%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Finite-Difference Temperature Prediction}
\label{chap:finite_diff}

This appendix shows, in more detail, the numerical technic used in
Section~\ref{sec:temp_predict} to predict the temperatures at different points of the
tool for known tool-chip interface mean temperature and dimensions; it uses the
finite-difference method applied for heat conduction. Information on finite-difference
theory applied to heat transfer problems can be found in published
material~\cite{holman02}.

The cutting tool geometry was simplified into a rectangular block with dimensions $X$,
$Y$, and $Z$, and this block was further divided into smaller rectangles, forming a
grid of $m + 1$, $n + 1$, and $o + 1$ elements in the $x$, $y$, and $z$ direction,
respectively (Figure~\ref{fig:finite_diff}).
\begin{figure}[h]
   \center{\includegraphics{finite_diff.eps}}
   \caption{Three dimension geometry of cutting tool for the
   finite-difference prediction of the temperature distribution.}
   \label{fig:finite_diff}
\end{figure}
To start formulating the finite difference equations, one must take into account the
heat that flows into each element $(i, j, k)$ ($i = 1\ldots m + 1$, $j = 1\ldots n +
1$, and $k = 1\ldots o + 1$): $\dot Q_1$ and $\dot Q_2$, with positive and negative
directions along the $x$-axis, respectively; $\dot Q_3$ and $\dot Q_4$, with positive
and negative directions along the $y$-axis, respectively; $\dot Q_5$ and $\dot Q_6$,
with positive and negative directions along the $z$-axis, respectively; i.e.\ the heat
flowing into element $(i, j, k)$ from the 6 adjacent elements $(i - 1, j, k)$, $(i + 1,
j, k)$, $(i, j - 1, k)$, $(i, j + 1, k)$, $(i, j, k - 1)$, and $(i, j, k + 1)$. If no
heat is generated inside the element, the following energy balance applies:
\begin{eqnarray}
   \dot Q_1 + \dot Q_2 + \dot Q_3 + \dot Q_4 + \dot Q_5 + \dot Q_6 = 0
   \label{eq:energy_balance}
\end{eqnarray}
Depending on the location of the element, two mechanisms oh heat transfer can be
considered: heat conduction between neighbouring elements ($\dot Q_\mr{cond}$) and heat
convection between the peripheral elements and the air surrounding the block ($\dot
Q_\mr{conv}$). The equation for $\dot Q_\mr{cond}$ is as follows:
\begin{eqnarray}
   \dot Q_\mr{cond} = \frac{A}{d}\kappa(T - T')
   \label{eq:q_cond}
\end{eqnarray}
where $A$ is the common heat transfer area between elements, $d$ is the distance between
elements, $\kappa$ is the thermal conductivity of the material, $T$ is the temperature of
the element of interest, and $T'$ is the temperature of the neighbouring element. The
equation for $\dot Q_\mr{conv}$ is defined as follows:
\begin{eqnarray}
   \dot Q_\mr{conv} = Ah(T - T_\infty)
   \label{eq:q_conv}
\end{eqnarray}
where $A$ is the area of contact between the element and the surrounding air, $h$ is the
convective heat transfer coefficient, $T$ is the temperature of the element, and
$T_\infty$ is the air temperature far away from the element. Moreover, by using
Equation~\ref{eq:energy_balance}, a general relationship that relates the temperature of
any element $(i, j, k)$ with the temperatures from the six neighbouring elements can be
