In an older post about Couette Poiseuille Flow Using Aerodynamic Navier Stokes Equations I derived a theoretical equation to simulate the behavior of airflow underneath a highly idealized race car underfloor. In this writeup I intend to validate that theory by checking the Boundary Conditions of the equation representing the velocity profile of the airflow, and checking the mass flow rate using the Continuity Equation.

Here is the equation I derived last time:

$u(y)=\frac{1}{2} [\frac{1}{\mu} \frac{dP}{dx}] y^{2} + \frac{V}{2h} y + \frac{\mu V - \frac{dP}{dx} h^{2}}{2\mu}$

What the equation represents is the velocity of the air flow in the x direction as a function of the y direction. Let me explain that briefly, I can make a more complete post on velocity profiles, boundary layers, and shear later. The air directly touching the ground is stationary since it is in contact with the ground. The air just above the ground has “friction” with the air below it. The same principle is in effect for the bottom of the race car. The air touching the bottom of the car is stationary with respect to the car, and “friction” effects the nearby air molecules. I say “friction” because in reality the air molecules are bouncing off each other. This creates a velocity profile.

So in saying all that jibber jabber above, it simply means the air at different heights between the ground and underfloor of the race car has different speeds, and we are taking that in to account with the equation. So if this article is over your head for now, bookmark it, and come back. Like I said, I’ll make a more detailed post on this later. Promise.

So the first method I will use to validate the theory, or mostly to make sure it isn’t dead wrong, is to double check the Boundary Conditions. I said in a quick sentence in my last post that I had done it, but I didn’t describe how the method works.

We know the air speed on the ground, and on the underbody. We can take the derived equation and make sure the speed is in agreement.

Lets assume we have a race car and atmosphere with the following specs:

• Ride height =  .1m
• Speed          = 40m/s
• Density        = 1.225kg/m^3
• Viscosity      = 1.8*10^-5 kg/(s*m)
• dP/dx=0

Lets plug and chug.

$u(y)=\frac{1}{2} [\frac{1}{\mu} \frac{dP}{dx}] y^{2} + \frac{V}{2h} y + \frac{\mu V - \frac{dP}{dx} h^{2}}{2\mu}$

$u(y)=\frac{1}{2} [\frac{1}{1.8*10^3} *0]*[-.05]^{2} + \frac{40}{2*.05}*[-.05] + \frac{1.8*10^3 40 - 0}{2*1.8*10^3}=0$

$u(y)=\frac{1}{2} [\frac{1}{1.8*10^3} *0]*[.05]^{2} + \frac{40}{2*.05}*[.05] + \frac{1.8*10^3 40 - 0}{2*1.8*10^3}=40$

So at the very least, we solved correctly for the Boundary Conditions.

Now, lets check the mass flow rate using the Continuity Equation.

$\int \int \rho \vec{V} \cdot \vec{n} dA = 0$

$\vec{V}=u(y)\vec{i}$

Since we are treating this as a 2 dimensional problem we can replace the first integral with a unit depth since there is no velocity change in the z direction due to the assumptions made in the previous article. Also, since the velocity only has an x component we don’t have any extra mess from the dot product.

$\int_{-h}^{h}\rho[\frac{1}{2} [\frac{1}{\mu} \frac{dP}{dx}] y^{2} + \frac{V}{2h} y + \frac{\mu V - \frac{dP}{dx} h^{2}}{2\mu}]dy=\frac{h\rho[3\mu V-2h^2*\frac{dp}{dx}}{3\mu}]$ 102

$\frac{h\rho[3\mu V-2h^2*\frac{dp}{dx}]}{3\mu}$

Since I am slowly transitioning to my new site, please read the rest here…