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Chapter 1 Introduction to Fluid Mechanics
1.1 What is Fluid Mechanics?
Fluid mechanics deals with the study of all fluids under static and dynamic situations.
Fluid mechanics is a branch of continuous mechanics which deals with
a relationship between forces, motions, and statical conditions in a continuous material.
This study area deals with many and diversified problems such as
surface tension, fluid statics, flow in enclose bodies, or flow round
bodies (solid or otherwise), flow stability, etc.
In fact, almost any action a person is doing involves some kind of a fluid mechanics problem.
Furthermore, the boundary between the solid mechanics and fluid
mechanics is some kind of gray shed and not a sharp distinction
(see Figure 1.1 for the complex relationships between the different branches
which only part of it should be drawn in the same time.).
For example, glass appears as a solid material, but a closer
look reveals that the glass is a liquid with a large viscosity.
A proof of the glass ``liquidity'' is the change of the glass
thickness in high windows in European Churches after hundred years.
The bottom part of the glass is thicker than the top part.
Materials like sand (some call it quick sand) and grains should be treated as liquids.
It is known that these materials have the ability to drown people.
Even material such as aluminum just below the mushy zone
also behaves as a liquid similarly to butter.
Furthermore, material particles that ``behaves'' as solid mixed with liquid creates a mixture
After it was established that the boundaries of fluid mechanics
aren't sharp, most of the discussion in this book is limited to simple
and (mostly) Newtonian (sometimes power fluids) fluids which will be defined later.
Fig. 1.1 Diagram to explain part of relationships of fluid mechanics branches.
The fluid mechanics study involve many fields that have no clear boundaries between them.
Researchers distinguish between orderly flow and chaotic flow
as the laminar flow and the turbulent flow.
The fluid mechanics can also be distinguish between a single phase
flow and multiphase flow (flow made more than one phase or
single distinguishable material).
The last boundary (as all the boundaries in fluid mechanics) isn't
sharp because fluid can go through a phase change
(condensation or evaporation) in the middle or during
the flow and switch from a single phase flow to a multi phase flow.
Moreover, flow with two phases (or materials) can
be treated as a single phase (for example, air with dust particle).
After it was made clear that the boundaries of fluid mechanics
aren't sharp, the study must make arbitrary boundaries between fields.
Then the dimensional analysis can be used explain why in certain cases one distinguish
area/principle is more relevant than the other and some effects can be neglected.
Or, when a general model is need because more parameters are effecting the situation.
It is this author's personal experience that the knowledge and ability
to know in what area the situation lay is one of the main problems.
For example, engineers in software company EKK Inc, ) analyzed
a flow of a complete still liquid assuming a complex turbulent flow model.
Such absurd analysis are common among engineers who do not know which model can be applied.
Thus, one of the main goals of this book is to explain what model should be applied.
Before dealing with the boundaries, the simplified private cases must be explained.
There are two main approaches of presenting an introduction of fluid mechanics materials.
The first approach introduces the fluid kinematic and then the basic governing equations,
to be followed by stability, turbulence, boundary layer
The second approach deals with the Integral Analysis
to be followed with Differential Analysis, and continue with Empirical Analysis.
These two approaches pose a dilemma to anyone who writes an
introductory book for the fluid mechanics.
These two approaches have justifications and positive points.
Reviewing many books on fluid mechanics made it clear, there isn't a clear winner.
This book attempts to find a hybrid approach in which
the kinematic is presented first (aside to standard initial four
chapters) follow by Integral analysis and continued by Differential analysis.
The ideal flow (frictionless flow) should be
expanded compared to the regular treatment.
This book is unique in providing chapter on multiphase flow.
Naturally, chapters on open channel flow (as a sub class of the
multiphase flow) and compressible flow (with the latest
developments) are provided.
1.2 Brief History
The need to have some understanding of fluid mechanics started with the need to obtain water supply.
For example, people realized that wells have to be dug and crude pumping devices need to be constructed.
Later, a large population created a need to solve waste (sewage) and some basic understanding was created.
At some point, people realized that water can be used to move things and provide power.
When cities increased to a larger size, aqueducts were constructed.
These aqueducts reached their greatest size and grandeur in those
of the City of Rome and China.
Yet, almost all knowledge of the ancients can be summarized as
application of instincts, with the exception Archimedes (250 B.C.) on
the principles of buoyancy.
For example, larger tunnels built for a larger water supply, etc.
There were no calculations even with the great need for water supply and transportation.
The first progress in fluid mechanics was made by Leonardo Da Vinci
(14521519) who built the first chambered canal lock near Milan.
He also made several attempts to study the flight (birds) and
developed some concepts on the origin of the forces.
After his initial work, the knowledge of fluid mechanics (hydraulic)
increasingly gained speed by the contributions of Galileo, Torricelli,
Euler, Newton, Bernoulli family, and D'Alembert.
At that stage theory and experiments had some discrepancy.
This fact was acknowledged by D'Alembert who stated that,
``The theory of fluids must necessarily be based upon experiment.''
For example the concept of ideal liquid that leads to motion with
no resistance, conflicts with the reality.
This discrepancy between theory and practice is called the
``D'Alembert paradox'' and serves to demonstrate the limitations
of theory alone in solving fluid problems.
As in thermodynamics, two different of school of thoughts were created:
the first believed that the solution will come from
theoretical aspect alone, and the second believed that solution
is the pure practical (experimental) aspect of fluid mechanics.
On the theoretical side, considerable contribution were made by Euler,
La Grange, Helmholtz, Kirchhoff, Rayleigh, Rankine, and Kelvin.
On the ``experimental'' side, mainly in pipes and open channels
area, were Brahms, Bossut, Chezy, Dubuat, Fabre, Coulomb, Dupuit,
d'Aubisson, Hagen, and Poisseuille.
In the middle of the nineteen century, first Navier in the molecular
level and later Stokes from continuous point of view succeeded
in creating governing equations for real fluid motion.
Thus, creating a matching between the two school of thoughts:
experimental and theoretical.
But, as in thermodynamics, people cannot relinquish control.
As results it created today ``strange'' names: Hydrodynamics,
Hydraulics, Gas Dynamics, and Aeronautics.
The NavierStokes equations, which describes the flow
(or even Euler equations), were considered unsolvable during the
mid nineteen century because of the high complexity.
This problem led to two consequences.
Theoreticians tried to simplify the equations and arrive at
approximated solutions representing specific cases.
Examples of such work are Hermann von Helmholtz's concept
of vortexes (1858), Lanchester's concept of circulatory
flow (1894), and the Kutta–Joukowski circulation theory
of lift (1906).
The experimentalists, at the same time proposed many correlations
to many fluid mechanics problems, for example, resistance by Darcy, Weisbach,
Fanning, Ganguillet, and Manning.
The obvious happened without theoretical guidance, the empirical
formulas generated by fitting curves to experimental data
(even sometime merely presenting the results in tabular form)
resulting in formulas that the relationship between the physics
and properties made very little sense.
At the end of the twenty century, the demand for vigorous scientific
knowledge that can be applied to various liquids as opposed to formula
for every fluid was created by the expansion of many industries.
This demand coupled with new several novel concepts like
the theoretical and experimental researches of Reynolds,
the development of dimensional analysis by Rayleigh,
and Froude's idea of the use of models change the
science of the fluid mechanics.
Perhaps the most radical concept that effects
the fluid mechanics is of Prandtl's idea of boundary layer
which is a combination of the modeling and dimensional analysis
that leads to modern fluid mechanics.
Therefore, many call Prandtl as the father of modern fluid mechanics.
This concept leads to mathematical basis for many approximations.
Thus, Prandtl and his students Blasius, von Karman, Meyer, and Blasius
and several other individuals as Nikuradse, Rose, Taylor, Bhuckingham,
Stanton, and many others, transformed the fluid mechanics to today modern science.
While the understanding of the fundamentals did not change much,
after World War Two, the way how it was calculated changed.
The introduction of the computers during the 60s and much more
powerful personal computer has changed the field.
There are many open source programs that can analyze many fluid mechanics situations.
Today many problems can be analyzed by using the numerical tools and provide reasonable results.
These programs in many cases can capture all the appropriate parameters
and adequately provide a reasonable description of the physics.
However, there are many other cases that numerical analysis cannot
provide any meaningful result (trends).
For example, no weather prediction program can produce good engineering quality results (where
the snow will fall within 50 kilometers accuracy.
Building a car with this accuracy is a disaster).
In the best scenario, these programs are as good as the input provided.
Thus, assuming turbulent flow for still flow simply provides
erroneous results (see for example, EKK, Inc).
1.3 Kinds of Fluids
Some differentiate fluid from solid by the reaction to shear stress.
The fluid continuously and permanently deformed under shear stress while the solid exhibits a
finite deformation which does not change with time.
It is also said that fluid cannot return to their original state after the deformation.
This differentiation leads to three groups of materials: solids and liquids and all
material between them.
This test creates a new material group that shows dual behaviors;
under certain limits; it behaves like solid and under others
it behaves like fluid (see Figure 1.1).
The study of this kind of material called rheology and it will (almost) not be discussed in this book.
It is evident from this discussion that when a fluid is at rest, no shear stress is applied.
The fluid is mainly divided into two categories: liquids and gases.
The main difference between the liquids and gases state is that
gas will occupy the whole volume while liquids has an almost fix volume.
This difference can be, for most practical purposes considered,
sharp even though in reality this difference isn't sharp.
The difference between a gas phase to a liquid phase above the critical point are practically minor.
But below the critical point, the change of water pressure by 1000% only change the
volume by less than 1 percent.
For example, a change in the volume by more 5% will required
tens of thousands percent change of the pressure.
So, if the change of pressure is significantly less than that,
Hence, the pressure will not affect the volume.
In gaseous phase, any change in pressure directly affects the volume.
The gas fills the volume and liquid cannot.
Gas has no free interface/surface (since it does fill the entire volume).
There are several quantities that have to be addressed in this discussion.
The first is force which was reviewed in physics.
The unit used to measure is [$N$].
It must be remember that force is a vector, e.g it has a direction.
The second quantity discussed here is the area.
This quantity was discussed in physics class but here it has an additional meaning,
and it is referred to the direction of the area.
The direction of area is perpendicular to the area.
The area is measured in [$m^2$].
Area of three–dimensional object has no single direction.
Thus, these kinds of areas should be addressed infinitesimally and locally.
The traditional quantity, which is force per area has a new meaning.
This is a result of division of a vector by a vector and it is referred to as tensor.
In this book, the emphasis is on the physics, so at this stage the tensor will have to be broken
into its components.
Later, the discussion on the mathematical meaning is presented (later version).
For the discussion here, the pressure has three components,
one in the area direction and two perpendicular to the area.
The pressure component in the area direction is called pressure
(great way to confuse, isn't it?).
The other two components are referred as the shear stresses.
