Formula One Car Refuel Tube Orifice Calculation

1. Defining the Problem

We want to calculate the diameter of the orifice in the refuel tube of a Formula One car. The orifice should avoid ignition risks due to:

Minimizing turbulent flow, which can cause frictional heating.
Ensuring fuel flow remains in the laminar or manageable transitional flow regime.

Let's define the following parameters:

Desired fuel flow rate $ Q $ in liters per second (let's assume $ Q = 12 \, \text{L/s} $ for a rapid refuel).
Dynamic viscosity of fuel, $ \mu \approx 0.001 \, \text{Pa}\cdot\text{s} $.
Density of fuel, $ \rho \approx 750 \, \text{kg/m}^3 $.
Acceptable Reynolds number $ \text{Re} $ for avoiding turbulent flow (for a manageable transitional flow, let’s set $ \text{Re} \approx 2000 $ as the upper limit).

The Reynolds number $ \text{Re} $ is given by:

$$ \text{Re} = \frac{\rho v D}{\mu} $$

where:

We can express the velocity $ v $ in terms of the flow rate $ Q $ and orifice diameter $ D $:

$$ v = \frac{Q}{A} = \frac{Q}{\frac{\pi D^2}{4}} = \frac{4Q}{\pi D^2} $$

Substituting $ v = \frac{4Q}{\pi D^2} $ into the Reynolds number formula:

$$ \text{Re} = \frac{\rho \cdot \left( \frac{4Q}{\pi D^2} \right) \cdot D}{\mu} = \frac{4 \rho Q}{\pi \mu D} $$

Rearrange to solve for $ D $ such that $ \text{Re} \leq 2000 $:

$$ D \geq \frac{4 \rho Q}{\pi \mu \cdot 2000} $$

Substitute $ \rho = 750 \, \text{kg/m}^3 $, $ Q = 0.012 \, \text{m}^3/\text{s} $, and $ \mu = 0.001 \, \text{Pa} \cdot \text{s} $:

$$ D \geq \frac{4 \times 750 \times 0.012}{\pi \times 0.001 \times 2000} \approx 0.0143 \, \text{m} \approx 14.3 \, \text{mm} $$

To maintain manageable flow without excessive turbulence and frictional heating, the orifice diameter should be approximately 14.3 mm or larger.