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\textbf{Question}

An incompressible viscous heat-conduction fluid on constant density
$\rho$ and constant kinematic viscosity $\nu$ flows past a flat plate
at $y=0$. The flow is two-dimensional. Far away from the plate, the
speed of the fluid is $(U,0)^T$ where $U$ is a constant. YOU MAY
ASSUME that for high Reynolds number steady boundary layer flow (with
no body forces) in the region close to the plate the horizontal and
vertical velocity components $u$ and $v$ satisfy the dimensional
equations
\begin{eqnarray*}
uu_x + vu_y & = & \un u_{yy}\\
u_x + v_y & = & 0
\end{eqnarray*}

The fluid in the mainstream flow has temperature $T_0$, and the flat
plate has temperature $T-1 > T_0$. YOU MAY ALSO ASSUME that the
temperature in the fluid obeys the energy equation
$$\rho c_p (uT_x +vT_y) = k(T_{xx} +T_{yy}) + \nu \rho\Phi$$
where
$$\Phi = 2u_x^2+2v_y^2+v_x^2+u_y^2+2v_xu_y$$
$k$ and $c_p$ are constants. By setting $T=T_0 + \overline{T}(T_1 -
T_0)$, where $\overline{T}$ is a dimensionless temperature, and
suitably scaling the other variables, show that the temperature in the
boundary layer is determined by the dimensional equation
$$\rho c_p (uT_x + vT_y) = kT_{yy} + \rho \nu u_y^2.$$

(NOTE: you may assume that the quantities $k/(\mu c_p)$ and
$U^2/(c_p(T_1-T_0))$ are both $O(1)$.)

By seeking a similarity solution to the equations for $u$, $v$ and $T$
of the form
\begin{eqnarray*}
\psi & = & \sqrt{\nu U x} f(\eta)\\
T & = & T_0 + (T_1-T_0)g(\eta)
\end{eqnarray*}
where $\psi$ is the stream function so that $u=\psi_y$, $v=-\psi_x$
and the similarity variable $\eta$ is given by
$$\eta = y \sqrt{\frac{U}{\nu x}},$$
show that $f$ and $g$ satisfy the ordinary differntial equations
\begin{eqnarray*}
f''' + \frac{1}{2} f f'' & = & 0\\
g'' + c_1fg'+c_2f''^2 & = & 0
\end{eqnarray*}
where $'=d/d\eta$ and $c_1$ and $c_2$ are constants that you should
specify. Given suitable boundary conditions for $f$ and $g$.

\newpage
\textbf{Answer}

ASSUME $\begin{array}{lcr} uu_x+vu_y & = & \nu u_{yy}\\ u_x+v_y &
= & 0 \end{array}$

And $\displaystyle \rho c_{\rho}(uT_x+vT_y)=k(T_{xx}+T_{yy})+\nu
\rho \frac{\partial q_i}{\partial x_j}\left ( \frac{\partial
q_j}{\partial x_i} + \frac{\partial q_i}{\partial x_j} \right )$

Now put $x=L\overline{x}$, $y=\epsilon L\overline{y}$,
$u=U\overline{u}$, $ v=\epsilon U\overline{v}$, ($p=\rho U^2
\overline{p}$) and $T=T_0+\overline{T}(T_1-T_0)$

$\displaystyle \Rightarrow \frac{\rho c_{\rho}
U(T_1-T_0)}{L}(\overline{u}\overline{T}_{\overline{x}}+\overline{v}\overline{T}_{\overline{y}})
= \frac{k(T_1-T_0)}{L^2}\left ( \overline{T}_{\overline{x}
\overline{x}} +
\frac{1}{\epsilon^2}\overline{T}_{\overline{y}\overline{y}} \right
) + \nu\rho\Phi$

$\Phi=2u_x^2+2v_y^2+v_x^2+u_y^2+2v_xu_y$

\begin{eqnarray*}
& & \frac{\rho c_{\rho}U(T_1-T_0)}{L}\left (
\overline{u}\overline{T}_{\overline{x}}+\overline{v}\overline{T}_{\overline{y}}
\right )\\
& = & \frac{k(T_1-T_0)}{L^2}\left (
\overline{T}_{\overline{x}
\overline{x}}+\frac{1}{\epsilon^2}\overline{T}_{\overline{y}
\overline{y}} \right )\\
& & + \frac{\nu\rho U^2}{L^2} \left (
2\overline{u}_{\overline{x}}^2+2\overline{v}_{\overline{y}}^2+
\epsilon^2\overline{v}_{\overline{x}}^2 +\frac{1}{\epsilon^2}
\overline{u}_{\overline{y}}^2+ 2\overline{v}_{\overline{x}}
\overline{u}_{\overline{y}} \right )\\
\overline{u}\overline{T}_{\overline{x}}+\overline{v}\overline{T}_{\overline{y}}
& = & \frac{k}{L\rho Uc_{\rho}}\left ( \overline{T}_{\overline{x}
\overline{x}} +\frac{\overline{T}_{\overline{y}
\overline{y}}}{\epsilon^2} + \frac{\nu U}{Lc_{\rho}(T_1-T_0)} \left
( 2\overline{u}_{\overline{x}}^2+2\overline{v}_{\overline{y}}^2
\right. \right.\\
& & \left. \left. +\epsilon^2\overline{v}_{\overline{x}}^2+
\frac{1}{\epsilon^2}\overline{u}_{\overline{y}}^2
+2\overline{v}_{\overline{x}} \overline{u}_{\overline{y}} \right
) \right )
\end{eqnarray*}

