Continuing on with tube flow, what happens when the fluid moving through the tube is low density and thus the no-slip boundary condition breaks down? (from 2B.9 Transport Phenomena)

I want the mass flowrate for a low density gas moving through a tube in slip flow.

First off we are going to make a pack of assumptions:that while the fluid is a gas it is *approximately* incompressible, and that flow is laminar etc. We can then jump ahead to:

$$ -\mu \frac{d v_z}{d r} = \frac{r^2}{2} \left( - \frac{d p}{d z} \right) + C_1 $$

Here I've shrunk the cylindrical shell of consideration from being of length *L* to a differential length, hence the pressure gradient has turned into a derivative.

By the normal arguments *C _{1} = 0*

$$ \frac{d v_z}{d r} = \frac{-r}{2 \mu} \left( - \frac{d p}{d z} \right) $$

Integrating with respect to *r*:

$$ v_z = \frac{-r^2}{4 \mu} \left( - \frac{d p}{d z} \right) + C_2 $$

Now we introduce a wrinkle, suppose that instead of the no-slip boundary condition, there is some *slip coefficient*, *ζ*, such that:
$$ v_z = - \zeta \frac{d v_z}{d r} \vert_{r=R}$$

Plugging that in...

$$ \frac{\zeta R}{2 \mu} \left( - \frac{d p}{d z} \right) = \frac{-R^2}{4 \mu} \left( - \frac{d p}{d z} \right) + C_2 $$ $$ C_2 = \left( \frac{\zeta R}{2 \mu} + \frac{R^2}{4 \mu} \right) \left( - \frac{d p}{d z} \right) $$

The velocity distribution is then: $$ v_z = \frac{1}{4 \mu} \left( - \frac{d p}{d z} \right) \left( R^2 - r^2 + 2 \zeta R \right) $$

The mass flow-rate is: $$ w = 2 \pi \rho \int \limits_0^R v_z r dr = \frac{\pi \rho}{2 \mu} \left( - \frac{d p}{d z} \right) \int \limits_0^R R^2 r + 2 \zeta R r - r^3 dr $$

Integrating and simplifying: $$ w = \frac{\pi R^4 \rho}{8 \mu} \left( 1 + \frac{4\zeta}{R} \right) \left( - \frac{d p}{d z} \right) $$

Now we want the mass flow-rate overall, not for particular point along the tube with a particular pressure gradient. So let's integrate over the whole tube from *z = 0* to *z = L*

$$ \int \limits_0^L w dz = \int \limits_{p_0}^{p_L} - \frac{\pi R^4 \rho}{8 \mu} \left( 1 + \frac{4\zeta}{R} \right) dp $$

Before we dive into this, note that the density depends upon the pressure in some way. Suppose it is merely an ideal gas: $$ \frac{\rho_0}{p_0} = \frac{\rho}{p} $$

But the slip coefficient *also* depends upon pressure:
$$ \zeta = \frac{\zeta_0}{p} $$

Plugging these in:

$$ \int \limits_0^L w dz = \frac{\pi R^4}{8 \mu} \frac{\rho_0}{p_0} \int \limits_{p_0}^{p_L} - p - \frac{4\zeta_0}{R} dp $$

$$ w L = \frac{\pi R^4}{8 \mu} \frac{\rho_0}{p_0} \left( - \frac{1}{2} \left( p_L^2 - p_0^2 \right) - \frac{4\zeta_0}{R} \left( p_L - p_0 \right) \right) $$

We can take an algebraic interlude by looking at the definition of the *average pressure* and *average density*
$$ p_{avg} = \frac{1}{2} \left( p_L + p_0 \right) $$

$$ \rho_{avg} = \frac{\rho_0}{p_0} p_{avg} $$

Now we notice that: $$ \frac{1}{2} \left( p_L^2 - p_0^2 \right) = \frac{1}{2} \left( p_L + p_0 \right) \left( p_L - p_0 \right) = p_{avg} \left( p_L - p_0 \right) $$

We can put this in, and re-arrange stuff to get: $$ w = \frac{\pi R^4 \rho_{avg}}{8 \mu} \frac{\left( p_0 - p_L \right)}{L} \left( 1 + \frac{4\zeta_0}{R p_{avg}} \right) $$

Which is where we wanted to be.

We can compare this to the no-slip boundary condition, the Hagen-Poiseuille law: $$ w = \frac{\pi R^4 \rho}{8 \mu} \frac{\left( p_0 - p_L \right)}{L} $$

Since there is less energy dissipated to the wall, slip flow gives you some additional mass flow for a given pressure drop. $$ w = \frac{\pi R^4 \rho_{avg}}{8 \mu} \frac{\left( p_0 - p_L \right)}{L} \frac{4\zeta_0}{R p_{avg}} $$