The units used for the pressure components is [$N/m^2$].
Fig. 1.2 Density as a function of the size of sample.}
The density is a property which requires that liquid to be continuous.
The density can be changed and it is a function of time and space
(location) but must have a continues property.
It doesn't mean that a sharp and abrupt change in the density cannot occur.
It referred to the fact that density is independent of the sampling size.
Figure 1.2 shows the density as a function of the sample size.
After certain sample size, the density remains constant.
Thus, the density is defined as
\begin{align}
\rho = \lim_{\Delta V \longrightarrow \varepsilon}
\dfrac{\Delta m} {\Delta V}
\label{intro:eq:rho}
\end{align}
It must be noted that $\varepsilon$ is chosen so that the continuous
assumption is not broken, that is, it did not reach/reduced to
the size where the atoms or molecular statistical calculations
are significant (see Figure 1.2 for point
where the green lines converge to constant density).
When this assumption is broken, then, the principles of
statistical mechanics must be utilized.
1.4 Shear Stress
Fig. 1.3 Schematics to describe the shear stress in fluid mechanics.
The shear stress is part of the pressure tensor.
However, here, and many parts of the book, it will be treated as a separate issue.
In solid mechanics, the shear stress is considered as the ratio of the force acting on area
in the direction of the forces perpendicular to area.
Different from solid, fluid cannot pull directly but through a solid surface.
Consider liquid that undergoes a shear stress between a short
distance of two plates as shown in Figure 1.3.
The upper plate velocity generally will be
\begin{align}
U = f ( A, F, h)
\label{intro:eq:uPlate}
\end{align}
Where $A$ is the area, the $F$ denotes the force, $h$ is the distance between the plates.
From solid mechanics study, it was shown that when the force per
area increases, the velocity of the plate increases also.
Experiments show that the increase of height will increase the velocity up to a certain range.
Consider moving the plate with a zero lubricant ($h \sim 0$) (results in large force) or a large amount
of lubricant (smaller force).
In this discussion, the aim is to develop differential equation,
thus the small distance analysis is applicable.
For cases where the dependency is linear, the following can be written
\begin{align}
U \propto \dfrac{h\,F}{A}
\label{intro:eq:Upropto}
\end{align}
Equations qref{intro:eq:Upropto} can be rearranged to be
\begin{align}
\dfrac{U}{h} \propto \dfrac{F}{A}
\label{intro:eq:UproptoA}
\end{align}
Shear stress was defined as
\begin{align}
\tau_{xy} = \dfrac {F}{A}
\label{intro:eq:tauDef}
\end{align}
The index $x$ represent the ``direction of the shear stress while
the $y$ represent the direction of the area(perpendicular to the area).
From equations qref{intro:eq:UproptoA} and
qref{intro:eq:tauDef} it follows that
ratio of the velocity to height is proportional to shear stress.
Hence, applying the coefficient to obtain a new equality as
\begin{align}
\tau_{xy} = \mu \,\dfrac{U}{h}
\label{intro:eq:shearS}
\end{align}
Where $\mu$ is called the absolute viscosity or dynamic viscosity
which will be discussed later in this chapter in a great length.
Fig. 1.4 The deformation of fluid due to shear stress as progression of time.
In steady state, the distance the upper plate moves after
small amount of time, $\delta t$ is
\begin{align}
dll = U \, \delta t
\label{intro:eq:delta_t}
\end{align}
From Figure 1.4 it can be noticed that
for a small angle, $\delta\beta \cong \sin\beta$, the regular approximation provides
\begin{align}
dll = U \, \delta t =
\overbrace{h \, \delta\beta}^{geometry}
\label{intro:eq:dUdx}
\end{align}
From equation qref{intro:eq:dUdx} it follows that
\begin{align}
U = h\, \dfrac{\delta\beta}{\delta t}
\label{intro:eq:dbdt}
\end{align}
Combining equation qref{intro:eq:dbdt} with equation
qref{intro:eq:shearS} yields
\begin{align}
\tau_{xy} = \mu \, \dfrac{\delta \beta}{\delta t}
\label{intro:eq:tau_dbdt}
\end{align}
If the velocity profile is linear between the plate (it will be
shown later that it is consistent with derivations of velocity),
then it can be written for small a angel that
\begin{align}
\dfrac{\delta \beta}{\delta t} =
\dfrac{dU}{dy}
\label{intro:eq:velocityTime}
\end{align}
Materials which obey equation qref{intro:eq:tau_dbdt} referred to
as Newtonian fluid.
For this kind of substance
\begin{align}
\tau_{xy} = \mu\,\dfrac{dU}{dy}
\label{intro:eq:tau_xy}
\end{align}
Newtonian fluids are fluids which the ratio is constant.
Many fluids fall into this category such as air, water etc.
This approximation is appropriate for many other fluids but
only within some ranges.
Equation qref{intro:eq:dbdt} can be interpreted as momentum
in the $x$ direction transferred into the $y$ direction.
Thus, the viscosity is the resistance to the flow (flux) or the movement.
The property of viscosity, which is exhibited by all
fluids, is due to the existence of cohesion and
interaction between fluid molecules.
These cohesion and interactions hamper the flux in y–direction.
Some referred to shear stress as viscous flux of
x–momentum in the y–direction.
The units of shear stress are the same as flux per time
as following
\[
\dfrac{F}{A} \,\left[ \dfrac {kg\, m}{sec^2}\; \dfrac{1}{m^2} \right] =
\dfrac{\dot{m}\,U} {A} \left[ \dfrac{kg}{sec}\;
\dfrac{m}{sec} \,\, \dfrac{1}{m^2} \right]
\]
Thus, the notation of $\tau_{xy}$ is easier to understand and
visualize.
In fact, this interpretation is more suitable to explain the
molecular mechanism of the viscosity.
The units of absolute viscosity are [$N\,sec/m^2$].
A space of 1 [cm] width between two large plane surfaces is filled
with glycerin.
Calculate the force that is required to drag a very thin plate of
1 [$m^2$] at a speed of 0.5 m/sec.
It can be assumed that the plates remains in equidistant
from each other and steady state is achieved instantly.
Solution
Assuming Newtonian flow, the following can be written (see equation
qref{intro:eq:shearS})
\[
F = \dfrac{A\,\mu U}{h} \sim
\dfrac{1\times 1.069 \times 0.5}{0.01} =
53.45 [N]
\]
Castor oil at $25^{\circ}C$ fills the space between two concentric
cylinders of 0.2[m] and 0.1[m] diameters with height of 0.1 [m].
Calculate the torque required to rotate the inner
cylinder at 12 rpm, when the outer cylinder remains stationary.
Assume steady state conditions.
Solution
The velocity is
\[
U = r\,\dot{\theta} = 2\,\pi\,r_i \,\mbox{rps}=
2\times \pi\times 0.1 \times
\overbrace{12 / 60}^{rps} = 0.4\,\pi\,r_i
\]
Where $rps$ is revolution per second.
The same way as in Example 1.1, the
moment can be calculated as the force times the distance as
\begin{align*}
M = F\,ll = \dfrac{\overbrace{ll}^{r_i}\,
\overbrace{A}^{2\,\pi\,r_i\,h} \,\mu U}{r_or_i}
\end{align*}
In this case ${r_or_i} = h$ thus,
\begin{align*}
M = \dfrac{2\,\pi^2\,\overbrace{{0.1}^{3}}^{r_i}\, \cancel{h} \,
\overbrace{0.986}^\mu \, 0.4 }
{\cancel{h} } \sim .0078 [N\,m]
\end{align*}
1.5 Viscosity
1.5.1 General Discussion
Fig. 1.5 The different of power fluids families.
Viscosity varies widely with temperature.
However, temperature variation has an opposite effect
on the viscosities of liquids and gases.
The difference is due to their fundamentally different mechanism
creating viscosity characteristics.
In gases, molecules are sparse and cohesion is negligible,
while in the liquids, the molecules are more compact and cohesion is more dominate.
Thus, in gases, the exchange of momentum between layers brought
as a result of molecular movement normal to the general direction
of flow, and it resists the flow.
This molecular activity is known to increase with temperature,
thus, the viscosity of gases will increase with temperature.
This reasoning is a result of the considerations of the kinetic theory.
This theory indicates that gas viscosities vary directly with
the square root of temperature.
In liquids, the momentum exchange due to molecular movement
is small compared to the cohesive forces between the molecules.
Thus, the viscosity is primarily dependent on the magnitude of these cohesive forces.
Since these forces decrease rapidly with increases of temperature, liquid viscosities
decrease as temperature increases.
Figure 1.6 Nitrogen (left) and Argon (right) viscosity as a function of the temperature and
pressure after Lemmon and Jacobsen.
Figure 1.6(a) demonstrates that
viscosity increases slightly with pressure, but this
variation is negligible for most engineering problems.
Well above the critical point, both phases are only a function of the temperature.
On the liquid side below the critical point, the pressure has minor effect on the viscosity.
It must be stress that the viscosity in the dome is meaningless.
There is no such a thing of viscosity at 30% liquid.
It simply depends on the structure of the flow as will be discussed
in the chapter on multi phase flow.
The lines in the above diagrams are only to show constant pressure lines.
Oils have the greatest increase of viscosity with pressure
which is a good thing for many engineering purposes.
1.5.2 Non–Newtonian Fluids
Fig. 1.7 The shear stress as a function of the shear rate.
In equation qref{intro:eq:tauDef}, the relationship between
the velocity and the shear stress was assumed to be linear.
Not all the materials obey this relationship.
There is a large class of materials which shows a nonlinear
relationship with velocity for any shear stress.
This class of materials can be approximated by a single polynomial
term that is $a = b x^{n}$.
From the physical point of view, the coefficient depends on
the velocity gradient.
This relationship is referred to as power relationship
and it can be written as
\begin{align}
\tau = \overbrace{K\, \left( \dfrac{d U}{dx} \right)^{n1}}^{viscosity}
\left( \dfrac{d U}{dx} \right)
\label{intro:eq:powerFluids}
\end{align}
The new coefficients ($n,K$) in equation qref{intro:eq:powerFluids} are constant.
When $n=1$ equation represent Newtonian fluid and $K$ becomes the familiar $\mu$.
The viscosity coefficient is always positive.
When $n$, is above one, the liquid is dilettante.
When $n$ is below one, the fluid is pseudoplastic.
The liquids which satisfy equation qref{intro:eq:powerFluids}
are referred to as purely viscous fluids.
Many fluids satisfy the above equation.
Fluids that show increase in the viscosity (with increase of the shear) referred
to as thixotropic and those that show decrease
are called rheopectic fluids (see Figure 1.5).
Materials which behave up to a certain shear stress as a solid and
above it as a liquid are referred as Bingham liquids.
In the simple case, the ``liquid side'' is
like Newtonian fluid for large shear stress.
