re VL VL (1.24) where V and L are characteristic velocity and length scales of the flow. The second form
of Re illustrates that the ratio of to has its own name, the kinematic viscosity:
(1.25)  v v It is called kinematic because the mass units cancel, leaving only the dimensions {L2/T}.  eText Main Menu  Textbook Table of Contents  Study Guide 1.7 Viscosity and Other Secondary Properties 10
9
8
7
6 25 Liquid 5
4
Dense gas
3
Twophase
region µ
µr = µ 25 2 10 c 5 Critical
point
1
0.9
0.8
0.7
0.6 3
2
1 0.5 0.5
pr = 0.2 0.4 Fig. 1.5 Fluid viscosity nondimensionalized by criticalpoint properties. This generalized chart is characteristic of all fluids but is only
accurate to 20 percent. (From
Ref. 14.) Lowdensity limit
0 0.3 0.2
0.4 0.6 0.8 1 2 3 4 5 6 7 8 9 10 Tr = T
Tc Generally, the first thing a fluids engineer should do is estimate the Reynolds number range of the flow under study. Very low Re indicates viscous creeping motion,
where inertia effects are negligible. Moderate Re implies a smoothly varying laminar
flow. High Re probably spells turbulent flow, which is slowly varying in the timemean
but has superimposed strong random highfrequency fluctuations. Explicit numerical
values for low, moderate, and high Reynolds numbers cannot be stated here. They depend upon flow geometry and will be discussed in Chaps. 5 through 7.
Table 1.4 also lists values of for the same eight fluids. The pecking order changes
considerably, and mercury, the heaviest, has the smallest viscosity relative to its own
weight. All gases have high relative to thin liquids such as gasoline, water, and alcohol. Oil and glycerin still have the highest , but the ratio is smaller. For a given
value of V and L in a flow, these fluids exhibit a spread of four orders of magnitude
in the Reynolds number.  v v Flow between Plates A classic problem is the flow induced between a fixed lower plate and an upper plate
moving steadily at velocity V, as shown in Fig. 1.6. The clearance between plates is
h, and the fluid is newtonian and does not slip at either plate. If the plates are large,  eText Main Menu  Textbook Table of Contents  Study Guide 26 Chapter 1 Introduction
y
Moving
plate:
u=V u=V
V h Fig. 1.6 Viscous flow induced by
relative motion between two parallel plates. Viscous
fluid u( y) Fixed plate u=0 this steady shearing motion will set up a velocity distribution u(y), as shown, with v
w 0. The fluid acceleration is zero everywhere.
With zero acceleration and assuming no pressure variation in the flow direction, you
should show that a force balance on a small fluid element leads to the result that the
shear stress is constant throughout the fluid. Then Eq. (1.23) becomes
du
dy const which we can integrate to obtain
u a by The velocity distribution is linear, as shown in Fig. 1.6, and the constants a and b can
be evaluated from the noslip condition at the upper and lower walls:
u
Hence a 0 and b 0
V a
a b(0)
b(h) at y
at y 0
h V/h. Then the velocity profile between the plates is given by
u V y
h (1.26) as indicated in Fig. 1.6. Turbulent flow (Chap. 6) does not have this shape.
Although viscosity has a profound effect on fluid motion, the actual viscous stresses
are quite small in magnitude even for oils, as shown in the following example. EXAMPLE 1.8
Suppose that the fluid being sheared in Fig. 1.6 is SAE 30 oil at 20°C. Compute the shear stress
in the oil if V 3 m/s and h 2 cm. Solution
The shear stress is found from Eq. (1.23) by differentiating Eq. (1.26):  v v du
dy  eText Main Menu  V
h Textbook Table of Contents (1)  Study Guide 1.7 Viscosity and Other Secondary Properties From Table 1.4 for SAE 30 oil,
Eq. (1) predicts 27 0.29 kg/(m s). Then, for the given values of V and h, [0.29 kg/(m s)](3 m/s)
0.02 m
43 N/m2 43 kg/(m s2) 43 Pa Ans. Although oil is very viscous, this is a modest shear stress, about 2400 times less than atmospheric pressure. Viscous stresses in gases and thin liquids are even smaller. Variation of Viscosity with
Temperature Temperature has a strong effect and pressure a moderate effect on viscosity. The viscosity of gases and most liquids increases slowly with pressure. Water is anomalous
in showing a very slight decrease below 30°C. Since the change in viscosity is only a
few percent up to 100 atm, we shall neglect pressure effects in this book.
Gas viscosity increases with temperature. Two common approximations are the
power law and the Sutherland law: T
T0 n power law (1.27)
(T/T0)3/2(T0 S)
Sutherland law
TS
where 0 is a known viscosity at a known absolute temperature T0 (usually 273 K). The
constants n and S are fit to the data, and both formulas are adequate over a wide range of
temperatures. For air, n 0.7 and S 110 K 199°R. Other values are given in Ref. 3.
Liquid viscosity decreases with temperature and is roughly exponential,
ae bT;
but a better fit is the empirical result that ln is quadratic in 1/T, where T is absolute
temperature
0 ln a
0 b T0
T c T0
T 2 (1.28) For water, with T0 273.16 K, 0 0.001792 kg/(m s), suggested values are a
1.94, b
4.80, and c 6.74, with accuracy about 1 percent. The viscosity of
water is tabulated in Table A.1. Curvefit viscosity formulas for 355 organic liquids are
given by Yaws et al. [34]. For further viscosity data, see Refs. 28 and 36. Thermal Conductivity Just as viscosity relates applied stress to resulting strain rate, there is a property called
thermal conductivity k which relates the vector rate of heat flow per unit area q to the
vector gradient of temperature T. This proportionality, observed experimentally for
fluids and solids, is known as Fourier’s law of heat conduction
q kT (1.29a) which can also be written as three scalar equations
k T
y Textbook Table of Contents   v v qx  eText Main Menu  k T
x qy qz k T
z Study Guide (1.29b) 28 Chapter 1 Introduction The minus sign satisfies the convention that heat flux is positive in the direction of decreasing temperature. Fourier’s law is dimensionally consistent, and k has SI units of
joules per secondmeterkelvin. Thermal conductivity k is a thermodynamic property
and varies with temperature and pressure in much the same way as viscosity. The ratio k/k0 can be correlated with T/T0 in the same manner as Eqs. (1.27) and (1.28) for
gases and liquids, respectively.
Further data on viscosity and thermalconductivity variations can be found in
Ref. 11. Nonnewtonian Fluids Fluids which do not follow the linear law of Eq. (1.23) are called nonnewtonian and
are treated in books on rheology [6]. Figure 1.7a compares four examples with a newtonian fluid. A dilatant, or shearthickening, fluid increases resistance with increasing
applied stress. Alternately, a pseudoplastic, or shearthinning, fluid decreases resistance
with increasing stress. If the thinning effect is very strong, as with the dashedline
curve, the fluid is termed plastic. The limiting case of a plastic substance is one which
requires a finite yield stress before it begins to flow. The linearflow Bingham plastic
idealization is shown, but the flow behavior after yield may also be nonlinear. An example of a yielding fluid is toothpaste, which will not flow out of the tube until a finite stress is applied by squeezing.
A further complication of nonnewtonian behavior is the transient effect shown in
Fig. 1.7b. Some fluids require a gradually increasing shear stress to maintain a constant strain rate and are called rheopectic. The opposite case of a fluid which thins out
with time and requires decreasing stress is termed thixotropic. We neglect nonnewtonian effects in this book; see Ref. 6 for further study. Shear
stress
τ Ideal Bingham
plastic Dilatant Plastic Shear
stress
τ Rheopectic Newtonian Yield
stress Common
fluids Pseudoplastic
Thixotropic Constant
strain rate  v v Fig. 1.7 Rheological behavior of
various viscous materials: (a) stress
versus strain rate; (b) effect of time
on applied stress. 0 Shear strain rate dθ
dt Time 0 (a)  eText Main Menu  (b) Textbook Table of Contents  Study Guide 1.7 Viscosity and Other Secondary Properties 29 A liquid, being unable to expand freely, will form an interface with a second liquid or
gas. The physical chemistry of such interfacial surfaces is quite complex, and whole
textbooks are devoted to this specialty [15]. Molecules deep within the liquid repel
each other because of their close packing. Molecules at the surface are less dense and
attract each other. Since half of their neighbors are missing, the mechanical effect is
that the surface is in tension. We can account adequately for surface effects in fluid
mechanics with the concept of surface tension.
If a cut of length dL is made in an interfacial surface, equal and opposite forces of
magnitude dL are exposed normal to the cut and parallel to the surface, where is
called the coefficient of surface tension. The dimensions of are {F/L}, with SI units
of newtons per meter and BG units of poundsforce per foot. An alternate concept is
to open up the cut to an area dA; this requires work to be done of amount dA. Thus
the coefficient can also be regarded as the surface energy per unit area of the interface, in N m/m2 or ft lbf/ft2.
The two most common interfaces are waterair and mercuryair. For a clean surface
at 20°C 68°F, the measured surface tension is Surface Tension 0.0050 lbf/ft 0.073 N/m
0.033 lbf/ft 0.48 N/m airwater
airmercury (1.30) These are design values and can change considerably if the surface contains contaminants like detergents or slicks. Generally decreases with liquid temperature and is
zero at the critical point. Values of for water are given in Fig. 1.8.
If the interface is curved, a mechanical balance shows that there is a pressure difference across the interface, the pressure being higher on the concave side, as illustrated in Fig. 1.9. In Fig. 1.9a, the pressure increase in the interior of a liquid cylinder
is balanced by two surfacetension forces
2RL p
or 2L p (1.31) R We are not considering the weight of the liquid in this calculation. In Fig. 1.9b, the pressure increase in the interior of a spherical droplet balances a ring of surfacetension force
R2 p
or p 2R
2
R (1.32) We can use this result to predict the pressure increase inside a soap bubble, which has
two interfaces with air, an inner and outer surface of nearly the same radius R:
pbubble 2 pdroplet 4
R (1.33) Figure 1.9c shows the general case of an arbitrarily curved interface whose principal
radii of curvature are R1 and R2. A force balance normal to the surface will show that
the pressure increase on the concave side is  v v p  eText Main Menu  (R1 1 Textbook Table of Contents R2 1)  Study Guide (1.34) 30 Chapter 1 Introduction 0.080 ϒ, N/m 0.070 0.060 0.050 Fig. 1.8 Surface tension of a clean
airwater interface. Data from Table
A.5. 0 10 20 30 40 50 60 70 80 90 100 T, ° C πR2 ∆p 2 RL ∆ p ∆ p dA L dL1
2π R dL2 L R2
R1
dL2
L dL1 2R (a) (b) (c) Fig. 1.9 Pressure change across a curved interface due to surface tension: (a) interior of a liquid cylinder; (b) interior of a spherical
droplet; (c) general curved interface.  v v Equations (1.31) to (1.33) can all be derived from this general relation; e.g., in
Eq. (1.31), R1 R and R2
.
