Next:
IdealFlow1
Previous:
Dimensional2
Chapter 9 Dim (continue)
9.3 Nusselt's Technique
The Nusselt's method is a bit more labor intensive, in that the governing equations
with the boundary and initial conditions are used to determine the dimensionless parameters.
In this method, the boundary conditions together with the governing equations
are taken into account as opposed to Buckingham's method.
A common mistake is to ignore the boundary conditions or initial conditions.
The parameters that results from this process are the dimensional parameters
which control the problems.
An example comparing the Buckingham's method with Nusselt's method is presented.
In this method, the governing equations, initial condition and boundary conditions are
normalized resulting in a creation of dimensionless parameters which govern the solution.
It is recommended, when the reader is out in the real world to simply abandon Buckingham's
method all together.
This point can be illustrated by example of flow over inclined plane.
For comparison reasons Buckingham's method presented and later the
results are compared with the results from Nusselt's method.
Utilize the Buckingham's method to analyze a two dimensional flow in incline plane.
Assume that the flow infinitely long and thus flow can be analyzed per width
which is a function of several parameters.
The potential parameters are the angle of inclination, $\theta$, liquid viscosity, $\nu$,
gravity, $g$, the height of the liquid, $h$, the density, $\rho$, and liquid velocity, $U$.
Assume that the flow is not affected by the surface tension (liquid), $\sigma$.
You furthermore are to assume that the flow is stable.
Develop the relationship between the flow to the other parameters.
Solution
Under the assumptions in the example presentation leads to following
\begin{align}
\label{dim:eq:buckinghamNusseltCompar}
\dot{m} = f\left( \theta, \nu, g, \rho, U\right)
\end{align}
The number of basic units is three while the number of the parameters is six
thus the difference is $63=3$.
Those groups (or the work on the groups creation) further can be reduced the because
angle $\theta$ is dimensionless.
The units of parameters can be obtained in Table 9.3 and summarized in
the following table.
Parameter 
Units 
Parameter 
Units 
$\nu$ 
$L^2t^{1}$ 
$g$ 
$L^{1}t{2}$ 
$\dot{m}$ 
$M\,t^{1}L^{1}$ 
$\theta$ 
none 
$U$ 
$L^{1}t^{1}$ 
$\rho$ 
$M\,L^{3}$ 
The basic units are chosen as for the time, $U$, for the mass, $\rho$,
and for the length $g$.
Utilizing the building blocks technique provides
\begin{align}
\label{inclidePlane:ini}
\overbrace{\dfrac{M}{t\,L} }^{\dot{m}}
= \left( \overbrace{\dfrac{M}{L^3}}^{\rho} \right)^a
\left( \overbrace{\dfrac{L}{t^2} }^{g} \right)^b
\left( \overbrace{\dfrac{L}{t} }^{U} \right)^c
\end{align}
The equations obtained from equation \eqref{inclidePlane:ini} are
\begin{align}
\label{inclidePlane:gov0}
\left.
\begin{array}{rrl}
\text{Mass}, M & a =& 1 \\
\text{Length}, L & 3a + b +c =& 1 \\
\text{time}, t & 2b c =&  1
\end{array}
\right\} \Longrightarrow
\pi_1 = \dfrac{\dot{m} \,g}{\rho\,\,U^3}
\end{align}
\begin{align}
\label{inclidePlane:ini1}
\overbrace{\dfrac{L^2}{t} }^{\nu}
= \left( \overbrace{\dfrac{M}{L^3}}^{\rho} \right)^a
\left( \overbrace{\dfrac{L}{t^2} }^{g} \right)^b
\left( \overbrace{\dfrac{L}{t} }^{U} \right)^c
\end{align}
The equations obtained from equation \eqref{inclidePlane:ini} are
\begin{align}
\label{inclidePlane:gov}
\left.
\begin{array}{rrl}
\text{Mass}, M & a =& 0 \\
\text{Length}, L & 3a + b +c =& 2 \\
\text{time}, t & 2b c =&  1
\end{array}
\right\} \Longrightarrow
\pi_2 = \dfrac{\nu \,g}{U^3}
\end{align}
Thus governing equation and adding the angle can be written as
\begin{align}
\label{inclidePlane:}
0 = f\left(\dfrac{\dot{m} \,g}{\rho\,\,U^3} , \dfrac{\nu \,g}{U^3} ,\theta\right)
\end{align}
The conclusion from this analysis are that the number of controlling parameters
totaled in three and that the initial conditions and boundaries are irrelevant.
A small note, it is well established that the combination of angle gravity or effective body force
is significant to the results.
Hence, this analysis misses, at the very least, the issue of the combination of the angle gravity.
Nusselt's analysis requires that the governing equations along with the boundary and
initial conditions to be written.
While the analytical solution for this situation exist, the parameters that effect the
problem are the focus of this discussion.
In Chapter 8, the Navier–Stokes equations were developed.
These equations along with the energy, mass or the chemical species of the system,
and second laws governed almost all cases in thermo–fluid mechanics.
This author is not aware of a compelling reason that this fact should be used in this chapter.
The two dimensional NS equation can obtained from equation \eqref{gravity:gov} as
\begin{align}
\label{dim:eq:twoDNSx}
\begin{array} {rll}
\rho \left(\dfrac{\partial U_x}{\partial t} +
\right. & U_x\dfrac{\partial U_x}{\partial x} +
U_y \dfrac{\partial U_x}{\partial y} + \left. U_z
\dfrac{\partial U_x}{\partial z}\right) = &\\
&\dfrac{\partial P}{\partial x} + \mu \left(\dfrac{\partial^2 U_x}{\partial x^2} +
\dfrac{\partial^2 U_x}{\partial y^2} + \dfrac{\partial^2 U_x}{\partial z^2}\right) +
\rho g\,\sin\theta
\end{array}
\end{align}
and
\begin{align}
\label{dim:eq:twoDNSy}
\begin{array} {rll}
\rho \left(\dfrac{\partial U_y}{\partial t} +
\right. & U_x\dfrac{\partial U_y}{\partial x} +
U_y \dfrac{\partial U_y}{\partial y} + \left. U_z
\dfrac{\partial U_y}{\partial z}\right) = &\\
&\dfrac{\partial P}{\partial x} + \mu \left(\dfrac{\partial^2 U_y}{\partial x^2} +
\dfrac{\partial^2 U_y}{\partial y^2} + \dfrac{\partial^2 U_y}{\partial z^2}\right) +
\rho g\,\sin\theta
\end{array}
\end{align}
With boundary conditions
\begin{align}
\label{dim:eq:twoDNSbcBotton}
\begin{array}{l}
U_x (y=0) = U_{0x} f(x) \\
\dfrac{\partial U_x}{\partial x} (y=h) = \tau_0 f(x)
\end{array}
\end{align}
The value $U_0x$ and $\tau_0$ are the characteristic and maximum values of the velocity
or the shear stress, respectively.
and the initial condition of
\begin{align}
\label{dim:eq:twoDNSic}
U_x (x=0) = U_{0y}\, f(y)
\end{align}
where $ U_{0y}$ is characteristic initial velocity.
These sets of equations \eqref{dim:eq:twoDNSx}–qref{dim:eq:twoDNSic} need to be converted to
dimensionless equations.
It can be noticed that the boundary and initial conditions are provided in a special form
were the representative velocity multiply a function.
Any function can be presented by this form.
In the process of transforming the equations into a dimensionless form associated with
some intelligent guess work.
However, no assumption is made or required about whether or not the velocity, in the $y$ direction.
The only exception is that the $y$ component of the velocity vanished on the boundary.
No assumption is required about the acceleration or the pressure gradient etc.
The boundary conditions have typical velocities which can be used.
The velocity is selected according to the situation or the needed velocity.
For example, if the effect of the initial condition is under investigation than
the characteristic of that velocity should be used.
Otherwise the velocity at the bottom should be used.
In that case, the boundary conditions are
\begin{align}
\label{dim:eq:twoDNScbles}
\begin{array}{l}
\dfrac{ U_x (y=0)}{ U_{0x}} = f(x) \\
\mu \dfrac{\partial U_x}{\partial x} (y=h) = \tau_0 \,g(x)
\end{array}
\end{align}
Now it is very convenient to define several new variables:
\begin{align}
\label{dim:eq:twoDNSdefVer1}
\begin{array}{rlcrl}
\overline{U} = & \dfrac{ U_x (\overline{x} ) }{ U_{0x}}\\
where:\
\overline{x} = & \dfrac{x}{h} &\qquad& \overline{y} &= \dfrac{y}{h} \\
\end{array}
\end{align}
The length $h$ is chosen as the characteristic length since no other length is provided.
