Combined Poiseuille-Couette flow

I stumbled upon this exercise in James Fay "Fluid Mechanics" book, which I'm using to learn fluid dynamics by my own, and I am struggling a bit with it, any help will be appreciated:

The figure shows a pig of length $$L$$ inside a pipe of radius $$a$$ and having a radial clearance $$h< between its surface and the inner surface of the pipe. When the pressure $$P_1$$ at $$1$$ exceeds the pressure $$P_2$$ at $$2$$, the pig will move to the right at a constant velocity $$V$$. Assuming that the flow between the pig and the pipe wall can be con sidered to be a steady plane Couette plus Poiseuille flow in a reference frame attached to the pig.

(a) Derive an expression for the pig velocity V in terms of the parameters $$P_1$$ , $$P_2$$, $$L$$, $$h$$, $$a$$ and the fluid viscosity $$\nu$$. (b) If $$Q$$ is the volume flow rate of fluid leaking through the clearance gap, relative to the pig, derive an expression for the ratio $$Q/\pi a^2V$$, which is the ratio of leakage rate to the flow rate of fluid through the pipe. I have tried solving $$a)$$ following the tip of using a Couette plus Poiseuille type flow. If I attach a reference frame to the pig (its x axis located at the surface of the pig and parallel the cylinder axis, and the y axis perpendicular to it), we see that the no-slip condition means that $$u(y=0)=0$$. Moreover, we see that at $$y=h$$, $$u(h)=V$$, where $$V$$ is the magnitude of the velocity of the pig in the laboratory frame. It can be proved that the velocity of the fluid within the pig-wall gap is:

$$u(y)=V\frac{y}{h}+\frac{1}{2\mu}(-\frac{dp}{dx})y(h-y)$$

Now, if I understand correctly, the problem asks me to obtain $$V$$ as a function of the aforementioned parameters. However, I am stuck in this step as after isolating $$V$$ I have no idea on how to get rid of the $$u$$ dependence of it. I can eliminate the $$\frac{dP}{dx}$$ term by noticing that this is just $$\frac{P_2-P_1}{L}$$ but that's it.

Integrating in y to get the ratio $$\frac{Q}{W}$$ yields:

$$\frac{Q}{W}=\frac{Vh}{2}+\frac{h^3}{12\mu}(\frac{P_2-P_1}{L})$$

So $$V=\frac{2Q}{hW}-\frac{h^2}{6\mu}(\frac{P_2-P_1}{L})=\frac{Q}{h\pi a}-\frac{h^2}{6\mu}(\frac{P_2-P_1}{L})$$

• The next step is to integrate with respect to y to get the volumetric flow rate Q. – Chet Miller Dec 5 '18 at 22:44
• I did that but then the variable W appears (the width) in this case what would it be? It varies with the distance from the axis. – mobzopi Dec 5 '18 at 22:49
• The width W is $2\pi R$. Because the gap is so small compared to the radius, you can treat the geometry as flat, and look at the channel in so-called "developed view" in which the width would then be $2\pi R$. – Chet Miller Dec 5 '18 at 23:22
• Is there a way to obtain Q in terms of the geometry in this case? – mobzopi Dec 5 '18 at 23:32
• You can do a force balance on the plug, including the pressure forces at the two ends and the shear force on the surface and see what you get. This should give you another relationship to work with. – Chet Miller Dec 6 '18 at 0:01

As reckoned from the frame of reference of the plug, your equations should read:

$$u(y)=-V\frac{y}{h}+\frac{1}{2\mu}(-\frac{dp}{dx})y(h-y)\tag{1}$$ $$\frac{Q}{W}=-\frac{Vh}{2}-\frac{h^3}{12\mu}(\frac{P_2-P_1}{L})\tag{2}$$where y = 0 is at the surface of the plug. If we differentiate Eqn. 1 with respect to y and evaluate the velocity gradient at y = 0, we obtain: $$\frac{du}{dy}(0)=-\frac{V}{h}-\frac{h}{2\mu}\frac{P_2-P_1}{L}$$So the shear stress on the plug is $$\tau=-\mu\frac{V}{h}-\frac{h}{2}\frac{(P_2-P_1)}{L}$$So, the force balance on the plug should be:$$\pi a^2(P_1-P_2)-2\pi aL\left(\mu\frac{V}{h}+\frac{h}{2}\frac{(P_2-P_1)}{L}\right)=0$$This leads to:$$(P_1-P_2)\left(1+\frac{h}{a}\right)=\frac{2L}{a}\mu\frac{V}{h}$$Since $$h<, this reduces to:$$\frac{(P_1-P_2)}{L}=\frac{2}{a}\mu \frac{V}{h}$$

Solving for the axial velocity in the channel in cylindrical coordinates with boundary conditions $$u(a)=0$$ and $$u(a-h)=V$$ yields:

$$u\left(r\right)=-\frac{1}{4}\frac{\Delta p}{\mu L}r^{2}+\frac{K_{1}}{\mu}\ln r+K_{2}$$

where: $$K_{1}=\frac{\mu V-\frac{\Delta p a^{2}}{L}\left(\frac{1}{2}\frac{h}{a}\right)\left(1-\frac{1}{2}\frac{h}{a}\right)}{\ln\left(1-\frac{h}{a}\right)} \qquad K_2=\frac{1}{4}\frac{\Delta p}{\mu L}a^{2}-\frac{K_{1}}{\mu}\ln a$$

The shear stress at the inner surface of the pipe is:

$$\tau(a)=-\mu\frac{d u}{d r}(a)=\frac{1}{2}\frac{\Delta p}{L}a+\frac{K_{1}}{a}$$

Following the analysis of Chester, at steady state the pressure drop across the pig must equal the frictional force at the wall, i.e.:

$$\pi a^{2}\Delta p=2\pi aL\tau\left(a\right)$$

Substituting the expression for the shear stress at the wall and rearranging for $$V$$, we find: $$V=\frac{\Delta pa^{2}}{\mu L}\left(\frac{1}{2}\frac{h}{a}\right)\left(1-\frac{1}{2}\frac{h}{a}\right)$$

Clearly, if we apply the small channel approximation $$\frac{h}{a}\ll 1$$ we end up with the same equation as Chester. Notice also that the constants simplify to:

$$K_{1}=0 \qquad K_2=\frac{1}{4}\frac{\Delta p}{\mu L}a^{2}$$

which indicates we can completely ignore any effects due to curvature in the channel.

We can now go ahead and solve for the fractional leakage flow. First, let's subtitute in the constants and simplify:

$$u\left(r\right)=-\frac{1}{4}\frac{\Delta p}{\mu L}\left(r^{2}-a^{2}\right)$$

The flow rate through the gap is: $$Q = \int_A u dA = 2\pi\int_{a-h}^a rudr = 2\pi\frac{\Delta p a^{4}}{\mu L}\left(\frac{1}{2}\frac{h}{a}\right)^{2}\left(1-\frac{1}{2}\frac{h}{a}\right)^{2}$$

The fractional leakage is then: $$\frac{Q}{\pi a^2 V} = \frac{h}{a}\left(1-\frac{1}{2}\frac{h}{a}\right) \approx \frac{h}{a}$$

which is completely determined by the relative channel height between the pig and inner surface of the pipe (perhaps unsurprisingly).