established:
\begin{eqnarray}
   \begin{array}{l}
      T^{N + 1, i, j, k} = \frac{
      C_1^{i, j, k}T^{N, i - 1, j, k} + C_2^{i, j, k}T^{N, i + 1, j, k} +
      C_3^{i, j, k}T^{N, i, j - 1, k} + C_4^{i, j, k}T^{N, i, j + 1, k} +
      C_5^{i, j, k}T^{N, i, j, k - 1} + C_6^{i, j, k}T^{N, i, j, k + 1}}
      {C_1^{i, j, k} + C_2^{i, j, k} + C_3^{i, j, k} +
      C_4^{i, j, k} + C_5^{i, j, k} + C_6^{i, j, k}}
   \end{array}
    \label{eq:finite_diff}
\end{eqnarray}
Before starting the numerical process, and since the temperatures of the elements are
still unknown, some initial value must be assigned to the temperatures of each element
$(i, j, k)$. Subsequently, Equation~\ref{eq:finite_diff} can be used to predict the
values of $T^{N + 1, i, j, k}$ one iteration ahead ($N + 1$) from the temperature of
its neighbouring elements. This procedure may be repeated until the temperatures start
to converge to its final values, i.e.\ when the convergence criterion that follows is
met:
\begin{eqnarray}
   \left| \frac{T^{N + 1, i, j, k} - T^{N, i, j, k}}
   {T^{N, i, j, k}}\right| \leq e
   \label{eq:conv_criterion}
\end{eqnarray}
where $e$ is the convergence error criterion. Moreover, since the values of $C_1^{i, j,
k}$, $C_2^{i, j, k}$, $C_3^{i, j, k}$, $C_4^{i, j, k}$, $C_5^{i, j, k}$, and $C_6^{i,
j, k}$ are associated with the heat flows $\dot Q_1$, $\dot Q_2$, $\dot Q_3$, $\dot
Q_4$, $\dot Q_5$, and $\dot Q_6$, their values are dependent on the location of the
element $(i, j, k)$ in the block. Next, the values of $C_1$ and $C_2$ that involve
convection are stated:
\begin{eqnarray}
   \begin{array}{lll}
      C_1^{1, n + 1, 1} =
      C_2^{m + 1, 1, 1} =
      C_2^{m + 1, n + 1, 1} & = &
      \frac{\Delta y \Delta z_1}{4}h
      \\
      C_1^{1, 1, o + 1} =
      C_1^{1, n + 1, o + 1} =
      C_2^{m + 1, 1, o + 1} =
      C_2^{m + 1, n + 1, o + 1} & = &
      \frac{\Delta y \Delta z}{4}h
      \\
      C_1^{1, 2\ldots n, 1} =
      C_2^{m + 1, 2\ldots n, 1} & = &
      \frac{\Delta y \Delta z_1}{2}h
      \\
      C_1^{1, 2\ldots n, o + 1} =
      C_1^{1, 1, 3\dots o} =
      C_1^{1, n + 1, 3\ldots o} & = & \\
      C_2^{m + 1, 2\ldots n, o + 1} =
      C_2^{m + 1, 1, 3\ldots o} =
      C_2^{m + 1, n + 1, 3\ldots o} & = &
      \frac{\Delta y \Delta z}{2}h
      \\
      C_1^{1, n + 1, 2} =
      C_2^{m + 1, 1, 2} =
      C_2^{m + 1, n + 1, 2} & = &
      \frac{\Delta y (\Delta z_1 + \Delta z)}{4}h
      \\
      C_1^{1, 2\ldots n, 3\ldots o} =
      C_2^{m + 1, 2\ldots n, 3\ldots o} & = &
      \Delta y \Delta z h
      \\
      C_1^{1, 2\ldots n, 2} =
      C_2^{m + 1, 2\ldots n, 2} & = &
      \frac{\Delta y (\Delta z_1 + \Delta z)}{2}h
   \end{array}
   \label{eq:}
\end{eqnarray}
The values of $C_1$ and $C_2$ that involve conduction:
\begin{eqnarray}
   \begin{array}{lll}
      C_1^{m_1 + 2\ldots m + 1, 1, 1} =
      C_1^{2\ldots m + 1, n + 1, 1} =
      C_2^{m_1 + 2\ldots m, 1, 1} =
      C_2^{1\ldots m, n + 1, 1} & = &
      \frac{\Delta y \Delta z_1}{4\Delta x}
      \\
      C_1^{2\ldots m + 1, 1, o + 1} =
      C_1^{2\ldots m + 1, n + 1, o + 1} =
      C_2{1\ldots m, 1, o + 1} =