It can be noticed that because the units consistency, the characteristic length can be
used for ``normalization'' (see Example 9.11).
Using these definitions the boundary and initial conditions becomes
\begin{align}
\label{dim:eq:twoDNScbles1}
\qquad
\begin{array}{l}
\dfrac{ \overline{U_x} (\overline{y}=0)}{ U_{0x}} = f^{'}(\overline{x}) \\
\dfrac{h \, \mu}{U_{0x}}\,
\dfrac{\partial \overline{U_x}}{\partial \overline{x}} (\overline{y}=1) = \tau_0\, g^{'}(\overline{x})
\end{array}
\end{align}
It commonly suggested to arrange the second part of equation \eqref{dim:eq:twoDNScbles1} as
\begin{align}
\label{dim:eq:twoDNScbles2}
\dfrac{\partial \overline{U_x}}{\partial \overline{x}} (\overline{y}=1) =
\dfrac{\tau_0\,U_{0x}}{h \, \mu}\,
g^{'}(\overline{x})
\end{align}
Where new dimensionless parameter, the shear stress number is defined as
\begin{align}
\label{dim:eq:tauDef}
\overline{\tau_0} = \dfrac{\tau_0\,U_{0x}}{h \, \mu}
\end{align}
With the new definition equation \eqref{dim:eq:twoDNScbles2} transformed into
\begin{align}
\label{dim:eq:twoDNScbles3}
\dfrac{\partial \overline{U_x}}{\partial \overline{x}} (\overline{y}=1) =
\overline{\tau_0} \, g^{'}(\overline{x})
\end{align}
Non–dimensionalize the following boundary condition.
What are the units of the coefficient in front of the variables, $x$.
What are relationship of the typical velocity, $U_0$ to $U_{max}$?
\begin{align}
\label{twoDNSbc:bc}
U_x (y = h) = U_0 \left( a\,x^2 + b\,xp(x) \right)
\end{align}
Solution
The coefficients $a$ and $b$ multiply different terms and therefore must have different units.
The results must be unitless thus $a$
\begin{align}
\label{twoDNSbc:aDef1}
L^0 = a \, \overbrace{ {L^2} }^{x^2}
\Longrightarrow a = \left[ \dfrac{1}{L^2} \right]
\end{align}
From equation \eqref{twoDNSbc:aDef1} it clear the conversion of the first term is
$U_x = a \, h^2 \overline{x}$.
The exponent appears a bit more complicated as
\begin{align}
\label{twoDNSbc:bDef1}
{L}^{0} = b \, xp\left( h\,\dfrac{x}{h}\right) =
b \, xp\left( h \right)
\, xp\left( \dfrac{x}{h}\right) =
b \, xp\left( h \right)
\, xp\left( \overline{x}\right)
\end{align}
Hence defining
\begin{align}
\label{twoDNSbc:bDef12}
\overline{b} = \dfrac{1}{xp{h}}
\end{align}
With the new coefficients for both terms and noticing that $y=h\longrightarrow \overline{y} =1$
now can be written as
\begin{align}
\label{twoDNSbc:}
\dfrac{ U_x (\overline{y} =1)}{U_{0}} = \overbrace{a\,h^2}^{\overline{a}}\,x^2 +
\overbrace{b \, xp\left( h \right) }^{\overline{b}}
\, xp\left( \overline{x}\right)
= \overline{a}\,\overline{x}^2 + \overline{b} xp{\overline{x}}
\end{align}
Where $\overline{a}$ and $\overline{b}$ are the transformed coefficients in the dimensionless
presentation.
After the boundary conditions the initial condition can undergo the non–dimensional process.
The initial condition \eqref{dim:eq:twoDNSic} utilizing the previous definitions transformed into
\begin{align}
\label{dim:eq:twoDNSicDless}
\dfrac{U_x(\overline{x}=0)}{U_{0x}} = \dfrac{U_{0y}}{U_{0x}}
f(\overline{y})
\end{align}
Notice the new dimensionless group of the velocity ratio as results of the boundary condition.
This dimensionless number was and cannot be obtained using the Buckingham's technique.
The physical significance of this number is an indication to the ``penetration'' of the
initial (condition) velocity.
The main part of the analysis if conversion of the governing equation into
a dimensionless form uses previous definition with additional definitions.
The dimensionless time is defined as $\overline{t} = t\,U_{0x}/h$.
This definition based on the characteristic time of $h/U_{0x}$.
Thus, the derivative with respect to time is
\begin{align}
\label{dim:eq:twoDNSdefDevT}
\dfrac{\partial U_x}{\partial t} =
\dfrac{\partial \overbrace{\overline{U_x}}^{\dfrac{U_x}{U_{0x}} } U_{0x}}
{\partial \underbrace{\overline{t}}_{ \dfrac{t\,U_{0x}}{ h } }\dfrac{h}{U_{0x}} }
=
\dfrac{{U_{0x}}^2 }{h} \dfrac{\partial \overline{U_x} }{\partial \overline{t} }
\end{align}
Notice that the coefficient has units of acceleration.
The second term
\begin{align}
\label{dim:eq:twoDNSdefDevT2}
U_x\dfrac{\partial U_x}{\partial x} =
\overbrace{\overline{U_x} }^{\dfrac{U_x}{U_{0x}}} U_{0x}
\dfrac{\partial \overbrace{\overline{U_x} }^{\dfrac{U_x}{U_{0x}}} U_{0x}}
{\partial \underbrace{\overline{x} }_{\dfrac{x}{h}} h }
=
\dfrac{{U_{0x}}^2}{h} \, {\overline{U_x} } \,
\dfrac{\partial \overline{U_x} } {\partial \overline{x} }
\end{align}
The pressure is normalized by the same initial pressure or the static pressure as
$\left(PP_\infty\right)/\left(P_0 P_\infty\right)$ and hence
\begin{align}
\label{dim:eq:twoDNSdefP}
\dfrac{\partial P}{\partial x} =
\dfrac{\partial \overbrace{\overline{P}}^{\dfrac{PP_\infty}{P_0P_\infty}} }
{\partial \overline{x} h} \left(P_0  P_\infty\right) =
\dfrac{\left(P_0  P_\infty\right)}{h} \dfrac{\partial \overline{P}}{\partial \overline{x} }
\end{align}
The second derivative of velocity looks like
\begin{align}
\label{dim:eq:twoDNSdef2Dir}
\dfrac{\partial^2 U_x}{\partial x^2} =
\dfrac{\partial}{\partial \left( \overline{x} h \right) }
\dfrac{\partial \left( \overline{U_x} U_{0x} \right) }{\partial \left( \overline{x} h \right) } =
\dfrac{U_{0x}}{h^2}
\dfrac{\partial^2 \overline{U_x} }{\partial \overline{x}^2}
\end{align}
The last term is the gravity $g$ which is left for the later stage.
