Question 2.8: Shear Stress in Simple COUETTE Flow Consider Couette-flow, i......
Shear Stress in Simple COUETTE Flow
Consider Couette-flow, i.e., a viscous fluid between two parallel plates a small gap h apart, where the upper plate moves with a constant velocity \mathrm{u}_o, in general due to a tangential force, F_{pull}.
Note: The experimentally observed boundary condition for a conventional fluid at any solid surface demands that:
\rm\vec{\mathrm{v}} _{fluid}=\vec{\mathrm{v}} _{wall} (E.2.8.1)
where in rectangular coordinate \vec{\mathrm{v}}=(\mathrm{u,v,w})\,\text{or}\,\vec{\mathrm{v}}=\mathrm{u\hat i+v\hat j +w \hat k.}
Applying Eq. (E.2.8.1) for the present case (see System Sketch), we have for a stationary solid wall (y = 0) :
{\mathrm{u}}_{fluid} = 0 < “no – slip” condition > (E.2.8.2a)
For the moving wall (y = h) :
{\mathrm{u}}_{fluid} = \mathrm{u}_0 <“no-slip” condition> (E.2.8.2b)
and the normal velocity component at both walls is:
{\mathrm{v}}_{fluid} = 0 <“no-penetration” condition> (E.2.8.2c)
| Approach | Assumptions | Sketch |
| • Reduced N-S equations based on assumptions and boundary conditions | • Steady laminar fully-developed (unidirectional) flow |  |
| • Constant fluid properties |
Translating the problem statement plus assumptions into mathematical shorthand, we have:
• Movement of the upper plate (\mathrm{u}_0 = constant) keeps the viscous fluid between the plates in motion via frictional effects propagating normal to the plate; hence, the usual “driving force” \frac{\partial \mathrm{p}}{\partial \mathrm{x}} \equiv 0.
• Steady flow⇒all time derivatives are zero, i.e., \frac{\partial}{\partial \mathrm{t}} \equiv 0.
• Laminar unidirectional flow⇒only one velocity component dependent on one dimension (1-D), is non-zero, i.e., \vec{\mathrm{v}} = (u,0,0) , where u = u(y) only. This implies parallel or fully-developed flow where \frac{\partial }{\partial \mathrm{x}} \equiv 0 and hence v = 0 .
In summary, we can postulate that
• u = u(y), v = w = 0;
• \frac{\partial \mathrm{u}}{\partial \mathrm{x}} = 0;\,\frac{\partial \mathrm{p}}{\partial \mathrm{x}}=0;\,\mathrm{g_x}=0
Checking Eqs. (2.23a–c) with these postulates, we realize the following:
(Continuity) \rm\frac{\partial u}{\partial x} +\frac{\partial v}{\partial y} =0
(x-momentum) \rm u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y} =-\frac{1}{\rho} \frac{\partial p}{\partial x} +\nu\left\lgroup\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \right\rgroup +g_x (2.23a–c)
(y-momentum) \rm u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y} =-\frac{1}{\rho} \frac{\partial p}{\partial y} +\nu\left\lgroup\frac{\partial^2v}{\partial x^2}+\frac{\partial^2v}{\partial y^2} \right\rgroup +g_y
• Continuity equation confirms: 0+\frac{\partial \mathrm{v}}{\partial \mathrm{y}}=0\,\succ \,\underline{\mathrm{v}=0}; (E.2.8.1)
or better, \mathrm{v}=0\Rightarrow \frac{\partial\mathrm{u}}{\partial \mathrm{x}} =0 ; i.e., fully-developed flow.
• x-momentum yields: 0+0=0+ \nu (0+\frac{\partial^2 {u}}{\partial {y}^2} )+0 (E.2.8.2)
and
• y-momentum collapses to: \frac{\partial \mathrm{p}}{\partial \mathrm{y}}=ρ\mathrm{g}_\mathrm{y} < fluid statics > (E.2.8.3)
Thus, Eq. (E.2.8.2) can be written as
\frac{\partial^2 \mathrm{u}}{\partial \mathrm{y}^2}=0 (E.2.8.4a)
subject to the “no-slip” conditions
u(y = 0) = 0 and u(y = h) = \mathrm{u}_0 (E.2.8.4b,c)
Double integration of (E.2.8.4a) and inserting the B.C.s (E.2.8.4b,c) yields:
\rm u(y)=u_0\frac{y}h (E.2.8.5)
Of the stress tensor (Eq. (2.19b)), \rm\tau _{ij}=\mu (\frac{\partial \mathrm{u}_i}{\partial \mathrm{x}_j}+\frac{\partial \mathrm{u}_j}{\partial \mathrm{x}_i} ) , only \tau _{\mathrm{yx}} is non-zero, i.e.,
\mathrm{\tau }_{\mathrm{{ij}}}=\mu({\frac{\partial{\mathrm{v}}_{\mathrm{{i}}}}{\partial\mathrm{x}_{\mathrm{{j}}}}}+{\frac{\partial\mathrm{v}_{\mathrm{{j}}}}{\partial\mathrm{x}_{\mathrm{{i}}}}})=\mu\dot{\gamma _{\mathrm{{ij}}}} (2.19b)
\tau _{\mathrm{xy}}=\mu (\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{v}}{\partial \mathrm{x}} ) (E.2.8.6)
With v ≡ 0 and Eq. (E.2.8.5)
\tau _{\mathrm{xy}}=\mu\,\mathrm{u}_0/h=¢Of the vorticity tensor \mathrm{\omega _{ij}}=\frac{\partial \mathrm{u}_i}{\partial \mathrm{x}_j}-\frac{\partial \mathrm{u}_j}{\partial \mathrm{x}_i} (App. A), only \mathrm{\omega _{yx}} is nonzero, i.e.,
\omega _{\mathrm{yx}}=\frac{\partial \mathrm{u}}{\partial \mathrm{y}}-\frac{\partial \mathrm{v}}{\partial \mathrm{x}} :\,=\mathrm{u}_0/h=¢ (E.2.8.7)
which implies that the fluid elements between the plates rotate with constant angular velocity \mathrm{\omega _{yx}}, while translating with u(y).
Note: The wall shear stress at the upper (moving) plate is also constant, i.e.,
so that
Profiles:
Comments: In the absence of a pressure gradient, only viscous effects set the fluid layer into (linear) motion. The necessary “pulling force” is inversely proportional to the gap height, i.e., the thinner the fluid layer the larger is the shear stress and hence \rm F_{pull}.
