The flow of water at room temperature (\mu =1.0\times 10^{-3} N s/m^{2}) between parallel plates need not be due to a pressure gradient; one can also generate a flow by moving the plates relative to each other while maintaining the gap distance at a constant value. Thus, consider the flow in Fig. 9.7, where the top plate is moving at a constant velocity U_{0}=0.1131 m/s with no pressure gradient in the x direction. The fluid layer is 2 mm thick and the plate is 1 m wide. Use the Navier–Stokes equation for Newtonian flows to find (a) the velocity field, (b) the volumetric flow rate, and (c) the shear stress field.
Question 9.2: The flow of water at room temperature (μ=1.0×10^-3 N s/m^2)b...
Given
U_{0}=0.1131\frac{m}{s}, h=2mm, w=1m, \frac{dp}{dx}=0\frac{Pa}{m}.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)
\hat{i}:-\frac{\partial p}{\partial x}+\mu \left(\frac{\partial^{2}\nu_{x}}{\partial x^{2}}+\frac{\partial^{2}\nu_{x}}{\partial y^{2}}+\frac{\partial^{2}\nu_{x}}{\partial z^{2}} \right)+\rho g_{x}
=\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)
\hat{k}:-\frac{\partial p}{\partial z}+\mu \left(\frac{\partial^{2}\nu_{z}}{\partial x^{2}}+\frac{\partial^{2}\nu_{z}}{\partial y^{2}}+\frac{\partial^{2}\nu_{z}}{\partial z^{2}} \right)+\rho g_{z}
=\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)
\mu \frac{\partial ^{2}\nu _{x}}{\partial y^{2}}=0, -\frac{\partial p}{\partial y}=0, -\frac{\partial p}{\partial z}=0.
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
\mu \frac{d^{2}\nu _{x}}{dy^{2}}=0.Integrating twice with respect to y, we obtain
\mu \nu _{x}=c_{1}y+c_{2}.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
0=c_{1}(0)+c_{2}\rightarrow c_{2}=0,\mu U_{0}=c_{1}(h)+0\rightarrow c_{1}=\frac{\mu U_{0}}{h}.
Thus, the velocity profile is
\nu _{x}(y)=\frac{U_{0}}{h}y,which is called a Couette flow. The volumetric flow rate Q, with n=\hat{i}, is thus given by
Some Exact Solutions
Q=\int_{A}^{}{\nu _{x}dA}=\int_{0}^{h}{}\int_{0}^{w}{\frac{U_{0}}{h} }y dz dy\rightarrow Q=\frac{U_{0}wh}{2}.To calculate the shear stress, recall again that for Newtonian fluids,
\sigma _{yx}=2\mu \left[\frac{1}{2}\left(\frac{\partial \nu _{x}}{\partial y}+\frac{\partial \nu _{y}}{\partial x} \right) \right],where, consistent with the assumptions, \nu_{y}\equiv 0. Hence,
\sigma _{yx}=\mu \frac{\partial \nu _{x} }{\partial y}\rightarrow \sigma _{yx}=\mu \frac{U_{0}}{h},or in terms of the volumetric flow rate Q,
\sigma _{yx}=\frac{2\mu Q}{wh^{2}}.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
Q=\frac{1}{2}\left(0.1131\frac{m}{s} \right)(1 m)(0.002 m)=0.000113\frac{m^{3}}{s}=113\frac{mL}{s},\left|\tau _{w}\right|=\frac{\mu U_{0}}{h}=\frac{(1.0\times 10^{-3}N s/m^{2})(0.1131 m/s)}{0.002 m}\approx 0.0566\frac{N}{m^{2}}=0.0566Pa.
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.
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