A second important surface effect is the contact angle which appears when a
liquid interface intersects with a solid surface, as in Fig. 1.10. The force balance
would then involve both and . If the contact angle is less than 90°, the liquid is
said to wet the solid; if
90°, the liquid is termed nonwetting. For example, water wets soap but does not wet wax. Water is extremely wetting to a clean glass surface, with
0°. Like , the contact angle is sensitive to the actual physicochemical conditions of the solidliquid interface. For a clean mercuryairglass
interface,
130°.
Example 1.9 illustrates how surface tension causes a fluid interface to rise or fall in
a capillary tube.  eText Main Menu  Textbook Table of Contents  Study Guide 1.7 Viscosity and Other Secondary Properties 31 Gas
Liquid Fig. 1.10 Contactangle effects at
liquidgassolid interface. If
90°, the liquid “wets” the solid; if
90°, the liquid is nonwetting. θ Nonwetting θ
Solid EXAMPLE 1.9
Derive an expression for the change in height h in a circular tube of a liquid with surface tension and contact angle , as in Fig. E1.9. Solution
θ The vertical component of the ring surfacetension force at the interface in the tube must balance the weight of the column of fluid of height h
2R R2h cos Solving for h, we have the desired result
h h 2R Fig. E1.9 2 cos
R Ans. Thus the capillary height increases inversely with tube radius R and is positive if
ting liquid) and negative (capillary depression) if
90°.
Suppose that R 1 mm. Then the capillary rise for a waterairglass interface,
0.073 N/m, and
1000 kg/m3 is
h 2(0.073 N/m)(cos 0°)
(1000 kg/m3)(9.81 m/s2)(0.001 m) For a mercuryairglass interface, with
illary rise is
h 0.015 (N s2)/kg 130°, 0.015 m 0.48 N/m, and 2(0.48)(cos 130°)
13,600(9.81)(0.001) 90° (wet0°, 1.5 cm 13,600 kg/m3, the cap 0.46 cm When a smalldiameter tube is used to make pressure measurements (Chap. 2), these capillary
effects must be corrected for. Vapor pressure is the pressure at which a liquid boils and is in equilibrium with its
own vapor. For example, the vapor pressure of water at 68°F is 49 lbf/ft2, while that
of mercury is only 0.0035 lbf/ft2. If the liquid pressure is greater than the vapor  v v Vapor Pressure  eText Main Menu  Textbook Table of Contents  Study Guide 32 Chapter 1 Introduction pressure, the only exchange between liquid and vapor is evaporation at the interface. If, however, the liquid pressure falls below the vapor pressure, vapor bubbles
begin to appear in the liquid. If water is heated to 212°F, its vapor pressure rises to
2116 lbf/ft2, and thus water at normal atmospheric pressure will boil. When the liquid pressure is dropped below the vapor pressure due to a flow phenomenon, we
call the process cavitation. As we shall see in Chap. 2, if water is accelerated from
rest to about 50 ft/s, its pressure drops by about 15 lbf/in2, or 1 atm. This can cause
cavitation.
The dimensionless parameter describing flowinduced boiling is the cavitation
number
Ca 1
2 where pa
pv
V pv pa (1.35) 2 V ambient pressure
vapor pressure
characteristic flow velocity Depending upon the geometry, a given flow has a critical value of Ca below which the
flow will begin to cavitate. Values of surface tension and vapor pressure of water are
given in Table A.5. The vapor pressure of water is plotted in Fig. 1.11.
Figure 1.12a shows cavitation bubbles being formed on the lowpressure surfaces
of a marine propeller. When these bubbles move into a higherpressure region, they
collapse implosively. Cavitation collapse can rapidly spall and erode metallic surfaces
and eventually destroy them, as shown in Fig. 1.12b. NoSlip and NoTemperatureJump Conditions When a fluid flow is bounded by a solid surface, molecular interactions cause the fluid
in contact with the surface to seek momentum and energy equilibrium with that surface.
All liquids essentially are in equilibrium with the surface they contact. All gases are, too, 100 80 pv , kPa 60 40 20 0
0  v v Fig. 1.11 Vapor pressure of water.
Data from Table A.5. 20 40 60 80 100 T, °C  eText Main Menu  Textbook Table of Contents  Study Guide 1.7 Viscosity and Other Secondary Properties  v v Fig. 1.12 Two aspects of cavitation
bubble formation in liquid flows:
(a) Beauty: spiral bubble sheets
form from the surface of a marine
propeller. (Courtesy of the Garfield
Thomas Water Tunnel, Pennsylvania State University); (b) ugliness:
collapsing bubbles erode a propeller surface. (Courtesy of Thomas
T. Huang, David Taylor Research
Center.)  eText Main Menu  Textbook Table of Contents  Study Guide 33 34 Chapter 1 Introduction except under the most rarefied conditions [8]. Excluding rarefied gases, then, all fluids
at a point of contact with a solid take on the velocity and temperature of that surface
Vfluid Vwall Tfluid Twall (1.36) These are called the noslip and notemperaturejump conditions, respectively. They
serve as boundary conditions for analysis of fluid flow past a solid surface (Chap. 6).
Figure 1.13 illustrates the noslip condition for water flow past the top and bottom surfaces of a fixed thin plate. The flow past the upper surface is disorderly, or turbulent,
while the lower surface flow is smooth, or laminar.7 In both cases there is clearly no
slip at the wall, where the water takes on the zero velocity of the fixed plate. The velocity profile is made visible by the discharge of a line of hydrogen bubbles from the
wire shown stretched across the flow.
To decrease the mathematical difficulty, the noslip condition is partially relaxed in
the analysis of inviscid flow (Chap. 8). The flow is allowed to “slip” past the surface
but not to permeate through the surface
Vnormal(fluid) Vnormal(solid) (1.37) while the tangential velocity Vt is allowed to be independent of the wall. The analysis
is much simpler, but the flow patterns are highly idealized. Fig. 1.13 The noslip condition in
water flow past a thin fixed plate.
The upper flow is turbulent; the
lower flow is laminar. The velocity
profile is made visible by a line of
hydrogen bubbles discharged from
the wire across the flow. [From Illustrated Experiments in Fluid Mechanics (The NCFMF Book of Film
Notes), National Committee for
Fluid Mechanics Films, Education
Development Center, Inc., copyright 1972.]
7  v v Laminar and turbulent flows are studied in Chaps. 6 and 7.  eText Main Menu  Textbook Table of Contents  Study Guide 1.8 Basic FlowAnalysis Techniques 35 In gas flow, one must be aware of compressibility effects (significant density changes
caused by the flow). We shall see in Sec. 4.2 and in Chap. 9 that compressibility becomes important when the flow velocity reaches a significant fraction of the speed of
sound of the fluid. The speed of sound a of a fluid is the rate of propagation of smalldisturbance pressure pulses (“sound waves”) through the fluid. In Chap. 9 we shall
show, from momentum and thermodynamic arguments, that the speed of sound is defined by Speed of Sound a2 p k p s k
T cp
cv (1.38) This is true for either a liquid or a gas, but it is for gases that the problem of compressibility occurs. For an ideal gas, Eq. (1.10), we obtain the simple formula
aideal gas (kRT)1/2 (1.39) where R is the gas constant, Eq. (1.11), and T the absolute temperature. For example,
for air at 20°C, a {(1.40)[287 m2/(s2 K)](293 K)}1/2 343 m/s (1126 ft/s 768
mi/h). If, in this case, the air velocity reaches a significant fraction of a, say, 100 m/s,
then we must account for compressibility effects (Chap. 9). Another way to state this
is to account for compressibility when the Mach number Ma V/a of the flow reaches
about 0.3.
The speed of sound of water is tabulated in Table A.5. The speed of sound of air
(or any approximately perfect gas) is simply calculated from Eq. (1.39). 1.8 Basic FlowAnalysis
Techniques There are three basic ways to attack a fluidflow problem. They are equally important
for a student learning the subject, and this book tries to give adequate coverage to each
method:
1. Controlvolume, or integral analysis (Chap. 3)
2. Infinitesimal system, or differential analysis (Chap. 4)
3. Experimental study, or dimensional analysis (Chap. 5)
In all cases, the flow must satisfy the three basic laws of mechanics8 plus a thermodynamic state relation and associated boundary conditions:
1.
2.
3.
4.
5. Conservation of mass (continuity)
Linear momentum (Newton’s second law)
First law of thermodynamics (conservation of energy)
A state relation like
( p, T)
Appropriate boundary conditions at solid surfaces, interfaces, inlets, and exits In integral and differential analyses, these five relations are modeled mathematically
and solved by computational methods. In an experimental study, the fluid itself performs this task without the use of any mathematics. In other words, these laws are believed to be fundamental to physics, and no fluid flow is known to violate them.  v v 8
In fluids which are variable mixtures of components, such as seawater, a fourth basic law is required,
conservation of species. For an example of salt conservation analysis, see Chap. 4, Ref. 16.  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 1 Introduction A control volume is a finite region, chosen carefully by the analyst, with open boundaries through which mass, momentum, and energy are allowed to cross. The analyst
makes a budget, or balance, between the incoming and outgoing fluid and the resultant changes within the control volume. The result is a powerful tool but a crude one.
Details of the flow are normally washed out or ignored in controlvolume analyses.
Nevertheless, the controlvolume technique of Chap. 3 never fails to yield useful and
quantitative information to the engineering analyst.
When the conservation laws are written for an infinitesimal system of fluid in motion,
they become the basic differential equations of fluid flow. To apply them to a specific
problem, one must integrate these equations mathematically subject to the boundary conditions of the particular problem. Exact analytic solutions are often possible only for very
simple geometries and boundary conditions (Chap. 4). Otherwise, one attempts numerical integration on a digital computer, i.e., a summing procedure for finitesized systems
which one hopes will approximate the exact integral calculus [1]. Even computer analysis often fails to provide an accurate simulation, because of either inadequate storage or
inability to model the finely detailed flow structure characteristic of irregular geometries
or turbulentflow patterns. Thus differential analysis sometimes promises more than it
delivers, although we can successfully study a number of classic and useful solutions.
