root/trunk/matml/transport/resources/buckpi/buckpi.tex

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\begin{document}
\title{A Quick Guide to Dimensional Analysis}
\date{June 19, 2007}
\author{Adam Powell}
\maketitle

Dimensional analysis is a powerful tool for reducing complex engineering
problems to their most basic components.  Typically, a functional relationship
involving five to eight parameters can be reduced to just two to four
dimensionless groups.  This allows one to more concisely represent the
solution, whether analytical, numerical or empirical.  Its result is akin to
the phase rule in its simplicity and power.

For example, suppose a flat solid sheet with thermal conductivity $k$ and
thickness $\delta$ is in contact with a fluid with heat transfer coefficient
$h$.  The fluid temperature is $T_{fl}$, the temperature on the side of the
sheet away from the fluid is $T_0$, and the temperature at the interface is
unknown.  The heat flux through this system is given by the driving force
divided by the sum of resistances:
\begin{equation}
  \label{eq:flux}
  q = \frac{T_0-T_{fl}}{\frac{\delta}{k}+\frac{1}{h}}.
\end{equation}
This is certainly not a complicated equation, but as the flux is a function of
four parameters (the temperature difference, heat transfer coefficient, sheet
thickness and conductivity), visualizing the five-dimensional space would be a
great challenge.  Dimensional analysis reduces this to just two parameters, as
described below.

\paragraph{Procedure}

Before beginning, it is necessary to define a couple of terms:
\begin{itemize}
\item {\bf Dimensions} are material properties, geometric dimensions, etc.  The
  above example has dimensions of heat flux $q$, temperature $T_0$ and
  $T_{fl}$, length $\delta$, conductivity $k$, and heat transfer coefficient
  $h$.
\item {\bf Base units} are the smallest set of units from which all units of
  all dimensions can be derived.  In the example above, those units can be
  expressed in terms of watts, meters, and degrees Kelvin or Celsius.
\item {\bf Derived units} are the products and quotients of the base units,
  such as W/m$^2$ for heat flux.
\end{itemize}
Based on these definitions, dimensional analysis proceeds as follows.

\begin{enumerate}
\item Postulate desired behavior as a function of the other variables.  In this
  example, the goal is to calculate heat flux $q$, and it is postulated to be a
  function of $T_0-T_{fl}$, $\delta$, $k$ and $h$.  The number of parameters is
  the number of dimensions $n$, in this case $n=5$, for heat flux and the four
  dimensions of which it is considered a function.  This is done by intuition,
  and is very often the hardest step in the process.

\item Find the number of independent units in the system $r$.  As identified
  above, there are four base units: m, s, kg, K; but kg and s are always linked
  in watts, so they're not independent for this problem.  In this case, all of
  the units can be expressed in terms of: W, m and K, so there are three
  independent units.

\item Calculate the number of dimensionless groups which completely
  characterize the system.  The Buckingham Pi theorem states that the number of
  dimensionless groups is $n-r$, in this case $5-3=2$ dimensionless groups.

\item Choose $r$ dimensionally-independent variables to eliminate, which will
  make the others dimensionless.  Here it is convenient to choose $T_0-T_{fl}$,
  $k$ and $\delta$.  As a counterexample, one can't use $h$, $k$ and $\delta$
  because they are not independent: the units of $h$ (W/m$^2\cdot$K) can be
  multiplied by those of $\delta$ (m) to give those of $k$ (W/m$\cdot$K).  Very
  often there are multiple ``right answers'', and one should choose the one
  which is most convenient, as will be discussed further below.

\item Form $\pi$ groups from what's left, these are unit-less versions of the
  remaining parameters.  Dimensionless flux $q$, called $\pi_q$, is:
  \begin{equation}
    \label{eq:piq}
    \pi_q = q \cdot [T_0-T_{fl}]^a \cdot [k]^b \cdot [\delta]^c.
  \end{equation}
  The exponents $a$, $b$ and $c$ must make the resulting $\pi_q$ dimensionless.
  To calculate these exponents, one can construct a table of unit exponents,
  whose columns must add to zero.  The table on the left has the unknown
  exponents, and on the right, the equations bringing totals to zero have been
  solved to give $a=-1$, $b=-1$ and $c=1$:
  \begin{center}
    \begin{tabular}{l|ccc|}
      Dimension & W & m & K \\ \hline
      $q$ & 1 & -2 & 0 \\
      $[T_0-T_{fl}]^a$ & 0 & 0 & $a$ \\
      $k^b$ & $b$ & $-b$ & $-b$ \\
      $\delta^c$ & 0 & $c$ & 0 \\ \hline
      Total & $1+b$ & $-2-b+c$ & $a-b$ \\ \hline
    \end{tabular} $\ \ \ \ $
    \begin{tabular}{l|ccc|}
      Dimension & W & m & K \\ \hline
      $q$ & 1 & -2 & 0 \\
      $[T_0-T_{fl}]^{-1}$ & 0 & 0 & -1 \\
      $k^{-1}$ & -1 & 1 & 1 \\
      $\delta^1$ & 0 & 1 & 0 \\ \hline
      Total & 0 & 0 & 0 \\ \hline
    \end{tabular}
  \end{center}
  This gives the result:
  \begin{equation}
    \label{eq:piqfinal}
    \pi_q=\frac{q\delta}{k(T_0-T_{fl})}.
  \end{equation}
  This is readily recognized as the ratio of total flux to flux in the absence
  of a fluid layer.  Likewise there is a dimensionless $\pi_h$:
  \begin{equation}
    \label{eq:pih}
    \pi_h = h \cdot [T_0-T_{fl}]^d \cdot [k]^e \cdot [\delta]^f,
  \end{equation}
  the exponents which make this dimensionless are $d=0$, $e=-1$, $f=1$ giving
  $\pi_h = h\delta/k$, the Biot number.