      C_2^{1\ldots m, n + 1, o + 1} & = &
      \frac{\Delta y \Delta z}{4\Delta x}\kappa
      \\
      C_1^{2\ldots m + 1, 2\ldots n, 1} =
      C_2^{1\ldots m, 2\ldots n, 1} & = &
      \frac{\Delta y \Delta z_1}{2\Delta x}\kappa
      \\
      C_1^{2\ldots m + 1, 2\ldots n, o + 1} =
      C_1^{2\ldots m + 1, 1, 3\ldots o} =
      C_1^{2\ldots m + 1, n + 1,3\ldots o} & = & \\
      C_2^{1\ldots m, 2\ldots n, o + 1} =
      C_2^{1\ldots m, 1, 3\ldots o} =
      C_2^{1\ldots m, n + 1, 3\ldots o} & = &
      \frac{\Delta y \Delta z}{2\Delta x}\kappa
      \\
      C_1^{m_1 + 2\ldots m + 1, 1, 2} =
      C_1^{2\ldots m + 1, n + 1, 2} =
      C_2^{m_1 + 2\ldots m, 1, 2} =
      C_2^{1\ldots m, n + 1, 2} & = &
      \frac{\Delta y (\Delta z_1 + \Delta z)}{4\Delta x}\kappa
      \\
      C_1^{2\ldots m + 1, 2\ldots n, 3\ldots o} =
      C_2^{1\ldots m, 2\ldots n, 3\ldots o} & = &
      \frac{\Delta y \Delta z}{\Delta x}\kappa
      \\
      C_1^{2\ldots m + 1, 2\ldots n, 2} =
      C_2^{1\ldots m, 2\ldots n, 2} & = &
      \frac{\Delta y (\Delta z_1 \Delta z)}{2\Delta x}\kappa
   \end{array}
\end{eqnarray}
The values of $C_3$ and $C_4$ that involve convection:
\begin{eqnarray}
   \begin{array}{lll}
      C_3^{m + 1, 1, 1} =
      C_4^{1, n + 1, 1} =
      C_4^{m + 1, n + 1} & = &
      \frac{\Delta x \Delta z_1}{4}h
      \\
      C_3^{1, 1, o + 1} =
      C_3^{m + 1, 1, o + 1} =
      C_4^{1, n + 1, o + 1} =
      C_4^{m + 1, n + 1, o + 1} & = &
      \frac{\Delta x \Delta z}{4}h
      \\
      C_3^{m_1 + 2\ldots m, 1, 1} =
      C_4^{2\ldots m, n + 1, 1} & = &
      \frac{\Delta x \Delta z_1}{2}h
      \\
      C_3^{2\ldots m, 1, o + 1} =
      C_3^{1, 1, 3\ldots o} =
      C_3^{m + 1, 1, 3\ldots o} & = & \\
      C_4^{2\ldots m, n + 1, o + 1} =
      C_4^{1, n + 1, 3\ldots o} =
      C_4^{m + 1, n + 1, 3\ldots o} & = &
      \frac{\Delta x \Delta z}{2}h
      \\
      C_3^{m + 1, 1, 2} =
      C_4^{1, n + 1, 2} =
      C_4^{m + 1, n + 1, 2} & = &
      \frac{\Delta x (\Delta z_1 + \Delta z)}{4}h
      \\
      C_3^{2\ldots m, 1, 3\ldots o} =
      C_4^{2\ldots m, n + 1, 3\ldots o} & = &
      \Delta x \Delta x h
      \\
      C_3^{m_1 + 2\ldots m, 1, 2} =
      C_4^{2\ldots m, n + 1, 2} & = &
      \frac{\Delta x (\Delta z_1 + \Delta z)}{2}h
   \end{array}
\end{eqnarray}
The values of $C_3$ and $C_4$ that involve conduction:
\begin{eqnarray}
   \begin{array}{lll}
      C_3^{1, 2\ldots n + 1, 1} =
      C_3^{m + 1, 2\ldots n + 1, 1} =
      C_4^{1, 2\ldots n, 1} =
      C_4^{m + 1, 1\ldots n, 1} & = &
      \frac{\Delta x \Delta z_1}{4\Delta}
      \\
      C_3^{1, 2\ldots  n + 1, o + 1} =
      C_3^{m + 1, 2\ldots  n + 1, o + 1} =
      C_4^{1, 1\ldots n, o + 1} =
      C_4^{m + 1, 1\ldots n, o + 1} & = &
      \frac{\Delta x \Delta z}{4\Delta y}\kappa
      \\
      C_3^{2\ldots m, 2\ldots n + 1, 1} =
      C_4^{2\ldots m_1 + 1, 2\ldots n, 1} =
      C_4^{m_1 + 2\ldots m, 1\ldots n, 1} & = &
      \frac{\Delta x \Delta z_1}{\Delta y}\kappa
      \\
      C_3^{2\ldots m, 2\ldots n + 1, o + 1} =
      C_3^{1, 2\ldots n + 1, 3\ldots o} =
      C_3^{m + 1, 2\ldots n + 1,   3\ldots o} & = & \\