A properly planned experiment is very often the best way to study a practical engineering flow problem. Guidelines for planning flow experiments are given in Chap.
5. For example, no theory presently available, whether differential or integral, calculus or computer, is able to make an accurate computation of the aerodynamic drag and
side force of an automobile moving down a highway with crosswinds. One must solve
the problem by experiment. The experiment may be fullscale: One can test a real automobile on a real highway in real crosswinds. For that matter, there are wind tunnels
in existence large enough to hold a fullscale car without significant blockage effects.
Normally, however, in the design stage, one tests a smallmodel automobile in a small
wind tunnel. Without proper interpretation, the model results may be poor and mislead
the designer (Chap. 5). For example, the model may lack important details such as surface finish or underbody protuberances. The “wind” produced by the tunnel propellers
may lack the turbulent gustiness of real winds. It is the job of the fluidflow analyst,
using such techniques as dimensional analysis, to plan an experiment which gives an
accurate estimate of fullscale or prototype results expected in the final product.
It is possible to classify flows, but there is no general agreement on how to do it.
Most classifications deal with the assumptions made in the proposed flow analysis.
They come in pairs, and we normally assume that a given flow is either
Steady or unsteady (1.40a) Inviscid or viscous (1.40b) Incompressible or compressible (1.40c) Gas or liquid (1.40d) As Fig. 1.14 indicates, we choose one assumption from each pair. We may have a steady
viscous compressible gas flow or an unsteady inviscid (
0) incompressible liquid
flow. Although there is no such thing as a truly inviscid fluid, the assumption
0
gives adequate results in many analyses (Chap. 8). Often the assumptions overlap: A
flow may be viscous in the boundary layer near a solid surface (Fig. 1.13) and effec  v v 36  eText Main Menu  Textbook Table of Contents  Study Guide 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 37
Fig. 1.14 Ready for a flow analysis? Then choose one assumption
from each box. Steady
Unsteady Inviscid
Viscous Incompressible
Compressible Gas
Liquid tively inviscid away from the surface. The viscous part of the flow may be laminar or
transitional or turbulent or combine patches of all three types of viscous flow. A flow
may involve both a gas and a liquid and the free surface, or interface, between them
(Chap. 10). A flow may be compressible in one region and have nearly constant density in another. Nevertheless, Eq. (1.40) and Fig. 1.14 give the basic binary assumptions of flow analysis, and Chaps. 6 to 10 try to separate them and isolate the basic effect of each assumption. 1.9 Flow Patterns: Streamlines,
Streaklines, and Pathlines Fluid mechanics is a highly visual subject. The patterns of flow can be visualized in a
dozen different ways, and you can view these sketches or photographs and learn a great
deal qualitatively and often quantitatively about the flow.
Four basic types of line patterns are used to visualize flows:
1. A streamline is a line everywhere tangent to the velocity vector at a given instant.
2. A pathline is the actual path traversed by a given fluid particle.
3. A streakline is the locus of particles which have earlier passed through a prescribed point.
4. A timeline is a set of fluid particles that form a line at a given instant.
The streamline is convenient to calculate mathematically, while the other three are easier to generate experimentally. Note that a streamline and a timeline are instantaneous
lines, while the pathline and the streakline are generated by the passage of time. The
velocity profile shown in Fig. 1.13 is really a timeline generated earlier by a single discharge of bubbles from the wire. A pathline can be found by a time exposure of a single marked particle moving through the flow. Streamlines are difficult to generate experimentally in unsteady flow unless one marks a great many particles and notes their
direction of motion during a very short time interval [17, p. 35]. In steady flow the situation simplifies greatly:
Streamlines, pathlines, and streaklines are identical in steady flow.
In fluid mechanics the most common mathematical result for visualization purposes
is the streamline pattern. Figure 1.15a shows a typical set of streamlines, and Fig. 1.15b
shows a closed pattern called a streamtube. By definition the fluid within a streamtube
is confined there because it cannot cross the streamlines; thus the streamtube walls
need not be solid but may be fluid surfaces.
Figure 1.16 shows an arbitrary velocity vector. If the elemental arc length dr of a
streamline is to be parallel to V, their respective components must be in proportion:
dx
u  v v Streamline:  eText Main Menu  dy
v dz
w Textbook Table of Contents dr
V  Study Guide (1.41) 38 Chapter 1 Introduction V
No flow across
streamtube walls Fig. 1.15 The most common
method of flowpattern presentation: (a) Streamlines are everywhere tangent to the local velocity
vector; (b) a streamtube is formed
by a closed collection of streamlines. Individual
streamline (a) (b) If the velocities (u, v, w) are known functions of position and time, Eq. (1.41) can be
integrated to find the streamline passing through the initial point (x0, y0, z0, t0). The
method is straightforward for steady flows (Example 1.10) but may be laborious for
unsteady flow.
The pathline, or displacement of a particle, is defined by integration of the velocity
components, as mentioned in Sec. 1.5:
Pathline: x u dt y v dt z w dt (1.42) Given (u, v, w) as known functions of position and time, the integration is begun at a
specified initial position (x0, y0, z0, t0). Again the integration may be laborious.
Streaklines, easily generated experimentally with smoke, dye, or bubble releases,
are very difficult to compute analytically. See Ref. 18 for mathematical details. z V
V
w
dr
dz dx
u y  Textbook Table of Contents dy v  v v Fig. 1.16 Geometric relations for
defining a streamline. x  eText Main Menu  Study Guide 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 39 EXAMPLE 1.10
Given the steady twodimensional velocity distribution
u Kx v Ky w 0 (1) where K is a positive constant, compute and plot the streamlines of the flow, including directions, and give some possible interpretations of the pattern. Solution
Since time does not appear explicitly in Eq. (1), the motion is steady, so that streamlines, pathlines, and streaklines will coincide. Since w 0 everywhere, the motion is two dimensional, in
the xy plane. The streamlines can be computed by substituting the expressions for u and v into
Eq. (1.41):
dx
Kx dy
Ky dx
x or
Integrating, we obtain ln x ln y dy
y ln C, or
xy C Ans. (2) This is the general expression for the streamlines, which are hyperbolas. The complete pattern is plotted in Fig. E1.10 by assigning various values to the constant C. The arrowheads
can be determined only by returning to Eqs. (1) to ascertain the velocity component directions, assuming K is positive. For example, in the upper right quadrant (x 0, y 0), u is
positive and v is negative; hence the flow moves down and to the right, establishing the arrowheads as shown. y C = –3 +3
+2 0 –2
–1 +1 C=0 C=0 x 0 +1
+2
C=+3 –1 0 –2
–3  v v Fig. E1.10 Streamlines for the velocity distribution given by
Eq. (1), for K 0.  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 1 Introduction
Note that the streamline pattern is entirely independent of constant K. It could represent the
impingement of two opposing streams, or the upper half could simulate the flow of a single downward stream against a flat wall. Taken in isolation, the upper right quadrant is similar to the flow
in a 90° corner. This is definitely a realistic flow pattern and is discussed again in Chap. 8.
Finally note the peculiarity that the two streamlines (C 0) have opposite directions and intersect. This is possible only at a point where u v w 0, which occurs at the origin in this
case. Such a point of zero velocity is called a stagnation point. A streakline can be produced experimentally by the continuous release of marked particles (dye, smoke, or bubbles) from a given point. Figure 1.17 shows two examples. The
flow in Fig. 1.17b is unsteady and periodic due to the flapping of the plate against the
oncoming stream. We see that the dashdot streakline does not coincide with either the
streamline or the pathline passing through the same release point. This is characteristic
of unsteady flow, but in Fig. 1.17a the smoke filaments form streaklines which are identical to the streamlines and pathlines. We noted earlier that this coincidence of lines is
always true of steady flow: Since the velocity never changes magnitude or direction at
any point, every particle which comes along repeats the behavior of its earlier neighbors.
Methods of experimental flow visualization include the following:
1.
2.
3.
4. Dye, smoke, or bubble discharges
Surface powder or flakes on liquid flows
Floating or neutraldensity particles
Optical techniques which detect density changes in gas flows: shadowgraph,
schlieren, and interferometer
5. Tufts of yarn attached to boundary surfaces
6. Evaporative coatings on boundary surfaces
7. Luminescent fluids or additives
Uniform
approach
flow Periodic
flapping
plate Release
point
Streamline
Pathline
Streakline
(b) (a) Fig. 1.17 Experimental visualization of steady and unsteady flow: (a) steady flow past an airfoil
visualized by smoke filaments (C. A. A. SCIENTIFIC — Prime Movers Laboratory Systems);
(b) unsteady flow past an oscillating plate with a point bubble release (from an experiment in
Ref. 17).  v v 40  eText Main Menu  Textbook Table of Contents  Study Guide 1.10 The Engineering Equation Solver 41 The mathematical implications of flowpattern analysis are discussed in detail in Ref.
18. References 19 and 20 are beautiful albums of photographs. References 21 and 22
are monographs on flow visualization. 1.10 The Engineering Equation
Solver EES Most of the examples and exercises in this text are amenable to direct calculation without guessing or iteration or looping. Until recently, only such direct problem assignments, whether “plugandchug” or more subtle, were appropriate for undergraduate
engineering courses. However, the recent introduction of computer software solvers
makes almost any set of algebraic relations viable for analysis and solution. The solver
recommended here is the Engineering Equation Solver (EES) developed by Klein and
Beckman [33] and described in Appendix E.
Any software solver should handle a purely mathematical set of relations, such as
the one posed in Ref. 33: X ln (X) Y3, X1/2 1/Y. Submit that pair to any commercial solver and you will no doubt receive the answer: X 1.467, Y 0.826. However,
for engineers, in the author’s opinion, EES is superior to most solvers because (1) equations can be entered in any order; (2) scores of mathematical formulas are builtin, such
as the Bessel functions; and (3) thermophysical properties of many fluids are builtin,
such as the steam tables [13]. Both metric and English units are allowed. Equations
need not be written in the traditional BASIC or FORTRAN style. For example, X
Y 1 0 is perfectly satisfactory; there is no need to retype this as X Y 1.
For example, reconsider Example 1.7 as an EES exercise. One would first enter the
reference properties p0 and 0 plus the curvefit constants B and n:
Pz 1.0 Rhoz 2.0 B 3000 n 7 Then specify the given pressure ratio and the curvefit relation, Eq. (1.19), for the equation of state of water:
P
P/Pz (B 1100*Pz
1) (Rho/Rhoz)
^n
* B  v v If you request an initial opinion from the CHECK/FORMAT menu, EES states that
there are six equations in six unknowns and there are no obvious difficulties. Then
request SOLVE from the menu and EES quickly prints out Rho 2.091, the correct
answer as seen already in Ex. 1.7. It also prints out values of the other five variables.