Substituting all terms and dividing by density, $\rho$ result in
\begin{align}
\label{dim:eq:twoDNSxx}
\begin{array} {rll}
\dfrac{{U_{0x}}^2} {h} \left(\dfrac{\partial \overline{U_x} }{\partial \overline{t} } +
\right. & \overline{U_x}\dfrac{\partial \overline{U_x} }{\partial \overline{x}} +
\overline{U_y} \dfrac{\partial \overline{U_x}}{\partial \overline{y}} +
\left. \overline{U_z} \dfrac{\partial \overline{U_x}}{\partial \overline{z}}\right) = &\\
& \dfrac{P_0P_\infty}{h\,\rho}
\dfrac{\partial \overline{P}}{\partial \overline{x}} +
\dfrac{{U_{0x}}\mu}{h^2\,\rho} \left(\dfrac{\partial^2 U_x}{\partial x^2} +
\dfrac{\partial^2 U_x}{\partial y^2} + \dfrac{\partial^2 U_x}{\partial z^2}\right) +
\dfrac{\cancel{\rho} g}{\cancel{\rho}}\,\sin\theta
\end{array}
\end{align}
Dividing equation \eqref{dim:eq:twoDNSxx} by ${U_{0x}}^2/ {h}$ yields
\begin{align}
\label{dim:eq:twoDNSx1i}
\begin{array} {rll}
\left(\dfrac{\partial \overline{U_x} }{\partial \overline{t} } +
\right. & \overline{U_x}\dfrac{\partial \overline{U_x} }{\partial \overline{x}} +
\overline{U_y} \dfrac{\partial \overline{U_x}}{\partial \overline{y}} +
\left. \overline{U_z} \dfrac{\partial \overline{U_x}}{\partial \overline{z}}\right) = &\\
& \dfrac{P_0P_\infty}{{U_{0x}}^2\,\rho}
\dfrac{\partial \overline{P}}{\partial \overline{x}} +
\dfrac{\mu}{U_{0x}\,h\,\rho} \left(\dfrac{\partial^2 U_x}{\partial x^2} +
\dfrac{\partial^2 U_x}{\partial y^2} + \dfrac{\partial^2 U_x}{\partial z^2}\right) +
\dfrac{ g \,h }{{U_{0x}}^2}\,\sin\theta
\end{array}
\end{align}
Defining ``initial'' dimensionless parameters as
\begin{align}
\label{dim:eq:iniDimlessPara}
\begin{array}{lcccr}
Re = \dfrac{U_{0x}\,h\,\rho}{\mu} &&
Fr = \dfrac{{U_{0x}}}{\sqrt{g \,h}} &&
Eu = \dfrac{P_0P_\infty}{{U_{0x}}^2\,\rho}
\end{array}
\end{align}
Substituting definition of equation \eqref{dim:eq:iniDimlessPara} into equation \eqref{dim:eq:twoDNSx1i} yields
\begin{align}
\label{dim:eq:twoDNSx1}
\begin{array} {rll}
\left(\dfrac{\partial \overline{U_x} }{\partial \overline{t} } +
\right. & \overline{U_x}\dfrac{\partial \overline{U_x} }{\partial \overline{x}} +
\overline{U_y} \dfrac{\partial \overline{U_x}}{\partial \overline{y}} +
\left. \overline{U_z} \dfrac{\partial \overline{U_x}}{\partial \overline{z}}\right) = &\\
& Eu\,
\dfrac{\partial \overline{P}}{\partial \overline{x}} +
\dfrac{1}{Re} \left(\dfrac{\partial^2 U_x}{\partial x^2} +
\dfrac{\partial^2 U_x}{\partial y^2} + \dfrac{\partial^2 U_x}{\partial z^2}\right) +
\dfrac{ 1 }{Fr^2}\,\sin\theta
\end{array}
\end{align}
Equation \eqref{dim:eq:twoDNSx1} show one common possibility of a dimensionless presentation of governing equation.
The significance of the large and small value of the dimensionless parameters will be discuss
later in the book.
Without actually solving the problem, Nusselt's method provides several more parameters that
were not obtained by the block method.
The solution of the governing equation is a function of all the parameters
present in that equation and boundaries condition as well the initial condition.
Thus, the solution is
\begin{align}
\label{dim:eq:twoDNSx2}
U_x = f \left( \overline{x}, \overline{y}, Eu, Re, Fr, \theta, \overline{\tau_0}, f_u, f_{\tau},
\dfrac{U_{0y}}{U_{0x}} \right)
\end{align}
The values of $ \overline{x},\, \overline{y}$ depend on $h$ and hence the value of $h$ is
an important parameter.
It can be noticed with Buckingham's method, the number of parameters obtained
was only three (3) while Nusselt's method yields 12 dimensionless parameters.
This is a very significant difference between the two methods.
In fact, there are numerous examples in the literature that showing people doing experiments
based on Buckingham's methods.
In these experiments, major parameters are ignored rendering these experiments useless
in many cases and deceiving.
Common Transformations
Fluid mechanics in particular and Thermo–Fluid field in general have several common
transformations that appear in boundary conditions, initial conditions and
equations.
It recognized that not all the possibilities can presented in the example shown above.
Several common boundary conditions which were not discussed in the above example
are presented below.
As an initial matter, the results of the non dimensional transformation depends on the
selection of what and how is nondimensionalization carried.
This section of these parameters depends on what is investigated.
Thus, one of the general nondimensionalization of the Navier–Stokes and energy
equations will be discussed at end of this chapter.
Boundary conditions are divided into several categories such as a given
given derivative (Neumann b.c.), mixed condition, and complex conditions.
The first and second categories were discussed to some degree earlier
and will be expanded later.
The third and fourth categories were not discussed previously.
The non–dimensionalization of the boundary conditions of the first category requires
finding and diving the boundary conditions by a typical or a characteristic value.
The second category involves the nondimensionalization of the derivative.
In general, this process involve dividing the function by a typical value
and the same for length variable (e.g. $x$) as
\begin{align}
\label{dim:eq:nondimDirative1}
\dfrac{\partial U}{\partial x} =
\dfrac{\ell}{U_0}
\dfrac{\partial \left( \dfrac{U}{U_0} \right) }{ \partial \left( \dfrac{x}{\ell} \right)} =
\dfrac{\ell}{U_0}
\dfrac{\partial \overline{U}}{\partial \overline{x}}
\end{align}
In the Thermo–Fluid field and others, the governing equation can be of higher order than second
order.
It can be noticed that the degree of the derivative boundary condition cannot exceed the
derivative degree of the governing equation (e.g. second order equation has at most the second
order differential boundary condition.).
In general ``nth'' order differential equation leads to
\begin{align}
\label{dim:eq:nondimDirative}
\dfrac{\partial^n U}{\partial x^n} =
\dfrac{U_0}{\ell^n}
\dfrac{\partial^n \left( \dfrac{U}{U_0} \right) }
{ \partial \left(\dfrac{x}{\ell}\right)^n } =
\dfrac{U_0}{\ell^n}
\dfrac{\partial^n \overline{U}}{\partial \overline{x}^n}
\end{align}
The third kind of boundary condition is the mix condition.
This category includes combination of the function with its derivative.
For example a typical heat balance at liquid solid interface reads
\begin{align}
\label{dim:eq:mixBCex}
h(T_0  T) =  k \dfrac{\partial T}{ \partial x}
\end{align}
This kind of boundary condition, since derivative of constant is zero, translated to
\begin{align}
\label{dim:eq:mixBCexDim1}
h\, \cancel{(T_0 T_{max})}\, \left( \dfrac{T_0  T }{ T_0 T_{max}} \right) =
 \dfrac{ k \,\cancel{\left( T_0 T_{max} \right)} }{ \ell} \,
\dfrac{  \partial \left( \dfrac{ T  T_0} { T_0 T_{max} } \right) }
{ \partial \left(\dfrac{x}{\ell} \right) }
\end{align}
or
\begin{align}
\label{dim:eq:mixBCexDim}
\left( \dfrac{T_0  T }{ T_0 T_{max}} \right) =
\dfrac{ k }{h\, \ell}
\dfrac{\partial \left( \dfrac{ T  T_0} { T_0 T_{max} } \right) }
{ \partial \left(\dfrac{x}{\ell} \right) }
\Longrightarrow
\Theta = \dfrac{1}{Nu} \dfrac{\partial \Theta}{\partial \overline{x}}
\end{align}
temperature are defined as
\begin{align}
\label{dim:eq:NuDef}
Nu = \dfrac{h\,\ell}{k} && \Theta = \dfrac{ T  T_0} { T_0 T_{max} }
\end{align}
and $T_{max}$ is the maximum or reference temperature of the system.
The last category is dealing with some non–linear conditions of the function with its
derivative.
For example,
\begin{align}
\label{dim:eq:surfaceTensionBCi}
\Delta P \approx {\sigma}\left({ \dfrac{1}{r_1} + \dfrac{1}{r_2} } \right) =
\dfrac{\sigma}{ r_1}\, \dfrac{r_1+ r_2}{r_2}
\end{align}
Where $r_1$ and $r_2$ are the typical principal radii of the free surface curvature, and,
$\sigma$, is the surface tension between the gas (or liquid) and the other phase.
The surface geometry (or the radii) is determined by several factors which include
the liquid movement instabilities etc chapters of the problem at hand.
This boundary condition \eqref{dim:eq:surfaceTensionBCi} can be rearranged to be
\begin{align}
\label{dim:eq:surfaceTensionBC}
\dfrac{\Delta P \, r_1}{\sigma} \approx \dfrac{r_1+ r_2}{r_2} \Longrightarrow
Av \approx \dfrac{r_1+ r_2}{r_2}
\end{align}
The Avi number represents the geometrical characteristics combined with the material properties.
The boundary condition \eqref{dim:eq:surfaceTensionBC} can be transferred into
\begin{align}
\label{dim:eq:surfaceTensionBCdim}
\dfrac{ \Delta P\,r_1}{\sigma} = Av
\end{align}
Where $\Delta P$ is the pressure difference between the two phases (normally between
the liquid and gas phase).
