Could the Bernoulli equation be explained by kinetic theory? I have watched on the Youtube channel of Eugene Khutoryansky an explanation of the Bernoulli equation on the molecular level.
It states that" An atom is more likely to enter the narrow section of a pipe if it has a larger component of its velocity parallel to the pipe walls.If it has a larger component of of its velocity perpendicular to the pipe walls, the atom is more likely to bounce back.As atoms leave the narrow section, their collision with other atoms cause them to again have a large component of their velocity perpendicular to the pipe walls"
Is this statement valid? If not, could there be an explanation of the principle on molecular level?
 A: The Bernoulli principle can be derived from the Navier-Stokes equations, which in turn can be derived from Newton's laws applied to small volume elements of a fluid (not usually to individual particles of fluid). It is a direct consequence of Newton's second law that the fluid will accelerate from a region of high pressure to a region of low pressure, and that is essentially what is stated by the Bernoulli principle. 
The explanation as given does not make much sense, either in terms of the Navier-Stokes equations or in terms of the flow of individual molecules. A molecule will enter the narrow pipe depending on its position, not its velocity. Pressure is "omnidirectional" because it is transmitted by collisions between random motions of individual particles. The actual distances travelled by particles of fluid between collisions don't have much bearing, but they are smaller than the size of the volume elements used to describe streamline flow (kinetic theory does not explain Bernoulli's principle directly, but it does explain pressure, and pressure explains Bernoulli's principle).
In the case of a constriction, pressure is applied by the narrowing of the walls, but this is balanced by the pressure originating the flow (if pressure applied by the walls were not equal to the applied pressure, there would have to be an acceleration according to Newton's second law, but if the pipe is uniform cross section area, there is no acceleration). 
At the position of the narrow tube itself there is no barrier applying pressure. Fluid accelerates because the volume flow must be the same through a narrower opening. This can only happen if pressure before the constriction is greater than pressure in the constriction. Thus the fluid is being squeezed through faster because of the pressure differential.
(Sorry if this sounds a bit circular, but Bernoulli's principle works both ways, meaning it is a bit circular. A pressure differential from high to low pressure accelerates the fluid, and the fluid can only be accelerated by a pressure differential). 
A: While it is not a great idea trying to describe fluid mechanics using their constituent particles, I think that this explanation holds up pretty well. As Tomas Brauner commented, when you talk about this explanation in the context of single atoms or molecules, it does not make sense practically, unless the size of the narrow opening is comparable to the size of the molecules. But in order for this explanation to work, we need to consider a group of molecules. First , I will complete the explanation and then move on to the validity of the statement
When you take a random bunch of molecules, some of them have velocities in the direction of the narrow pipe while some have motion perpendicular to the opening or towards the wall:

Now what this mean is that not all of the particles have velocities towards the narrow opening, but a large number of them do. Some particles also have velocity toward the wall(away from the opening of the narrow section). These particles are the ones that mainly contribute to the pressure. Pressure on the walls of the pipe is due to the motion of these particles. Also, some particles have a bit of parallel velocity and a bit of perpendicular velocity. These contribute a little to the pressure and also to the forward flow of the fluid. (NOTE: when I talk about particles having perpendicular of parallel velocities, I am simplifying a bit and mean most of the velocity is perpendicular to the direction of flow or most of the velocity is parallel to the direction of the flow. The direction of the flow is where the narrow opening is located.)
Now, let's say the group of particles enters the narrow section. Take $m$ to be the total mass of the particles. Now, the Mass Flow Rate($\frac{m}{t}$) of the particles must be the same for both the wider section of the pipe and the narrower section. The narrow section has a smaller volume, so the only way for this to be true is that the velocity of the particles(towards the right) must increase. so now, even a larger portion of the particles has parallel velocity and a smaller number of particles have perpendicular velocity. That corresponds to a lowering of the pressure on the wall(which was due to the perpendicular velocities of some of the particles)

Now, why does this happen? Why does the perpendicular velocity of some particles become parallel? Here is where I consider the statement. What Eugene says is somewhat correct. The particles having perpendicular velocities could actually traverse the narrow pipe, because their sizes are too small compared to the diameter of the pipe. But the second part of the statement gives us something to consider. 

As atoms leave the narrow section, their collision with other atoms cause them to again have a large component of their velocity perpendicular to the pipe walls

When the atoms(which I have been referring to as particles this whole time), enter the narrow section, they can collide with other atoms, which have a large component of their velocity parallel to the narrow opening. When this happens, their velocity changes an is now aligned parallel to the pipe. They still have some perpendicular velocity, which is why the narrow section will have some pressure, instead of zero pressure. 
Okay. So TL;DR: Particles or atoms of the fluid, have some random velocity, with a large portion of them having velocity parallel to the narrow opening(that is why the fluid flows to the right). When particles enter the narrow section of the pipe, the fluid's speed increases. Some of the particles  collide with each other, causing even a larger number of them to have velocities parallel to the pipe. But even then, some of the particles have perpendicular velocity, which is why there is some pressure on the walls(pressure being caused mainly by the perpendicular component of velocity of the atoms).Thus, we get a molecular explanation of Bernoulli's principle. Please note that this is a heuristic view.
(PS: Ignore my bad drawing/doodling. I am a newbie to using Paint for drawing.)
A: 
Is this statement valid? If not, could there be an explanation of the principle on molecular level?

