First of all, I am slowly migrating to my new site about F1 Race Car Technology.

In my previous Pitot Tube Example Using Bernoulli’s Equation I made some simplifying assumptions and did not explain things as clearly as I could have.

Before  you can really understand how Pitot Tubes work, whether it’s for aircraft or a race car, you need to understand the different “types” of pressure. You may have seen words like Dynamic Pressure, Static, Total, Absolute, Stagnation, Gage,  and others.

First, I will describe Static Pressure. For a more official description you can read a book, but the way I think of static pressure is the pressure I feel in stationary air. At the bottom of a pool that pressure goes up, up in a mountain that pressure goes down. What happens though when you are in a race car traveling at speed? The static pressure in this case would be the pressure of the air measured by an observer traveling with the air, not against it.

Static Pressure is created when the random motions of air molecules interact with a body. As the particles bounce off the surface, they transfer momentum to the surface and subjects it to a force.

In the picture above, the random motion paths are shown with the red lines. A few collisions are observed and they are imparting small forces on the body shown with black arrows.

Dynamic Pressure is used in finding lift and drag. It takes in to account the density and velocity of the air, while basically ignoring the little random motions of the air that make up the static pressure.

The image above shows that Dynamic Pressure doesn’t care about the random motions of the particles, only the momentum transfer due to the density of the air and the freestream velocity.

The image above shows how an object with more mass will impart a larger force than a smaller object with equal velocity.

Additionally, the faster of the two objects with equal mass will impart a larger force on the body.

$q = \frac{1}{2} \rho V^{2}$

Based on the equation above, for Dynamic Pressure, it is clear that changes in freestream velocity have a greater effect than does the density of the free stream.

Stagnation Pressure and Total Pressure are redundant terms. Stagnation pressure is the pressure measured if the moving airflow is brought to a stop Isentropically. Intuitively it makes sense. If air flow is hitting the front of an F1 car the pressure is higher since that flow had to slow down. Basically stagnation pressure adds up the static pressure from random molecular motion, and the dynamic pressure made up of the actual freestream velocity of the air.

$P_0 = P + \frac{1}{2} \rho V^{2} = P + q$

That’s all great. How does this apply to Pitot Tubes? A Pitot Tube is used to find the velocity of a fluid, in the case of a race car that fluid is air. To do this, a Pitot Tube compares the Static Pressure to the Stagnation Pressure.

Once both the Static Pressure and the Stagnation Pressures are known, the velocity of the race car relative to the air can be determined. This can be done using Bernoulli’s Equation.

$\frac{P_1}{\rho} + \frac{1}{2} V_1^{2} + g h_1 = \frac{P_2}{\rho} + \frac{1}{2} V_2^{2} + g h_2$

Since the Formula 1 car speed is much less than the speed of sound, we can make some assumptions such as constant air density. We can also assume that the difference in height of the fluid is negligible. Therefore we can rewrite the equation.

$\frac{P_1}{\rho} + \frac{1}{2} V_1^{2} = \frac{P_2}{\rho} + \frac{1}{2} V_2^{2}$

This is better, but still a little confusing. It can be made much simpler when we multiply both sides by the density.

$P_1 + \frac{1}{2} \rho V_1^{2} = P_2 + \frac{1}{2} \rho V_2^{2}$

Now let’s designate the variables. Let’s have the first values relate to the Stagnation portion and the second set of values relate to the Static values. We know the velocity at the Pitot Tube inlet is zero, hence the Stagnation Pressure. The Stagnation Pressure is measured, and therefore known. Next, we have also measured the Static Pressure. Lastly, we should know the air density. The only thing we don’t know is the Velocity of the freestream as it passes by the hole, or port. Since the Velocity is the only unknown, we can solve.

$P_{stag} = P_{stat} + \frac{1}{2} \rho V_2^{2}$

With a little bit of algebra, we arrive at the equation for the velocity.

$V = \sqrt{\frac{[2 [P_{stag} - P_{stat}]]}{\rho}}$

F1 Car Pitot Tube Example, Pikes Peak International Hill Climb:

But let’s make it a little more difficult than that. An F1 car runs up Pikes Peak. Toward the top of the mountain, what Velocity is required to make the Stagnation/Static Pressure Ratio equal to 1.0082?

According to Wolfram Alpha:

Height of Pikes Peak: 4,300m

Air Density at 4,300m: .79kg/m^3

Air Pressure at 4,300m: 59kPa

$P_1 + \frac{1}{2} \rho V_1^{2} = P_2 + \frac{1}{2} \rho V_2^{2}$

$P_{stag} = P_{stat} + \frac{1}{2} \rho V_2^{2}$

$\frac{P_{stag}}{P_{stat}} = 1 + \frac{\frac{1}{2} \rho V_2^{2}}{P_{stat}}$

$V_2 = \sqrt{2 \frac{P_{stat}}{\rho} [\frac{P_{stag}}{P_{stat}}-1]}$

$V_2 = \sqrt{2 \frac{59000}{.79} [1.0082-1]} = 35 \frac{m}{s}$

$P_{Static} = 59{kPa}$

$P_{Dynamic} = \frac{1}{2} \rho V^{2} = 0.484{kPa}$

$P_{Stagnation} = P_{Static} + P_{Dynamic} = P + \frac{1}{2} \rho V^{2} = 59.484{kPa}$

Hope this helped you understand these things a little better. This article tried to explain the differences between Static Pressure, Dynamic Pressure, and Stagnation Pressure aka Total Pressure. It also showed how to use Bernoulli’s Equation to find the velocity of the freestream by using a Pitot Tube. Please comment and rate. I am fairly new to writing and need to know if the general reader thinks this material and example was too complicated or too simple.

Think you get it? What Stagnation/Static Pressure ratio is required when the racing car is at ground level at the same speed? You can see if you got it correct in my next article.