One of advantage of Nusselt's method is the Object–Oriented nature which allows
one to add
additional dimensionless parameters for addition ``degree of freedom.''
It is common assumption, to initially assume, that liquid is incompressible.
If greater accuracy is needed than this assumption is removed.
In that case, a new dimensionless parameters is introduced as the ratio of
the density to a reference density as
\begin{align}
\label{dim:eq:refereceRho}
\overline{\rho} = \dfrac{\rho}{\rho_0}
\end{align}
In case of ideal gas model with isentropic flow this assumption becomes
\begin{align}
\label{dim:eq:gasDensity}
\bar{\rho} = {\rho \over \rho_0} = \left(\dfrac{ P_{0}}{ P } \right)^{\dfrac{1}{n}}
\end{align}
The power $n$ depends on the gas properties.
Characteristics Values
Normally, the characteristics values are determined by physical values e.g.
The diameter of cylinder as a typical length.
There are several situations where the characteristic length, velocity, for example, are determined
by the physical properties of the fluid(s).
The characteristic velocity can determined from $U_0 =\sqrt{2P_{0} / \rho}$.
The characteristic length can be determined from ratio of $\ell = \Delta P/\sigma$.
One idea of renewable energy is to use and to utilize the high concentration of
of brine water such as in the Salt Lake and the Salt Sea (in Israel).
This process requires analysis the mass transfer process.
The governing equation is non–linear and this example provides opportunity
to study nondimensionalizing of this kind of equation.
The conversion of the species yields a governing nonlinear equationootnotemark{} for such process is
\begin{align}
\label{highMass:gov}
U_0 \dfrac{\partial C_A}{\partial x} =
\dfrac{\partial }{ \partial y }
\dfrac {D_{AB}}{ \left( 1  X_A\right) } \dfrac{\partial C_A}{ \partial y }
\end{align}
Where the concentration, $C_A$ is defended as the molar density i.e. the number of
moles per volume.
The molar fraction, $X_A$ is defined as the molar fraction of species $A$ divide by
the total amount of material (in moles).
The diffusivity coefficient, $D_{AB}$ is defined as penetration of species $A$ into
the material.
What are the units of the diffusivity coefficient?
The boundary conditions of this partial differential equation are given by
\begin{align}
\label{highMass:BCy}
\dfrac{\partial C_A}{\partial y} \left(y=\infty\right) = 0
\end{align}
\begin{align}
\label{highMass:BCy2i}
C_A (y=0) = C_e
\end{align}
Where $C_e$ is the equilibrium concentration.
The initial condition is
\begin{align}
\label{highMass:BCy2}
C_A (x=0) = C_0
\end{align}
Select dimensionless parameters so that the governing equation and boundary and initial
condition can be presented in a dimensionless form.
There is no need to discuss the physical significance of the problem.
ootnotetext{More information how this equation was derived can be found
in Bar–Meir (Meyerson), Genick
``Hygroscopic absorption to falling films: The effects of the concentration level''
M.S. Thesis TelAviv Univ. (Israel). Dept. of Fluid Mechanics and Heat Transfer
12/1991.}
Solution
This governing equation requires to work with dimension associated with mass transfer
and chemical reactions, the ``mole.''
However, the units should not cause confusion or fear since it appear on both
sides of the governing equation.
Hence, this unit will be canceled.
Now the units are compared to make sure that diffusion coefficient is kept the units
on both sides the same.
From units point of view, equation \eqref{highMass:gov} can be written
(when the concentration is simply ignored) as
\begin{align}
\label{highMass:govDim}
\overbrace{\dfrac{L}{t}} ^{U}
\overbrace{\dfrac{\cancel{C}}{L} }^{\dfrac{\partial C}{\partial x}} =
\overbrace{\dfrac{1}{L}}^{\dfrac{\partial }{\partial y}}
\overbrace{\dfrac{D_{AB}}{1}}^{\dfrac{D_{AB} }{ \left(1  X \right)}}
\overbrace{\dfrac{\cancel{C}}{L} }^{\dfrac{\partial C}{\partial y}}
\end{align}
It can be noticed that $X$ is unitless parameter because two same quantities are divided.
\begin{align}
\label{highMass:govGov}
\dfrac{1}{t} = \dfrac{1}{L^2} D_{AB} \Longrightarrow D_{AB} = \dfrac{L^2}{t}
\end{align}
Hence the units of diffusion coefficient are typically given by $\left[m^2/sec\right]$
(it also can be observed that based on Fick's laws of diffusion it has the same units).
The potential of possibilities of dimensionless parameter is large.
Typically, dimensionless parameters are presented as ratio of two
quantities.
In addition to that, in heat and mass transfer (also in pressure driven flow etc.)
the relative or reference to certain point has to accounted for.
The boundary and initial conditions here provides the potential of the ``driving force'' for the mass
flow or mass transfer.
Hence, the potential definition is
\begin{align}
\label{highMass:phi}
\Phi = \dfrac{C_A  C_0 }{C_e  C_0}
\end{align}
With almost ``standard'' transformation
\begin{align}
\label{highMass:coordinates}
\overline{x} = \dfrac{x}{\ell} &\qquad & \overline{y} = \dfrac{y}{\ell}
\end{align}
Hence the derivative of $\Phi$ with respect to time is
\begin{align}
\label{highMass:PhiD}
\dfrac{\partial \Phi}{\partial \overline{x}} =
\dfrac{\partial \dfrac{C_A  C_0 }{C_e  C_0} }{\partial \dfrac{x}{\ell} }
= \dfrac{\ell}{C_e  C_0}
\dfrac{\partial \left(C_A  \cancelto{0}{C_0} \,\,\,\,\,\right)} {\partial {x} }
= \dfrac{\ell}{C_e  C_0}
\dfrac{\partial C_A } {\partial {x} }
\end{align}
In general a derivative with respect to $\overline{x}$ or $\overline{y}$ leave yields
multiplication of $\ell$.
Hence, equation \eqref{highMass:gov} transformed into
\begin{align}
\label{highMass:totalDim}
\begin{array}{rl}
U_0\dfrac{\cancel{\left(C_e  C_0\right)}} {\ell}\dfrac{\partial \Phi}{\partial \overline{x}} &=
\dfrac{1}{\ell}\, \dfrac{\partial }{ \partial \overline{y} }
\dfrac {D_{AB}}{ \left( 1  X_A\right) }
\dfrac{\cancel{\left(C_e  C_0\right)}} {\ell}\dfrac{\partial \Phi}{\partial \overline{y}} \
\mbox{\Huge $\displaystyle\leadsto$}
\dfrac{U_0} {\ell}\dfrac{\partial \Phi}{\partial \overline{x}} &=
\dfrac{1}{\ell^2}\, \dfrac{\partial }{ \partial \overline{y} }
\dfrac {D_{AB}}{ \left( 1  X_A\right) }
\dfrac{\partial \Phi}{\partial \overline{y}}
\end{array}
\end{align}
Equation \eqref{highMass:totalDim} like non–dimensionalized and proper version.
However, the term $X_A$, while is dimensionless, is not proper.
Yet, $X_A$ is a function of $\Phi$ because it contains $C_A$.
Hence, this term, $X_A$ has to be converted or presented by $\Phi$.
Using the definition of $X_A$ it can be written as
\begin{align}
\label{highMass:X}
X_A = \dfrac{C_A} {C} = \left( C_e C_0 \right)\dfrac{C_A  C_0}{ C_e C_0} \dfrac{1}{C}
\end{align}
Thus the transformation in equation \eqref{highMass:X} another unexpected dimensionless
parameter as
\begin{align}
\label{highMass:Xf}
X_A = \Phi \, \dfrac{C_e C_0}{C}
\end{align}
Thus number, $\dfrac{C_e C_0}{C}$ was not expected and it represent ratio of the driving
force to the height of the concentration which was not possible to attend by Buckingham's
method.
9.4 Summary of Dimensionless Numbers
This section summarizes all the major dimensionless parameters which are commonly used in
the fluid mechanics field.