\item Rewrite Step 1 in dimensionless terms, and we're done:
  \begin{equation}
    \label{eq:piqfunc}
    \pi_q = f(\pi_h).
  \end{equation}
\end{enumerate}

\paragraph{Interpreting the result}

What's this?  So simple?  Can't be.  Check it against the analytical equation
above:
$$q=\frac{T_0-T_{fl}}{\frac{1}{h} + \frac{\delta}{k}};$$
\begin{equation}
  \label{eq:piqsolved}
  \pi_q=\frac{q\delta}{k(T_0-T_{fl})} =
  \frac{T_0-T_{fl}}{\frac{1}{h} + \frac{\delta}{k}}
  \frac{\delta}{k(T_0-T_{fl})} =
  \frac{1}{\frac{k}{h\delta} + 1} =
  \frac{1}{\frac{1}{\pi_h}+1} = 1-\frac{1}{\pi_h+1}.
\end{equation}
So the dimensional analysis works, the dimensionless flux is indeed only a
simple function of the dimensionless heat transfer coefficient; that function
looks like:
\begin{center}
  \PSbox{buckpi-graph.ps}{355pt}{128pt}
\end{center}

The limiting cases are instructive here.  For a large Biot number $\pi_h$, the
dimensionless flux $\pi_q$ approaches one, so $q\simeq k(T_0-T_{fl})/\delta$,
the pure conduction result.  For a small Biot number, dimensionless flux is
approximately equal to the Biot number $\pi_q\simeq\pi_h$, so $q\simeq
h(T_0-T_{fl})$, the pure convection result.

Again, the purpose of this process is to simplify the functional relationship
and characterize it in as few parameters as possible.  If the equation has no
analytical solution, the functional relationship could be obtained from a
series of experiments, whose results would then apply to any other
conduction-convection problem of the same nature.

\paragraph{Physical modeling}

Furthermore, this principle forms the basis of physical modeling.  For example,
in a wind tunnel model of flow past an aircraft, the dimensionless drag force
is a function of the Reynolds number (inertial/viscous drag), the Mach number
(velocity/speed of sound) and the shape.  If the model aircraft shape is right,
and the pressure and velocity are adjusted to give the correct Reynolds and
Mach numbers, then every detail of flow past the model in the wind tunnel will
be identical to flow past the full-scale aircraft at its cruising altitude, and
the dimensionless drag force will be identical as well.

A materials processing analogue is water modeling: in many processes, one can
use a water tank with the same Reynolds number to model flow in the real
process.

\paragraph{Resolving ambiguity}

There are very often multiple possible choices of variables to eliminate, and
choices of parameter forms.  For example, in the above case, though one {\em
  could not} keep only $q$ and $T_0-T_{fl}$ and eliminate $h$, $k$ and $\delta$
because they are not dimensionally independent, one {\em could} keep $q$ and
$h$ as above, or $q$ and $k$, or $q$ and $\delta$.  The latter two would have
yielded dimensionless pi groups as follows:
\begin{equation}
  \label{eq:q-k}
  {\rm Keep\ }q,k: \pi_q = \frac{q}{h(T_0-T_{fl})}, \pi_k = \frac{k}{h\delta};
\end{equation}
\begin{equation}
  \label{eq:q-d}
  {\rm Keep\ }q,\delta: \pi_q = \frac{q}{h(T_0-T_{fl})},
  \pi_\delta = \frac{h\delta}{k}.
\end{equation}
Note that these give the same $\pi_q$.  The $\pi_\delta$ expression is more
attractive than $\pi_k$ because it has more parameters in the numerator; this
is why the Biot number is written this way.  But which $\pi_q$ is better: this
or the one obtained above?
\begin{equation}
  \label{eq:piq2}
  \pi_q = \frac{q}{h(T_0-T_{fl})} =
  \frac{T_0-T_{fl}}{\frac{1}{h} + \frac{\delta}{k}} \frac{1}{h(T_0-T_{fl})} =
  \frac{1}{1+\pi_\delta},
\end{equation}
which looks like:
\begin{center}
  \PSbox{buckpi-graph2.ps}{355pt}{128pt}
\end{center}
Looking at the asymptotics, it is clear here that as $\pi_\delta$ (the Biot
number) approaches zero, $\pi_q\rightarrow1$ so $q\simeq h(T_0-T_{fl})$, the
convection-limited case.  What is less clear is the asymptotic behavior for
large Biot number: it clearly goes to zero, but that it approaches
$1/\pi_\delta$, so $q\simeq k(T_0-T_{fl})/\delta$, is not immediately apparent
from the graph.  Thus the former nondimensionalization is preferred, as its
asymptotic behavior at both ends is more readily apparent.

