再论飞机的升力--牛顿第二定律还是纳维尔-斯托克斯方程

$F = 2 \rho A v \sin\theta \int_0^{\pi/2} \frac{2}{\pi} 2 v\sin(2\phi) d\phi = \frac{8}{\pi} \rho A v^2\sin\theta$

F =  8* 1.225 * 47 * (54*1000/3600)^2 * sin (5*pi/180) /3.14 ~ 2876（牛顿）

Yesterday, I saw a popular science video on Douyin, saying that the lift generated by the aircraft wing cannot be simply explained by Newton's law, nor can it be simply explained by Bernoulli's principle, but the Navier–Stokes equation must be used and one must consider the viscous force of air, and so on. The reason given is that the direct use of Bernoulli's principle cannot explain why the air above the wing flows faster. When the lift is calculated using Newton's law, the result is proportional to the square of the sine of the angle of attack. When the angle of attack is very small, this is a second order small quantity, not enough to provide enough lift. It is said that the Wright brothers succeeded because they used the correct lift formula. The conclusion is that many things in science cannot give simple explanations, and laymen can only watch the excitement. This question aroused my strong interest, and I searched the Internet and found that Michael Merrifield, a British physics professor on YouTube, held a similar view. But this claim is of course unconvincing. Newton's laws are first principles. Both Bernoulli's principle and Navier-Stokes equation are derived from Newton's law and are the direct application of Newton's equations. The viscosity coefficient of air can also be derived from the mean free path of rigid-ball molecular collisions. If the interpretation made by Newton's law is wrong, it can only be an error or omission in the application of Newton's laws. Newton's second law states that the force on an object is equal to the rate of change of its momentum. In physics, air pressure is the transfer of momentum. For an aircraft in flight, the system is the aircraft and the air molecules that collide with the aircraft. In order for such a system to maintain its height under gravity, there must be a corresponding momentum transfer. Specifically, objects must move downward to take away the impulse generated by gravity. This conservation of momentum is an irresistible fundamental law of physics that cannot possibly err in our analysis based on first principles. The lift force of an airplane must certainly be explained by Newton's laws. Let us first review the usual Newtonian mechanics model. In this model, the horizontal airflow encounters the downward deflection of the wing to generate lift, the angle of attack of the wing (which is the angle between the wing and the horizontal direction in horizontal flight) is set as θ, the speed of the aircraft is v, and the air goes down along the wing after encountering the wing, and the air mass hitting the wing per unit time is ρ A sinθ(A is the wing area, \rho is the air density), and the vertical component of the air speed changes from 0 to v sinθ. According to Newton's second law, lift is dp / dt = ρ A v sinθ v sinθ = ρ Asin2θ v2. Let's use this formula to calculate the "Newton's Theory" lift for the Wright Brothers' aircraft. The total wing area of ​​this plane is about 47 square meters (this plane has two upper and lower wings, the wingspan is about 12.2 meters, and the width is 2 meters), the speed is 54 kilometers per hour, and the angle of attack is 5 degrees. The density of air is 1.225 kg/m^3. The sine of 5 degrees is about 0.0872. Bringing in the formula, the lift force is F = 1.225 * 47 * (54*1000/3600)^2 * sin^2 (5*pi/180) ~ 12954 * 0.0872*0.087 ~ 98 (N). The calculated lift was only 10 kilograms, while the Wright brothers' aircraft weighed more than 300 kilograms. More than 30 times less. Indeed, the above formula with the square of the sine of the angle of attack cannot explain the flight of the Wright brothers. ( See https://www.thewrightbrothersusa.com/products/1903-flyer-full-scale-replica for Wright Brothers aircraft data , , . ) So what is the problem with the above explanation? If it can't be explained by air fluid plus Newtonian mechanics, it's probably because we haven't really used first principles. Air is made up of molecules, so we have to add molecular theory to airflow. Suppose a molecule of mass m in the air is flying towards the front of the wing with a horizontal velocity V, and the wing velocity is v, the molecule's velocity relative to the wing is V + v. According to the elastic rigid sphere model, the molecule is reflect on the wing. With ϕ the angle of incidence , the reflection obtains the momentum change in the vertical direction asm ( V+ v ) sin( 2 ϕ ). At the same time we consider another air molecule flying towards the back of the wing with a horizontal velocity V, this molecule has a velocity V - v relative to the wing, and after bouncing on the wing, its momentum changes as m ( − V+ v ) sin( 2 ϕ ). (In the latter case, there must be a condition for the rebound to occur, that is, the horizontal speed of the molecule must be greater than the speed of the wing, otherwise it cannot catch up with the wing.) The sum of the two is 2 m v sin( 2 ϕ ). Simple calculations show that, even when considering the case where the molecular velocities are not horizontal, the vertical transfer of momentum after a pair of molecules on opposite sides of the wing of the ground reference frame after bouncing off the wing is the same 2 m v sin( 2 ϕ ), that is, twice the momentum of the molecule due to the overall motion of the airflow times the sine of twice the angle of reflection. Therefore, even if we consider the momentum transfer and dispersion due to collisions among molecules, the overall result does not change. Why did I not use the angle of attack θ in the above calculation to represent the angle between the molecular motion and the rebound surface ϕ? When I consider the picture at the molecular level, the wing surface will not be flat, but with small bumps all over the surface. From a microscopic point of view, the angle of attack varies continuously between 0 and 90 degrees. When the flight velocity v is much smaller than the average molecular velocity, the angular distribution of the number of incident molecules can be approximated as uniform, so the lift force F is F= 2 ρ A v sinθ∫pi/ 202pi2 v sin( 2 ϕ ) dϕ =8piρ Av2sinθ .  Using this formula to calculate the lift of the Wright Brothers aircraft, I have : F = 8* 1.225 * 47 * (54*1000/3600)^2 * sin (5*pi/180) /3.14 ~ 2876 (Newton).  This value is very close to the take-off weight of 3300 Newtons for the first successful test flight of the Wright Brothers. The difference may be explained by the difference in the actual angle of attack vs. the recoreded 5 degree angle of attack of the Wright Brothers aircraft, which was only a rough estimate. In the above formula, if the angle of attack is 6 degrees, then the lift is 8* 1.225 * 47 * (54*1000/3600)^2 * sin (6*pi/180) /3.14 ~ 3449 Newtons, and the plane can leave the ground. In addition, our analysis also ignores a considerable portion of the calculation of the contribution of the back of the aircraft wing. At this point, we should restore confidence in first principles under classical mechanics.
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