Now we were told to assume that $|2|\mu c_{\rho}$,
$U^2/c_p(T_1-T_0)$ were $O(1)$. So TAKE THEM TO BE 1.

\begin{eqnarray*}
\Rightarrow
\overline{u}\overline{T}_{\overline{x}}+\overline{v}\overline{T}_{\overline{y}}
& = & \left ( \frac{\nu}{LU} \right ) \left ( \overline{T}_{\overline{x}
\overline{x}}+\frac{1}{\epsilon^2}\overline{T}_{\overline{y}\overline{y}}
\right )\\
& & + \left ( \frac{\nu}{LU} \right ) \left (
2\overline{u}_{\overline{x}}^2+2\overline{v}_{\overline{x}}^2+\epsilon^2
\overline{v}_{\overline{y}}^2 +
\frac{1}{\epsilon^2}\overline{u}_{\overline{y}}^2+2\overline{v}_{\overline{x}}\overline{u}_{\overline{y}}
\right )
\end{eqnarray*}

Now $\nu/LU = 1/Re$ and the assumption that was used to derive the
momentum boundary layer equations in the fluid was
$\epsilon^2Re=1$.

Thus for $\epsilon \ll 1$, $Re \gg 1$, $\epsilon^2Re=1$ we get

$\overline{u}\overline{T}_{\overline{x}}+\overline{v}\overline{T}_{\overline{y}}
=
\overline{T}_{\overline{y}\overline{y}}+\overline{u}_{\overline{y}}^2$.
Redimensionalising $\Rightarrow$

$\rho c_p(uT_x+vT_y)=kT_{yy}+\rho\nu u_y^2$.

Now seek $\psi=\sqrt{vUx}f(\eta)$, $T=T_0+(T_1-T_0)g(\eta)$, \ \ \
$\left ( \eta=y\sqrt{\frac{U}{\nu x}} \right )$.

$\displaystyle u=\psi_y=Uf'(\eta)$, $\displaystyle
u_y=\psi_{yy}=\frac{u^{3/2}}{\sqrt{\nu x}}f''$, $\displaystyle
u_{yy}=\frac{U^2}{\nu x}f'''$

\begin{eqnarray*}
u_x & = & -\frac{1}{2}x^{-\frac{3}{2}}Uy\sqrt{\frac{U}{\nu}}f''\\
T_x & = & (T_1-T_0)y\sqrt{\frac{U}{\nu}}\left (
-\frac{1}{2} \right )x^{-\frac{3}{2}}g'\\
T_y & = & (T_1-T_0)\sqrt{\frac{U}{\nu x}}g\\
T_{yy} & = & (T_1-T_0)\frac{U}{\nu x}g''\\
v & = & -\psi_x=-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{U\nu}f- \sqrt{\nu
Ux}y(-\frac{1}{2}x^{-\frac{3}{2}})y\sqrt{\frac{U}{\nu}}f'
\end{eqnarray*}

\begin{eqnarray*}
\Rightarrow
& & Uf' \left (
-\frac{1}{2}x^{-\frac{3}{2}}U^{\frac{3}{2}}\nu^{-\frac{1}{2}}f''
\right )\\
& & +U^{
\frac{3}{2}}\nu^{-\frac{1}{2}}x^{-\frac{1}{2}}f'' \left
( -\frac{1}{2}fx^{-\frac{1}{2}}U^{\frac{1}{2}}
\nu^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{3}{2}}yx^{\frac{1}{2}}Uf'
\right )\\
& = & \frac{U^2}{x}f'''
\end{eqnarray*}

\begin{eqnarray*}
& & \rho c_p \left (
Uf'(T_1-T_0)yU^{\frac{1}{2}}\nu^{-\frac{1}{2}}\left
( -\frac{1}{2} \right ) x^{-\frac{3}{2}}g' \right. \\
& & +(T_1-T_0)\sqrt{\frac{U}{\nu x}}g'
\left. \left ( -\frac{1}{2}x^{-\frac{1}{2}}\sqrt{U}\nu
f+\frac{1}{2}x^{-1}yUf' \right ) \right )\\
& = & (T_1-T_0)\frac{U}{\nu x}g''k+\rho\frac{\nu
U^3}{vx}f''^2
\end{eqnarray*}

$\Rightarrow$

$$U^2x^{-1}\left ( -\frac{1}{2}
\right )ff''=\frac{U^2}{x}f'''$$
$$\rho c_{\rho}(T_1-T_0)\sqrt{\frac{U}{\nu x}}g' \left ( -\frac{1}{2}
\right ) x^{-\frac{1}{2}}\sqrt{U\nu}f$$
$$=(T_1-T_0)\left (
\frac{U}{\nu x}\right ) kg''+\frac{\rho U^3}{x}f''^2$$

i.e $\displaystyle \left. \begin{array}{l}
f'''+\frac{1}{2}ff''=0\\ \displaystyle g''+c_1fg'+c_2f''^2
\end{array} \right \}
\begin{array}{l}  c_1+\frac{1}{2}\frac{\mu c_{\rho}}{k}\\ c_2=\mu
U^2/k(T_1-T_0)
\end{array} $

B/C's:- $f(0)=f'(0)=)$ (no slip), $f'(\infty)=1$ (MATCHING)

$g(0)=1$, $g(\infty)=0$


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