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再论飞机的升力--牛顿第二定律还是纳维尔-斯托克斯方程

热度 1已有 893 次阅读2022-5-16 08:49 |个人分类:科普|系统分类:科技

昨天在抖音上看到一段科普视频,说是飞机机翼产生的升力不能用空气被机翼转向下方而用牛顿定律简单解释,也不能用伯努利原理简单解释,必须用 Navier–Stokes 方程,考虑空气的粘滞力,云云。其给出的理由是,直接用伯努利原理,不能解释为什么机翼上方的空气流得快;用牛顿定律计算出来,升力正比于攻角正弦的平方,攻角很小时,这是个二阶小量,不足以提供足够的升力。还说乔治-凯利给出的升力公式里,升力正比于攻角的正弦,这才是对的。莱特兄弟之所以成功,就是因为用了正确的升力公式。其结论是,科学里很多东西无法给出简单的解释,外行只能看看热闹。这个问题引起了我强烈的兴趣,在网上搜索,发现 YouTube 上英国物理教授 Michael Merrifield 也持类似的观点。

但这种说法当然不能令人信服。牛顿定律是第一原理。伯努利原理与 Navier–Stokes 方程都是从牛顿定律推导出来,是牛顿方程的直接运用。空气的粘滞系数也可以通过刚球分子碰撞的平均自由程推导。如果用牛顿定律做出的解释不对,只能是在牛顿定律的运用中出现了错误或者遗漏。牛顿第二定律就是说物体受到的力等于其动量变化率。空气压力在物理里面那都是动量的传递。对于飞行中的飞机,这个系统就是飞机与飞机碰撞的空气分子,要这样一个系统在重力下保持高度,必须有对应的动量传递,具体来说必须有物体向下运动,带走重力产生的冲量。这种动量的守恒,是不可抗拒的基本物理规律,不可能影响我们基于第一原理的分析。用牛顿定律肯定能够解释飞机的升力。

先让我们回顾通常的牛顿力学模型。在这个水平气流遇到机翼向下偏转而产生升力的模型里,设机翼攻角 (这是水平飞行时机翼与水平方向的角度)为 $\theta$, 飞机速度为 v, 设空气遇到机翼后沿着机翼下行,单位时间内撞到机翼上的空气质量为 $\rho A \sin \theta$ (A 为机翼面积,\rho 为空气密度), 而空气速度转向后其垂直方向分量从0变为 $v \sin\theta$。根据牛顿第二定律,升力就是 $dp/dt = \rho A v \sin\theta v \sin\theta = \rho A \sin^2\theta\ v^2$。

且让我们用这个公式对莱特兄弟的飞机计算其“牛顿理论”升力。这架飞机的机翼总面积约 47 平方米(这款飞机有上下两个机翼、翼展约12.2米、宽2米),速度 54 公里每小时,攻角 5 度。空气密度  1.225 kg/m^3 。5度正弦约为 0.0872 。带入公式,升力为 F = 1.225 * 47 * (54*1000/3600)^2 * sin^2 (5*pi/180) ~ 12954 * 0.0872*0.087 ~ 98 (N)。计算出来升力只有 10 公斤,而莱特兄弟的飞机重量有300多公斤。差了30多倍。确实,上面带着攻角正弦平方的公式无法解释莱特兄弟的飞行。(莱特兄弟飞机数据 参见 https://www.thewrightbrothersusa.com/products/1903-flyer-full-scale-replica https://wrightstories.com/wright-brothers-get-a-lift/,https://corescholar.libraries.wright.edu/cgi/viewcontent.cgi?article=1001&context=following https://en.wikipedia.org/wiki/Wright_Flyer, http://www.wrightflyer.org/wp-content/uploads/2012/10/Flight-Mechanics-In-Modern-Terms.pdf。

那么上面的解释问题在哪呢?如果用空气流体加上牛顿力学不能解释,这可能是因为我们还没有真正使用第一原理。空气是分子组成,所以我们得在气流之上加上分子理论。假设空气中一个质量为 m 的分子以 水平速度 V 向机翼正面飞来,而机翼速度为 v ,该分子相对于机翼的速度为 V + v。按照弹性刚球模型,该分子在机翼上以与机翼成$\phi$角度入射,反射获得垂直方向的动量变化为$m (V+v) \sin(2\phi) $。同时我们考虑另一个空气分子以水平速度 V 朝机翼背面飞来,该分子相对于机翼的速度为 V - v,在机翼上反弹之后,其动量变化为 $m (-V+v) \sin(2\phi) $。(后一种情况要发生反弹必须有一个条件,那就是分子水平速度必须大于机翼速度,否则它就追不上机翼。)两者之和为 $2 m v \sin(2\phi) $。简单的计算表明,即便是考虑分子速度不是水平的情况,地面参考系速度相反的机翼两面的一对分子在机翼上反弹之后产生的垂直动量传递也是  $2 m v \sin(2\phi) $,也就是分子因气流整体运动的动量的两倍乘以反射角二倍的正弦。因此,即使我们考虑分子之间的碰撞发生动量传递分散,总的结果也不会变。

为什么上面的计算里,我不用 攻角$\theta$表示分子运动与反弹面的夹角$\phi$呢?这是我考虑到在分子级的微观角度,机翼不会是平的,而是一个个的小鼓包。或者说,从微观的观点,攻角是0到90度之间连续变化的。在飞行速度 v 远小于分子平均速度时,入射分子数量的角分布可以近似为均匀的,因此升力 F 为

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

用这个公式计算莱特兄弟飞机的升力为:

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

这个数值已经非常接近莱特兄弟第一次成功试飞 3300 牛顿的起飞重量了。其中的差异也许可以用攻角与的差别解释,莱特兄弟飞机的5度攻角只是一个大概估计。上面公式中如果攻角为6度,那么升力就可以达到 8* 1.225 * 47 * (54*1000/3600)^2 * sin (6*pi/180) /3.14  ~ 3449 牛顿,就飞起来了。另外,我们的分析对飞机机翼背面的贡献的计算也忽略了相当一部分。

在之前的“牛顿力学”计算中, \rho A \sin \theta v,这是单位时间入射到机翼上的空气质量,这是对的。但是以前的计算假设空气沿着机翼下行而引入了又一个攻角正弦 \sin \theta 的因子,这在飞行速度远低于分子速度时是错误的。实际上从分子微观角度,机翼是不平整的,大部分分子其实是几乎垂直于机翼面反弹,也就是将水平空气的转向下方。同时飞机背面的分子碰撞也产生同样级别的升力。考虑到这些细节,牛顿定律就给出了近乎实际的结果。

至此,我们应该恢复了经典力学下对第一原理的信心。


lift.png



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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