Occasionally EES reports “unable to converge” and states what went wrong (division
by zero, square root of a negative number, etc.). One needs only to improve the guesses
and ranges of the unknowns in Variable Information to assist EES to the solution.
In subsequent chapters we will illustrate some implicit (iterative) examples by using EES and will also assign some advanced problem exercises for which EES is an
ideal approach. The use of an engineering solver, notably EES, is recommended to all
engineers in this era of the personal computer.  eText Main Menu  Textbook Table of Contents  Study Guide 42 Chapter 1 Introduction Earlier in this chapter we referred to the uncertainty of the principle of corresponding
states in discussing Fig. 1.5. Uncertainty is a fact of life in engineering. We rarely know
any engineering properties or variables to an extreme degree of accuracy. Therefore,
we need to know the uncertainty U of our data, usually defined as the band within
which the experimenter is 95 percent confident that the true value lies (Refs. 30, 31).
In Fig. 1.5, we were given that the uncertainty of / c is U
20 percent.
Fluid mechanics is heavily dependent upon experimentation, and the data uncertainty is needed before we can use it for prediction or design purposes. Sometimes uncertainty completely changes our viewpoint. As an offbeat example, suppose that astronomers reported that the length of the earth year was 365.25 days “give or take a
couple of months.” First, that would make the fivefigure accuracy ridiculous, and the
year would better be stated as Y 365 60 days. Second, we could no longer plan
confidently or put together accurate calendars. Scheduling Christmas vacation would
be chancy.
Multiple variables make uncertainty estimates cumulative. Suppose a given result P
depends upon N variables, P P(x1, x2, x3, . . . , xN), each with its own uncertainty;
for example, x1 has uncertainty x1. Then, by common agreement among experimenters,
the total uncertainty of P is calculated as a rootmeansquare average of all effects: 1.11 Uncertainty of
Experimental Data P
x1
x1 P 2 P
x2
x2 2 P
xN
xN 2 1/2 (1.43) This calculation is statistically much more probable than simply adding linearly the
various uncertainties xi, thereby making the unlikely assumption that all variables
simultaneously attain maximum error. Note that it is the responsibility of the experimenter to establish and report accurate estimates of all the relevant uncertainties xi.
If the quantity P is a simple powerlaw expression of the other variables, for example, P Const x1n x2n x3n . . . , then each derivative in Eq. (1.43) is proportional to
P and the relevant powerlaw exponent and is inversely proportional to that variable.
If P Const x1n x2n x3n . . . , then
1 2 1 2 3 3 P
x1 n1P P
,
x1
x2 n2P P
,
x2
x3 n3P
,
x3 Thus, from Eq. (1.43),
P
P n1 x1
x1 2 n2 x2
x2 2 n3 x3
x3 2 1/2 (1.44) Evaluation of P is then a straightforward procedure, as in the following example.
EXAMPLE 1.11
The socalled dimensionless Moody pipefriction factor f, plotted in Fig. 6.13, is calculated in
experiments from the following formula involving pipe diameter D, pressure drop p, density
, volume flow rate Q, and pipe length L:
2  v v f  eText Main Menu  D5 p
32 Q2L Textbook Table of Contents  Study Guide 1.12 The Fundamentals of Engineering (FE) Examination 43 Measurement uncertainties are given for a certain experiment: for D: 0.5 percent, p: 2.0 percent, : 1.0 percent, Q: 3.5 percent, and L: 0.4 percent. Estimate the overall uncertainty of the
friction factor f. Solution
The coefficient 2/32 is assumed to be a pure theoretical number, with no uncertainty. The other
variables may be collected using Eqs. (1.43) and (1.44):
U f
f 5 D
D 2 [{5(0.5%)}2 1 p
p 2 (2.0%)2 2 1
(1.0%)2 2 Q
Q 2 1 {2(3.5%)}2 L
L 2 1/2 (0.4%)2]1/2 7.8% Ans. By far the dominant effect in this particular calculation is the 3.5 percent error in Q, which is
amplified by doubling, due to the power of 2 on flow rate. The diameter uncertainty, which is
quintupled, would have contributed more had D been larger than 0.5 percent. 1.12 The Fundamentals of
Engineering (FE) Examination The road toward a professional engineer’s license has a first stop, the Fundamentals
of Engineering Examination, known as the FE exam. It was formerly known as
the EngineerinTraining (EIT) Examination. This 8h national test will probably
soon be required of all engineering graduates, not just for licensure, but as a studentassessment tool. The 120problem morning session covers many general studies:
Chemistry
Electric circuits
Materials science
Statics Computers
Engineering economics
Mathematics
Thermodynamics Dynamics
Fluid Mechanics
Mechanics of materials
Ethics For the 60problem afternoon session you may choose chemical, civil, electrical, industrial, or mechanical engineering or take more generalengineering problems for remaining disciplines. As you can see, fluid mechanics is central to the FE exam. Therefore, this text includes a number of endofchapter FE problems where appropriate.
The format for the FE exam questions is multiplechoice, usually with five selections, chosen carefully to tempt you with plausible answers if you used incorrect units
or forgot to double or halve something or are missing a factor of , etc. In some cases,
the selections are unintentionally ambiguous, such as the following example from a
previous exam:
Transition from laminar to turbulent flow occurs at a Reynolds number of
(A) 900
(B) 1200
(C) 1500
(D) 2100
(E) 3000  v v The “correct” answer was graded as (D), Re 2100. Clearly the examiner was thinking, but forgot to specify, Red for flow in a smooth circular pipe, since (see Chaps. 6
and 7) transition is highly dependent upon geometry, surface roughness, and the length
scale used in the definition of Re. The moral is not to get peevish about the exam but
simply to go with the flow (pun intended) and decide which answer best fits an  eText Main Menu  Textbook Table of Contents  Study Guide 44 Chapter 1 Introduction undergraduatetraining situation. Every effort has been made to keep the FE exam questions in this text unambiguous. 1.13 ProblemSolving
Techniques Fluid flow analysis generates a plethora of problems, 1500 in this text alone! To solve
these problems, one must deal with various equations, data, tables, assumptions, unit
systems, and numbers. The writer recommends these problemsolving steps:
1. Gather all the given system parameters and data in one place.
2. Find, from tables or charts, all needed fluid property data: , , cp, k, , etc.
3. Use SI units (N, s, kg, m) if possible, and no conversion factors will be necessary.
4. Make sure what is asked. It is all too common for students to answer the wrong
question, for example, reporting mass flow instead of volume flow, pressure instead of pressure gradient, drag force instead of lift force. Engineers are expected to read carefully.
5. Make a detailed sketch of the system, with everything clearly labeled.
6. Think carefully and then list your assumptions. Here knowledge is power; you
should not guess the answer. You must be able to decide correctly if the flow
can be considered steady or unsteady, compressible or incompressible, onedimensional, or multidimensional, viscous or inviscid, and whether a control volume or partial differential equations are needed.
7. Based on steps 1 to 6 above, write out the appropriate equations, data correlations, and fluid state relations for your problem. If the algebra is straightforward,
solve for what is asked. If the equations are complicated, e.g., nonlinear or too
plentiful, use the Engineering Equation Solver (EES).
8. Report your solution clearly, with proper units listed and to the proper number
of significant figures (usually two or three) that the overall uncertainty of the
data will allow.  Like most scientific disciplines, fluid mechanics has a history of erratically occurring
early achievements, then an intermediate era of steady fundamental discoveries in the
eighteenth and nineteenth centuries, leading to the twentiethcentury era of “modern
practice,” as we selfcenteredly term our limited but uptodate knowledge. Ancient
civilizations had enough knowledge to solve certain flow problems. Sailing ships with
oars and irrigation systems were both known in prehistoric times. The Greeks produced
quantitative information. Archimedes and Hero of Alexandria both postulated the parallelogram law for addition of vectors in the third century B.C. Archimedes (285 – 212
B.C.) formulated the laws of buoyancy and applied them to floating and submerged
bodies, actually deriving a form of the differential calculus as part of the analysis. The
Romans built extensive aqueduct systems in the fourth century B.C. but left no records
showing any quantitative knowledge of design principles.
From the birth of Christ to the Renaissance there was a steady improvement in the
design of such flow systems as ships and canals and water conduits but no recorded
evidence of fundamental improvements in flow analysis. Then Leonardo da Vinci
(1452 – 1519) derived the equation of conservation of mass in onedimensional steady v v 1.14 History and Scope of
Fluid Mechanics  eText Main Menu  Textbook Table of Contents  Study Guide 1.14 History and Scope of Fluid Mechanics 45  v v flow. Leonardo was an excellent experimentalist, and his notes contain accurate descriptions of waves, jets, hydraulic jumps, eddy formation, and both lowdrag (streamlined) and highdrag (parachute) designs. A Frenchman, Edme Mariotte (1620 – 1684),
built the first wind tunnel and tested models in it.
Problems involving the momentum of fluids could finally be analyzed after Isaac Newton (1642 – 1727) postulated his laws of motion and the law of viscosity of the linear fluids now called newtonian. The theory first yielded to the assumption of a “perfect” or
frictionless fluid, and eighteenthcentury mathematicians (Daniel Bernoulli, Leonhard
Euler, Jean d’Alembert, JosephLouis Lagrange, and PierreSimon Laplace) produced
many beautiful solutions of frictionlessflow problems. Euler developed both the differential equations of motion and their integrated form, now called the Bernoulli equation.
D’Alembert used them to show his famous paradox: that a body immersed in a frictionless fluid has zero drag. These beautiful results amounted to overkill, since perfectfluid
assumptions have very limited application in practice and most engineering flows are
dominated by the effects of viscosity. Engineers began to reject what they regarded as a
totally unrealistic theory and developed the science of hydraulics, relying almost entirely
on experiment. Such experimentalists as Chézy, Pitot, Borda, Weber, Francis, Hagen,
Poiseuille, Darcy, Manning, Bazin, and Weisbach produced data on a variety of flows
such as open channels, ship resistance, pipe flows, waves, and turbines. All too often the
data were used in raw form without regard to the fundamental physics of flow.