\paragraph{Second example: drag force}

In flow past an object with circular cross section, such as a sphere,
ellipsoid, or disc perpendicular to flow, drag force on the object $F_d$ is a
function of fluid density $\rho$ and viscosity $\mu$, far stream flow velocity
$U_\infty$, cross section diameter $d$,\footnote{For some reason, diameter is
  nearly always used as the length dimension of a sphere, cylinder, etc. in
  fluid flow, whereas radius is nearly always used in heat transfer.} and
object shape:
\begin{equation}
  \label{eq:dragfunc}
  F_d = f (U_\infty, \rho, \mu, d, {\rm shape}).
\end{equation}
Putting aside shape for the moment, which we can assume is dimensionless
(aspect ratio, etc.), one can say {\em a priori} that force will probably be
proportional to either inertial force $\frac{1}{2}\rho U^2$ (a.k.a. dynamic
pressure), or viscous shear $\mu U/d$, so the best dimensionless force will be
a ratio of $F_d$ to one of those two.  Dimensional analysis results in one of
three pairs of dimensionless groups:
\begin{center}
  \begin{tabular}{|c|c|c|} \hline
    Variables kept & Dimensionless $F_d$ & Dimensionless other \\ \hline
    $F_d, \mu$     & $\pi_F = \frac{F_d}{\rho U^2d^2}$ &
    $\pi_\mu = \frac{\mu}{\rho Ud}$ \\
    $F_d, \rho$    & $\pi_F = \frac{F_d}{\mu Ud}$ &
    $\pi_\rho = \frac{\rho Ud}{\mu}$ \\
    $F_d, U$ or $F_d, d$ & $\pi_F = \frac{F_d\rho}{\mu^2}$ &
    $\pi_{U/d} = \frac{\rho Ud}{\mu}$ \\ \hline
  \end{tabular}
\end{center}
Based on our initial understanding of the physics, the third dimensionless
force $\frac{F_d\rho}{\mu^2}$ is clearly not helpful.  And the best
dimensionless independent group is clearly $\frac{\rho Ud}{\mu}$, also known as
the Reynolds number, the ratio of inertial to viscous force.  But the best
expression for $\pi_F$, the dimensionless force, is less clear: should it be
the ratio with respect to dynamic pressure and area $\rho U^2L^2$, or viscous
drag $\mu UL$?

As it turns out, at low Reynolds number, viscous drag has an analytical
expression which is not sensitive to shape (or roughness), whereas at high
Reynolds number, boundary layer separation and turbulent flow are very
sensitive to them.  So using $\pi_F = F_d/(\rho U^2d^2)$ results in a single
curve at low Reynolds number, and multiple nearly flat lines for different
shapes at high Reynolds number.  In contrast, $\pi_F = \frac{F_d}{\mu Ud}$
results in a single flat line at low Reynolds number, and multiple curves at
high Reynolds number.  Multiple flat lines are easier to communicate than
multiple curves, so dynamic pressure is the better quantity for normalizing the
force.  This dimensionless force scaled to the dynamic pressure and cross
section area is called the {\em friction factor} given by:
\begin{equation}
  \label{eq:frictfact}
  f = \frac{F_d}{\frac{1}{2}\rho U_\infty^2 \cdot \frac{1}{4}\pi d^2} = f({\rm
    Re,\ shape}).
\end{equation}

Thus for low Reynolds number (less than 0.1), $f=24/{\rm Re}$, the analytical
result with no dependence on shape.  For high Reynolds number ($10^3-10^5$),
the friction factor is roughly independent of Reynolds number, and defines the
{\em drag coefficient} $c_d$: $f=c_d({\rm shape})$.  For a sphere, $c_d=0.44$;
for a disc perpendicular to the flow direction, $c_d\simeq 1$.

We can draw two lessons from this.  First, a brief assessment of the physics
beforehand can inform the process of dimensional analysis.  Second, where there
are still ambiguities after dimensional analysis, plotting the dimensionless
data can help to determine which is the most useful choice in the long run.

\end{document}