The general relationship for simple Bingham flow is
\begin{align}
\tau_{xy} =  \mu \dfrac{}{} \pm \tau_0 \qquad
if \left \tau_{yx} \right > \tau_{0}
\label{intro:eq:binghamA}
\end{align}
\begin{align}
\dfrac{dU_x}{dy} = 0 \qquad
if \left \tau_{yx} \right < \tau_{0}
\label{intro:eq:binghamB}
\end{align}
There are materials that simple Bingham model does not provide
adequate explanation and a more sophisticate model is required.
The Newtonian part of the model has to be replaced by power liquid.
For example, according to Ferraris at el
concrete behaves as shown in Figure 1.7.
However, for most practical purposes, this kind of figures
isn't used in regular engineering practice.
1.5.3 Kinematic Viscosity
Fig. 1.8 Air viscosity as a function of the temperature.}
The kinematic viscosity is another way to look at the viscosity.
The reason for this new definition is that some experimental data are given in this form.
These results also explained better using the new definition.
The kinematic viscosity embraces both the viscosity and density
properties of a fluid.
The above equation shows that the dimensions
of $\nu$ to be square meter per second, [$m^2/sec]$, which
are acceleration units (a combination of kinematic terms).
This fact explains the name ``kinematic'' viscosity.
The kinematic viscosity is defined as
\begin{align}
\nu = \dfrac{\mu}{\rho}
\label{intro:eq:nu}
\end{align}
The gas density decreases with the temperature.
However, The increase of the absolute viscosity with the temperature
is enough to overcome the increase of density and thus,
the kinematic viscosity also increase with the temperature for many materials.
Fig. 1.9 Water viscosity as a function temperature.}
1.5.4 Estimation of The Viscosity
The absolute viscosity of many fluids relatively doesn't change with the
pressure but very sensitive to temperature.
For isothermal flow, the viscosity can be considered constant in many cases.
The variations of air and water as a function of the temperature
at atmospheric pressure are plotted in Figures 1.8
and 1.9.
Some common materials (pure and mixture) have expressions that provide an estimate.
For many gases, Sutherland's equation is used and according to the literature, provides reasonable
of $40^{\circ}C$ to $1600^{\circ}C$.
\begin{align}
\mu = \mu_0 \; \dfrac{0.555\;T_{i0} + Suth }
{0.555\;T_{in} + Suth } \left(\dfrac{T}{T_0}\right)^{\dfrac{3}{2}}
\label{intro:eq:sutherland}
\end{align}
Where
$\mu$  viscosity at input temperature, T 
$\mu_0$  reference viscosity at reference temperature, $T_{i0}$ 
$T_{in}$  input temperature in degrees Kelvin 
$T_{i0}$  reference temperature in degrees Kelvin 
$Suth$  Suth is Sutherland's constant and it is presented in the Table 1.1 
Calculate the viscosity of air at 800K based on Sutherland's equation.
Use the data provide in Table 1.1.
Solution
Applying the constants from Suthelnd's table provides
\begin{multline*}
\mu = 0.00001827 \times
\dfrac{ 0.555\times524.07+120}{0.555\times800+120}
\times \left( \dfrac{800}{524.07}\right)^{\dfrac{3}{2}} \
\sim 2.51\,{10}^{5} \left[\dfrac{N\, sec}{m^2}\right]
nd{multline*}
The observed viscosity is about $\sim 3.7{10}^{5}\left[\dfrac{N\, sec}{m^2}\right]$.
Table 1.2 Viscosity of selected gases.
Substance 
Chemical formula 
Temperature, $T\,[^{\circ}C]$ 
Viscosity, $\left[\dfrac{N\, sec}{m^2} \right]$ 

$iC_4\,H_{10}$ 
23 
0.0000076 

$CH_4$ 
20 
0.0000109 
oxygen 
$O_2$ 
20 
0.0000203 
Mercury vapor 
$Hg$ 
380 
0.0000654 
Table 1.3 Viscosity of selected liquids.
Substance 
Chemical formula 
Temperature, $T\,[^{\circ}C]$ 
Viscosity, $\left[\dfrac{N\, sec}{m^2} \right]$ 
 $(C_2H_5)O$ 
20

0.000245

 $C_6H_6$ 
20

0.000647

 $Br_2$  26  0.000946 
 $C_2H_5OH$  20  0.001194 
 $Hg$  25  0.001547 
 $H_2SO_4$  25  0.01915 
Olive Oil  $ $  25  0.084 
Castor Oil  $ $  25  0.986 
Clucuse  $ $  25  520 
Corn Oil  $ $  20  0.072 
SAE 30  $ $    0.150.200 
SAE 50  $ $  $\sim25^{\circ}C$  0.54 
SAE 70  $ $  $\sim25^{\circ}C$  1.6 
Ketchup  $ $  $\sim20^{\circ}C$  0,05 
Ketchup  $ $  $\sim25^{\circ}C$  0,098 
Benzene  $ $  $\sim20^{\circ}C$  0.000652 
Firm glass  $ $    $\sim 1\times10^7$ 
Glycerol  $ $  20  1.069 
Fig. 1.10 Liquid metals viscosity as a function of the temperature.
Liquid Metals
Liquid metal can be considered as a Newtonian fluid for many applications.
Furthermore, many aluminum alloys are behaving as a Newtonian
liquid until the first solidification appears (assuming
steady state thermodynamics properties).
Even when there is a solidification (mushy zone), the metal
behavior can be estimated as a Newtonian material (further reading
can be done in this author's book ``Fundamentals of Die Casting Design'').
Figure 1.10 exhibits several liquid
metals (from The Reactor Handbook, Vol. Atomic Energy Commission
AECD3646 U.S. Government Printing Office, Washington D.C. May 1995 p. 258.)
The General Viscosity Graphs
In case ``ordinary'' fluids where information is limit,
Hougen et al suggested to use graph similar to compressibility chart.
In this graph, if one point is well documented, other points can be estimated.
Furthermore, this graph also shows the trends.
In Figure 1.11 the relative viscosity
$\mu_r = \mu / \mu_c$ is plotted as a function of relative temperature, $T_r$.
$\mu_c$ is the viscosity at critical condition and
$\mu$ is the viscosity at any given condition.
The lines of constant relative pressure, $P_r = P/P_c$ are drawn.
The lower pressure is, for practical purpose, $\sim 1[bar]$.
Table 1.3 Viscosity of selected liquids.
Chemical component 
Molecular Weight 
$T_c$ [K] 
$P_c$ [Bar] 
$\mu_c$ $ \left[\dfrac{N\,sec}{m^2}\right]$ 
$H_2$  2.016  33.3  12.9696  3.47 
$He$  4.003  5.26  2.289945  2.54 
$Ne$  20.183  44.5  27.256425  15.6 
$Ar$  39.944  151  48.636  26.4 
$Xe$  131.3  289.8  58.7685  49. 
Air ``mixed''  28.97  132  36.8823  19.3 
$CO_2$  44.01  304.2  73.865925  19.0 
$O_2$  32.00  154.4  50.358525  18.0 
$C_2H_6$  30.07  305.4  48.83865  21.0 
$CH_4$  16.04  190.7  46.40685  15.9 
Water  18.01528  647.096 K  22.064 [MPa]  $\sim$ 11. 
The critical pressure can be evaluated in the following three ways.
The simplest way is by obtaining the data from Table
1.4 or similar information.
The second way, if the information is available and is close enough to the critical point,
then the critical viscosity is obtained as
\begin{align}
\mu_c = \dfrac{\overbrace{\mu}^{given} }
{\underbrace{\mu_r}_{ Figure 1.11 } }
\label{intro:eq:critical}
\end{align}
The third way, when none is available, is by utilizing the following approximation
\begin{align}
\mu_c = \sqrt{M\,T_c} \tilde{v_c}^{2/3}
\label{intro:eq:muC}
\end{align}
Where $\tilde{v_c}$ is the critical molecular volume and $M$ is molecular weight.
Or
\begin{align}
\mu_c = \sqrt{M} {P_c}^{2/3} {T_c}^{1/6}
\label{intro:eq:muC2}
\end{align}
Calculate the reduced pressure and the reduced temperature
and from the Figure 1.11 obtain the reduced viscosity.
Estimate the viscosity of oxygen, $O_2$ at $100^{\circ}C$ and 20[Bar].
Solution
$P_c = 50.35[Bar]\,$, $\,T_c = 154.4$ and therefor
$\mu_c = 18 \left[ \dfrac{N\,sec}{m^2}\right]$ The value of the reduced temperature is
\[
T_r \sim \dfrac{373.15}{154.4} \sim 2.41
\]
The value of the reduced pressure is
\[
P_r \sim \dfrac{20}{50.35} \sim 0.4
\]
From Figure 1.11 it can be obtained $\mu_r\sim 1.2$
and the predicted viscosity is
\[
\mu = \mu_c \, \overbrace{\left( \dfrac{\mu}{\mu_c}\right)}
^{Table } = 18 \times 1.2 = 21.6[N sec/m^2]
\]
Fig. 1.11 Reduced viscosity as function of the reduced temperature.
Fig. 1.12 Reduced viscosity as function of the reduced temperature.
Viscosity of Mixtures
In general the viscosity of liquid mixture has to be evaluated experimentally.
Even for homogeneous mixture, there isn't silver bullet to estimate the viscosity.
In this book, only the mixture of low density gases is discussed for analytical expression.
For most cases, the following Wilke's correlation for gas at low
density provides a result in a reasonable range.
\begin{align}
\mu_{mix} =
\sum_{i=1}^{n} \dfrac{x_i\,\mu_i}{\sum_{j=1}^{n} x_i\,\Phi_{ij}}
\label{intro:eq:muMix}
\end{align}
where $\Phi_ij$ is defined as
\begin{align}
\Phi_{ij} = \dfrac{1}{\sqrt{8}}
\sqrt{ 1+ \dfrac{M_i}{M_j}} \left(
1+ \sqrt{\dfrac{\mu_i}{\mu_j}}
\sqrt[4]{\dfrac{M_j}{M_i}}
\right)^2
\label{intro:eq:Phiij}
\end{align}
Here, $n$ is the number of the chemical components in the mixture.
$x_i$ is the mole fraction of component $i$, and
$\mu_i$ is the viscosity of component $i$.
The subscript $i$ should be used for the $j$ index.
The dimensionless parameter $\Phi_{ij}$ is equal to one when $i = j$.
The mixture viscosity is highly nonlinear function of the fractions of the components.
Calculate the viscosity of a mixture (air)
made of 20% oxygen, $O_2$ and 80% nitrogen $N_2$
for the temperature of $20^{\circ}C$.
Solution
The following table summarizes the known details
Table summary 1.