At the end of the nineteenth century, unification between experimental hydraulics
and theoretical hydrodynamics finally began. William Froude (1810 – 1879) and his son
Robert (1846 – 1924) developed laws of model testing, Lord Rayleigh (1842 – 1919) proposed the technique of dimensional analysis, and Osborne Reynolds (1842 – 1912) published the classic pipe experiment in 1883 which showed the importance of the dimensionless Reynolds number named after him. Meanwhile, viscousflow theory was
available but unexploited, since Navier (1785 – 1836) and Stokes (1819 – 1903) had successfully added newtonian viscous terms to the equations of motion. The resulting
NavierStokes equations were too difficult to analyze for arbitrary flows. Then, in 1904,
a German engineer, Ludwig Prandtl (1875 – 1953), published perhaps the most important paper ever written on fluid mechanics. Prandtl pointed out that fluid flows with
small viscosity, e.g., water flows and airflows, can be divided into a thin viscous layer,
or boundary layer, near solid surfaces and interfaces, patched onto a nearly inviscid
outer layer, where the Euler and Bernoulli equations apply. Boundarylayer theory has
proved to be the single most important tool in modern flow analysis. The twentiethcentury foundations for the present state of the art in fluid mechanics were laid in a
series of broadbased experiments and theories by Prandtl and his two chief friendly
competitors, Theodore von Kármán (1881 – 1963) and Sir Geoffrey I. Taylor (1886 –
1975). Many of the results sketched here from a historical point of view will, of course,
be discussed in this textbook. More historical details can be found in Refs. 23 to 25.
Since the earth is 75 percent covered with water and 100 percent covered with air,
the scope of fluid mechanics is vast and touches nearly every human endeavor. The
sciences of meteorology, physical oceanography, and hydrology are concerned with
naturally occurring fluid flows, as are medical studies of breathing and blood circulation. All transportation problems involve fluid motion, with welldeveloped specialties
in aerodynamics of aircraft and rockets and in naval hydrodynamics of ships and submarines. Almost all our electric energy is developed either from water flow or from  eText Main Menu  Textbook Table of Contents  Study Guide 46 Chapter 1 Introduction steam flow through turbine generators. All combustion problems involve fluid motion,
as do the more classic problems of irrigation, flood control, water supply, sewage disposal, projectile motion, and oil and gas pipelines. The aim of this book is to present
enough fundamental concepts and practical applications in fluid mechanics to prepare
you to move smoothly into any of these specialized fields of the science of flow — and
then be prepared to move out again as new technologies develop. Problems
Most of the problems herein are fairly straightforward. More difficult or openended assignments are labeled with an asterisk as in
Prob. 1.18. Problems labeled with an EES icon (for example, Prob.
2.62), will benefit from the use of the Engineering Equation Solver
(EES), while problems labeled with a computer disk may require
the use of a computer. The standard endofchapter problems 1.1 to
1.85 (categorized in the problem list below) are followed by fundamentals of engineering (FE) exam problems FE3.1 to FE3.10,
and comprehensive problems C1.1 to C1.4. 1.1, 1.2, 1.3
1.4
1.5
1.6
1.7
1.7
1.7
1.7
1.8,9
1.10 Topic Problems Fluidcontinuum concept
Dimensions, units, dynamics
Velocity field
Thermodynamic properties
Viscosity; noslip condition
Surface tension
Vapor pressure; cavitation
Speed of sound; Mach number
Flow patterns, streamlines, pathlines
History of fluid mechanics 1.1 – 1.3
1.4 – 1.20
1.21 – 1.23
1.24 – 1.37
1.38 – 1.61
1.62 – 1.71
1.72 – 1.75
1.76 – 1.78
1.79 – 1.84
1.85 v v  eText Main Menu P1.3 1.26 P1.1 A gas at 20°C may be considered rarefied, deviating from
the continuum concept, when it contains less than 1012 molecules per cubic millimeter. If Avogadro’s number is 6.023
E23 molecules per mole, what absolute pressure (in Pa) for
air does this represent?
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.
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.
Hint: Account for the weight of the fluid and show that a
noshear condition will cause horizontal forces to be out of
balance.  Fluid density P1.4 A beaker approximates a right circular cone of diameter 7 in
and height 9 in. When filled with liquid, it weighs 70 oz.
When empty, it weighs 14 oz. Estimate the density of this
liquid in both SI and BG units.
P1.5 The mean free path of a gas, , is defined as the average
distance traveled by molecules between collisions. A proposed formula for estimating of an ideal gas is Problem Distribution
Section pa  RT What are the dimensions of the constant 1.26? Use the formula to estimate the mean free path of air at 20°C and 7 kPa.
Would you consider air rarefied at this condition?
P1.6 In the {MLT } system, what is the dimensional representation of (a) enthalpy, (b) mass rate of flow, (c) bending moment, (d ) angular velocity, (e) modulus of elasticity;
( f ) Poisson’s ratio?
P1.7 A small village draws 1.5 acre ft/day of water from its
reservoir. Convert this average water usage to (a) gallons
per minute and (b) liters per second.
P1.8 Suppose we know little about the strength of materials
but are told that the bending stress in a beam is proportional to the beam halfthickness y and also depends
upon the bending moment M and the beam area moment
of inertia I. We also learn that, for the particular case
M 2900 in lbf, y 1.5 in, and I 0.4 in4, the predicted stress is 75 MPa. Using this information and dimensional reasoning only, find, to three significant figures, the only possible dimensionally homogeneous
formula
y f (M, I ). Textbook Table of Contents  Study Guide Problems 47
P1.9 The kinematic viscosity of a fluid is the ratio of viscosity
where Q is the volume rate of flow and p is the pressure
to density,
/ . What is the only possible dimensionrise produced by the pump. Suppose that a certain pump
less group combining with velocity V and length L? What
develops a pressure rise of 35 lbf/in2 when its flow rate is
is the name of this grouping? (More information on this
40 L/s. If the input power is 16 hp, what is the efficiency?
will be given in Chap. 5.)
*P1.14 Figure P1.14 shows the flow of water over a dam. The volP1.10 The StokesOseen formula [18] for drag force F on a
ume flow Q is known to depend only upon crest width B,
sphere of diameter D in a fluid stream of low velocity V,
acceleration of gravity g, and upstream water height H
density , and viscosity , is
above the dam crest. It is further known that Q is proportional to B. What is the form of the only possible dimen9
F 3 DV
V2D2
sionally homogeneous relation for this flow rate?
16
Is this formula dimensionally homogeneous?
P1.11 Engineers sometimes use the following formula for the
volume rate of flow Q of a liquid flowing through a hole
of diameter D in the side of a tank:
Q 0.68 D2 gh Water level where g is the acceleration of gravity and h is the height
of the liquid surface above the hole. What are the dimensions of the constant 0.68?
P1.12 For lowspeed (laminar) steady flow through a circular
pipe, as shown in Fig. P1.12, the velocity u varies with radius and takes the form
u B p 2
(r0 Q H Dam
B r2)
P1.14 where is the fluid viscosity and p is the pressure drop
from entrance to exit. What are the dimensions of the constant B?
Pipe wall
r = r0 r
u (r) r=0 P1.15 As a practical application of Fig. P1.14, often termed a
sharpcrested weir, civil engineers use the following formula for flow rate: Q 3.3BH3/2, with Q in ft3/s and B
and H in feet. Is this formula dimensionally homogeneous?
If not, try to explain the difficulty and how it might be converted to a more homogeneous form.
P1.16 Algebraic equations such as Bernoulli’s relation, Eq. (1)
of Ex. 1.3, are dimensionally consistent, but what about
differential equations? Consider, for example, the boundarylayer xmomentum equation, first derived by Ludwig
Prandtl in 1904:
u P1.12
P1.13 The efficiency of a pump is defined as the (dimensionless) ratio of the power developed by the flow to the power
required to drive the pump: u
x v v  eText Main Menu Q  u
y p
x gx y where is the boundarylayer shear stress and gx is the
component of gravity in the x direction. Is this equation
dimensionally consistent? Can you draw a general conclusion?
P1.17 The HazenWilliams hydraulics formula for volume rate of
flow Q through a pipe of diameter D and length L is given by Qp
input power  v Textbook Table of Contents  61.9 D 2.63 Study Guide p
L 0.54 48 Chapter 1 Introduction *P1.18 *P1.19 P1.20 P1.21 *P1.22 where p is the pressure drop required to drive the flow.
What are the dimensions of the constant 61.9? Can this formula be used with confidence for various liquids and gases?
For small particles at low velocities, the first term in the
StokesOseen drag law, Prob. 1.10, is dominant; hence,
F KV, where K is a constant. Suppose a particle of mass
m is constrained to move horizontally from the initial position x 0 with initial velocity V0. Show (a) that its velocity will decrease exponentially with time and (b) that it
will stop after traveling a distance x mV0 /K.
For larger particles at higher velocities, the quadratic term
in the StokesOseen drag law, Prob. 1.10, is dominant;
hence, F CV2, where C is a constant. Repeat Prob. 1.18
to show that (a) its velocity will decrease as 1/(1 CV0t/m)
and (b) it will never quite stop in a finite time span.
A baseball, with m 145 g, is thrown directly upward from
the initial position z 0 and V0 45 m/s. The air drag on
the ball is CV 2, as in Prob. 1.19, where C 0.0013 N
s2/m2. Set up a differential equation for the ball motion, and
solve for the instantaneous velocity V(t) and position z(t).
Find the maximum height zmax reached by the ball, and
compare your results with the classical case of zero air drag.
A velocity field is given by V Kxti Kytj 0k, where
K is a positive constant. Evaluate (a)
and (b)
V.
According to the theory of Chap. 8, as a uniform stream
approaches a cylinder of radius R along the symmetry line
AB in Fig. P1.22, the velocity has only one component:
u U1 R2
for
x2 x R P1.24 P1.25 P1.26 P1.27 Q/Acrosssection and the dimensionless Reynolds number of the
flow, Re
VavgD/ . Comment on your results.
Air at 1 atm and 20°C has an internal energy of approximately 2.1 E5 J/kg. If this air moves at 150 m/s at an altitude z 8 m, what is its total energy, in J/kg, relative to
the datum z 0? Are any energy contributions negligible?
A tank contains 0.9 m3 of helium at 200 kPa and 20°C.
Estimate the total mass of this gas, in kg, (a) on earth and
(b) on the moon. Also, (c) how much heat transfer, in MJ,
is required to expand this gas at constant temperature to a
new volume of 1.5 m3?
When we in the United States say a car’s tire is filled “to
32 lb,” we mean that its internal pressure is 32 lbf/in2 above
the ambient atmosphere. If the tire is at sea level, has a
volume of 3.0 ft3, and is at 75°F, estimate the total weight
of air, in lbf, inside the tire.
