Question 9.2: The flow of water at room temperature (μ=1.0×10^-3 N s/m^2)b...

Given

Assume
1. Newtonian fluid ($\mu =$constant)
2. Incompressible flow $(\triangledown \cdot \nu =0)$

FIGURE 9.7 Couette flow induced by the relative motion of an upper plate, at constant velocity $U_{0},$ relative to a fixed rigid plate at the bottom. The associated velocity profile in the fluid is linear and, consequently, the fluid shear stress is uniform.

3. Steady flow $(\partial \nu /\partial t =0 )$
4. Unidirectional flow $(\nu _{y}=\nu _{z}=0)$
5. Negligible body forces $(g=0)$
6. Fully developed flow $(\partial \nu /\partial x =0 )$
7. 1-D flow $(\partial \nu _{x}/\partial z =0,  \partial \nu _{x}/\partial x =0 )$
8. No pressure gradient in $x(\partial p/\partial x =0 )$

Mass balance is given by Eq. (9.3); it is again satisfied identically because $\nu =\nu _{x}(y)\hat{i}$   only. The balance of linear momentum in Cartesian coordinates is given by Eqs. (9.4)–(9.6). Eliminate the terms that disappear given the above assumptions and show that we have

$\frac{\partial \nu _{x}}{\partial x}+\frac{\partial \nu _{y}}{\partial y}+\frac{\partial \nu _{z}}{\partial z}=0,$            (9.3)

$=\rho \left(\frac{\partial \nu _{x}}{\partial t}+\nu _{x}\frac{\partial \nu _{x}}{\partial x}+\nu _{y}\frac{\partial \nu _{x}}{\partial y}+\nu _{z}\frac{\partial \nu _{x}}{\partial z} \right),$                            (9.4)

$=\rho \left(\frac{\partial \nu _{z}}{\partial t}+\nu _{x}\frac{\partial \nu _{z}}{\partial x}+\nu _{y}\frac{\partial \nu _{z}}{\partial y}+\nu _{z}\frac{\partial \nu _{z}}{\partial z} \right).$          (9.6)

The second and third equations, together with assumption 8, reveal that p=constant. Hence, in the absence of a pressure-driven flow, the governing differential equation of motion is

Integrating twice with respect to y, we obtain

Invoking the no-slip condition at the bottom plate, $\nu_{x}(y=0)=0,$ and the no-slip condition at the top plate, $\nu_{x}(y=h)=U_{0},$ we find that

Thus, the velocity profile is

which is called a Couette flow. The volumetric flow rate Q, with $n=\hat{i},$  is thus given by

Some Exact Solutions

To calculate the shear stress, recall again that for Newtonian fluids,

where, consistent with the assumptions, $\nu_{y}\equiv 0.$ Hence,

or in terms of the volumetric flow rate Q,

In contrast to the pressure-driven flow wherein $\sigma _{yx}$ varied with position y and indeed went to zero at $y=h/2,$ we see that $\sigma _{yx}$ is constant in this Couette flow. Moreover, because the computed value of $\sigma _{yx}$ is everywhere positive, each fluid element experiences a simple shear (cf. Fig. 7.7). The wall shear stress $\tau _{w}$ is equal and opposite the fluid shear stress at y=0 and h. As we emphasize throughout, although the problem statement requires a specific computation, it is always better to first solve the problem generally. Now that we have the general relations, we can substitute the numerical values given in the problem statement into our equations for the volumetric flow rate and the shear stress. They are respectively

We will discover in Sect. 9.3.3 that this simple (general) solution has important implications in various real-world problems, including determination of the viscosity of a fluid.