Component  Molecular Weight, $M$  Fraction, $x$  Viscosity, $\mu$ 
$O_2$  32.  0.2  0.0000203 
$N_2$  28.  0.8  0.00001754 
Table summary 2.
i  j  $M_i/M_j$  $\mu_i/\mu_j$  $\Phi_{ij}$ 
1  1  1.0  1.0  1.0 
1  2  1.143  1.157  1.0024 
2  1  0.875  .86  0.996 
2  2  1.0  1.0  1. 
\begin{multline*}
\mu_{mix} \sim \dfrac{0.2\times 0.0000203}{0.2\times1.0 +
0.8\times 1.0024} + \\
\dfrac{0.8\times 0.00001754}{0.2\times0.996 +
0.8\times 1.0} \sim 0.0000181 \left[\dfrac{N\,sec}{m^2}\right]
nd{multline*}
The observed value is $\sim0.0000182 \left[\dfrac{N\,sec}{m^2}\right]$.
In very low pressure, in theory, the viscosity is only a function of
the temperature with a ``simple'' molecular structure.
For gases with very long molecular structure or complexity structure
these formulas cannot be applied.
For some mixtures of two liquids it was observed that at a low
shear stress, the viscosity is dominated by a liquid with
high viscosity and at high shear stress to be dominated by a
liquid with the low viscosity liquid.
The higher viscosity is more dominate at low shear stress.
Reiner and Phillippoff suggested the following formula
\begin{align}
\dfrac{dU_x}{dy} = \left(
\dfrac{1}
{\mu_{\infty} + \dfrac{\mu_0  \mu_{\infty}}
{1+ \left( \dfrac{\tau_{xy} }{\tau_s} \right)^2 } }
\right)\; \tau_{xy}
\label{intro:eq:reinerPhillippoff}
\end{align}
Where the term $\mu_{\infty}$ is the experimental value at high shear stress.
The term $\mu_{0}$ is the experimental viscosity at shear stress approaching zero.
The term $\tau_s$ is the characteristic shear stress of the mixture.
An example for values for this formula, for Molten Sulfur
at temperature $120^{\circ}C$ are
$\mu_{\infty} = 0.0215 \left({N\,sec}/{m^2}\right) $,
$\mu_{0} = 0.00105 \left({N\,sec}/{m^2}\right)$,
and $\tau_s = 0.0000073 \left({kN}/{m^2}\right)$.
This equation qref{intro:eq:reinerPhillippoff} provides reasonable
value only up to $\tau = 0.001\left({kN}/{m^2}\right)$.
Figure 1.12 can be used for a crude estimate of dense gases mixture.
To estimate the viscosity of the mixture with $n$ component
Hougen and Watson's method for pseudocritial properties is adapted.
In this method the following are defined as mixed critical pressure as
\begin{align}
{P_c}_{mix} = \sum_{i=1}^{n} \, x_i \,{P_c}_i
\label{intro:eq:mixPc}
\end{align}
the mixed critical temperature is
\begin{align}
{T_c}_{mix} = \sum_{i=1}^{n} \,x_i\, {T_c}_i
\label{intro:eq:mixTc}
\end{align}
and the mixed critical viscosity is
\begin{align}
{\mu_c}_{mix} = \sum_{i=1}^{n} \,x_i\, {\mu_c}_i
\label{intro:eq:mixMUc}
\end{align}
Fig. 1.13 Concentrating cylinders with the rotating inner cylinder.
of 0.101 [m] radius and the cylinders length is 0.2 [m].
It is given that a moment of 1 [$N\times m$] is required to maintain an angular velocity
of 31.4 revolution per second (these number represent only academic question not real value
of actual liquid).
Estimate the liquid viscosity used between the cylinders.}
Solution
The moment or the torque is transmitted through the liquid to the outer cylinder.
Control volume around the inner cylinder shows that moment is a function of the area and shear stress.
The shear stress calculations can be estimated as a linear between the two concentric cylinders.
The velocity at the inner cylinders surface is
\begin{align}
\label{concentricCylinders:Ui}
U_i = r\,\omega = 0.1\times 31.4[rad/second] = 3.14 [m/s]
nd{align}
The velocity at the outer cylinder surface is zero.
The velocity gradient may be assumed to be linear, hence,
\begin{align}
\label{concentricCylinders:dUdr}
\dfrac{dU}{dr} \cong \dfrac{0.1 0}{0.101  0.1} = 100 sec^{1}
nd{align}
The used moment is
\begin{align}
\label{concentricCylinders:M1}
M = \overbrace{2\,\pi\,r_i\,h}^{A} \overbrace{\mu \dfrac{dU}{dr}}^{\tau} \,\overbrace{r_i}^{ll}
nd{align}
or the viscosity is
\begin{align}
\label{concentricCylinders:M}
\mu = \dfrac{M}{ {2\,\pi\,{r_i}^2\,h} { \dfrac{dU}{dr}} } =
\dfrac{1}{2\times\pi\times{0.1}^2 \times 0.2 \times 100} =
nd{align}
A square block weighing 1.0 [kN] with a side surfaces area of 0.1 [$m^2$] slides down an incline
surface with an angle of 20\^{0}C.
The surface is covered with oil film.
The oil creates a distance between the block and the inclined surface of $1\times10^{6}[m]$.
What is the speed of the block at steady state?
Assuming a linear velocity profile in the oil and that the whole oil is under steady state.
The viscosity of the oil is $3 \times 10^{5} [m^2/sec]$.
Solution
The shear stress at the surface is estimated for steady state by
\begin{align}
\label{slidingBlock:shear}
\tau = \mu \dfrac{dU}{dx} = 3 \times 10^{5} \times \dfrac{U}{1\times10^{6}} = 30 \, U
nd{align}
The total fiction force is then
\begin{align}
\label{slidingBlock:frictionForce}
f = \tau\, A = 0.1 \times 30\,U = 3\,U
nd{align}
The gravity force that acting against the friction is equal to the friction hence
\begin{align}
\label{slidingBlock:solPre1}
F_g = f = 3\,U\Longrightarrow U = \dfrac{m\,g\,\sin\,20^{\circ}}{3}
nd{align}
Or the solution is
\begin{align}
\label{slidingBlock:solPre}
U = \dfrac{1\times 9.8\times\sin\,20^{\circ}}{3}
nd{align}
Fig. 1.14 Rotating disc in a steady state.
The edge effects can be neglected.
The gap is given and equal to $\delta$ and
the rotation speed is $\omega$.
The shear stress can be assumed to be linear.}
Solution
In this cases the shear stress is a function of the radius, $r$ and an expression has to be
developed.
Additionally, the differential area also increases and is a function of $r$.
The shear stress can be estimated as
\begin{align}
\label{discRotating:tau}
\tau \cong \mu \,\dfrac{U}{\delta} = \mu\,\dfrac{\omega \, r}{\delta}
nd{align}
This torque can be integrated for the entire area as
\begin{align}
\label{discRotating:F}
T = \int_0^R r\, \tau \,dA =
\int_0^R \overbrace{r}^{ll} \, \overbrace{\mu\, \dfrac{\omega \, r}{\delta}}^{\tau} \,
\overbrace{2\,\pi\,r\,dr}^{dA}
nd{align}
The results of the integration is
\begin{align}
\label{discRotating:I}
T = \dfrac{\pi\,\mu\,\omega\,R^4}{2\,\delta}
nd{align}
1.6 Fluid Properties
The fluids have many properties which are similar to solid.
A discussion of viscosity and surface tension should be part of this section
but because special importance these topics have separate sections.
The rest of the properties lumped into this section.
1.6.1 Fluid Density
\img{intro/drawing/waterDensityTemperaturePressure}
{Water Density}
{0.7}
{intro:fig:waterDensityTemperaturePressure}
{Water density as a function of temperature}
{Water density as a function of temperature for various pressure.
This figure illustrates the typical situations like the one that appear in Example 1.9}
The density is a property that is simple to analyzed and understand.
The density is related to the other state properties such temperature and pressure
through the equation of state or similar.
Examples to describe the usage of property are provided.
A steel tank filled with water undergoes heating from $10^{\circ}C$ to $50^{\circ}C$.
The initial pressure can be assumed to atmospheric.
Due to the change temperature the tank, (strong steel structure) undergoes linear expansion of
$8\times10^{6}$ {per} \^{0}C.
Calculate the pressure at the end of the process.
$E$ denotes the Young's modulus.
Assume that the Young modulus of the water is
State your assumptions.
Solution
The expansion of the steel tank will be due to two contributions: one due to the thermal expansion
and one due to the pressure increase in the tank.
For this example, it is assumed that the expansion due to pressure change is negligible.
The tank volume change under the assumptions state here but in the same time
the tank walls remain straight.
The new density is
\begin{align}
\label{rhoSwater:tankExp}
\rho_{2} = \dfrac{\rho_1}\,{\underbrace{\left( 1 + \alpha \, \Delta\,T \right)^3}
_{\text{thermal expansion}}}
\end{align}
The more accurate calculations require looking into the steam tables.
As estimated value of the density using Young's modulus and $V_2 \propto
\left(L_2\right)^3$.
\begin{align}
\label{rhoSwater:secondSide}
\rho_2 \propto \dfrac{1}{\left(L_2\right)^3} \Longrightarrow
\rho_2 \cong \dfrac{m}{\left(L_1 \left( 1  \dfrac{\Delta P}{E} \right) \right)^3 }
nd{align}
It can be noticed that $\rho_1 \cong m/{L_1}^3$ and thus
\begin{align}
\label{rhoSwater:pRho}
\dfrac{\rho_1}{\left( 1 + \alpha \Delta\,T \right)^3} =
\dfrac{\rho_1}{ \left( 1  \dfrac{\Delta P}{E} \right)^3 }
nd{align}
The change is then
\begin{align}
\label{rhoSwater:tankTp}
1 + \alpha \Delta\,T = 1  \dfrac{\Delta P}{E}
nd{align}
Thus the final pressure is
\begin{align}
\label{rhoSwater:tankTp1}
P_2 = P_1  E\,\alpha\, \Delta\,T
nd{align}
In this case, what happen when the value of $P_1  E\,\alpha\, \Delta\,T$ becomes negative or very
very small? The basic assumption falls and the water evaporates.
If the expansion of the water is taken into account then the change (increase) of
water volume has to be taken into account.
The tank volume was calculated earlier and since the claim of ``strong'' steel the volume of the
tank is only effected by the temperature.
\begin{align}
\label{rhoSwater:tankT}
\left. \dfrac{V_{2} } {V_{1} } \right_{tank} = \left( 1+ \alpha\, \Delta T\right)^3
nd{align}
The volume of the water undergoes also a change and is a function of the temperature and pressure.
The water pressure at the end of the process is unknown but the volume is known.
