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 $h<<a$ 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 $V=\frac{2Q}{hW}-\frac{h^2}{6\mu}(\frac{P_2-P_1}{L})=\frac{Q}{h\pi a}-\frac{h^2}{6\mu}(\frac{P_2-P_1}{L})$
 A: 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 $h<<R$, this reduces to:$$\frac{(P_1-P_2)}{L}=\frac{2}{a}\mu \frac{V}{h}$$
A: 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).
