First order flow correction for inertia

It is convenient to decompose the first order flow into the parts due to inertia and buoyancy:

$$\mathbf{u}_1 = \mathbf{u}_{1I} + \mathbf{u}_{1B},$$ (8.51)

where the parts are both solenoidal and vanish at the boundary and satisfy

$$\mathbf{I}_1 = -\mbox{\boldmath$\nabla$}p_{1I} + \nabla^2\mathbf{u}_{1I}$$ (8.52)

$$-\mathbf{B}_1 = -\mbox{\boldmath$\nabla$}p_{1B} + \nabla^2\mathbf{u}_{1B}.$$ (8.53)

The inertia force, $\mathbf{I}_1$, contains the single term $\mathbf{u}_0\cdot\mbox{\boldmath$\nabla$}\mathbf{u}_0$, which simplifies greatly from the general form (8.26). Since only the azimuthal component of $\mathbf{u}$ is nonzero, and since it is independent of $\phi$, $\mathbf{I}_{1}$ reduces to

$$\mathbf{I}_1 = \mathbf{u}_0\cdot\mbox{\boldmath$\nabla$}\mat... ...\mathbf{u}_0)^2\cot\theta}{r}\;\mbox{\boldmath$\hat{\theta}$}.$$ (8.54)

These are precisely the terms of the equation of motion referred to in §8.1.1: they are proportional to the square of the component of velocity about the axis and inversely proportional to the distance from the axis. Bird et al. (1960, p. 85) called them the `centrifugal force'.

Again, $\mathbf{u}_{1I}$ is found by the method of §B.4. The inertia force, $\mathbf{I}_1$, is found to have no toroidal part; the scalars defining its decomposition are:

$$\mbox{$\mathcal S$}[\mathbf{I}_1] = \frac{r^2(112r^4-72r^2+15)}{604\,800}P_2^0(\cos\theta)$$

$$-{}\frac{r^2(16r^4-12r^2+3)}{57\,600}$$ (8.55)

$$\mbox{$\mathcal P$}[\mathbf{I}_1-\mathbf{I}_1^{(S)}] = \frac{r^2(4r^2-1)(28r^2-11)}{1\,209\,600}P_2^0(\cos\theta)$$ (8.56)

$$\mbox{$\mathcal T$}[\mathbf{I}_1-\mathbf{I}_1^{(S)}] = 0.$$ (8.57)

As a result, the velocity is purely poloidal:

$$\mbox{$\mathcal P$}[\mathbf{u}_{1I}] = \frac{r^2(1-4r^2)^2(28r^2-19)}{319\,334\,400}P_2^0(\cos\theta)$$ (8.58)

$$\mbox{$\mathcal T$}[\mathbf{u}_{1I}] = 0.$$ (8.59)

Its spherical components are:

$$\mathbf{u}_{1I} = \frac{r(1-4r^2)^2(28r^2-19)(3\cos^2\theta-1)}{106\,444\,800}\;\mbox{\boldmath$\hat{r}$}$$

$$+ \frac{r(1-4r^2)(336r^4-224r^2+19)\sin\theta\cos\theta}{35\,481\,600}\; \mbox{\boldmath$\hat{\theta}$}.$$ (8.60)

Since, in addition, $\mbox{$\mathcal P$}[\mathbf{u}_{1I}]$ is independent of $\phi$, the flow field in any plane passing through the $z$-axis can be represented by a Stokes's stream-function (see §B.5):

$$\psi_{1I} = \frac{r^3(1-4r^2)^2(19-28r^2)\sin^2\theta\cos\theta} {106\,444\,800},$$ (8.61)

the contours of which are plotted in figure 8.7.

Figure 8.7: Stream-lines of the first order flow due to inertia (8.61) in any plane of constant $\phi$. $\psi_{1I}$ is nonnegative for $z\geq 0$. Contour levels at 0.01, 0.1(0.1)0.4, 0.6(0.1)0.9, 0.99 of range.

The pressure,

$$p_{1I} = \frac{r^2[(3696r^4-2376r^2+567)\sin^2\theta+1232r^4-1188r^2+315]} {13\,305\,600},$$ (8.62)

is illustrated in figure 8.8.

Figure 8.8: First order pressure due to inertia, $p_{1I}$, (8.62) in any plane of constant $\phi$. The maxima occur in the circle $r=1/2$, $\theta=\pi/2$. Contour levels at 0.01, 0.1(0.1)0.4, 0.6(0.1)0.9, 0.99 of range.

Since the circulation is positive around maxima of the stream-function, the flow is radially inward along the $z$-axis and outward in the $xy$-plane, and so is analogous to secondary flow in curved pipes and rivers, circular aquaculture tanks and stirred tea cups, as discussed in §8.1.1. The correction, $\mathbf{u}_{I1}$, is independent of all the parameters of the problem, but its effect on the velocity field vanishes with $\mbox{\textit{Gr}}(1+N)$.