Lift me up, lift

We are dealing with fluids in movement. Hence, we need an equation of fluidynamics that relates both pressure difference and speed of the fluid. The Bernoulli Equation could help us. As we could neglect the hydrostatic pressure, we have:

$P_{1} + \frac{d \cdot v_{1}^2}{2} = P_{2} + \frac{d \cdot v_{2}^2}{2}$

Where P stands for the hydrodynami pressure, d for for the density of the fluid and v for the speed of the fluid.

Hence, we should calculate the speed of the fluid in both areas: below and above the wing. In order to do that, we should notice that the air takes 1s to go from one edge to edge of the wing to another and meet again. Hence, the speed of the air below (the displacement of the fluid in this area is igual to the diameter of the circular section) is $v_{1}= \frac {R \cdot 2 m}{1s} = 2 \frac {m}{s}$ . And the speed of the air above (the displacement of the fluid in this area is igual to the lenght of the semi-circular section) is $v_{2}= \frac {R \cdot \pi m}{1s} = \pi \frac {m}{s}$ .

Thus, applying the equation:

$P_{1}-P{2} = \frac {d \cdot (\pi^2-2^2)}{2}=\frac {1.23 \cdot (\pi^2-2^2)}{2}=3.61$

The bottom of the semicircle has length $2 \ m$ (the diameter of the circle), so air molecules must travel at a speed of $v_1 = \dfrac{2 \ m}{1 \ s} = 2 \ m/s$ to go from one edge of the wing to the other edge in $1 \ s$ .

The bottom of the semicircle has length $\pi \ m$ (half the circumference of the circle), so air molecules must travel at a speed of $v_2 = \dfrac{\pi \ m}{1 \ s} = \pi \ m/s$ to go from one edge of the wing to the other edge in $1 \ s$ .

Since the density of air is $\rho = 1.23 \ kg/m^3$ , by the Bernoulii Equation. the pressure difference is $\Delta P = \dfrac{1}{2}\rho (v^2_2 - v^2_1) = 3.61 \ N/m^2$ .

According to Bernoulli's equation, air speed above the wing must be greater than that below the wing to produce a lift. Therefore, we can assume that the semi-circular wing is placed such that the diameter is the bottom side of the wing, while the semi-circular arc is the upper side of the wing. In that position, air must travel longer distance on the upper side of the wing than on the bottom side, for the same time interval.

If $v_1$ is the speed of the air above the wing, and $d$ is the diameter :

$v_1 = \frac {distance}{time} = \frac {0.5\pi d}{1 s} = \frac {0.5\pi\times 2 m}{1 s} = \pi \frac ms$

If $v_2$ is the speed of the air below the wing :

$v_2 = \frac {distance}{time} = \frac {d}{1 s} = \frac {2 m}{1 s} = 2 \frac ms$

Applying Bernoulli's equation :

$P_1 + \frac {1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac {1}{2}\rho v_2^2 + \rho gh_2$

Assuming that $h_1 = h_2$ :

$P_1 + \frac 12\rho v_1^2 = P_2 + \frac 12\rho v_2^2$

$P_2 - P_1 = \frac 12\rho(v_1^2 - v_2^2)$

$P_2 - P_1 = \frac 12\times 1.23 \frac {kg}{m^3}\times[(\pi \frac ms)^2 - (2 \frac ms)^2] = 3.6098 \frac {N}{m^2}$

It is obvious that $P_2 > P_1$ , hence the pressure difference produces a lift.

According to the Bernoulli equation, the sum of the static pressure and dynamic pressure is constant in a fluid line. Therefore, $\Delta p = \frac{\rho}{2} (V_1^2 - V_2^2)$ . We can calculate $V_1$ and $V_2$ from the geometry of the wings: $V_1 = \frac{r\pi}{t}$ and $V_2 = \frac{2r}{t}$ . Substituting into $\Delta p$ , we get $\Delta p = 3.604~\mbox{N/m}^2$ .