For steam at 40 lbf/in2, some values of temperature and
specific volume are as follows, from Ref. 13: T, °F
3 v, ft /lbm 400 500 600 700 800 12.624 14.165 15.685 17.195 18.699 Is steam, for these conditions, nearly a perfect gas, or is it
wildly nonideal? If reasonably perfect, find a leastsquares†
value for the gas constant R, in m2/(s2 K), estimate the percent error in this approximation, and compare with Table A.4.
P1.28 Wet atmospheric air at 100 percent relative humidity contains saturated water vapor and, by Dalton’s law of partial
pressures,
patm pdry air pwater vapor where U is the stream velocity far from the cylinder. Using the concepts from Ex. 1.5, find (a) the maximum flow
deceleration along AB and (b) its location.  v v Suppose this wet atmosphere is at 40°C and 1 atm. Calculate the density of this 100 percent humid air, and compare it with the density of dry air at the same conditions.
P1.29 A compressedair tank holds 5 ft3 of air at 120 lbf/in2
“gage,” that is, above atmospheric pressure. Estimate the
y
energy, in ftlbf, required to compress this air from the atmosphere, assuming an ideal isothermal process.
P1.30 Repeat Prob. 1.29 if the tank is filled with compressed wau
x
ter instead of air. Why is the result thousands of times less
A
B
R
than the result of 215,000 ft lbf in Prob. 1.29?
*P1.31 The density of (fresh) water at 1 atm, over the temperature range 0 to 100°C, is given in Table A.1. Fit these valEES
P1.22
ues to a leastsquares† equation of the form
a bT
cT 2, with T in °C, and estimate its accuracy. Use your forP1.23 Experiment with a faucet (kitchen or otherwise) to determine
mula to compute the density of water at 45°C, and comtypical flow rates Q in m3/h, perhaps timing the discharge of
pare your result with the accepted experimental value of
a known volume. Try to achieve an exit jet condition which
990.1 kg/m3.
is (a) smooth and round and (b) disorderly and fluctuating.
†
Measure the supplypipe diameter (look under the sink). For
The concept of “leastsquares” error is very important and should be
both cases, calculate the average flow velocity, Vavg
learned by everyone.  eText Main Menu  Textbook Table of Contents  Study Guide Problems 49
P1.32 A blimp is approximated by a prolate spheroid 90 m long
and 30 m in diameter. Estimate the weight of 20°C gas
within the blimp for (a) helium at 1.1 atm and (b) air at
1.0 atm. What might the difference between these two values represent (see Chap. 2)?
*P1.33 Experimental data for the density of mercury versus pressure at 20°C are as follows:
p, atm 1
3 , kg/m 500 1,000 1,500 2,000 13,545 13,573 13,600 13,625 T, K 300 400 500 600 700 800 , kg/(m s) 2.27 E5 2.85 E5 3.37 E5 3.83 E5 4.25 E5 4.64 E5 Fit these value to either (a) a power law or (b) the Sutherland law, Eq. (1.30).
P1.42 Experimental values for the viscosity of helium at 1 atm
are as follows: 13,653 T, K 200 400 600 800 1000 1200 , kg/(m s) 1.50 E5 2.43 E5 3.20 E5 3.88 E5 4.50 E5 5.08 E5  v v Fit this data to the empirical state relation for liquids,
Fit these values to either (a) a power law or (b) the SutherEq. (1.22), to find the best values of B and n for mercury.
land law, Eq. (1.30).
Then, assuming the data are nearly isentropic, use these
*P1.43 Yaws et al. [34] suggest the following curvefit formula
values to estimate the speed of sound of mercury at 1 atm
for viscosity versus temperature of organic liquids:
and compare with Table 9.1.
3
P1.34 If water occupies 1 m at 1 atm pressure, estimate the presB
log10
A
CT DT2
sure required to reduce its volume by 5 percent.
T
P1.35 In Table A.4, most common gases (air, nitrogen, oxygen,
hydrogen) have a specific heat ratio k 1.40. Why do arwith T in absolute units. (a) Can this formula be criticized
gon and helium have such high values? Why does NH3
on dimensional grounds? (b) Disregarding (a), indicate anhave such a low value? What is the lowest k for any gas
alytically how the curvefit constants A, B, C, D could be
that you know of?
found from N data points ( i, Ti) using the method of least
P1.36 The isentropic bulk modulus B of a fluid is defined as the
squares. Do not actually carry out a calculation.
isentropic change in pressure per fractional change in den P1.44 The values for SAE 30 oil in Table 1.4 are strictly “repsity:
resentative,” not exact, because lubricating oils vary considerably according to the type of crude oil from which
p
B
they are refined. The Society of Automotive Engineers [26]
s
allows certain kinematic viscosity ranges for all lubricatWhat are the dimensions of B? Using theoretical p( ) reing oils: for SAE 30, 9.3
12.5 mm2/s at 100°C. SAE
lations, estimate the bulk modulus of (a) N2O, assumed to
30 oil density can also vary 2 percent from the tabulated
be an ideal gas, and (b) water, at 20°C and 1 atm.
value of 891 kg/m3. Consider the following data for an acP1.37 A nearideal gas has a molecular weight of 44 and a speceptable grade of SAE 30 oil:
cific heat cv 610 J/(kg K). What are (a) its specific heat
T, °C
0
20
40
60
80
100
ratio, k, and (b) its speed of sound at 100°C?
P1.38 In Fig. 1.6, if the fluid is glycerin at 20°C and the width be, kg/(m s)
2.00
0.40
0.11
0.042
0.017
0.0095
tween plates is 6 mm, what shear stress (in Pa) is required
How does this oil compare with the plot in Appendix Fig.
to move the upper plate at 5.5 m/s? What is the Reynolds
A.1? How well does the data fit Andrade’s equation in
number if L is taken to be the distance between plates?
Prob. 1.40?
P1.39 Knowing for air at 20°C from Table 1.4, estimate its viscosity at 500°C by (a) the power law and (b) the Suther P1.45 A block of weight W slides down an inclined plane while
lubricated by a thin film of oil, as in Fig. P1.45. The film
land law. Also make an estimate from (c) Fig. 1.5. Comcontact area is A and its thickness is h. Assuming a linear
pare with the accepted value of
3.58 E5 kg/m s.
velocity distribution in the film, derive an expression for
*P1.40 For liquid viscosity as a function of temperature, a simthe “terminal” (zeroacceleration) velocity V of the block.
plification of the logquadratic law of Eq. (1.31) is Andrade’s equation [11],
A exp (B/T), where (A, B) are P1.46 Find the terminal velocity of the block in Fig. P1.45 if the
block mass is 6 kg, A 35 cm2,
15°, and the film is
curvefit constants and T is absolute temperature. Fit this
1mmthick SAE 30 oil at 20°C.
relation to the data for water in Table A.1 and estimate the
P1.47 A shaft 6.00 cm in diameter is being pushed axially
percent error of the approximation.
through a bearing sleeve 6.02 cm in diameter and 40 cm
P1.41 Some experimental values of the viscosity of argon gas at
long. The clearance, assumed uniform, is filled with oil
1 atm are as follows:
EES  eText Main Menu  Textbook Table of Contents  Study Guide 50 Chapter 1 Introduction
sured torque is 0.293 N m, what is the fluid viscosity?
Suppose that the uncertainties of the experiment are as follows: L ( 0.5 mm), M ( 0.003 N m), ( 1 percent),
and ri or ro ( 0.02 mm). What is the uncertainty in the
measured viscosity?
P1.52 The belt in Fig. P1.52 moves at a steady velocity V and
skims the top of a tank of oil of viscosity , as shown. Assuming a linear velocity profile in the oil, develop a simple formula for the required beltdrive power P as a function of (h, L, V, b, ). What beltdrive power P, in watts,
is required if the belt moves at 2.5 m/s over SAE 30W oil
at 20°C, with L 2 m, b 60 cm, and h 3 cm? Liquid film of
thickness h
W
V Block contact
area A θ P1.45 whose properties are
0.003 m2/s and SG 0.88. Estimate the force required to pull the shaft at a steady veL
locity of 0.4 m/s.
V
P1.48 A thin plate is separated from two fixed plates by very visMoving belt, width b
cous liquids 1 and 2, respectively, as in Fig. P1.48. The
plate spacings h1 and h2 are unequal, as shown. The conOil, depth h
tact area is A between the center plate and each fluid.
(a) Assuming a linear velocity distribution in each fluid,
P1.52
derive the force F required to pull the plate at velocity V.
(b) Is there a necessary relation between the two viscosities, 1 and 2?
*P1.53 A solid cone of angle 2 , base r0, and density c is rotating with initial angular velocity 0 inside a conical seat,
as shown in Fig. P1.53. The clearance h is filled with oil
of viscosity . Neglecting air drag, derive an analytical exh1
µ1
pression for the cone’s angular velocity (t) if there is no
applied torque.
F, V
µ2 h2 ω (t) Base
radius r0 Oil P1.48  v v P1.49 The shaft in Prob. 1.47 is now fixed axially and rotated in2θ
side the sleeve at 1500 r/min. Estimate (a) the torque
(N m) and (b) the power (kW) required to rotate the shaft.
h
P1.50 An amazing number of commercial and laboratory devices
have been developed to measure the viscosity of fluids, as
described in Ref. 27. The concentric rotating shaft of Prob.
1.49 is an example of a rotational viscometer. Let the inner and outer cylinders have radii ri and ro, respectively,
P1.53
with total sleeve length L. Let the rotational rate be
(rad/s) and the applied torque be M. Derive a theoretical
relation for the viscosity of the clearance fluid, , in terms *P1.54 A disk of radius R rotates at an angular velocity inside
of these parameters.
a diskshaped container filled with oil of viscosity , as
P1.51 Use the theory of Prob. 1.50 (or derive an ad hoc expresshown in Fig. P1.54. Assuming a linear velocity profile
sion if you like) for a shaft 8 cm long, rotating at 1200
and neglecting shear stress on the outer disk edges, derive
r/min, with ri 2.00 cm and ro 2.05 cm. If the meaa formula for the viscous torque on the disk.  eText Main Menu  Textbook Table of Contents  Study Guide Problems 51
r4 p
0
8LQ Ω
Clearance
h Pipe end effects are neglected [27]. Suppose our capillary
has r0 2 mm and L 25 cm. The following flow rate
and pressure drop data are obtained for a certain fluid: Oil
Q, m3/h
R 0.72 1.08 1.44 1.80 p, kPa R 0.36
159 318 477 1274 1851 P1.54 *P1.55 The device in Fig. P1.54 is called a rotating disk viscometer
[27]. Suppose that R 5 cm and h 1 mm. If the torque
required to rotate the disk at 900 r/min is 0.537 N m,
what is the viscosity of the fluid? If the uncertainty in each
parameter (M, R, h, ) is 1 percent, what is the overall
uncertainty in the viscosity?
