What amount of lift would be created in the fallacious "lift from equal transit time" theory?

Question

Context

A wrong --but simple to understand-- theory for lift states:

• Air molecules that were close before being separated at the leading edge must be close again after they join back at the trailing edge. For this to happen, molecules must travel the two paths in the same time.
  
  

• As the path is longer on the upper side, due to the camber of the wing, air is accelerated on the upper side. By Bernoulli's equation, static pressure decreases when velocity increases, so the pressure is lower on the upper side, and higher on the lower side.

• Lift is the force created by this pressure differential on the wing sides.

(What is wrong is not Bernoulli's equation, but the assumed relationship between path length and speed.)

Such theory is, for instance, presented as correct in this book.

More at Nasa.

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Question

We know this theory is wrong and lift created by path length difference would be a very small percentage of the lift actually measured (and even null for symmetrical airfoils).

Let's assume a rectangular wing with a simple cambered airfoil, at the critical angle of attack:

• What would be the pressure differential according to equal transit time difference? What percentage of actual lift does it represent?

Answer 1

The theoretical lift coefficient of a NACA 4412 at the critical angle of attack (about 16 degrees) is approximately 1.7. \begin{equation} c_\mathrm{l}=1.7 \end{equation}

For our purposes, that same airfoil has its upper "surface" about 1.03 times as long as its lower, which can be seen by using the distance formula on the coordinates or integrating the cambered NACA airfoil equation. Here is the distance-formula method applied to the Airfoil Tools numbers (thanks to ROIMaison for making me check this calculation):

On average, therefore, the fallacy would predict the air moving along the upper surface at about 1.03 times the velocity as along the lower in order to join up properly.

Making all the assumptions from the NASA link, Bernoulli's equation says \begin{equation} P_1+\frac{\rho V_1^2}{2}=P_2+\frac{\rho V_2^2}{2} \end{equation} Let's represent the lower surface with subscript 1 and the upper with subscript 2. The respective velocities are averages across the surfaces. We have \begin{align} P_1-P_2&=\frac{\rho (1.03V_1)^2}{2}-\frac{\rho V_1^2}{2}\\ &=\frac{609}{20000}\rho V_1^2\\ \end{align} So the average difference in pressure coefficient (i.e., the lift coefficient) is given by \begin{equation} \mathrm{avg}(c_\mathrm{p_{ETT}})=\frac{609}{10000}\frac{V_1^2}{V^2}=c_\mathrm{l_{ETT}} \end{equation} where $V$ is the freestream velocity and ETT stands for equal transit time.

We have another expression for pressure coefficient that we can apply to get a final value. We end up with \begin{equation} c_\mathrm{l_{ETT}}=\frac{609}{10000}\frac{V_1^2}{V^2}=1-\frac{V_1^2}{V^2} \end{equation} or \begin{equation} \frac{V_1^2}{V^2}=\frac{10000}{10609} \end{equation} and therefore \begin{equation} c_\mathrm{l_{ETT}}=\frac{609}{10609}\approx.0574 \approx .034c_\mathrm{l} \end{equation} which is less than 5% of the actual value. However, it is important to keep in mind that the fallacy predicts no lift change with angle of attack (just one of the many reasons it is silly), and so the relationship between $c_\mathrm{l_{ETT}}$ and $c_\mathrm{l}$ varies with angle of attack for a given airfoil. For the 4412, they meet at just over a -5° AOA. For a symmetric airfoil, they meet at zero AOA. For any other airfoil, the calculation must be redone with the correct path lengths and actual lift coefficients, although magnitude of the result is not likely to change dramatically.