Common Dimensionless Parameters of Thermo–Fluid Field 
Name 
Symbol 
Equation 
Interpretation 
Application 
Archimedes Number 
$Ar$ 
$\dfrac {g\, {\ell}^3 \rho_f (\rho  \rho_f)}{\mu^2} $ 
$\dfrac{\text{buoyancy forces}}{\text{viscous forces}}$ 
in nature and force convection 
Atwood Number 
$A$ 
$\dfrac {(\rho_a  \rho_b)}{\rho_a + \rho_b}$ 
$\dfrac{\text{buoyancy forces}}{\text{``penetration'' force}}$ 
in stability of liquid layer $a$ over $b$ Rayleighâ€“Taylor instability etc 
Bond Number 
$Bo$ 
$\dfrac{\rho\,g\,\ell^2}{\sigma}$ 
$\dfrac{\text{gravity forces}}{\text{surface tension force}}$ 
in open channel flow, thin film flow 
Brinkman Number 
$Br$ 
$\dfrac {\mu U^2}{k\,\Delta T}$ 
$\dfrac{\text{heat forces}}{\text{heat force}}$ 
in open channel flow, thin film flow 
Capillary Number 
$Ca$ 
$\dfrac {\mu U}{\sigma}$ 
$\dfrac{\text{viscous forces}}{\text{surface tension force}}$ 
For small $Re$ and surface tension involve problem 
Cauchy Number 
$Cau$ 
$\dfrac{\rho\,U^2}{E}$ 
$\dfrac{\text{inertia forces}}{\text{elastic forces}}$ 
For large $Re$ and surface tension involve problem 
Cavitation Number 
$\sigma$ 
$\dfrac{P_l  P_v}{\dfrac{1}{2} \rho U^2}$ 
$\dfrac{\text{pressure difference}}{\text{inertia energy}}$ 
pressure difference to vapor pressure to the potential of phase change (mostly to gas) 
Courant Number 
$Co$ 
$\dfrac{\Delta t\, U}{\Delta x}$ 
$\dfrac{\text{wave distance }}{\text{typical distance}}$ 
A requirement in numerical schematic to achieve stability 
Dean Number 
$D$ 
$\dfrac{Re}{\sqrt{R / h}}$ 
$\dfrac{\text{inertia forces}}{\text{viscous deviation forces}}$ 
related to radius of channel with width $h$ stability 
Deborah Number 
$De$ 
$\dfrac{t_c}{t_p}$ 
$\dfrac{\text{stress relaxation time}}{\text{observation time}}$ 
the ratio of the fluidity of material primary used in rheology 
Drag Coefficient 
$C_D$ 
$\dfrac{D}{\dfrac{1}{2}\,\rho\,U^2\,A }$ 
$\dfrac{\text{drag force}}{\text{inertia effects }}$ 
Aerodynamics, hydrodynamics, note this coefficient has many definitions 
Eckert Number 
$Ec$ 
$\dfrac{U^2}{C_p\,\Delta T}$ 
$\dfrac{\text{inertia effects}}{\text{thermal effects }}$ 
during dissipation processes 
Ekman Number 
$Ek$ 
$\dfrac{\nu}{2\ell^2\,\omega}$ 
$\dfrac{\text{viscous forces}}{\text{Coriolis forces }}$ 
geophysical flow like atmospheric flow 
Ekman Number 
$Ek$ 
$\dfrac{\nu}{2\ell^2\,\omega}$ 
$\dfrac{\text{viscous forces}}{\text{Coriolis forces }}$ 
geophysical flow like atmospheric flow 
Euler Number 
$Eu$ 
$\dfrac{P_0P_{\infty}}{\dfrac{1}{2}\,\rho\,U^2} $ 
$\dfrac{\text{pressure effects}}{\text{inertia effects }}$ 
potential of resistance problems 
Froude Number 
$Fr$ 
$ \dfrac{U}{\sqrt{g\,\ell}} $ 
$\dfrac{\text{inertia effects}}{\text{gravitational effects }}$ 
open channel flow and two phase flow 
Galileo Number 
$Ga$ 
$\dfrac{\rho\, g\,\ell^3}{\mu^2} $ 
$\dfrac{\text{gravitational effects}}{\text{viscous effects }}$ 
open channel flow and Stokes flow 
Grashof Number 
$Gr$ 
$\dfrac{\beta\,\Delta T \,g\,\ell^3\,\rho^2}{\mu^2 } $ 
$\dfrac{\text{buoyancy effects}}{\text{viscous effects }}$ 
natural convection 
Knudsen Number 
$Kn$ 
$\dfrac{\lambda}{\ell} $ 
$\dfrac{\text{LMFP }}{\text{characteristic length }}$ 
length of mean free path, LMFP, to characteristic length 
Laplace Constant 
$La$ 
$ \sqrt{\dfrac{2\,\sigma}{g(\rho_1\rho_2)}} $ 
$\dfrac{\text{surface force }}{\text{gravity effects }}$ 
liquid raise, surface tension problem, (also ref Capillary constant) 
Lift Coefficient 
$C_L$ 
$ \dfrac{L}{\dfrac{1}{2}\,\rho\,U^2\,A } $ 
$\dfrac{\text{lift force }}{\text{inertia effects }}$ 
Aerodynamics, hydrodynamics, note this coefficient has many definitions 
Mach Number 
$M$ 
$ \dfrac{U}{c} $ 
$\dfrac{\text{velocity }}{\text{sound speed }}$ 
Compressibility and propagation of disturbance 
Marangoni Number 
$Ma$ 
$ {\dfrac{d\sigma}{dT}}\dfrac{\ell \, \Delta T}{\nu \alpha} $ 
$\dfrac{\text{ `thermal' s. tension }}{\text{viscous force }}$ 
Compressibility and propagation of disturbance 
Morton Number 
$Mo$ 
$\dfrac{g \mu_c^4 \, \Delta \rho}{\rho_c^2 \sigma^3} $ 
$\dfrac{\text{ viscous force }}{\text{surface tension }}$ 
bubble and drop flow 
Ozer Number 
$Oz$ 
$ \dfrac{\dfrac{{C_{D}}^2\,P_{max}}{\rho} }{\left(\dfrac{Q_{max}}{A}\right)^2 } $ 
$\dfrac{\text{ `maximum' supply }}{\text{`maximum' demand }}$ 
supply and demand analysis such pump and pipe system, economy 
Prandtl Number 
$Pr$ 
$ \dfrac{\nu}{\alpha}$ 
$\dfrac{\text{ viscous diffusion}}{\text{thermal diffusion }}$ 
Prandtl number is fluid property important in flow due to thermal forces 
Reynolds Number 
$Re$ 
$ \dfrac{\rho\,U\,\ell}{\mu} $ 
$\dfrac{\text{ inertia force}}{\text{viscous force }}$ 
in most fluid mechanics issues 
Rossby Number 
$Ro$ 
$ \dfrac{U}{\omega\,ll_{0}} $ 
$\dfrac{\text{ inertia force}}{\text{Coriolis force }}$ 
in rotating fluids 
Shear Number 
$Sn$ 
$ \dfrac{\tau_c\,ll_c}{\mu_c\,U_c} $ 
$\dfrac{\text{actual shear}}{\text{`potential' shear }}$ 
shear flow 
Stokes Number 
$Stk$ 
$ \dfrac{\rho\, g\,\ell^3}{\mu^2} $ 
$\dfrac{\text{particle relaxation time }}{\text{Kolmogorov time }}$ 
in aerosol flow dealing with penetration of particles 
Strouhal Number 
$St$ 
$\dfrac{\omega\,\ell}{U} $ 
$\dfrac{\text{`unsteady' effects }}{\text{inertia effect }}$ 
The effects of natural or forced frequency in all the field that is how
much the `unsteadiness' of the flow is 
Taylor Number 
$Ta$ 
$ \dfrac{\rho^2\,{\omega_i}^2\,\ell^4}{\mu^4} $ 
$\dfrac{\text{centrifugal force}}{\text{viscous force }}$ 
Stability of rotating cylinders Notice $\ell$ has special definition 
Weber Number 
$We$ 
$ \dfrac{\rho\,U^2\,\ell}{\sigma} $ 
$\dfrac{\text{inertia force}}{\text{viscous force }}$ 
For large $Re$ and surface tension involve problem 
The dimensional parameters that were used in the construction of the dimensionless parameters
in Table 9.8 are the characteristics of the system.
Therefore there are several definition of Reynolds number.
In fact, in the study of the physical situations often people refers to local $Re$ number and the
global $Re$ number.
Keeping this point in mind, there several typical dimensions which need to be mentioned.
The typical body force is the gravity $g$ which has a direction to center of Earth.
The typical length is denoted as $\ell$ and in many cases it is referred to as the diameter or
the radius.
