What exactly does mixture control do?

Carburetors are supposedly volumetric (i.e. they deliver a constant air:fuel ratio by volume), so as the air density decreases we need to decrease the amount of fuel delivered in order to maintain the same mass flow ratio. This is how nearly every pilot is taught. But that explanation is not quite correct. Carburetors do not maintain a constant volumetric ratio of fuel and air. The amount of fuel delivered actually declines as the air density declines. The problem is that it declines slower than the air density, and this is what makes the mixture richer as we climb. Hence we need to restrict the flow of fuel to maintain the proper air:fuel ratio.

Float style carburetor (source: FAA Pilot’s Handbook of Aeronautical Knowledge)

As air flows through the carburetor’s constriction, the drop in pressure can be written as

\delta P = \frac{1}{2}\rho_a {v_a}^2

where \delta P is the pressure difference (compared to ambient), \rho_a is the density of air, and v_a is the velocity in the constriction. This pressure difference \delta P is what forces the fuel from the bowl into the venturi.

Next, we need to relate \delta P to the fuel draw rate. For this, we need to use the orifice plate equation, which states

Q_f = K\sqrt{2\delta P \rho_f}

where Q_f is the fuel flow rate, \rho_f is the fuel density and K is a function of the geometry of the orifice.

Combining the above two equations, we can get

Q_f = K\sqrt{\rho_a \rho_f}v_a.

The velocity of air v_a is a function of engine rpm, engine displacement and the carburetor’s diameter. This is

v_a = \frac{V f}{2D}

where f is the engine rpm, V is the engine displacement volume, and D is the carburetor’s diameter. As a result, we can get

Q_f = \frac{KV f}{2D}\sqrt{\rho_a \rho_f}.

The mass flow rate of air is

Q_a= \rho_a v_a =\frac{V f\rho_a}{2D}.

Therefore, the ratio between the fuel flow rate (in mass) and the air flow rate (in mass) is

\frac{Q_f}{Q_a}= K \sqrt{\frac{\rho_f}{\rho_a}}.

We can assume the fuel density to remain more or less constant (although it does change slightly with temperature). Therefore, the only two variables here are the air density and the orifice geometry.

  • If the aircraft climbs to 8000 ft, where the density is 75% that of sea level, assuming you leave the mixture unchanged (K remaining unchanged), we can see that the mixture will become richer by 15% (which comes from \sqrt{\frac{1}{0.75}}). One way to compensate for this is by making the orifice smaller (which would make K smaller). This is exactly what the mixture knob does.
  • Here we are assuming that we want to maintain a fixed air:fuel ratio. This is not always the case. The stoichiometric mass flow ratio for complete combustion is 1:14.7. That is, we need 14.7 grams of air for each gram of fuel. However, due to other factors such as cooling and combustion efficiency, the ideal mass flow ratio is not always 1:14.7. At very high power outputs, it is better to use something like 1:12 (which is known as rich of peak) and at low power settings it is more efficiency to use something like 1:16 (known as lean of peak). As a result, even if there is no change in air density, the mixture should be adjusted whenever there is a change in power. On takeoff, the mixture should be slightly enriched. Once the power is reduced after take-off, the mixture can be leaned.

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