*P1.56 The device in Fig. P1.56 is called a coneplate viscometer
[27]. The angle of the cone is very small, so that sin
, and the gap is filled with the test liquid. The torque M
to rotate the cone at a rate is measured. Assuming a linear velocity profile in the fluid film, derive an expression
for fluid viscosity as a function of (M, R, , ). P1.59 P1.60 *P1.61
Ω R
Fluid
θ P1.62 θ P1.56
P1.63  v v *P1.57 For the coneplate viscometer of Fig. P1.56, suppose that
R 6 cm and
3°. If the torque required to rotate the
cone at 600 r/min is 0.157 N m, what is the viscosity of
the fluid? If the uncertainty in each parameter (M, R, ,
) is 1 percent, what is the overall uncertainty in the viscosity?
*P1.58 The laminarpipeflow example of Prob. 1.12 can be used
to design a capillary viscometer [27]. If Q is the volume
flow rate, L is the pipe length, and p is the pressure drop
from entrance to exit, the theory of Chap. 6 yields a formula for viscosity:  eText Main Menu  P1.64 P1.65 What is the viscosity of the fluid? Note: Only the first three
points give the proper viscosity. What is peculiar about the
last two points, which were measured accurately?
A solid cylinder of diameter D, length L, and density s
falls due to gravity inside a tube of diameter D0. The clearance, D0 D
D, is filled with fluid of density and
viscosity . Neglect the air above and below the cylinder.
Derive a formula for the terminal fall velocity of the cylinder. Apply your formula to the case of a steel cylinder,
D 2 cm, D0 2.04 cm, L 15 cm, with a film of SAE
30 oil at 20°C.
For Prob. 1.52 suppose that P 0.1 hp when V 6 ft/s,
L 4.5 ft, b 22 in, and h 7/8 in. Estimate the viscosity of the oil, in kg/(m s). If the uncertainty in each
parameter (P, L, b, h, V) is 1 percent, what is the overall uncertainty in the viscosity?
An airhockey puck has a mass of 50 g and is 9 cm in diameter. When placed on the air table, a 20°C air film, of
0.12mm thickness, forms under the puck. The puck is
struck with an initial velocity of 10 m/s. Assuming a linear velocity distribution in the air film, how long will it
take the puck to (a) slow down to 1 m/s and (b) stop completely? Also, (c) how far along this extremely long table
will the puck have traveled for condition (a)?
The hydrogen bubbles which produced the velocity profiles in Fig. 1.13 are quite small, D 0.01 mm. If the hydrogenwater interface is comparable to airwater and the
water temperature is 30°C estimate the excess pressure
within the bubble.
Derive Eq. (1.37) by making a force balance on the fluid
interface in Fig. 1.9c.
At 60°C the surface tension of mercury and water is 0.47
and 0.0662 N/m, respectively. What capillary height
changes will occur in these two fluids when they are in
contact with air in a clean glass tube of diameter 0.4 mm?
The system in Fig. P1.65 is used to calculate the pressure
p1 in the tank by measuring the 15cm height of liquid in
the 1mmdiameter tube. The fluid is at 60°C (see Prob.
1.64). Calculate the true fluid height in the tube and the
percent error due to capillarity if the fluid is (a) water and
(b) mercury. Textbook Table of Contents  Study Guide 52 Chapter 1 Introduction
mula for the maximum diameter dmax able to float in the
liquid. Calculate dmax for a steel needle (SG 7.84) in
water at 20°C.
P1.70 Derive an expression for the capillary height change h for
a fluid of surface tension Y and contact angle between
two vertical parallel plates a distance W apart, as in Fig.
P1.70. What will h be for water at 20°C if W 0.5 mm? 15 cm
p1 θ P1.65
P1.66 A thin wire ring, 3 cm in diameter, is lifted from a water
surface at 20°C. Neglecting the wire weight, what is the
force required to lift the ring? Is this a good way to meah
sure surface tension? Should the wire be made of any particular material?
P1.67 Experiment with a capillary tube, perhaps borrowed from
W
the chemistry department, to verify, in clean water, the rise
due to surface tension predicted by Example 1.9. Add small
P1.70
amounts of liquid soap to the water, and report to the class
whether detergents significantly lower the surface tension.
*P1.71 A soap bubble of diameter D1 coalesces with another bubWhat practical difficulties do detergents present?
ble of diameter D2 to form a single bubble D3 with the
*P1.68 Make an analysis of the shape (x) of the waterair intersame amount of air. Assuming an isothermal process, deface near a plane wall, as in Fig. P1.68, assuming that the
rive an expression for finding D3 as a function of D1, D2,
slope is small, R 1 d2 /dx2. Also assume that the prespatm, and Y.
sure difference across the interface is balanced by the specific weight and the interface height, p
g . The P1.72 Early mountaineers boiled water to estimate their altitude.
If they reach the top and find that water boils at 84°C, apboundary conditions are a wetting contact angle at x 0 EES
proximately how high is the mountain?
and a horizontal surface
0 as x → . What is the maxP1.73 A small submersible moves at velocity V, in fresh water
imum height h at the wall?
at 20°C, at a 2m depth, where ambient pressure is 131
y
kPa. Its critical cavitation number is known to be Ca
0.25. At what velocity will cavitation bubbles begin to form
y=h
on the body? Will the body cavitate if V 30 m/s and the
water is cold (5°C)?
P1.74 A propeller is tested in a water tunnel at 20°C as in Fig.
1.12a. The lowest pressure on the blade can be estimated
by a form of Bernoulli’s equation (Ex. 1.3):
pmin θ
η (x)
x
x=0 P1.68  v v P1.69 A solid cylindrical needle of diameter d, length L, and density n may float in liquid of surface tension Y. Neglect
buoyancy and assume a contact angle of 0°. Derive a for  eText Main Menu  p0 1
2 V2 where p0 1.5 atm and V tunnel velocity. If we run the
tunnel at V 18 m/s, can we be sure that there will be no
cavitation? If not, can we change the water temperature
and avoid cavitation?
P1.75 Oil, with a vapor pressure of 20 kPa, is delivered through
a pipeline by equally spaced pumps, each of which increases the oil pressure by 1.3 MPa. Friction losses in the
pipe are 150 Pa per meter of pipe. What is the maximum
possible pump spacing to avoid cavitation of the oil? Textbook Table of Contents  Study Guide Fundamentals of Engineering Exam Problems 53 P1.76 An airplane flies at 555 mi/h. At what altitude in the stan P1.82 A velocity field is given by u V cos , v V sin , and
dard atmosphere will the airplane’s Mach number be exw 0, where V and are constants. Derive a formula for
actly 0.8?
the streamlines of this flow.
*P1.77 The density of 20°C gasoline varies with pressure ap *P1.83 A twodimensional unsteady velocity field is given by u
proximately as follows:
x(1 2t), v y. Find the equation of the timevarying
EES
streamlines which all pass through the point (x0, y0) at
p, atm
1
500
1000
1500
some time t. Sketch a few of these.
, lbm/ft3
42.45
44.85
46.60
47.98 *P1.84 Repeat Prob. 1.83 to find and sketch the equation of the
pathline which passes through (x0, y0) at time t 0.
Use these data to estimate (a) the speed of sound (m/s) P1.85 Do some reading and report to the class on the life and
and (b) the bulk modulus (MPa) of gasoline at 1 atm.
achievements, especially visàvis fluid mechanics, of
P1.78 Sir Isaac Newton measured the speed of sound by timing
(a) Evangelista Torricelli (1608–1647)
the difference between seeing a cannon’s puff of smoke
(b) Henri de Pitot (1695–1771)
and hearing its boom. If the cannon is on a mountain 5.2 mi
away, estimate the air temperature in degrees Celsius if the
(c) Antoine Chézy (1718–1798)
time difference is (a) 24.2 s and (b) 25.1 s.
(d) Gotthilf Heinrich Ludwig Hagen (1797–1884)
P1.79 Examine the photographs in Figs. 1.12a, 1.13, 5.2a, 7.14a,
(e) Julius Weisbach (1806–1871)
and 9.10b and classify them according to the boxes in Fig.
(f) George Gabriel Stokes (1819–1903)
1.14.
(g) Moritz Weber (1871–1951)
*P1.80 A twodimensional steady velocity field is given by u
2
2
(h) Theodor von Kármán (1881–1963)
x
y, v
2xy. Derive the streamline pattern and
sketch a few streamlines in the upper half plane. Hint: The
(i) Paul Richard Heinrich Blasius (1883–1970)
differential equation is exact.
(j) Ludwig Prandtl (1875–1953)
P1.81 Repeat Ex. 1.10 by letting the velocity components in(k) Osborne Reynolds (1842–1912)
crease linearly with time:
(l) John William Strutt, Lord Rayleigh (1842–1919)
V Kxt i Kyt j 0k
(m) Daniel Bernoulli (1700–1782)
(n) Leonhard Euler (1707–1783)
Find and sketch, for a few representative times, the instantaneous streamlines. How do they differ from the
steady flow lines in Ex. 1.10? Fundamentals of Engineering Exam Problems  v v FE1.1 The absolute viscosity of a fluid is primarily a function of
(a) Density, (b) Temperature, (c) Pressure, (d) Velocity,
(e) Surface tension
FE1.2 If a uniform solid body weighs 50 N in air and 30 N in
water, its specific gravity is
(a) 1.5, (b) 1.67, (c) 2.5, (d) 3.0, (e) 5.0
FE1.3 Helium has a molecular weight of 4.003. What is the
weight of 2 m3 of helium at 1 atm and 20°C?
(a) 3.3 N, (b) 6.5 N, (c) 11.8 N, (d) 23.5 N, (e) 94.2 N
FE1.4 An oil has a kinematic viscosity of 1.25 E4 m2/s and a
specific gravity of 0.80. What is its dynamic (absolute)
viscosity in kg/(m s)?
(a) 0.08, (b) 0.10, (c) 0.125, (d) 1.0, (e) 1.25
FE1.5 Consider a soap bubble of diameter 3 mm. If the surface
tension coefficient is 0.072 N/m and external pressure is
0 Pa gage, what is the bubble’s internal gage pressure?  eText Main Menu  (a) 24 Pa, (b) 48 Pa, (c) 96 Pa, (d) 192 Pa,
(e) 192 Pa
FE1.6 The only possible dimensionless group which combines
velocity V, body size L, fluid density , and surface tension coefficient is
(a) L /V, (b) VL2/ , (c) V2/L, (d) LV2/ ,
(e) LV2/
FE1.7 Two parallel plates, one moving at 4 m/s and the other
fixed, are separated by a 5mmthick layer of oil of specific gravity 0.80 and kinematic viscosity 1.25 E4 m2/s.