Thus, the density at end is also known
\begin{align}
\label{rhoSwater:rhoTwo}
\rho_2 = \dfrac{m_w} {\left. T_2 \right_{tank}}
nd{align}
The pressure is a function volume and the temperature $P=P(v,T)$ thus
\begin{align}
\label{rhoSwater:PvTg}
dP = \overbrace{\left( \dfrac{\partial P }{ \partial v }\right)}^{\sim \beta_v} dv
+ \overbrace{\left( \dfrac{\partial P }{ \partial T }\right)}^{\sim E} dT
nd{align}
As approximation it can written as
\begin{align}
\label{rhoSwater:PvT}
\Delta P = \beta_v \, \Delta v + E \, \Delta T
nd{align}
Substituting the values results for
\begin{align}
\label{rhoSwater:PvTn}
\Delta P = \dfrac{0.0002} { \Delta \rho}
+ 2.15\times10^9 \, \Delta T
nd{align}
Notice that density change, $\Delta\rho< 0$.
Caution: advance matherial can be skipped
1.6.2 Bulk Modulus
Similar to solids (hook's law), liquids have a property that describes the volume
change as results of pressure change for constant temperature.
It can be noted that this property is not the result of the equation
of state but related to it.
Bulk modulus is usually obtained from experimental or theoretical or semi theoretical
(theory with experimental work) to fit energy–volume data.
Most (theoretical) studies are obtained by uniformly changing the unit cells in global energy
variations especially for isotropic systems ( where the molecules has a structure with cubic symmetries).
The bulk modulus is a measure of the energy can be stored in the liquid.
This coefficient is analogous to the coefficient of spring.
The reason that liquid has different coefficient is because it is three dimensional verse one
dimension that appear in regular spring.
The bulk modulus is defined as
\begin{align}
B_T =  v \, \left( \dfrac{\partial P}{\partial v} \right)_T
\label{intro:eq:bulkModulus_v}
\end{align}
Using the identity of $v=1/\rho$ transfers equation
qref{intro:eq:bulkModulus_v} into
\begin{align}
B_T = \rho \,\left( \dfrac{\partial P}{\partial \rho} \right)_T
\label{intro:eq:bulkModulus_r}
\end{align}
The bulk modulus for several selected liquids is presented in Table
1.5.
Table 1.5 The bulk modulus for selected material
Chemical component 
Bulk Modoulus $10^{9}\dfrac{N}{m}$ 
$T_c$ 
$P_c$ 
Acetic Acid  2.49  593K  57.8 [Bar] 
Acetone  0.80  508 K  48 [Bar] 
Benzene  1.10  562 K  4.74 [MPa] 
Carbon Tetrachloride  1.32  556.4 K  4.49 [MPa] 
Ethyl Alcohol  1.06  514 K  6.3 [Mpa] 
Gasoline  1.3  nf  nf 
Glycerol  4.034.52  850 K  7.5 [Bar] 
Mercury  26.228.5  1750 K  172.00 [MPa] 
Methyl Alcohol  0.97  Est 513  Est 78.5 [Bar] 
Nitrobenzene  2.20  nf  nf 
Olive Oil  1.60  nf  nf 
Paraffin Oil  1.62  nf  nf 
SAE 30 Oil  1.5  na  na 
Seawater  2.34  na  na 
Toluene  1.09  591.79 K  4.109 [MPa] 
Turpentine  1.28  na  na 
Water  2.152.174  647.096 K  22.064 [MPa] 
In the literature, additional expansions for similar parameters are defined.
The thermal expansion is defined as
\begin{align}
\beta_P = \dfrac{1 }{v} \, \left( \dfrac{\partial v}{\partial T} \right) _P
\label{intro:eq:betaP}
\end{align}
This parameter indicates the change of volume due to temperature
change when the pressure is constant.
Another definition is referred as coefficient of tension and it is defined as
\begin{align}
\beta_v = \dfrac{1 }{P} \,\left( \dfrac{\partial P}{\partial T} \right)_v
\label{eq:beta_v}
\end{align}
This parameter indicates the change of the pressure due to the
change of temperature (where $v=constant$).
These definitions are related to each other.
This relationship is obtained by the observation that the pressure as a function of
the temperature and specific volume as
\begin{align}
P = f (T,\,v)
\label{intro:eq:PfTv}
\end{align}
The full pressure derivative is
\begin{align}
dP = \left( \dfrac{\partial P}{\partial T} \right) _v dT
+ \left( \dfrac{\partial P}{\partial v} \right) _T dv
\label{intro:eq:dPfTv}
\end{align}
On constant pressure lines, $dP =0$, and therefore
equation qref{intro:eq:dPfTv} reduces
\begin{align}
0 = \left( \dfrac{\partial P}{\partial T} \right) _v dT
+ \left( \dfrac{\partial P}{\partial v} \right) _T dv
\label{intro:eq:dPfTv0}
\end{align}
From equation qref{intro:eq:dPfTv0} follows that
\begin{align}
\left. \dfrac{dv}{dT} \right_{P=const}
= \, \dfrac{\left( \dfrac{\partial P}{\partial T} \right) _v}
{\left( \dfrac{\partial P}{\partial v} \right) _T}
\label{intro:eq:Pconsbeta}
\end{align}
Equation qref{intro:eq:Pconsbeta} indicates that relationship
for these three coefficients is
\begin{align}
\beta_{T} = \dfrac{\beta_v}{\beta_{P}}
\label{intro:eq:genealRbeta}
\end{align}
The last equation qref{intro:eq:genealRbeta} sometimes is used
in measurement of the bulk modulus.
The increase of the pressure increases the bulk modulus
due to the molecules increase of the rejecting forces
between each other when they are closer.
In contrast, the temperature increase results in reduction of the
bulk of modulus because the molecular are further away.
Calculate the modulus of liquid elasticity
that reduced 0.035 per cent of its volume
by applying a pressure of 5[Bar] in a s slow process.
Solution
Using the definition for the bulk modulus
\[
\beta_T =  v \, \dfrac{\partial P }{\partial v} \simeq
\dfrac{v} {\Delta v} \, \Delta P = \dfrac {5}{0.00035} \simeq 14285.714 [Bar]
\]
Calculate the pressure needed to apply on water to reduce its volume
by 1 per cent. Assume the temperature to be $20^{\circ}C$.
Solution
Using the definition for the bulk modulus
\[
\Delta P \sim \beta_T \dfrac{\Delta v}{v} \sim
2.15\,10^{9} ime 0.01 = 2.15\,10^{7} [N/m^2]
= 215 [Bar]
\]
Fig. 1.16 Two liquid layers under pressure.
Initially the pressure in the tank is $P_0$.
The liquids are compressed due to the pressure increases.
The new pressure is $P_1$.
The area of the tank is $A$ and liquid A height is $h_1$
and liquid B height is $h_2$.
Estimate the change of the heights of the liquids depicted in the Figure 1.16.
State your assumptions.}
Solution
The volume change in a liquid is
\begin{align}
\label{twoLiquid:1liquid}
B_T \cong \dfrac{\Delta P }{ \Delta V /V}
nd{align}
Hence the change for the any liquid is
\begin{align}
\label{twoLiquid:DeltaH}
\Delta h = \dfrac{\Delta P }{ A\, B_T/V} = \dfrac{h\, \Delta P}{B_T}
nd{align}
The total change when the hydrostatic pressure is ignored.
\begin{align}
\label{twoLiquid:totalH}
\Delta h_{1+2} = \Delta P \left( \dfrac{h_1 }{{B_T}_1} + \dfrac{h_2 }{{B_T}_2} \right)
nd{align}
In the Internet the following problem ( here with
$\LaTeX{}$
modification) was posted which related to Pushka equation.\
A cylindrical steel pressure vessel with volume 1.31 $m^3$ is to be tested.
The vessel is entirely filled with water, then a piston at one end of the cylinder
is pushed in until the pressure inside the vessel has increased by 1000 kPa.
Suddenly, a safety plug on the top bursts. How many liters of water come out?
Relevant equations and data suggested by the user were:
$B_T=0.2\times10^10N/m^2$, $P_1=P_0+\rho\,g\,h$, $P_1=B_T\Delta V/V$
with the suggested solution of
``I am assuming that I have to look for $\Delta V$ as that would be the water that comes
out causing the change in volume.''
\begin{align*}
\Delta V=\dfrac{V\,\Delta P}{B_T}
=1.31(1000)/(0.2\times10^{10})
\Delta V= 6.55*10^{7}
\end{align*}
Another user suggest that:\
We are supposed to use the bulk modulus from our textbook, and that one is
$0.2\times10^{10}$.
Anything else would give a wrong answer in the system. So with this bulk modulus, is 0.655L right?
In this post several assumptions were made.
What is a better way to solve this problem.
Solution
It is assumed that this process can be between two extremes: one isothermal and one isentropic.
The assumption of isentropic process is applicable after a shock wave that travel in the tank.
If the shock wave is ignored (too advance material for this book.)
the process is isentropic.
The process involve some thermodynamics identities to be connected.
Since the pressure is related or a function of density and temperature it follows that
\begin{align}
\label{physicsForum:PrhoT}
P = P \left( \rho, T \right)
nd{align}
Hence the full differential is
\begin{align}
\label{physicsForum:dP}
dP = \left.\dfrac{\partial P }{ \partial \rho} \right_T d\rho
+ \left.\dfrac{\partial P }{ \partial T} \right_{\rho} dT
nd{align}
Equation qref{physicsForum:dP} can be multiplied by $\rho/P$ to be
\begin{align}
\label{physicsForum:dPrhoP}
\dfrac{ \rho\,dP}{P} = \dfrac{1}{P}\,\left( \overbrace{\rho\left.\dfrac{\partial P }{ \partial \rho} \right_T}^{B_T} d\rho \right)
+ \rho\, \left( \overbrace{ \dfrac{1}{P} \left.\dfrac{\partial P }{ \partial T} \right_{\rho} }^{\beta_v} dT \right)
nd{align}
The definitions that were provided before can be used to write
\begin{align}
\label{physicsForum:dPrhoP1}
\dfrac{ \rho\,dP}{P} = \dfrac{1}{P}\, {B_T} \,d\rho + \rho\, \beta_v \, dT
nd{align}
The infinitesimal change of density will be then
\begin{align}
\label{physicsForum:drho1}
\dfrac{1}{P}\, {B_T} \,d\rho = \dfrac{ \rho\,dP}{P}  \rho\, \beta_v \, dT
nd{align}
or
\begin{align}
\label{physicsForum:drho}
d\rho = \dfrac{ \rho\,dP}{B_T}  \dfrac{\rho\,P\, \beta_v \, dT}{B_T}
nd{align}
Thus, the calculation that were provide on line need to have corrections by subtracting the second terms.
1.6.2.1 Bulk Modulus of Mixtures
In the discussion above it was assumed that the liquid is pure.
In this short section a discussion about the bulk modulus averaged is presented.
When more than one liquid are exposed to pressure the value of these two (or more liquids) can
have to be added in special way.