I think the statement is valid and to understand why; the following exercise might be helpful. In his video, the molecules enter the narrow tube from high pressure (wide) region 1, pass through the low pressure (narrow) region 2, and exits again at a wide region 3. In his explanation, region 3 had molecules which on impact with our original molecule lead to an increase in the wall pressure. But, for simplicity, imagine if there are no molecules in region 3. What will be the scenario? 
The absence of any molecules means region 3 is a vacuum and our original molecule after exiting region 2 will retain its velocity effectively forming a jet (basically gas is escaping from a high-pressure region to a very low-pressure region). Now let us add 5 molecules in region 3. Now, we have to introduce the concept of mean free path since there is an average distance travelled by a our original molecule without hitting the other 5 molecules. On collision, our original molecule will now have a normal velocity component which will lead to a slight increase in the wall pressure. However, the axial velocity component is the dominant vector but the magnitude is slightly reduced. Following this, if we add more molecules to region 3 the mean free path length keeps on decreasing thereby increasing the chances of a collision and the number of collisions. The tangential velocity component also keeps on decreasing as the normal velocity component keeps increasing thus effectively increasing the wall pressure.
So, I personally think the statement is valid and hope this helps in some way.
A: As others have mentioned, the mean free path of atoms is usually much smaller than the dimension of the pipe. As mentioned by PNS, the statement 

If it has a larger component of of its velocity perpendicular to the
  pipe walls, the atom is more likely to bounce back

means that as the atom enters the tube, close to the boundary, it will collide with atoms already present in the tube. Let's call "x" the direction in which the fluid flows, along the axis tubes.
If the atom enters the tube heads on, it has only a small chance of being reflected back when colliding with another atom. This is because the momentum along x of the center of mass of the two atoms is more likely to be directed towards the general direction of motion of the gas, and you are already going fast in this direction. Imagine a collision with a random atom already in the tube: The entering atom contributes a lot to forward total momentum along x. After the collision, the center of mass still has the same speed in the x direction (conservation of momentum), even if the atoms can change direction of motion. However, to get back into the large tube, you need to go faster in the -x direction than the center of mass. Otherwise, you keep on going. This is relatively unlikely in this situation, when compared to the whole ensemble of possible collisions. 
If the atom just glances the entrance of the tube, at a very shallow angle, its x momentum is small. Therefore, in the same interactions as before, you don't need much speed in the -x direction to get back out of the tube. The speed  of  the center of mass of any collision in the x direction will be smaller than in the previous case, and will in many cases be negative, ie directed towards the larger tube, so there are more particles in the tube that can give enough of a boost to the incoming particle to go back out.
This whole discussion of centers of mass may seem a bit complicated, but I think it is required to understand what happens as atoms can (and will most of the time) be deflected when colliding with each other. Also, it makes things easier to understand as we know that interactions in the center of mass reference frame are simple: to conserve energy, the two identical atoms must keep the same speed. 
The same effect will of course occur in the direction from the smaller to the larger tube. However, since there is a net movement of the gas in the x direction, the number of reflections will be smaller in the number of transitioning particles from small to large than from large to small, as the center of mass speed of any collision has a "drift" in the right direction.
This is far from a complete answer as to "how do you get the Bernouilli equation" from kinetic principles, but I think it clarifies at least the reflection of particles aspect. 
A: Im sorry for finding this post so late, but Ive seen her video recently as well and tried to put the thoughts and ideas from her video into mathematical terms. I was able to derive the Bernoulli equation with the kinetic gas theory. The only flaw I noticed in her video is that bernoulli's equation has nothing to do with probability. It's merely a direct consequence of energy conservation.
Since I dont want to post a too long anwer on this already well answered thread, Ill post a brief summary of my derivation:
From the kinetic gas theory we know that the to the surface normal speed component squared is directly proportional to the pressure $p$:
$$v_\perp^2=\frac p\rho$$
Energy conservation gives us
$$E_{kin}=E_{kin}'\rightarrow v_x^2+v_y^2+v_z^2=v_x'^2+v_y'^2+v_z'^2$$
Because the gas is isotropic, $v_y^2=v_z^2$ must be true. Suppose the gas is flowing parallel to $v_x=u$, then both $v_y$ and $v_z$ must be normal to the surface the pressure is acting on. We then get
$$u^2+\frac p\rho+\frac p\rho=u'^2+\frac{p'}\rho+\frac{p'}\rho$$
This ultimately gives us
$$\frac12\rho u^2+p=\frac12\rho u'^2+p'$$
or
$$\frac12\rho u^2+p=p_0$$
Edit: Strictly speaking, were only talking about average speeds, so substitute all $v$'s with $\bar{v}$. (note that in this case its $\bar{v}^2$ not $\bar{v^2}$)
Edit2: Furthemore we aren't actually allowed to plug $\bar{v}^2$ directly into the energy conservation formula, since $E_{kin}\propto \bar{v^2}$. However, $\bar{v^2}=\frac38\pi\cdot\bar{v}^2$. The factor cancels out in the derivation.
A: The former response in my opnion has a problem. The person says correctly that the pressure is given by:
\begin{equation}
P = v_{y}^{2}/\rho = v_{z}^{2}/\rho.
\end{equation}
But the point is that the equation the person wrote is not correct, because the mean square velocity in the X direction has two components and the energy conservation equation must be:
\begin{equation}
 0.5(\rho V^{2} + \rho v_{x}^{2} + \rho v_{y}^{2} + \rho v_{z}^{2}) = 0.5(\rho V^{'2} + \rho v_{x}^{'2} + \rho v_{y}^{'2} + \rho v_{z}^{'2})
\end{equation}
But we see in this last equation that the remain terms in both sides is not more just the pressure, but 3/2 of it.
So we miss the Bernoulli equation due to the presence of the x component in both sides.