The density, $\rho$ is referred to the characteristic density or density at infinity.
The area, $A$ in drag and lift coefficients is referred normally to projected area.
Fig. 9.4 Oscillating Von Karman Vortex Street.
The frequency $\omega$ or $f$ is referred to as the ``unsteadiness'' of the system.
Generally, the periodic effect is enforced by the boundary conditions or the initial conditions.
In other situations, the physics itself instores or forces periodic instability.
For example, flow around cylinder at first looks like symmetrical situation.
And indeed in a low Reynolds number it is a steady state.
However after a certain value of Reynolds number, vortexes are created in an infinite parade
and this phenomenon is called Von Karman vortex street (see Figure 9.4)
These vortexes are created in a non–symmetrical way and hence create an unsteady situation.
When Reynolds number increases, these vortexes are mixed and the flow becomes turbulent
which, can be considered a steady state.
The pressure $P$ is the pressure at infinity or when the velocity is at rest.
$c$ is the speed of sound of the fluid at rest or characteristic value.
The value of the viscosity, $\mu$ is typically some kind averaged value.
The inability to define a fix value leads also to new dimensionless numbers
which represent the deviations of these properties.
9.4.1 The Significance of these Dimensionless Numbers
Reynolds number, named in the honor of Reynolds, represents the ratio of the momentum forces
Historically, this number was one of the first numbers to be introduced to fluid mechanics.
This number determines, in many cases, the flow regime.
Eckert number^{*} determines whether the role of the momentum energy is transferred
to thermal energy is significant to affect the flow.
This effect is important in situations where high speed is involved.
This fact suggests that Eckert number is related to Mach number.
Determine this relationship and under what circumstances this relationship is true.
*This example is based on Bird, Lightfoot and Stuart ``Transport Phenomena''.
Solution
In Table 9.8 Mach and Eckert numbers are defined as
\begin{align}
\label{eckert:definition}
Ec = \dfrac{U^2}{C_p\,\Delta T} &\qquad&\qquad&\qquad& M = \dfrac{U}{\sqrt{\dfrac{P}{\rho}} }
\end{align}
The material which obeys the ideal flow model ($P/\rho = R\,T$ and $P = C_1\,\rho^k$)
can be written that
\begin{align}
\label{eckert:MachDefIdealGas}
M = U \left/ \sqrt{\dfrac{P}{\rho}} \right.= \dfrac{U}{\sqrt{k\,R\,T}}
\end{align}
For the comparison, the reference temperature used to be equal to zero.
Thus Eckert number can be written as
\begin{align}
\label{eckert:expendated}
\sqrt{Ec} = \dfrac{U}{\sqrt{C_p\,T} } =
\dfrac{U}{\sqrt{\underbrace{\left(\dfrac{R \, k}{k1} \right)}_{C_p}\,T} } =
\dfrac{\sqrt{k1}\,U}{\sqrt{k\,R\,T} } = \sqrt{k 1}\, M
\end{align}
The Eckert number and Mach number are related under ideal gas model and isentropic relationship.
Brinkman number measures of the importance of the viscous heating relative
the conductive heat transfer.
This number is important in cases when a large velocity change occurs over
short distances such as lubricant, supersonic flow in rocket mechanics creating large
heat effect in the head due to large velocity (in many place it is a combination of Eckert
number with Brinkman number.
The Mach number is based on different equations depending on the property of the medium
in which pressure disturbance moves through.
Cauchy number and Mach number are related as well and see
Example 9.15 for explanation.
For historical reason some fields prefer to use certain numbers and not other ones.
For example in Mechanical engineers prefer to use the combination $Re$ and $We$ number
while Chemical engineers prefers to use the combination of $Re$ and the Capillary number.
While in some instances this combination is justified, other cases it is arbitrary.
Show what the relationship between these dimensionless numbers.
Solution
The definitions of these number in Table 9.8
\begin{align}
\label{CaWeRe:we}
We = \dfrac{\rho\, U^2\, \ell}{\sigma} &\qquad &
Re = \dfrac{\rho\,U\,\ell}{\mu} & \qquad &
Ca = \dfrac{\mu\,U}{\sigma} = \dfrac{U}{\dfrac{\sigma}{\mu} }
\end{align}
Dividing Weber number by Reynolds number yields
\begin{align}
\label{CaWeRe:WedRe}
\dfrac{We}{Re} = \dfrac{\dfrac{\rho\, U^2\, \ell}{\sigma} } {\dfrac{\rho\,U\,\ell}{\mu} } =
\dfrac{U}{\dfrac{\sigma}{\mu} } = Ca
\end{align}
Physicist who pioneered so many fields that it is hard to say what and where are his greatest contributions.
Eulerâ€™s number and Cavitation number are essentially the same with the exception that these numbers
represent different driving pressure differences.
This difference from dimensional analysis is minimal.
Furthermore, Euler number is referred to as the pressure coefficient, $C_p$.
This confusion arises in dimensional analysis because historical reasons and the main focus area.
The cavitation number is used in the study of cavitation phenomena while Euler number
is mainly used in calculation of resistances.
Explained under what conditions and what are relationship between the
Mach number and Cauchy number?
Solution
Cauchy number is defined as
\begin{align}
\label{M2Ca:CaDef}
Cau = \dfrac{\rho\,\pmb{U}^2 } {E}
\end{align}
The square root of Cauchy number is
\begin{align}
\label{M2Ca:Cau12}
\sqrt{Cau} = \dfrac{U}{\sqrt{\dfrac{E}{\rho}}}
\end{align}
In the liquid phase the speed of sound is approximated as
\begin{align}
\label{M2Ca:liquidSound}
c = \dfrac{E}{\rho}
\end{align}
Using equation \eqref{M2Ca:Cau12} transforms equation \eqref{M2Ca:CaDef} into
\begin{align}
\label{M2Ca:eq:Cau12f}
\sqrt{Cau} = \dfrac{U}{c} = M
\end{align}
Thus the square root of $Cau$ is equal to Mach number in the liquid phase.
In the solid phase equation \eqref{M2Ca:liquidSound} is less accurate and speed of sound
depends on the direction of the grains.
However, as first approximation, this analysis can be applied also to the solid phase.
9.4.2 Relationship Between Dimensionless Numbers
The Dimensionless numbers since many of them have formulated in a certain field tend to
be duplicated.
For example, the Bond number is referred in Europe as Eotvos number.
In addition to the above confusion, many dimensional numbers expressed the same things
under certain conditions.
For example, Mach number and Eckert Number under certain circumstances are same.
Galileo Number is a dimensionless number which represents the ratio of
\begin{align}
\label{GalileoNumber:def}
Ga = \dfrac{\rho^2\,g\,\ell^3}{\mu^2}
\end{align}
The definition of Reynolds number has viscous forces and the definition of Froude number has
gravitational forces.
What are the relation between these numbers?
Laplace Number is another dimensionless number that appears in fluid mechanics
which related to Capillary number.
The Laplace number definition is
\begin{align}
\label{Laplace:def}
La = \dfrac{\rho \, \sigma \, \ell }{\mu^2}
\end{align}
Show what are the relationships between Reynolds number, Weber number and Laplace number.
The Rotating Froude Number is a somewhat a similar number to the regular Froude number.
This number is defined as
\begin{align}
\label{RotatingFr:def}
Fr_R = \dfrac{\omega^2\,\ell}{g}
\end{align}
What is the relationship between two Froude numbers?
Ohnesorge Number is another dimensionless parameter that deals with
surface tension and is similar to Capillary number and it is defined as
\begin{align}
\label{ohnesorge:def}
Oh = \dfrac{\mu}{\sqrt{\rho\,\sigma\,\ell} }
\end{align}
Defined $Oh$ in term of $We$ and $Re$ numbers.
9.4.3 Examples for Dimensional Analysis
The similarity of pumps is determined by comparing several dimensional numbers
among them are Reynolds number, Euler number, Rossby number etc.
Assume that the only numbers which affect the flow are Reynolds and Euler number.
The flow rate of the imaginary pump is 0.25 [$m^3/sec$] and pressure increase for
this flow rate is 2 [Bar] with 2500 [kw].
Due to increase of demand, it is suggested to replace the pump with a
4 times larger pump.
What is the new estimated flow rate, pressure increase, and power consumption?
Solution
It provided that the Reynolds number controls the situation.
