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

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:

\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).