What is the average shear stress in the oil?
(a) 80 Pa, (b) 100 Pa, (c) 125 Pa, (d) 160 Pa, (e) 200 Pa
FE1.8 Carbon dioxide has a specific heat ratio of 1.30 and a
gas constant of 189 J/(kg °C). If its temperature rises
from 20 to 45°C, what is its internal energy rise?
(a) 12.6 kJ/kg, (b) 15.8 kJ/kg, (c) 17.6 kJ/kg, (d) 20.5
kJ/kg, (e) 25.1 kJ/kg Textbook Table of Contents  Study Guide 54 Chapter 1 Introduction FE1.9 A certain water flow at 20°C has a critical cavitation
number, where bubbles form, Ca 0.25, where Ca
2(pa pvap)/ V2. If pa 1 atm and the vapor pressure
is 0.34 pounds per square inch absolute (psia), for what
water velocity will bubbles form?
(a) 12 mi/h, (b) 28 mi/h, (c) 36 mi/h, (d) 55 mi/h,
(e) 63 mi/h FE1.10 A steady incompressible flow, moving through a contraction section of length L, has a onedimensional average velocity distribution given by u U0(1 2x/L).
What is its convective acceleration at the end of the contraction, x L?
2
2
2
2
2
(a) U0 /L, (b) 2U0 /L, (c) 3U0 /L, (d) 4U0 /L, (e) 6U0 /L Comprehensive Problems
Sometimes equations can be developed and practical problems can be solved by knowing nothing more than the dimensions of the key parameters in the problem. For example, consider the heat loss through a window in a
building. Window efficiency is rated in terms of “R value”
which has units of (ft2 h °F)/Btu. A certain manufacturer advertises a doublepane window with an R value of
2.5. The same company produces a triplepane window
with an R value of 3.4. In either case the window dimensions are 3 ft by 5 ft. On a given winter day, the temperature difference between the inside and outside of the
building is 45°F.
(a) Develop an equation for the amount of heat lost in a
given time period t, through a window of area A, with
R value R, and temperature difference T. How much
heat (in Btu) is lost through the doublepane window
in one 24h period?
(b) How much heat (in Btu) is lost through the triplepane
window in one 24h period?
(c) Suppose the building is heated with propane gas, which
costs $1.25 per gallon. The propane burner is 80 percent efficient. Propane has approximately 90,000 Btu
of available energy per gallon. In that same 24h period, how much money would a homeowner save per
window by installing triplepane rather than doublepane windows?
(d) Finally, suppose the homeowner buys 20 such triplepane windows for the house. A typical winter has the
equivalent of about 120 heating days at a temperature
difference of 45°F. Each triplepane window costs $85
more than the doublepane window. Ignoring interest
and inflation, how many years will it take the homeowner to make up the additional cost of the triplepane
windows from heating bill savings?
C1.2 When a person ice skates, the surface of the ice actually
melts beneath the blades, so that he or she skates on a thin
sheet of water between the blade and the ice.
(a) Find an expression for total friction force on the bottom of the blade as a function of skater velocity V, blade
length L, water thickness (between the blade and the
ice) h, water viscosity , and blade width W.  v v C1.1  eText Main Menu  (b) Suppose an ice skater of total mass m is skating along
at a constant speed of V0 when she suddenly stands stiff
with her skates pointed directly forward, allowing herself to coast to a stop. Neglecting friction due to air resistance, how far will she travel before she comes to a
stop? (Remember, she is coasting on two skate blades.)
Give your answer for the total distance traveled, x, as
a function of V0, m, L, h, , and W.
(c) Find x for the case where V0 4.0 m/s, m 100 kg, L
30 cm, W 5.0 mm, and h 0.10 mm. Do you think our
assumption of negligible air resistance is a good one?
C1.3 Two thin flat plates, tilted at an angle , are placed in a
tank of liquid of known surface tension and contact angle , as shown in Fig. C1.3. At the free surface of the liquid in the tank, the two plates are a distance L apart and
have width b into the page. The liquid rises a distance h
between the plates, as shown.
(a) What is the total upward (zdirected) force, due to surface tension, acting on the liquid column between the
plates?
(b) If the liquid density is , find an expression for surface
tension in terms of the other variables. z h g L C1.3 Textbook Table of Contents  Study Guide References 55 C1.4 that, for the coordinate system given, both w and
(dw/dx)wall are negative. Oil of viscosity and density drains steadily down the
side of a tall, wide vertical plate, as shown in Fig. C1.4.
In the region shown, fully developed conditions exist; that
is, the velocity profile shape and the film thickness are
independent of distance z along the plate. The vertical velocity w becomes a function only of x, and the shear resistance from the atmosphere is negligible.
(a) Sketch the approximate shape of the velocity profile
w(x), considering the boundary conditions at the wall and
at the film surface.
(b) Suppose film thickness , and the slope of the velocity profile at the wall, (dw/dx)wall, are measured by a laser
Doppler anemometer (to be discussed in Chap. 6). Find an
expression for the viscosity of the oil as a function of ,
, (dw/dx)wall, and the gravitational accleration g. Note Plate
Oil film
Air g
z x C1.4 References 2.
3.
4.
5.
6.
7.
8.
9.
10.
11. 12. 13. 14. J. C. Tannehill, D. A. Anderson, and R. H. Pletcher, Computational Fluid Mechanics and Heat Transfer, 2d ed., Taylor
and Francis, Bristol, PA, 1997.
S. V. Patankar, Numerical Heat Transfer and Fluid Flow,
McGrawHill, New York, 1980.
F. M. White, Viscous Fluid Flow, 2d ed., McGrawHill, New
York, 1991.
R. J. Goldstein (ed.), Fluid Mechanics Measurements, 2d ed.,
Taylor and Francis, Bristol, PA, 1997.
R. A. Granger, Experiments in Fluid Mechanics, Oxford University Press, 1995.
H. A. Barnes, J. F. Hutton, and K. Walters, An Introduction
to Rheology, Elsevier, New York, 1989.
A. E. Bergeles and S. Ishigai, TwoPhase Flow Dynamics and
Reactor Safety, McGrawHill, New York, 1981.
G. N. Patterson, Introduction to the Kinetic Theory of Gas
Flows, University of Toronto Press, Toronto, 1971.
ASME Orientation and Guide for Use of Metric Units, 9th ed.,
American Society of Mechanical Engineers, New York, 1982.
J. P. Holman, Heat Transfer, 8th ed., McGrawHill, New York,
1997.
R. C. Reid, J. M. Prausnitz, and T. K. Sherwood, The Properties of Gases and Liquids, 4th ed., McGrawHill, New York,
1987.
J. Hilsenrath et al., “Tables of Thermodynamic and Transport
Properties,” U. S. Nat. Bur. Stand. Circ. 564, 1955; reprinted
by Pergamon, New York, 1960.
R. A. Spencer et al., ASME Steam Tables with Mollier Chart,
6th ed., American Society of Mechanical Engineers, New
York, 1993.
O. A. Hougen and K. M. Watson, Chemical Process Principles Charts, Wiley, New York, 1960.  v v 1.  eText Main Menu  15.
16.
17. 18.
19.
20.
21.
22.
23. 24.
25.
26.
27.
28.
29. A. W. Adamson, Physical Chemistry of Surfaces, 5th ed., Interscience, New York, 1990.
J. A. Knauss, Introduction to Physical Oceanography, PrenticeHall, Englewood Cliffs, NJ, 1978.
National Committee for Fluid Mechanics Films, Illustrated
Experiments in Fluid Mechanics, M.I.T. Press, Cambridge,
MA, 1972.
I. G. Currie, Fundamental Mechanics of Fluids, 2d ed.,
McGrawHill, New York, 1993.
M. van Dyke, An Album of Fluid Motion, Parabolic Press,
Stanford, CA, 1982.
Y. Nakayama (ed.), Visualized Flow, Pergamon Press, Oxford, 1988.
W. J. Yang (ed.), Handbook of Flow Visualization, Hemisphere, New York, 1989.
W. Merzkirch, Flow Visualization, 2d ed., Academic, New
York, 1987.
H. Rouse and S. Ince, History of Hydraulics, Iowa Institute
of Hydraulic Research, Univ. of Iowa, Iowa City, 1957;
reprinted by Dover, New York, 1963.
H. Rouse, Hydraulics in the United States 1776 – 1976, Iowa
Institute of Hydraulic Research, Univ. of Iowa, Iowa City, 1976.
G. Garbrecht, Hydraulics and Hydraulic Research: An Historical Review, Gower Pub., Aldershot, UK, 1987.
1986 SAE Handbook, 4 vols., Society of Automotive Engineers, Warrendale, PA.
J. R. van Wazer, Viscosity and Flow Measurement, Interscience, New York, 1963.
SAE Fuels and Lubricants Standards Manual, Society of Automotive Engineers, Warrendale, PA, 1995.
John D. Anderson, Computational Fluid Dynamics: The Basics with Applications, McGrawHill, New York, 1995. Textbook Table of Contents  Study Guide 56 Chapter 1 Introduction H. W. Coleman and W. G. Steele, Experimentation and Uncertainty Analysis for Engineers, John Wiley, New York,
1989.
31. R. J. Moffatt, “Describing the Uncertainties in Experimental
Results,” Experimental Thermal and Fluid Science, vol., 1,
1988, pp. 3 – 17.
32. Paul A. Libby, An Introduction to Turbulence, Taylor and
Francis, Bristol, PA, 1996.
33. Sanford Klein and William Beckman, Engineering Equation
Solver (EES), FChart Software, Middleton, WI, 1997.  v v 30.  eText Main Menu  34. C. L. Yaws, X. Lin, and L. Bu, “Calculate Viscosities for 355
Compounds. An Equation Can Be Used to Calculate Liquid
Viscosity as a Function of Temperature,” Chemical Engineering, vol. 101, no. 4, April 1994, pp. 119 – 128.
35. Frank E. Jones, Techniques and Topics in Flow Measurement,
CRC Press, Boca Raton, FL, 1995.
36. Carl L. Yaws, Handbook of Viscosity, 3 vols., Gulf Publishing, Houston, TX, 1994. Textbook Table of Contents  Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
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