The density and viscosity remains the same and hence
\begin{align}
\label{pump1:Re}
Re_m = Re_p \Longrightarrow U_m\,D_m = U_p\, D_p \Longrightarrow U_p = \dfrac{D_m}{D_P} U_m
\end{align}
It can be noticed that initial situation is considered as the model and while the new
pump is the prototype.
The new flow rate, $Q$, depends on the ratio of the area and velocity as
\begin{align}
\label{pump1:Q}
\dfrac{Q_p}{Q_m} = \dfrac{A_p \, U_p}{A_m \, U_m} \Longrightarrow
Q_p = {Q_m}\dfrac{A_p \, U_p}{A_m \, U_m}
= {Q_m}\dfrac{{D_p}^2 \, U_p}{{D_m}^2 \, U_m}
\end{align}
Thus the prototype flow rate is
\begin{align}
\label{pump1:Qf}
Q_p = Q_m\, \left(\dfrac{D_p}{D_m}\right)^3 = 0.25\times 4^3 = 16 \left[ \dfrac{m^3}{sec}\right]
\end{align}
The new pressure is obtain by comparing the Euler number as
\begin{align}
\label{pump1:Eu}
Eu_p = Eu_m \Longrightarrow
\left(\dfrac{\Delta P} {\dfrac{1}{2}\rho\,U^2} \right)_p = \left(\dfrac{\Delta P} {\dfrac{1}{2}\rho\,U^2} \right)_m
\end{align}
Rearranging equation \eqref{pump1:Eu} provides
\begin{align}
\label{pump1:Euarranged}
\dfrac{\left(\Delta P\right)_p} {\left(\Delta P\right)_m}
= \dfrac{\left(\cancel{\rho}\,U^2 \right)_p} {\left(\cancel{\rho}\,U^2 \right)_m} =
\dfrac{\left(U^2 \right)_p} {\left(U^2 \right)_m}
\end{align}
Utilizing equation \eqref{pump1:Re}
\begin{align}
\label{pump1:DeltaP}
\Delta P_p = \Delta P_m \left( \dfrac{D_p}{D_m} \right)^2
\end{align}
The power can be obtained from the following
\begin{align}
\label{pump1:power}
\dot{W} = \dfrac{F\, \ell}{t} = F\, U = P \, A \, U
\end{align}
In this analysis, it is assumed that pressure is uniform in the cross section.
This assumption is appropriate because only the secondary flows in the radial direction
(to be discussed in this book section on pumps.).
Hence, the ratio of power between the two pump can be written as
\begin{align}
\label{pump1:powerRatio}
\dfrac{\dot{W}_p}{\dot{W}_m} =
\dfrac{\left( P \, A \, U \right)_p} {\left( P \, A \, U \right)_m}
\end{align}
Utilizing equations above in this ratio leads to
\begin{align}
\label{pump1:pRvalue}
\dfrac{\dot{W}_p}{\dot{W}_m} =
\overbrace{\left(\dfrac{D_p}{D_m}\right)^2}^{P_p/P_m}
\overbrace{\left(\dfrac{D_p}{D_m}\right)^2}^{A_p/A_m}
\overbrace{\left(\dfrac{D_p}{D_m}\right)}^{U_p/U_m} =
\left(\dfrac{D_p}{D_m}\right)^5
\end{align}
The flow resistance to flow of the water in a pipe is to be simulated by flow of air.
Estimate the pressure loss ratio if Reynolds number remains constant.
This kind of study appears in the industry in which the compressibility of
the air is ignored.
However, the air is a compressible substance that flows the ideal gas model.
Water is a substance that can be considered incompressible flow for relatively small
pressure change.
Estimate the error using the averaged properties of the air.
Solution
For the first part, the Reynolds number is the single controlling parameter
which affects the pressure loss.
Thus it can be written that the Euler number is function of the Reynolds number.
\begin{align}
\label{airWater:ReEu}
Eu = f (Re)
\end{align}
Thus, to have a similar situation the Reynolds and Euler have to be same.
\begin{align}
\label{airWater:ReEu}
Re_p = Re_m \qquad \qquad Eu_m = Eu_p
\end{align}
Hence,
\begin{align}
\label{airWater:Rer}
\dfrac{U_m}{U_p} =
\dfrac{ll_p}{ll_m}
\dfrac{\rho}{\rho_m}
\dfrac{\mu_p}{\mu_m}
\end{align}
and for Euler number
\begin{align}
\label{airWater:Eur}
\dfrac{\Delta P_m}{\Delta P_p} =
\dfrac{\rho_m}{\rho_p}
\dfrac{U_m}{U_p}
\end{align}
and utilizing equation \eqref{airWater:Rer} yields
\begin{align}
\label{airWater:ReEucombined}
\dfrac{\Delta P_m}{\Delta P_p} =
\left(\dfrac{ll_p}{ll_m}\right)^2
\left(\dfrac{\mu_m}{\mu_p}\right)^2
\left(\dfrac{\rho_p}{\rho_m}\right)
\end{align}
Inserting the numerical values results in
\begin{align}
\label{airWater:numerical}
\dfrac{\Delta P_m}{\Delta P_p} = 1 \times 1000 \times
\end{align}
It can be noticed that the density of the air changes considerably hence
the calculations produce a considerable error which can render the calculations
useless (a typical problem of Buckingham's method).
Assuming a new variable that effect the problem, air density variation.
If that variable is introduced into problem, air can be used to simulate water flow.
However as a first approximation, the air properties are calculated based on the
averaged values between the entrance and exit values.
If the pressure reduction is a function of pressure reduction (iterative process).
to be continue
A device operating on a surface of a liquid to study using a model with a ratio 1:20.
What should be ratio of kinematic viscosity between the model and prototype
so that Froude and Reynolds numbers remain the same.
Assume that body force remains the same and velocity is reduced by half.
Solution
The requirement is that Reynolds
\begin{align}
\label{kinimaticViscosity:Re}
Re_m = R_p \Longrightarrow
\left(\dfrac{U\,\ell}{\nu} \right)_p = \left(\dfrac{U\,\ell}{\nu} \right)_m
\end{align}
The Froude needs to be similar so
\begin{align}
\label{kinimaticViscosity:Fr}
Fr_m= Fr_p \Longrightarrow
\left(\dfrac{U}{\sqrt{g\,\ell}} \right)_p = \left(\dfrac{U\,\ell}{\nu} \right)_m
\end{align}
dividing equation \eqref{kinimaticViscosity:Re} by equation \eqref{kinimaticViscosity:Fr} results in
\begin{align}
\label{kinimaticViscosity:ReDRe1}
\left(\dfrac{U\,\ell}{\nu} \right)_p /
\left(\dfrac{U}{\sqrt{g\,\ell}} \right)_p
= \left(\dfrac{U\,\ell}{\nu} \right)_m /
\left(\dfrac{U}{\sqrt{g\,\ell}} \right)_m
\end{align}
or
\begin{align}
\label{kinimaticViscosity:ReDRe}
\left(\dfrac{\ell\, \sqrt{g\,\ell}} {\nu} \right)_p =
\left(\dfrac{\ell\, \sqrt{g\,\ell}} {\nu} \right)_m
\end{align}
The kinematic viscosity ratio is then
\begin{align}
\label{kinimaticViscosity:knRatio}
\dfrac{\nu_p}{\nu_m} = \left(\dfrac{ll_m} {ll_p} \right)^{3/2} = \left(1/20\right)^{3/2}
\end{align}
It can be noticed that this can be achieved using Ohnesorge Number like this
presentation.
In AP physics in 2005 the first question reads
``A ball of mass $M$ is thrown vertically upward with an initial speed of $U_0$.
Does it take longer for the ball to rise to its maximum height or to fall from its maximum height back to the
height from which it was thrown?
It also was mentioned that resistance is proportional to ball velocity (Stoke flow).
Justify your answer.''
Use the dimensional analysis to examine this situation.
Solution
The parameters that can effect the situation are (initial) velocity of the ball,
air resistance (assuming Stokes flow e.g. the resistance is function of the velocity), maximum height, and gravity.
Functionality of these parameters can be written as
\begin{align}
\label{dim:ex:APballBuckingham}
t = f ( U,\, k,\, H,\, m,\, g)
\end{align}
The time up and/or down must be written in the same fashion since fundamental principle of Buckingham's $\pi$ theorem
the functionally is unknown but only dimensionless parameters are dictated.
Hence, no relationship between the time up and down can be provided.
However, Nusselt's method provides first to written the governing equations.
The governing equation for the ball climbing up is
\begin{align}
\label{dim:eq:ballClimbing}
m\,\dfrac{dU}{dt} =  m\,g  k\,U
\end{align}
when the negative sign indicates that the positive direction is up.