The definition of the bulk modulus is given by equation qref{intro:eq:bulkModulus_v} or
qref{intro:eq:bulkModulus_r} and can be written (where the partial derivative can looks as
delta $\Delta$ as
\begin{align}
\label{intro:eq:bulkModulusi}
\partial V = \dfrac{V\,\partial P }{B_T} \cong \dfrac{V\,\Delta P }{B_T}
\end{align}
The total change is compromised by the change of individual liquids or phases if two materials are present.
Even in some cases of emulsion (a suspension of small globules of one liquid in a second liquid
with which the first will not mix) the total change is the summation of the individuals change.
In case the total change isn't, in special mixture, another approach with taking into account the
energyvolume is needed.
Thus, the total change is
\begin{align}
\label{intro:eq:DeltaV}
\partial V = \partial V_1 + \partial V_2 + \cdots \partial V_i \cong
\Delta V_1 + \Delta V_2 + \cdots \Delta V_i
\end{align}
Substituting equation qref{intro:eq:bulkModulusi} into equation qref{intro:eq:DeltaV} results in
\begin{multline}
\label{intro:eq:DeltaVexpli}
\partial V = \dfrac{V_1\,\partial P }{{B_T}_1} + \dfrac{V_2\,\partial P }{{B_T}_2} + \cdots +
\dfrac{V_i\,\partial P }{{B_T}_i} \cong \\
\dfrac{V_1\,\Delta P }{{B_T}_1} + \dfrac{V_2\,\Delta P }{{B_T}_2} + \cdots + \dfrac{V_i\,\Delta P }{{B_T}_i}
\,\,
nd{multline}
Under the main assumption in this model the total volume is comprised of the individual volume hence,
\begin{align}
\label{intro:eq:vvi}
V = x_1 \, V + x_1 \,V + \cdots + x_i \, V
\end{align}
Where $x_1$, $x_2$ and $x_i$ are the fraction volume such as $x_i= V_i/V$.
Hence, using this identity and the fact that the pressure is change for all the phase uniformly
equation qref{intro:eq:vvi} can be written as
\begin{multline}
\label{intro:eq:vvi1}
\partial V = V \,\partial P \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots + \dfrac{x_i}{{B_T}_i}
\right) \cong \\
V\, \Delta P \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots +\dfrac{x_i}{{B_T}_i} \right)
nd{multline}
Rearranging equation qref{intro:eq:vvi1} yields
\begin{align}
\label{intro:eq:newBTstart}
v\, \dfrac{\partial P}{\partial v} \cong
v\, \dfrac{\Delta P}{\Delta v} =
\dfrac{1}{ \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots + \dfrac{x_i}{{B_T}_i} \right) }
\end{align}
Equation qref{intro:eq:newBTstart} suggested an averaged new bulk modulus
\begin{align}
\label{intro:eq:BTmixDef}
{B_T}_{mix} = \dfrac{1}{ \left( \dfrac{x_1}{{B_T}_1} + \dfrac{x_2}{{B_T}_2} + \cdots + \dfrac{x_i}{{B_T}_i} \right) }
\end{align}
In that case the equation for mixture can be written as
\begin{align}
\label{intro:eq:BTmix}
v\, \dfrac{\partial P}{\partial v} = {B_T}_{mix}
\end{align}
End Caution: advance matherial
1.6.2.2 When the Bulk Modulus is Important? and Hydraulics System
There are only several situations in which the bulk modulus is important.
These situations include hydraulic systems, deep ocean (on several occasions), geology system like the Earth, Cosmology.
The Pushka equation normally can address the situations in deep ocean and geological system.
This author is not aware of any special issues that involve in Cosmology as opposed to geological system.
The only issue that was not addressed is the effect on hydraulic systems.
The hydraulic system normally refers to systems in which a liquid is used to transmit forces (pressure)
for surface of moving object (normally piston) to another object.
In theoretical or hypothetical liquids the moving one object (surface) results in movement of the
other object under the condition that liquid volume is fix.
The movement of the responsive object is unpredictable when the liquid volume or
density is a function of the pressure (and temperature due to the friction).
In very rapid systems the temperature and pressure varies during the operation significantly.
In practical situations, the commercial hydraulic fluid can change due to friction by $50^{\circ}C$.
The bulk modulus or the volume for the hydraulic oil changes by more 60%.
The change of the bulk modulus by this amount can change the response time significantly.
Hence the analysis has to take into account the above effects.
1.7 Surface Tension
Fig. 1.17 Surface tension control volume analysis describing principles radii.
The surface tension manifested itself by a rise or depression of the liquid at the free surface edge.
Surface tension is also responsible for the creation of the drops and bubbles.
It also responsible for the breakage of a liquid jet into other medium/phase to many drops (atomization).
The surface tension is force per length and is measured by [N/m] and is acting to stretch the surface.
Surface tension results from a sharp change in the density between two adjoined phases or materials.
There is a common misconception for the source of the surface tension.
In many (physics, surface tension, and fluid mechanics) books
explained that the surface tension is a result from unbalanced molecular cohesive forces.
This explanation is wrong since it is in conflict with Newton's second law (see Example
1.14).
This erroneous explanation can be traced to Adam's book but earlier source may be
found.
Fig. 1.18 Surface tension erroneous explanation.
``The cohesive forces between molecules down into a liquid are shared with all neighboring atoms.
Those on the surface have no neighboring atoms above, and exhibit stronger attractive forces upon their
nearest neighbors on the surface. This enhancement of the intermolecular attractive forces at the surface
is called.''
Explain the fundamental error of this explanation (see Figure 1.18).
Solution
It amazing that this erroneous explanation is prevalent in physics and chemistry (check the standard books
for general chemistry in any college).
In fact, even in Wikipedia this erroneous explanation appears.
The explanation based on the unbalance of the top layer of molecules.
Consider the control volume shown in Figure 1.18.
The control volume is made from a molecule thickness and larger width.
If this explanation was to be believed it must obey Newton's Laws.
However, as it will be shown, this explanation violates Newton's Laws and hence it is not valid.
The entire liquid domain is in a static equilibrium and hence every element is static equilibrium
including the control volume.
The pulling on the left of control volume is balanced with forcing that pulling to the right.
However, the control volume is pulled by the molecules below while there counter force to balance it.
There are no molecules about to balance it.
If this explanation was correct the top layer (control volume) was supposed to be balanced.
According to Newton second Law this layer should move down and the liquid cannot be at rest ever.
Obviously, the liquid is at rest and this explanation violates Newton second law.
In the Dimensional Analysis Chapter, provide another reason why this explanation violate all what is known
experimentally about the surface tension.
The relationship between the surface tension and the pressure on
the two sides of the surface is based on geometry.
Consider a small element of surface.
The pressure on one side is $P_i$ and the pressure on the other side is $P_o$.
When the surface tension is constant,
the horizontal forces cancel each other because symmetry.
In the vertical direction, the surface tension forces are puling the surface upward.
Thus, the pressure difference has to balance the surface tension.
The forces in the vertical direction reads
\begin{align}
\left(P_iP_o\right) dll_1\,dll_2 =
\Delta\,P dll_1\,dll_2 =
2 \,\sigma dll_1 \sin \beta_1 +
2 \,\sigma dll_2 \sin \beta_2
\label{intro:eq:STbalance}
\end{align}
For a very small area, the angles are very small and thus
$(\sin\beta \sim \beta)$.
Furthermore, it can be noticed that $dll_i\sim2\,R_i\,d\beta_i$.
Thus, the equation qref{intro:eq:STbalance} can be simplified as
\begin{align}
\Delta\,P = \sigma \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)
\label{intro:eq:STbSimple}
\end{align}
Equation qref{intro:eq:STbSimple} predicts that pressure
difference increase with inverse of the radius.
There are two extreme cases: one) radius of infinite and
radius of finite size.
The second with two equal radii.
The first case is for an infinite long cylinder for which
the equation qref{intro:eq:STbSimple} is reduced to
\begin{align}
\Delta\,P = \sigma \left( \dfrac{1}{R} \right)
\label{intro:eq:STbcylinder}
\end{align}
Other extreme is for a sphere for which the main radii are
the same and equation qref{intro:eq:STbSimple} is reduced to
\begin{align}
\Delta\,P = \dfrac{2\,\sigma }{R}
\label{intro:eq:STbspher}
\end{align}
Where $R\,$ is the radius of the sphere.
A soap bubble is made of two layers, inner and outer,
thus the pressure inside the bubble is
\begin{align}
\Delta\,P = \dfrac{4\,\sigma }{R}
\label{intro:eq:STbbubble}
\end{align}
A glass tube is inserted into bath of mercury.
It was observed that contact angle between the glass and mercury is $55^{\circ}$.
Fig. 1.19 Glass tube inserted into mercury.
Estimate the force due to the surface tension (tube is depicted in Figure 1.19).
It can be assume that the contact angle is the same for the inside and outside
part of the tube.
Estimate the depression size.
Assume that the surface tension for this combination of material is 0.5 [N/m]}
Solution
The mercury as free body that several forces act on it.
\begin{align}
\label{tubeMercury:inside}
F = \sigma 2\,\pi \cos55^\circ\,\left( {D_i} + {D_o} \right)
nd{align}
This force is upward and the horizontal force almost canceled.
However, if the inside and the outside diameters are considerable different the results is
\begin{align}
\label{tubeMercury:horizontal}
F = \sigma 2\,\pi \sin55\circ\,\left( {D_o}  {D_o} \right)
nd{align}
The balance of the forces on the meniscus show under the magnified glass are
\begin{align}
\label{tubeMercury:balance}
P \, \overbrace{\pi\,r^2}^{A} = \sigma\,2\,\pi\,r + \cancelto{\sim0}{W}
nd{align}
or
\begin{align}
\label{tubeMercury:balanceF}
g\,\rho\,h \, \pi\,r^2 = \sigma\,2\,\pi\,r + \cancelto{\sim0}{W}
nd{align}
Or after simplification
\begin{align}
\label{tubeMercury:balancea}
h = \dfrac{2\,\sigma}{ g\,\rho\,r}
nd{align}
A Tank filled with liquid, which contains $n$ bubbles with equal radii, $r$.
Calculate the minimum work required to increase the pressure in tank by $\Delta P$.
Assume that the liquid bulk modulus is infinity.
Solution
The work is due to the change of the bubbles volume.
The work is
\begin{align}
w = \int_{r_0}^{r_f} \Delta P(v) dv
\label{intro:eq:bubbleWork}
\end{align}
The minimum work will be for a reversible process.
The reversible process requires very slow compression.
It is worth noting that for very slow process, the temperature must
remain constant due to heat transfer.
The relationship between pressure difference and the radius is
described by equation qref{intro:eq:STbspher} for
reversible process.