      C_4^{2\ldots m, 1\ldots n, o + 1} =
      C_4^{1, 1\ldots n, 3\ldots o} =
      C_4^{m + 1, 1\ldots n, 3\ldots o} & = &
      \frac{\Delta x \Delta z}{2\Delta y}\kappa
      \\
      C_3^{1, 2\ldots n + 1, 2} =
      C_3^{m + 1, 2\ldots n + 1, 2} =
      C_4^{1, 2\ldots n, 2} =
      C_4^{m + 1, 1\ldots n, 2} & = &
      \frac{\Delta x (\Delta z_1 + \Delta z)}{4\Delta y}\kappa
      \\
      C_3^{2\ldots m, 2\ldots n + 1, 3\ldots o} =
      C_4^{2\ldots m, 1\ldots n, 3\ldots o} & = &
      \frac{\Delta x \Delta z}{\Delta y}\kappa
      \\
      C_3^{2\ldots m, 2\ldots n + 1, 2} =
      C_4^{2\ldots m1 + 1, 2\ldots n, 2} =
      C_4^{m1 + 2\ldots m, 1\ldots n, 2} & = &
      \frac{\Delta x (\Delta z_1 \Delta z)}{2\Delta y}\kappa
   \end{array}
\end{eqnarray}
The values of $C_5$ and $C_6$ that involve convection:
\begin{eqnarray}
   \begin{array}{lll}
       C_5^{m + 1, 1, 1} =
       C_5^{1, n + 1, 1} =
       C_5^{m + 1, n + 1, 1} & = & \\
       C_6^{1, 1, o + 1} =
       C_6^{m + 1, 1, o + 1} =
       C_6^{1, n + 1, o + 1} =
       C_6^{m + 1, n + 1, o + 1} & = &
       \frac{\Delta x \Delta y}{4}h
       \\
      C_5^{m_1 + 2\ldots m, 1, 1} =
      C_5^{2\ldots m, n + 1, 1} =
      C_5^{1, 2\ldots n, 1} =
      C_5^{m + 1, 2\ldots n, 1} & = & \\
      C_6^{2\ldots m, 1, o + 1} =
      C_6^{2\ldots m, n + 1, o + 1} =
      C_6^{1, 2\ldots n, o + 1} =
      C_6^{m + 1, 2\ldots n, o + 1} & = &
      \frac{\Delta x \Delta y}{2}h
      \\
      C_5^{2\ldots m, 2\ldots n, 1} =
      C_6^{2\ldots m, 2\ldots n, o + 1} & = &
      \Delta z \Delta y h
   \end{array}
\end{eqnarray}
Finally, the values of $C_5$ and $C_6$ that involve conduction:
\begin{eqnarray}
   \begin{array}{lll}
      C_5^{1, 1, 3\ldots o + 1} =
      C_5^{m + 1, 1, 3\ldots o + 1} =
      C_5^{1, n + 1, 3\ldots o + 1} =
      C_5^{m + 1, n + 1, 3\ldots o + 1} & = & \\
      C_6^{1, 1, 3\ldots o} =
      C_6^{m + 1, 1, 2\ldots o} =
      C_6^{1, n + 1, 2\ldots o} =
      C_6^{m + 1, n + 1, 2\ldots o} & = &
      \frac{\Delta x \Delta y}{4\Delta z}\kappa
      \\
      C_5^{m + 1, 1, 2} =
      C_5^{1, n + 1, 2} =
      C_5^{m + 1, n + 1, 2} & = & \\
      C_6^{m + 1, 1, 1} =
      C_6^{1, n + 1, 1} =
      C_6^{m + 1, n + 1, 1} & = &
      \frac{\Delta x \Delta y}{4 \Delta z_1}\kappa
      \\
      C_5^{2\ldots m, 1, 3\ldots o + 1} =
      C_5^{2\ldots m, n + 1, 3\ldots o + 1} =
      C_5^{1, 2\ldots n, 3\ldots o + 1} =
      C_5^{m + 1, 2\ldots n, 3\ldots o + 1} & = & \\
      C_6^{2\ldots m, 1, 3\ldots o} =
      C_6^{2\ldots m, n + 1, 2\ldots o} =
      C_6^{1, 2\ldots n, 2\ldots o} =
      C_6^{m + 1, 2\ldots n, 2\ldots o} & = &
      \frac{\Delta x \Delta y}{2\Delta z}\kappa
      \\
      C_5^{m_1 + 2\ldots m, 1, 2} =
      C_5^{2\ldots m, n + 1, 2} =
      C_5^{1, 2\ldots n, 2} =
      C_5^{m + 1, 2\ldots n, 2} & = & \\
      C_6^{m_1 + 2\ldots m, 1, 1\ldots 2} =
      C_6^{2\ldots m, n + 1, 1} =
      C_6^{1, 2\ldots n, 1} =
      C_6^{m + 1, 2\ldots n, 1} & = &
      \frac{\Delta x \Delta y}{2\Delta z_1}\kappa
      \\
      C_5^{2\ldots m, 2\ldots n, 3\ldots o + 1} =
      C_6^{2\ldots m, 2\ldots n, 2\ldots o} & = &
      \frac{\Delta x \Delta y}{\Delta z}\kappa
