For the planar shear flow in Figure 3.15, u = (u1(x2), 0), what is DΓ/Dt when Γ is computed from a small rectangular contour, ABCD − centered on (x1, x2) with sides Δx1 and Δx2 − when ρ and μ are constants?

Question

For the planar shear flow in Figure 3.15, [latex]\mathbf{u}=(u_{1}(x_{2}),0)[/latex], what is DΓ/Dt when Γ is computed from a small rectangular contour, ABCD − centered on [latex](x_{1},x_{2})[/latex] with sides [latex]\Delta x_{1}[/latex] and [latex]\Delta x_{2}[/latex] − when ρ and μ are constants?

Accepted Answer

Evaluate (5.11) to determine DΓ/Dt. The first step is to determine the viscous stress tensor [latex] au_{ij}[/latex] from the given velocity field. This is an incompressible flow (∇·u = 0), so [latex] au_{ij}=2\mu S_{ij}[/latex] and can be written in as a 2-by-2 matrix in terms of the one nonzero velocity gradient. Similarly the net viscous force on a fluid element, [latex]\partial au_{ij}=\partial x_{j}[/latex], can be written as a 2-by-1 row vector:

[latex]\frac{D\Gamma}{Dt}=\underset{C}{\int}\left(\frac{1}{ ho}\frac{\partial au_{ij}}{\partial x_{j}} ight)dx_{i}[/latex], (5.11)

[latex] au_{ij}=2\mu S_{ij}=\mu\begin{bmatrix}0&\partial u_{1/\partial x_{2}}\\partial u_{1/\partial x_{2}}&0\end{bmatrix}[/latex] and [latex]\frac{\partial au_{ij}}{\partial x_{j}}=\mu\begin{bmatrix}\partial^{2}u_{1}/\partial x^{2}_{2}&0\end{bmatrix}[/latex].

Here, a net horizontal viscous force is present when the velocity profile has a nonzero second derivative. For the rectangular contour ABCD, the integral in (5.11) can be written:

[latex]\frac{D\Gamma}{Dt}=\underset{C}{\int}\left(\frac{1}{ ho}\frac{\partial au_{ij}}{\partial x_{j}} ight)dx_{i}=\frac{1}{ ho}\left[\underset{AB}{\int}+\underset{BC}{\int}+\underset{CD}{\int}+\underset{DA}{\int} ight]\left(\frac{\partial au_{ij}}{\partial x_{j}} ight)dx_{i}[/latex].

On the vertical AB and CD portions of the contour [latex]d\chi=(0,dx_{2})[/latex], so the dot product with [latex]\partial au_{ij}/\partial x_{j}[/latex] will be zero. This leaves the horizontal BC and DA portions of the contour:

[latex]\frac{D\Gamma}{Dt}=\frac{1}{ ho}\left[\underset{BC}{\int}+\underset{DA}{\int} ight]\left(\frac{\partial au_{ij}}{\partial x_{j}} ight)dx_{i}=\frac{1}{ ho}\left[\overset{x_{1}+\Delta x_{1}/2}{\underset{x_{1}-\Delta x_{1}/2}{\int}}\left(\mu\frac{\partial^{2}u_{1}}{\partial u^{2}_{2}} ight)_{x_{2}-\Delta x_{2}/2}dx^{\prime}_{1}-\overset{x_{1}+\Delta x_{1}/2}{\underset{x_{1}-\Delta x_{1}/2}{\int}}\left(\mu\frac{\partial^{2}u_{1}}{\partial u^{2}_{2}} ight)_{x_{2}+\Delta x_{2}/2}dx^{\prime}_{1} ight][/latex],

where [latex]x^{\prime}_{1}[/latex] is an integration variable, and the minus sign arises in the DA integration because the direction of integration opposes the direction of increasing [latex]x_{1}[/latex]. Both integrands are independent of [latex]x_{1}[/latex], and, when [latex]\Delta x_{2}[/latex] is small enough, they may be Taylor-expanded to find:

[latex]\frac{D\Gamma}{Dt}=-\frac{\mu}{ ho}\frac{\partial^{3}u_{1}}{\partial x^{3}_{2}}\Delta x_{1}\Delta x_{2}[/latex].

Thus, the circulation around fluid elements in a simple shear flow will only change when the third derivative of the velocity profile is nonzero. This finding can be understood as follows. When the velocity profile is linear, [latex]\partial u_1/\partial x_2=\mathrm{const}[/latex]., fluid elements have vorticity and Γ computed on ABCD is not zero, but there is no net viscous force or torque on fluid elements so Γ does not change. When the velocity profile is quadratic, [latex]\partial^{2}u_1/\partial x^{2}_2=\mathrm{const}[/latex]., fluid elements again have vorticity, Γ computed on ABCD is not zero, and there is a net viscous force on fluid elements, but no net viscous torque. In this case, fluid elements may accelerate, or pressure or body forces may balance the net viscous force, but fluid elements feel no net torque so Γ computed on ABCD does not change. However, when the velocity profile is cubic (or higher) so that [latex]\partial^{3}u_1/\partial x^{3}_2 eq0[/latex], then there is a net viscous torque on fluid elements and Γ computed on ABCD will increase or decrease appropriately.
Interestingly, the final result of this example is a precursor to (5.13), the main result of the next section, since a limit allows it to be rewritten in terms [latex]\omega_{3}=-\partial u_{1}/\partial x_{2}[/latex], the plane-normal vorticity (circulation per unit area) in the simple shear flow:

[latex]\frac{D\omega}{Dt}=(\omega\cdot abla)\mathbf{u}+ u abla^2\omega[/latex] (5.13)

[latex]\underset{\Delta x_{1},\Delta x_{2} ightarrow0}{\lim}\frac{1}{\Delta x_{1}\Delta x_{2}}\frac{D\Gamma}{Dt}=\frac{D\omega_{3}}{Dt}=-\frac{\mu}{ ho}\frac{\partial^{3}u_{1}}{\partial x^{3}_{2}}= u\frac{\partial^{2}\omega_{3}}{\partial x^{2}_{2}}[/latex],

where [latex]\Delta x_{1}\Delta x_{2}[/latex] is the area inside the square ABCD contour, and [latex] u=\mu/ ho[/latex] is the kinematic viscosity.

Question 5.2: For the planar shear flow in Figure 3.15, u = (u1(x2), 0), w...

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