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\textbf{Question}

A viscous fluid of constant density $\rho$ and constant kinematic
viscosity $\nu$ flows steadily in a two-dimensional channel $-\infty <
x < \infty$, $0 \le z \le h$. There are no body fporces. Thwo solid
planes at $z=0$ and $z=h$ confine the fluid and flow is produced by a
pressure gradient $p_x=-A$ where $A$ is a constant. By assuming a
velocity of the form $\un{q}=(u(z),0,0)$ show that the velocity is
given by
$$u = -\frac{A}{2\mu}(z^2-zh)$$
and find the mass flow per unit width of the channel.

The EZflow ice cream company makes fancy ice cream products in a large
production room at the heart of the EZfood Corporation factory. Ice
cream is feed to the production room by two different channels, both
of which may be considered to be two-dimensional channels of depth
$h$. Ashok makes products from the output of channel $A$, whilst Betty
makes products from the output of channel $B$. The ice cream in
channel $A$ has constant density $\rho_A$ and constant kinematic
viscosity $\nu_A$ and is driven by a constant pressure gradient $-A$.

One wall of channel $B$ lies against a heat exchanger, so that though
the ice cream in channel $B$ has constant density $\rho_B$, its
kinematic viscosity is given by $\nu_B e^{-\alpha z}$ where $\nu_B$
and $\alpha$ are constants and $0 \le z le h$. By again assuming a
solution of the form $\un{q} = (u(z),0,0)$ find an exact solution to
the Navier-Stokes equations in channel $B$. Use your exact solution to
determine the (constant) pressure gradient $-B$ that must be imposed
in channel $B$ if Ashok and Betty are to be supplied with the same
amount of ice cream, thereby avoiding a damaging industrial dispute.

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\textbf{Answer}

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Seek $\underline{q}=(u(y),0,0)^T$. Then certainly
$div(\underline{q})=0$ OK.

Now we have steady flow, $\underline{q}_t=0$.
$\underline{q}=(u(y),0,0)^T$ $\Rightarrow$
$(\underline{q}.\nabla)\underline{q}=0$.

Also $p_x=-A$. So Navier-Stokes $\Rightarrow$

$0 = A/\rho + \nu(u_{xx}+u_{yy}+u_{zz}) = A/\rho + \nu u_{zz}$

$\displaystyle \left. \begin{array}{ll} 0 = -py/\rho + \nu(0)\\ 0
= -px/\rho + \nu(0) \end{array} \right \} $ $p=p(x)$ only

$\Rightarrow$  $\displaystyle \frac{-A}{\mu}=u_{zz}$
$\displaystyle \Rightarrow u_z=\frac{-Az}{\mu}+K_1$,
$\displaystyle u=\frac{-Az^2}{2\mu}+K_1z+K_2$


No slip $\Rightarrow$  $u(0)=u(h)=0$  $\Rightarrow$  $K_2=0$,
$\displaystyle 0=\frac{-Ah^2}{2\mu}+K_1h$

$\Rightarrow$  $\displaystyle u=\frac{-A}{2\mu}(z^2-zh)$

Volume flow in channel $\displaystyle = \rho\int_0^h u\,dz.$

$\displaystyle =\frac{-A\rho}{2\mu}\int_0^h (z^2-zh)\,dz =
\frac{-A\rho}{2\rho} \left [ \begin{array}{l} \displaystyle
\frac{z^3}{3}-\frac{z^2h}{2} \end{array} \right ] _0^h=\frac{A\rho
h^3}{2\mu b}=\frac{A\rho h^3}{12\mu}$

Now consider the Ice-cream factory:-

By the above, Ashok's tube has mass flow

$\displaystyle M_A=\frac{Ap_Ah^3}{12\mu _A}$

Now in channel B, we have, with $\underline{q}=(u(z),0,0)^T$ and
$p_A=-B$ say

$\displaystyle 0=\frac{B}{\rho_B}+\nu_Be^{-\alpha z}u_{zz}$

and of course $div(\underline{q})$

$\displaystyle \Rightarrow \frac{-B}{\rho_b \nu_B}e^{\alpha
z}=u_{zz}$

$u_z=\frac{-Be^{\alpha z}}{\alpha \rho_B \nu_B} + K_1$, \
$\displaystyle u=\frac{-Be^{\alpha z}}{\alpha^2 \rho_B \nu_B}$

$\left. \begin{array}{ll} \displaystyle u(0)=0 \Rightarrow
0=\frac{-B}{\alpha^2 \rho_B \nu_B}+K_2, &K_2= \frac{B}{\alpha^2
\rho_B \nu_B}\\ \displaystyle u(h)=0 \Rightarrow
0=\frac{-Be^{\alpha h}}{\alpha^2 \rho_B \nu_B}
+K_1h+\frac{B}{\alpha^2 \rho_B \nu_B}, &K_1= \frac{-B(1-e^{\alpha
h})}{\alpha^2 \rho_B \nu_B h} \end{array} \right.$

$\displaystyle u=\frac{-Be^{\alpha z}h}{\alpha^2 \rho_B \nu_B
h}-\frac{Bz(1-e^{\alpha h})}{\alpha^2 \rho_B \nu_B h} +
\frac{Bh}{\alpha^2 \rho_B \nu_B h}$

$\displaystyle \Rightarrow u=\frac{B}{\alpha^2 \rho_B \nu_B
h}(h-z-he^{\alpha z}+ze^{\alpha h})$

So $\displaystyle M_B=\rho_B\int_0^h u\,dz = \frac{B}{\alpha^2
\rho_B \nu_B h}(h^2-\frac{h^2}{2}-\frac{h}{\alpha}(e^{\alpha
h}-1)+\frac{h^2}{2}e^{\alpha h})$

\begin{eqnarray*} \displaystyle
M_B & = & \frac{B \rho_B}{\alpha^2 \mu_B
h}(\frac{h^2}{2}(1+e^{\alpha h})-\frac{h}{\alpha}(e^{\alpha
h}-1))\\ & = & \frac{B \rho_B}{\alpha^2
\mu_B}(\frac{h}{2}(1+e^{\alpha h})-\frac{1}{\alpha}(e^{\alpha
h}-1))
\end{eqnarray*}

So $\displaystyle M_B=M_A$ when $\displaystyle
\frac{B\rho_B}{\alpha^2 \mu_B}(\frac{h}{2}(1+e^{\alpha
h})-\frac{1}{\alpha}(e^{\alpha h}-1))=\frac{A\rho_A h^3}{12\mu_A}$

$\displaystyle B=\frac{A\rho_A h^3 \alpha^2 \mu_B}{\rho_B
12\mu_A}\frac{1}{(\frac{h}{2}(1+e^{\alpha
h})-\frac{1}{\alpha}(e^{\alpha h}-1))}$

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