Hence the work is
\begin{align}
w = \int_{r_0}^{r_f} \overbrace{\dfrac{2\,\sigma}{r}}^{\Delta P}
\overbrace{4\,\pi\,r^2\,dr}^{dv}
= 8\, \pi\,\sigma \int_{r_0}^{r_f} r dr
= 4\,\pi\,\sigma \left( {r_f}^2  {r_0}^2 \right)
\label{intro:eq:bubbleWork1}
\end{align}
Where, $r_0$ is the radius at the initial stage and
$r_f$ is the radius at the final stage.
The work for $n$ bubbles is then $4\,\pi\,\sigma\,n\,\left( {r_f}^2 
{r_0}^2 \right) $.
It can be noticed that the work is negative, that is the work is
done on the system.
Fig. 1.20 Capillary rise between two plates.
1.20.
Notice that previously a rise for circular tube was developed which different from
simple one dimensional case.
The distance between the two plates is $ll$ and the and surface tension is $\sigma$.
Assume that the contact angle is $0^{\circ}$ (the maximum possible force).
Compute the value for surface tension of $0.05[N/m]$, the density $1000[kg/m^3]$
and distance between the plates of $0.001[m]$.}
Solution
In Figure 1.20 exhibits the liquid under the current study.
The vertical forces acting on the body are the gravity, the pressure above and
below and surface tension.
It can be noted that the pressure and above are the same with the exception of the curvature
on the upper part.
Thus, the control volume is taken just above the liquid and the air part is neglected.
The question when the curvature should be answered in the Dimensional analysis and for simplification
this effect is neglected.
The net forces in the vertical direction (positive upwards) per unit length are
\begin{align}
\label{2DTubeRise:eq:gov}
2 \sigma \cos\,0^{\circ} = g\,h\,ll\,\rho \Longrightarrow h = \dfrac {2\,\sigma}{ll\,\rho\,g}
\end{align}
Inserting the values into equation qref{2DTubeRise:eq:gov} results in
\begin{align}
\label{2DTubeRise:eq:resutlts}
h = \dfrac {2\times 0.05}{0.001 \times 9.8\times \times 1000 } =
\end{align}
Develop expression for rise of the liquid due to surface tension in
concentric cylinders.
Solution
The difference lie in the fact that ``missing''cylinder add additional force and reduce
the amount of liquid that has to raise.
The balance between gravity and surface tension is
\begin{align}
\label{hConcentricTube:gov}
\sigma \, 2\,\pi\,\left(r_i\,\cos\theta_i + r_o\cos\theta_o\right) =
\rho\,g\,h\, \left( \pi (r_o)^2  \pi (r_i)^2 \right)
nd{align}
Which can be simplified as
\begin{align}
\label{hConcentricTube:results}
h = \dfrac{2\,\sigma \left(r_i\,\cos\theta_i + r_o\cos\theta_o\right) }
{ \rho\,g\, \left((r_o)^2  (r_i)^2 \right) }
nd{align}
The maximum is obtained when $\cos\theta_i = \cos\theta_o = 1$.
Thus, equation qref{hConcentricTube:results} can be simplified
\begin{align}
\label{hConcentricTube:resultsF}
h = \dfrac{2\,\sigma }
{ \rho\,g\, \left(r_o  r_i \right) }
nd{align}
Wetting of Surfaces
Fig. 1.21 Forces in Contact angle.
To explain the source of the contact angle, consider the point
where three phases became in contact.
This contact point occurs due to free surface reaching a solid
boundary.
The surface tension occurs between gas phase (G) to liquid phase (L) and
also occurs between the solid (S) and the liquid phases
as well as between the gas phase and the solid phase.
In Figure 1.21, forces diagram is shown
when control volume is chosen so that the masses of the
solid, liquid, and gas can be ignored.
Regardless to the magnitude of the surface tensions (except to
zero) the forces cannot be balanced for the description of
straight lines.
For example, forces balanced along the line of solid
boundary is
\begin{align}
\sigma_{gs}  \sigma_{ls}  \sigma_{lg} \cos\beta = 0
\label{intro:eq:contanctXy}
\end{align}
and in the tangent direction to the solid line the forces balance is
\begin{align}
F_{solid} = \sigma_{lg} \sin\beta
\label{intro:eq:contanctXx}
\end{align}
substituting equation qref{intro:eq:contanctXx} into
equation qref{intro:eq:contanctXy} yields
\begin{align}
\sigma_{gs}  \sigma_{ls} = \dfrac{ F_{solid}}{ \tan\beta}
\label{intro:eq:contactComb}
\end{align}
Fig. 1.22 Description of wetting and non–wetting fluids.
For $\beta= \pi/2 \Longrightarrow \tan\beta = \infty$.
Thus, the solid reaction force must be zero.
The gas solid surface tension is different from the liquid solid surface
tension and hence violating equation qref{intro:eq:contanctXy}.
The surface tension forces must be balanced,
thus, a contact angle is created to balance it.
The contact angle is determined by whether the surface tension
between the gas solid (gs) is larger or smaller then the
surface tension of liquid solid (ls) and the local geometry.
It must be noted that the solid boundary isn't straight.
The surface tension is a molecular phenomenon, thus depend on the
locale structure of the surface and it provides the balance for these local structures.
The connection of the three phases–materials–mediums
creates two situations which are categorized as wetting or non–wetting.
There is a common definition of wetting the surface.
If the angle of the contact between three materials is larger than $90^{\circ}$ then it is non–wetting.
On the other hand, if the angle is below than $90^{\circ}$ the material is
wetting the surface (see Figure 1.22).
The angle is determined by properties of the liquid, gas medium and the solid surface.
And a small change on the solid surface can change the wetting condition to non–wetting.
In fact there are commercial sprays that are intent to change the
surface from wetting to non wetting.
This fact is the reason that no reliable data can be provided
with the exception to pure substances and perfect geometries.
For example, water is described in many books as a wetting fluid.
This statement is correct in most cases, however, when solid
surface is made or cotted with certain materials, the water
is changed to be wetting (for example 3M selling product to ``change'' water to non–wetting).
So, the wetness of fluids is a function of the solid as well.
Table 1.6. The contact angle for air, distilled water with selected
materials to demonstrate the inconsistency.
Chemical component 
Contact angle 
Source 
Steel  $\pi/3.7$  [1] 
Steel, Nickel  $\pi/4.74$  [2] 
Nickel  $\pi/4.76$ to $\pi/3.83$  [1] 
Nickel  $\pi/4.76$ to $\pi/3.83$  [3] 
ChromeNickel Steel  $\pi/3.7$  [4] 
Silver  $\pi/6$ to $\pi/4.5 $  [5] 
Zink  $\pi/3.4$  [4] 
Bronze  $\pi/3.2$  [4] 
Copper  $\pi/4$  [4] 
Copper  $\pi/3$  [7] 
Copper  $\pi/2$  [8]


R. Siegel, E. G. Keshock (1975)
``Effects of reduced gravity on nucleate boiling bubble dynamics in
saturated water,'' AIChE Journal
Volume 10 Issue 4, Pages 509  517. 1975

Bergles A. E. and Rohsenow W. M.
'The determination of forced convection surface–boiling heat
transfer, ASME, J. Heat Transfer, vol 1 pp 365  372.

Tolubinsky, V.I. and Ostrovsky, Y.N. (1966) ``On the mechanism of
boiling heat transfer'',. International Journal of Heat and Mass
Transfer, Vol. 9, No 12, pages 14651470.

Arefeva E.I., Aladev O, I.T., (1958) ``wlijanii smatchivaemosti na
teploobmen pri kipenii,'' Injenerno Fizitcheskij Jurnal,
1117 1(7) In Russian.

Labuntsov D. A. (1939)
``Approximate theory of heat transfer by developed nucleate
boiling'' In Sussian Izvestiya An SSSR , Energetika I transport, No 1.

Basu, N., Warrier, G. R., and Dhir, V. K., (2002)
``Onset of Nucleate Boiling and Active Nucleation Site Density during
Subcooled Flow Boiling,''
ASME Journal of Heat Transfer, Vol. 124, papes 717 728.

Gaetner, R. F., and Westwater, J. W., (1960) ``Population of
Active Sites in Nucleate Boiling Heat Transfer,'' Chem. Eng.
Prog. Symp., Ser. 56.

Wang, C. H., and Dhir, V. K., (1993), ``Effect of Surface
Wettability on Active Nucleation Site Density During Pool Boiling
of Water on a Vertical Surface,''
J. Heat Transfer 115, pp. 659669
To explain the contour of the surface, and the contact angle
consider simple ``wetting'' liquid contacting a solid material
in two–dimensional shape as depicted in Figure 1.23.
To solve the shape of the liquid surface, the pressure difference
between the two sides of free surface has to be balanced by the surface tension.
In Figure 1.23\
\hbox to \textwidth{describes the raising of the liquid as results of the surface tension. The surface tension }
reduces the pressure in the liquid above the liquid line (the dotted line in the Figure
1.23).
The pressure just below the surface is $g\,h(x)\,\rho$
(this pressure difference will be explained in more details in
Chapter 4).
The pressure, on the gas side, is the atmospheric pressure.
This problem is a two dimensional problem and equation
qref{intro:eq:STbcylinder} is applicable to it.
Appalling equation qref{intro:eq:STbcylinder} and using the pressure difference yields
\begin{align}
g\,h(x)\,\rho = \dfrac{\sigma}{R(x)}
\label{intro:eq:hFsigma}
\end{align}
The radius of any continuous function, h = h(x), is
\begin{align}
R(x) = \dfrac{\left( 1+\left[\dot{h}(x)\right]^2 \right)^{3/2} }
{\ddot{h}(x)}
\label{intro:eq:radius}
\end{align}
Where $\dot{h}$ is the derivative of $h$ with respect to $x$.
Equation qref{intro:eq:radius} can be derived either by
forcing a circle at three points
at (x, x+dx, and x+2dx) and thus finding the the diameter or by
geometrical analysis of triangles build on points x and x+dx
(perpendicular to the tangent at these points).
Substituting equation qref{intro:eq:radius} into equation
qref{intro:eq:hFsigma} yields
\begin{align}
g\,h(x)\,\rho = \dfrac{\sigma}{\dfrac{
\left( 1+\left[\dot{h}(x)\right]^2 \right)^{3/2} }
{\ddot{h}(x)} }
\label{intro:eq:hFsigmaExF}
\end{align}
Equation qref{intro:eq:hFsigmaExF} is non–linear differential equation for
height and can be written as
1D Surface Due to Surface Tension
\begin{align}
\label{intro:eq:hFsigmaExS}
\dfrac{g\,h\,\rho } {\sigma}
\left( 1+\left[\dfrac{dh}{dx}\right]^2 \right)^{3/2}
 \dfrac{d^2h}{dx^2} = 0
\end{align}
With the boundary conditions that specify either the derivative
$\dot{h}(x=r)=0$ (symmetry) and the derivative at
$\dot{h}x=\beta$ or heights in two points or other combinations.
