Fluid Dynamical problems

Question

I'm just going through Batchelor's book on Fluid Dynamics, and I'm not too sure about a couple of questions in "Exercises for Chapter 4", which are the following:

Exercise 3 for Chapter 4

"A long circular tube has a cylindrical layer of liquid of un iform thickness adhering to its inner surface. In order to remove the liquid air is blown t hrough the tube by the application of a difference between the pressures in the air a t the two ends. Determine the ratio of the steady fluxes of volume of the air and the liquid from the end of the tube."

Now earlier on in the chapter Batchelor uses the following formula: $$Q=\int_{0}^{a}2u\pi r \space dr = \frac{\pi G a^4}{8\mu}=\frac{\pi a^4 (p_0 - p_1)}{8\mu L} , \quad u(r)=\frac{G}{4\mu}(a^2-r^2) $$

But this only applies when the flow fills the cylinder with only one fluid. Can we instead assume that the air is travelling with some speed $U_{air}(r)=\frac{G}{4\mu_{air}}(a^2-r^2)$ and has it's own flow, from $r=0$ to $a-\delta$ (where $\delta$ is the thickness of fluid).

Likewise the fluid has its own flow $U_{fluid}(r)=\frac{G}{4\mu_{fluid}}(a^2+\log{r}-r^2)$, workout their flux seperately then divide them?

Second question is

Exercise 4 for Chapter 4

"A thin layer of viscous fluid lies between two parallel rigi d planes, one of which is stationary and the other of which is in oscillatory translat ional motion with frequency n in its own plane. Determine the ratio of the magnitudes of the (oscillatory) frictional forces on the two planes, and examine the cases of large and sm all values of n"

I'm uncertain how to derive the Frictional forces from the Navier-Stokes equations.

Any help is greatly appreciated!

Answer 1

For the first question you should solve the Navier-Stokes equations for unidirectional, laminar flow -- obtaining general solutions that apply to the two regions occupied by fluid and air, respectively.

Assuming unidirectional flow, the the axial component of velocity $u$ depends only on the radial coordinate $r$ and the pressure $p$ decreases linearly with respect to the axial coordinate $z$. Ignoring body forces or absorbing them into pressure, the pressure depends only on $z$ because there is no non-zero component of velocity in either the radial or azimuthal directions. Hence, the pressure gradient in the air and the layer of fluid are identical.

The Navier-Stokes equations reduce to:

$$\frac{\mu}{r}\frac{d}{d r}\left(r\frac{d u}{d r}\right)= \frac{dp}{dz}= -G,$$

with general solution

$$u(r) = \begin{cases} \frac{G}{4\mu_{fluid}}r^2 + A_1 \log r+ A_2 &\mbox{ if } \,\, a - \delta \leqslant r \leqslant a,\\ \frac{G}{4\mu_{air}}r^2 + B_1 \log r+ B_2 &\mbox{ if } \,\,0 \leqslant r \leqslant a-\delta,\end{cases}.$$

To obtain a finite solution, we immediately find $B_1 = 0$. The remaining constants are determined using the no-slip condition,

$$u(a) = 0,$$

and continuity of velocity and shear stress at the fluid-air interface,

$$\lim_{r \rightarrow (a-\delta)+}u(r) = \lim_{r \rightarrow (a-\delta)-}u(r),\\\lim_{r \rightarrow (a-\delta)+}\mu_{fluid}\frac{du}{dr} = \lim_{r \rightarrow (a-\delta)-}\mu_{air}\frac{du}{dr}. $$

Application of the boundary conditions results in three equations that determine the constants:

$$ \frac{G}{4\mu_{fluid}}a^2 + \log (a- \delta)A_1+ A_2=0, \\ \frac{G}{4\mu_{fluid}}(a-\delta)^2 + A_1 \log(a-\delta)+ A_2=\frac{G}{4\mu_{air}}(a-\delta)^2 + B_2, \\ \frac{G}{2}(a-\delta) + \frac{\mu_{fluid}}{a - \delta}A_1= \frac{G}{2}(a-\delta).$$

After solving for $A_1$,$A_2$, and $B_2$, we can find the ratio of the volumetric flow rates for air and fluid as:

$$\frac{Q_{air}}{Q_{fluid}}= \frac{\int_0^{a - \delta}ru(r) \,dr}{\int_{a - \delta}^aru(r) \, dr}.$$

For the second question, again, assume unidirectional flow. The longitudinal component of velocity $u_x(y,t)$ depends on time $t$ and the coordinate $y$, where $y=0$ at the fixed plane and $y = \delta$ at the oscillating plane.

Solve the Navier-stokes equations assuming zero pressure gradient and the no-slip boundary conditions,

$$u_x(0,t) = 0, \\ u_x(\delta,t) = A\sin(nt).$$

For a Newtonian fluid, the shear stress is

$$\tau_{xy} = \mu \frac{\partial u_x}{ \partial y}.$$

The "frictional" force (per unit area) on a plane is just the shear stress evaluated at the location of the plane. For example, the force per unit area on the oscillating plane is

$$\hat{F}(t) = \left.\mu \frac{\partial u_x}{ \partial y}\right|_{y= \delta}.$$