Can someone else how the physics behind the water cyclone in

>>7762756
pressure

Can someone please explain how using this mixer causes the water to spin in a way where >>7762762
Care to elaborate?

eletrcomagnetic force produced by the crust of the Earth interacts with the ions in undistilled water and creates potential energy that is released when an exit is created at the bottom. That is why if you fill your bath with distilled water and let it drain, it won't create a cyclone.

>>7762799
But what causes this shape?
Sorry if my questions are a bit general

>>7762799
ayy lmao
>>7762805
youre getting rused m8

>>7762805
because of the shape of the Earth's electromagnetic field. You can imagine the field being like straight arrows perpendicular to the surface of the Earth, like a million of straight colons, and since the ions want to move in a direction that is perpendicular to these arrows, it results in an circular motion above the hole. Then gravity do its job.

File: 1139265064469.jpg (93KB, 700x700px) Image search: [Google]

93KB, 700x700px

>>7762756

Just Google "electro-static pressure in hydrocentric systems". Its a lil complex but it should still be understandable to you

File: Vortex-Bottle-Connector.jpg (30KB, 600x759px) Image search: [Google]

30KB, 600x759px

In the early 1930’s in Austria, Victor Schauberger, e.g. fabricated conical pipes of special materials, which contained a corkscrew turbine. Operated by an electric motor, the spiral turbines screwed water into a vortex flow and directed the water onto a conventional water turbine coupled to a generator.

Schauberger claimed, that as the water was screwed faster and faster, it suddenly began to produce enormous amounts of energy. Coupled to a dynamo, the turbine began to produce more electricity than the input motor was consuming. The system quickly went out of control as the apparatus tore itself away from its mountings and smashed itself against the ceiling.

>>7762756
dont waste your time here, youll have better luck asking on /b/

this place is full of people who call themselves physicists but have the intellectual capacity of a nematode

>>7762756
We actually had to derive this in our continuum mech. final.

Basically once you apply the IBVP to the Navier-Stokes you'll end up with a pressure gradient towards the centre.

>>7763839
Fuck you Anon. You really think we can't do this shit whenever we feel like it?

Fine I'll do it quickly. Here's an easy way to find the pressure gradient with minimal assumptions.

Approximate the fluid as inviscid ,the inviscid form of the Navier-Stokes is:
[math]\nabla P = \rho \bar{g} - \rho \frac{D \bar{v}}{Dt} [/math]

Using a cylindrical coord. system with [math]z[/math] the "vertical" direction orthogonal to the ground, [math]r[/math]is in the radial direction and [math]\theta[/math] is angular coord. I'll use the vector notation [math]e[/math] with subscript. The velocity of the fluid is thus [math]\bar{v} = \frac{A e_\theta}{r}[/math] for some constant A. With the no-slip condition on the center shaft of the fluid fully developed we have

[math]v(R)= \omega R = \frac{A}{R}[/math]
where [math]\omega[/math] is the angular velocity, R is the radius of the entire tube, so
[math]A = \omega R^2[/math] and
[math]v = \frac{\omega R^2}{r} e_\theta[/math]
Note, because
[math]\omega =\frac{|v| \sin{\theta}}{|r|} \Rightarrow \omega = \frac{A}{r} e^\theta [/math]
[math]\omega =\frac{d \theta}{dt} [/math]
[math]v_\perp =r \frac{d \theta}{dt} [/math]
[math]v_\perp =|v| \frac{d \theta}{dt} [/math]

The substantial derovatove can be evaluating by taking the total derivative and rearranging
[math]\frac{d \bar{v}}{dt} = - \frac{\omega R^2}{r^2} e_\theta \dot{r} + \frac{\omega R^2}{r} \frac{d e_\theta}{dt} [/math]
From [math]\frac{d e_\theta}{dt} = - \dot{\theta} e_r[/math], after some elementary manipulations and considering the fluid velocity in the radial direction is zero and [math] \dot{\theta} = \frac{v}{r} [/math],
[math]\frac{d \bar{v}}{dt} = - \frac{\omega R^2}{r} \dot{\theta} e_r [/math]
Now
[math]\frac{D \bar{v}}{Dt} = - \frac{\omega R^2}{r^2} v e_r = - \frac{\omega^2 R^4}{r^3} e_r [/math]

So the gradient becomes

(cont.)

>>7763839
>>7763931
[math]\nabla P = \rho \bar{g}e_z + \frac{\omega^2 R^4}{r^3} e_r [/math]

You can plot this out on a 2D, remembering that [math]z[/math] is the vertical direction (use that as the "y-axis" for convenience), the gravity vector is non-zero everywhere except downwards so [math]\bar{g} = -g [/math] the vector becomes [math]-\rho g e_z[/math] so on your plot draw a downward arrow. Now make the [math]r[/math] direction on your "x-axis" draw an arrow with the start connected to the end of the gravity vector. As with all vector addition connect another arrwo from hte start of the [math]z[/math] vector to the end of the [math]r[/math] vector to find a 2D direction for [math]\nabla P[/math].

You'll notice from the direction of the vector that the [math]\nabla P[/math] of the fluid is to the right and downwards which creates the vortex at a quasi-equilibrium.


>>7763839
Also Physics is a meme major for kids who weren't smart enough to get into ChemE. Prove me wrong.

>>7763821
Build one then.

>>7763839
nematodes are fantastic human beings

>>7763991
Agreed. Better than Abos anyway.