An alternative presentation of equation qref{intro:eq:hFsigmaExF} is
\begin{align}
g\,h\,\rho = \dfrac{\sigma \ddot{h} }
{ \left( 1+\dot{h}^2 \right)^{3/2} }
\label{intro:eq:hFsigmaExF1}
\end{align}
Integrating equation qref{intro:eq:hFsigmaExF1} transforms into
\begin{align}
\int \dfrac{ g\,\rho}{\sigma} h\, dh = \int \dfrac{\ddot{h} }
{ \left( 1+\dot{h}^2 \right)^{3/2} } dh
\label{intro:eq:hFsigmaExFInt}
\end{align}
The constant $Lp\,=\sigma/\rho\,g$ is referred to as Laplace's capillarity constant.
The units of this constant are meter squared.
The differential $dh$ is $\dot{h}$.
Using dummy variable and the identities $\dot{h} = \xi$ and hence,
$\ddot{h} =\dot{\xi} = d\xi$
transforms equation qref{intro:eq:hFsigmaExFInt} into
\begin{align}
\int \dfrac{ 1}{Lp} h\, dh =
\int \dfrac{ \xi d\xi }
{ \left( 1+\xi^2 \right)^{3/2} }
\label{intro:eq:hFsigmaExFIntm}
\end{align}
After the integration equation qref{intro:eq:hFsigmaExFIntm} becomes
\begin{align}
\dfrac{ h^2}{2\,Lp} + constant =
 \dfrac{1}{\left( 1+\dot{h}^2 \right)^{1/2} }
\label{intro:eq:hFsigmaExFIntmF}
\end{align}
At infinity, the height and the derivative of the height must by
zero so $constant + 0 = 1/1$ and hence, $constant=1$ .
\begin{align}
1  \dfrac{h^2}{2\,Lp} =
\dfrac{1}{\left( 1+\dot{h}^2 \right)^{1/2} }
\label{intro:eq:hFsigmaExFIntmFC}
\end{align}
Equation qref{intro:eq:hFsigmaExFIntmFC} is a first order
differential equation that can be solved by variables
separation.
Equation qref{intro:eq:hFsigmaExFIntmFC} can be rearranged to be
\begin{align}
{\left( 1+\dot{h}^2 \right)^{1/2} } =
\dfrac{1}
{ 1  \dfrac{h^2}{2\,Lp} }
\label{intro:eq:hFsigmaExFIntmFR}
\end{align}
Squaring both sides and moving the one to the right side yields
\begin{align}
\dot{h}^2 =
\left( \dfrac{1}{1  \dfrac{h^2}{2\,Lp} } \right)^2  1
\label{intro:eq:hFsigmaExFIntmS}
\end{align}
The last stage of the separation is taking the square root of both sides to be
\begin{align}
\dot{h} = \dfrac{dh}{dx} = \sqrt{
\left( \dfrac{1}{1  \dfrac{h^2}{2\,Lp} } \right)^2  1}
\label{intro:eq:hFsigmaExFIntmSR}
\end{align}
or
\begin{align}
\dfrac{dh}{\sqrt{
\left( \dfrac{1}{1  \dfrac{h^2}{2\,Lp} } \right)^2  1}}
= dx
\label{intro:eq:hFsigmaExFIntmSRi}
\end{align}
Equation qref{intro:eq:hFsigmaExFIntmSRi} can be integrated to yield
\begin{align}
\label{intro:eq:hFsigmaExFIntmSRi1}
\int \dfrac{dh}{\sqrt{
\left( \dfrac{1}{1  \dfrac{h^2}{2\,Lp} } \right)^2  1}}
= x + constant
\end{align}
The constant is determined by the boundary condition at $x=0$.
For example if $h(x0) = h_0$ then $constant = h_0$.
This equation is studied extensively in classes on surface tension.
Furthermore, this equation describes the dimensionless parameter
that affects this phenomenon and this parameter will be studied in Chapter 9.
This book is introductory, therefore this discussion on surface tension equation will be limited.
1.7.1.1 Capillarity
The capillary forces referred to the fact that surface tension causes
liquid to rise or penetrate into area (volume), otherwise it will
not be there.
It can be shown that the height that the liquid raised in a tube due to
the surface
tension is
\begin{align}
h = \dfrac{2\,\sigma \cos\beta}{g\,\Delta \rho\,r}
\label{intro:eq:capilaryH}
\end{align}
Where $\Delta \rho $ is the difference of liquid density to the gas
density and $r$ is the radius of tube.
Fig. 1.24 The raising height as a function of the radii.}
But this simplistic equation is unusable and useless unless the contact angle (assuming
that the contact angel is constant or a repressive average can be found or provided or
can be measured) is given.
However, in reality there is no readily information for contact
angle and therefore this equation is useful to show the treads.
The maximum that the contact angle can be obtained in equation
qref{intro:eq:capilaryH} when $\beta =0$ and thus $\cos\beta = 1$.
This angle is obtained when a perfect half a sphere shape exist
of the liquid surface.
In that case equation qref{intro:eq:capilaryH} becomes
\begin{align}
h_{max} = \dfrac{2\,\sigma }{g\,\Delta \rho\,r}
\label{intro:eq:capilaryHS}
\end{align}
Fig. 1.25 The raising height as a function of the radius.}
Figure 1.25 exhibits the height as a function of the radius of the tube.
The height based on equation qref{intro:eq:capilaryHS} is shown in
Figure 1.24 as blue line.
The actual height is shown in the red line.
Equation qref{intro:eq:capilaryHS} provides reasonable results only in a certain range.
For a small tube radius, equation qref{intro:eq:hFsigmaExS} proved better results because the
curve approaches hemispherical sphere (small gravity effect).
For large radii equation qref{intro:eq:hFsigmaExS} approaches the strait line (the liquid line)
strong gravity effect.
On the other hand, for extremely small radii equation qref{intro:eq:capilaryHS}
indicates that the high height which indicates a negative pressure.
The liquid at a certain pressure will be vaporized and will breakdown the model upon this equation was constructed.
Furthermore, the small scale indicates that the simplistic and
continuous approach is not appropriate and a different model is needed.
The conclusion of this discussion are shown in Figure 1.24.
The actual dimension for many liquids (even water) is about 15 [$mm$].
The discussion above was referred to ``wetting'' contact angle.
The depression of the liquid occurs in a ``negative'' contact angle similarly to ``wetting.''
The depression height, $h$ is similar to equation qref{intro:eq:capilaryHS} with a minus sign.
However, the gravity is working against the surface tension and
reducing the range and quality of the predictions of equation qref{intro:eq:capilaryHS}.
The measurements of the height of distilled water and mercury are
presented in Figure 1.25.
The experimental results of these materials are with agreement with
the discussion above.
The surface tension of a selected material is given in
Table 1.7.
In conclusion, the surface tension issue is important only in case
where the radius is very small and gravity is negligible.
The surface tension depends on the two materials or mediums
that it separates.
Calculate the diameter of a water droplet to attain
pressure difference of 1000[$N/m^2$]. You
can assume that temperature is $20^{\circ}C$.
Solution
The pressure inside the droplet is given by equation
qref{intro:eq:STbspher}.
\[
D = 2\,R = \dfrac{2\,2\,\sigma}{\Delta P} =
\dfrac{4\times 0.0728 }{1000} \sim 2.912\,{10}^{4} [m]
\]
Calculate the pressure difference between a droplet of water at
$20^{\circ}C$ when the droplet has a diameter of 0.02 cm.
Solution
using equation
\[
\Delta P = \dfrac{2\,\sigma}{r} \sim
\dfrac{2\times 0.0728}{0.0002} \sim
728.0 [N/m^2]
\]
Calculate the maximum force necessary to lift a thin wire ring of 0.04[m]
diameter from a water surface at $20^{\circ}C$.
Neglect the weight of the ring.
Solution
\[
F = 2(2\,\pi\,r\,\sigma)\cos\beta
\]
The actual force is unknown since the contact angle is
unknown.
However, the maximum Force is obtained when $\beta = 0$
and thus $\cos\beta = 1$.
Therefore,
\[
F = 4\,\pi\,r\,\sigma = 4\times \pi \times 0.04\times
0.0728 \sim .0366 [N]
\]
In this value the gravity is not accounted for.
A small liquid drop is surrounded with the air and has a diameter of 0.001 [m].
The pressure difference between the inside and outside droplet is 1[kPa].
Estimate the surface tension?
Table 1.7. The surface tension for selected materials
Chemical formula 
Temperature, $T\,[^{\circ}C]$ 
Surface Tension, $\left[\dfrac{ N}{m} \right]$ 
correction 
Acetic Acid  27.6  20  n/a 
Acetone  25.20    0.1120 
Aniline  43.4  22  0.1085 
enzene  28.88    0.1291 
Benzylalcohol  39.00    0.0920 
Benzylbenzoate  45.95    0.1066 
Bromobenzene  36.50    0.1160 
Bromobenzene  36.50    0.1160 
Bromoform  41.50    0.1308 
Butyronitrile  28.10    0.1037 
Carbon disulfid  32.30    0.1484 
Quinoline  43.12    0.1063 
Chloro benzene  33.60    0.1191 
Chloroform  27.50    0.1295 
Cyclohexane  24.95    0.1211 
Cyclohexanol  34.40  $25^{\circ}C$  0.0966 
Cyclopentanol  32.70    0.1011 
Carbon Tetrachloride  26.8    n/a 
Carbon disulfid  32.30    0.1484 
Chlorobutane  23.10    0.1117 
Ethyl Alcohol  22.3    n/a 
Ethanol  22.10    0.0832 
Ethylbenzene  29.20    0.1094 
Ethylbromide  24.20    0.1159 
Ethylene glycol  47.70    0.0890 
Formamide  58.20    0.0842 
Gasoline  $\sim$ 21    n/a 
Glycerol  64.0    0.0598 
Helium  0.12  $269^{\circ}C$  n/a 
Mercury  425465.0    0.2049 
Methanol  22.70    0.0773 
Methyl naphthalene  38.60    0.1118 
Methyl Alcohol  22.6    n/a 
Neon  5.15  $247^{\circ}C$  n/a 
Nitrobenzene  43.90    0.1177 
Olive Oil  43.048.0    0.067 
Perfluoroheptane  12.85    0.0972 
Perfluorohexane  11.91    0.0935 
Perfluorooctane  14.00    0.0902 
Phenylisothiocyanate  41.50    0.1172 
Propanol  23.70  $25^{\circ}C$  0.0777 
Pyridine  38.00    0.1372 
Pyrrol  36.60    0.1100 
SAE 30 Oil  n/a    n/a 
Seawater  5469    n/a 
Toluene  28.4    0.1189 
Turpentine  27    n/a 
Water  72.80    0.1514 
oXylene  30.10    0.1101 
mXylene  28.90    0.1104 
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