      \\
      C_5^{2\ldots m, 2\ldots n, 2} =
      C_6^{2\ldots m, 2\ldots n, 1} & = &
      \frac{\Delta x \Delta y}{\Delta z_1}\kappa
   \end{array}
\end{eqnarray}
where $l_\mr{int}$ is the tool-chip contact length, $\Delta x = \frac{X}{m}$, $\Delta y
= \frac{Y}{n}$, $\Delta z_1 = l_\mr{int}$, $\Delta z = \frac{Z - l_\mr{int}}{o - 1}$,
and $m_1 = \frac{w}{\Delta x}$. In addition, since the temperature of the tool-chip
contact area remains constant during the whole process, the following equation must be
combined with Equation~\ref{eq:finite_diff}:
\begin{eqnarray}
   \begin{array}{lll}
      T^{N + 1, i, j, k} = T_\mr{int} & \mbox{for} &
      i \leq \frac{w}{\Delta x} + 0.5 \mbox{ and } j = 1 \mbox{ and } k \leq 2
   \end{array}
   \label{eq:finite_diff_aux}
\end{eqnarray}
Finally, it must be noted that Equations~\ref{eq:finite_diff}
and~\ref{eq:finite_diff_aux} solely define $T^{i, j, k}$ for the rectangular block
shown in Figure~\ref{fig:finite_diff}, i.e.\ $i = 1\ldots m + 1$, $j = 1\ldots n +
1$, and $k = 1\ldots o + 1$. However, in order to complete the model, the convective
heat that flows between the air and the boundaries of the block must also be
specified; this is done as follows:
\begin{eqnarray}
   \begin{array}{lll}
      T^{0, 0\ldots n + 2, 0\ldots o + 2} =
      T^{m + 2, 0\ldots n + 2, 0\ldots o + 2} & = & \\
      T^{1\ldots m + 1, 0, 0\ldots o + 2} =
      T^{1\ldots m + 1, n + 2, 0\ldots o + 2} & = & \\
      T^{1\ldots m + 1, 1\ldots n + 1, 0} =
      T^{1\ldots m + 1, 1\ldots n + 1, o +2} & = & T_\infty
   \end{array}
   \label{eq:air_temp}
\end{eqnarray}

\begin{figure}[h]
   \center{\includegraphics{temp_example.eps}}
   \caption[Example of temperature predictions using the finite-difference method.]
   {Example of temperature predictions using the finite-difference method. (a)~Variation
   of temperature $T_1$ with $N$ until convergence is reached. (b)~Two-dimensional
   temperature distribution on plane $z = 0 \mbox{ mm}$ (temperature units: $^\circ$C).}
   \label{fig:finite_diff_example}
\end{figure}
Figure~\ref{fig:finite_diff_example} shows an example of the temperatures predicted by
the finite-difference method described above for $T_\mr{int} = 1000^\circ$C, $T_\infty
= 25^\circ$C, $w = 1$~mm, $l_\mr{int} = 0.1$~mm, $X = 20$~mm, $Y = 20$~mm, $Z = 20$~mm.
The values of $\kappa$ and $h$ were estimated: \mbox{50~W$/$m-$^\circ$C} and
\mbox{100~W$/$m$^2$-$^\circ$C}, respectively. Convergence was attained for $e =
10^{-7}$ (Equation~\ref{eq:conv_criterion}) for the successive calculated temperatures
obtained at location $x = 0\mbox{ mm}$, $y = 5\mbox{ mm}$, and $z = 0\mbox{ mm}$,
temperature $T_1$. Figure~\ref{fig:finite_diff_example}a shows the variation of $T_1$
for each iteration during the finite-difference computation procedure. It can also be
observed that $T_1$ rises initially quickly from the initial value given to the unknown
elements of the block (150$^\circ$C), converging later slowly to its final value
(234$^\circ$C). Figure~\ref{fig:finite_diff_example}b shows the temperature
distribution after convergence on the $xy$-face of the block that is under the
tool-chip interface.