The initial condition is that
\begin{align}
\label{dim::eq:ballInitialConidtion}
U(0) = U_0
\end{align}
The governing equation the way down is
\begin{align}
\label{dim:eq:ballDown}
m\,\dfrac{dU}{dt} =  m\,g + k\,U
\end{align}
with initial condition of
\begin{align}
\label{dim:eq:ballInitialConidtionDown}
U(0) = 0
\end{align}
Equation \eqref{dim:eq:ballDown} has no typical velocity (assuming at this stage that solution was not solved ever before).
Dividing equation \eqref{dim:eq:ballDown} by $m\,g$ and inserting the gravitation constant into the derivative
results in
\begin{align}
\label{dim:eq:boatGovDimless2}
\dfrac{dU}{d\,(g\,t)} = 1 + \dfrac{k\,U}{ m\,g}
\end{align}
The gravity constant, $g$, could be inserted because it is constant.
Equation suggests that velocity should be normalized by as dimensionless group, $\left.{k\,U}\right/{ m\,g}$.
Without solving the equations, it can be observed that value of dimensionless group above or below one
change the characteristic of the governing equation (positive slop or negative slop).
Non–dimensioning of initial condition \eqref{dim::eq:ballInitialConidtion} yields
\begin{align}
\label{dim:eq:ballInitialConidtionD}
\dfrac{k\,U(0)}{ m\,g} = \dfrac{k\,U_0}{ m\,g}
\end{align}
In this case, if the value $\left.{k\,U_0}\right/{ m\,g}$ is above one change the characteristic of the situation.
This exercise what not to solve this simple Physics mechanics problem but rather to demonstrate the power
of dimensional analysis power.
So, What this information tell us?
In the case the supper critical initial velocity, the ball can be above critical velocity $\dfrac{k\,U_0}{ m\,g} >1$ on the up.
However the ball never can be above the critical velocity and hence the time up will
shorter the time done.
For the initial velocity below the critical velocity, while it is know that the answer is the same, the dimensional analysis
does not provide a solution.
On the way up ball can start
Fig. Description of the boat crossing river.
Two boats sail from the opposite sides of river (see Figure ).
They meet at a distance $ll_1$ (for example 1000) meters from bank $\bbb{A}$.
The boats reach the opposite side respectively and continue back to their original bank.
The boats meet for the second time at $ll_2$ (for example 500) $[m]$ from bank $\bbb{B}$.
What is the river width?
What are the dimensional parameters that control the problem?
Solution
The original problem was constructed so it was suitable to the 11 years old author's daughter
who was doing her precalculus.
However, it appears that this question can be used to demonstrate some of the power of the
the dimensional analysis.
Using the Buckingham's method it is assumed that diameter is a function of the velocities and lengths.
Hence, the following can be written
\begin{align}
\label{dim:eq:govBuck}
D = f(ll_1, \, ll_2,\,U_A,\,U_B)
\end{align}
Where $D$ is the river width.
Hence, according basic idea the following can be written
\begin{align}
\label{dim:eq:govR}
D = {ll_1}^a\,{ll_1}^b\,{U_A}^c\,{U_B}^d
\end{align}
The solution of equation \eqref{dim:eq:govR} requires that
\begin{align}
D = [L]^a[L]^b\left[\dfrac{L}{T}\right]^c\left[\dfrac{L}{T}\right]^d
\end{align}
The time has to be zero hence it requires that
\begin{align}
\label{dim:eq:boatTime}
0 = c+d
\end{align}
The units length requires that
\begin{align}
\label{dim:eq:boatLength}
1 = a +b + c+ d
\end{align}
Combined equation \eqref{dim:eq:boatTime} with equation \eqref{dim:eq:boatLength} results in
\begin{align}
\label{dim:eq:boatCombined}
1=a+b
\end{align}
It can noticed that symmetry arguments require that $a$ and $b$ must be identical.
Hence, $a=b=\sqrt{1/2}$ and the solutions is of the form $D= f(\sqrt{ll_1\,ll_2})$.
From the analytical solution it was found that this solution is wrong.
Another approach utilizing the minimized Buckingham's approach reads
\begin{align}
\label{dim:eq:boatMinGov}
D = f (ll_1,\,U_A)
\end{align}
In the standard form this leads to
\begin{align}
\label{dim:eq:boadMinLead}
D = [L]^a\,\left[\dfrac{L}{T}\right]^b
\end{align}
Which leads to the requirements of $b=0$ and $a=1$.
Which again conflict with the actual analytical solution.
Using Nusselt's method requires to write the governing equation.
The governing equations are based equating the time traveled to first and second meeting as the following
\begin{align}
\label{dim:eq:boatNusseltGov1}
\dfrac{ll_1}{U_A} = \dfrac{Dll_1}{U_B}
\end{align}
At the second meeting the time is
\begin{align}
\label{dim:eq:boatNusseltGov2}
\dfrac{D+ll_2}{U_A} = \dfrac{2\,Dll_2}{U_B}
\end{align}
Equations \eqref{dim:eq:boatNusseltGov1} and \eqref{dim:eq:boatNusseltGov2} have three unknowns $D$, $U_A$ and $U_B$.
The non–dimensioning process can be carried by dividing governing equations by $D$ and multiply by $U_B$ to become
\begin{align}
\label{dim:eq:boatNusseltGov1d}
\overline{ll_1} = \left( 1  \overline{ll_1} \right) \,\dfrac{U_A}{U_B}
\end{align}
\begin{align}
\label{dim:eq:boatNusseltGov2d}
1+\overline{ll_2} = \left( 2  \overline{ll_2} \right) \,\dfrac{U_A}{U_B}
\end{align}
Equations \eqref{dim:eq:boatNusseltGov1d} and \eqref{dim:eq:boatNusseltGov2d} have three unknowns.
However, the velocity ratio is artificial parameter or dependent parameter.
Hence division of the dimensionless governing equations yield one equation with one unknown as
\begin{align}
\label{dim:eq:boatNusseltSsol}
\dfrac{\overline{ll_1}}{1+\overline{ll_2} } =
\dfrac{1  \overline{ll_1}}{ 2  \overline{ll_2}}
\end{align}
Equation determines that $\overline{ll_1}$ is a function of $\overline{ll_2}$.
It can be noticed that $D$, $ll_1$ and $ll_2$ are connected.
Hence, knowing two parameters leads to the solution of the missing parameter.
From dimensional analysis it can be written that the
\begin{align}
\label{dim:eq:boatSolNusselt}
\overline{ll_2} = f ( \overline{ll_1} ) = \dfrac{2\,\dfrac{\overline{ll_1} }{1\overline{ll_1} }  1 }
{ 1+ \dfrac{\overline{ll_1}}{1\overline{ll_1} } }
\end{align}
It can be concluded that river width is a function of implicit of $\overline{ll_1}$ and $\overline{ll_2}$.
Clearly the Nusselt's technique provided write based to obtain the dimensionless parameters.
A bit smarter selection of the normalizing parameter can provide explicit solution.
An alternative definition of dimensionless parameters of $\widetilde{D} = D/ll_1$
and $\widetilde{ll_2} = ll_2/ll_1$ can provide the need path.
Equation \eqref{dim:eq:boatNusseltSsol} can be converted quadratic equation for $D$ as
\begin{align}
\label{dim:eq:boatSolNusseltNew}
\dfrac{1}{\widetilde{D}  \widetilde{ll_2} } =
\dfrac{\widetilde{D}  1}{ 2\, \widetilde{D}  \widetilde{ll_2} }
\end{align}
Equation \eqref{dim:eq:boatSolNusseltNew} is quadratic which can be solved analytically.
The solution can be presented as
\begin{align}
\label{dim:eq:boatSolPresentation}
D = ll_1 \, f \left(\dfrac{ll_2}{ll_1} \right)
\end{align}
9.5 Summary
The two dimensional analysis methods or approaches were presented in this chapter.
Buckingham's $\pi$ technique is a quick ``fix approach'' which allow rough estimates
and relationship between model and prototype.
Nusselt's approach provides an heavy duties approach to examine what dimensionless
parameters effect the problem.
It can be shown that these two techniques in some situations provide almost similar solution.
In other cases, these technique proves different and even conflicting results.
The dimensional analysis technique provides a way to simplify models (solving the governing
equation by experimental means) and to predict effecting parameters.
Next:
IdealFlow1
Previous:
Dimensional2
