# Question 3.15: Simple Couette Flow with Viscous Dissipation As an example o......

Simple Couette Flow with Viscous Dissipation
As an example of planar lubrication with significant heat generation due to oil-film friction, consider simple thermal Couette flow with adiabatic wall and constant temperature of the moving plate.

 Approach Assumptions Sketch • Reduced N–S equations and HT eqs. • Steady laminar 1-D flow • Constant thermal wall cond. • $\rm ∇p = 0; u_0$ and d are constant

Step-by-Step
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Based on the postulates $\rm \vec{\bf v}=[u(y),0,0]$ and ∇p ≡ 0, the Navier– Stokes equations reduce to:

0 = 0 <continunity>

and

$\rm 0=\frac{d^2u}{dy^2}$ <x-momentum>                           (E.3.15.1a)

subject to u(y = 0) = 0 and u(y = d) = $\rm u_0$. Thus,

$\rm u(y)=u_0\frac{y}{d}$                       (E.3.15.1b)

The heat transfer equation (3.64) with $\rm\Phi =\left\lgroup\frac{\partial u}{\partial y} \right\rgroup ^2$ from Eq. (3.65) reduces to:

$\rm\frac{\partial T}{\partial t} +(\vec{\bf v}\cdot\nabla)T=\alpha \,\nabla^2\,T-\frac{\mu}{\rho c_p} \Phi\pm S_{heat}$                              (3.64)
$\Phi={\rm 2\left[\left\lgroup\frac{\partial u}{\partial x} \right\rgroup^2+\left\lgroup\frac{\partial v}{\partial y} \right\rgroup^2+\left\lgroup\frac{\partial w}{\partial z} \right\rgroup^2\right]} \\\qquad{\rm +\left[\left\lgroup\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} \right\rgroup^2 +\left\lgroup\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y} \right\rgroup^2+\left\lgroup\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} \right\rgroup^2\right] }\\\qquad{\rm -\frac{2}{3} \left\lgroup\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} \right\rgroup^2 }$                     (3.65)

$\mathrm{k}{\frac{\mathrm{d}^{2}\mathrm{T}}{\mathrm{d}\mathrm{y^{2}}}}=-\mathrm{\mu}\left\lgroup{\frac{\mathrm{u}_{0}}{\mathrm{d}}}\right\rgroup^2$                                (E.3.15.2)

subject to $\rm \frac{dT}{dy} \Big|_{y=0}=0 ~\text{and}~T(y=d)=T_0$

Double integration yields:

$\begin{array}{c c}\rm {\rm{{ T}({ y})=T_{0}+\frac{\mu{u}_{0}^{2}}{2\mathrm{k}}\left[1-\left\lgroup\frac{\mathrm{y}}{\mathrm{d}}\right\rgroup ^{2}\right]}}\end{array}$                          (E.3.15.3)

At the plate surface $\rm q_s=q(y=d)=-k\frac{\partial T}{\partial y} \Big|_{y=d}$ we have:

$\rm q_s=\mu \frac{u_0^2}{d}$                            (E.3.15.4)

Comments: Clearly, as μ and $\rm u_0$ increase and the spacing decreases, $\rm q_s$ shoots up. For simple Couette flow du/dy evaluated at y = d is equal to $\rm u_0 / d$ so that $\rm q_s = u_0 τ_{wall}$ here, which is a simple example of the heat transfer and momentum transfer relation (see Reynolds–Colburn analogy). Of interest would be the evaluation of the mean fluid temperature, $\rm T_m=\frac{1}{\dot m} \int_A\rho\,u\,T\,dA$, to estimate h from $\rm q_s=h(T_0-T_m).$

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