# Not no-slip -- low density gas flow

Posted in math on Monday, April 28 2014

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 C1 = 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. $$w = \frac{\pi R^4 \rho_{avg}}{8 \mu} \frac{\left( p_0 - p_L \right)}{L} \frac{4\zeta_0}{R p_{avg}}$$