Re: [Phys-l] question about Bernoulli

[William Robertson <wrobert9@ix.netcom.com>]

It seems that adding this component of velocity parallel to the pipe  
does not affect the component of velocity perpendicular to the pipe.  
You have a greater overall velocity, with the component perpendicular  
to the pipe remaining as it was when the air was "still." And if you  
are somehow reducing this perpendicular component of velocity, what is  
the mechanism for that? Sorry if I'm being dense on this.

Bill


William C. Robertson, Ph.D.


On Nov 17, 2010, at 6:02 PM, brian whatcott wrote:

> How about this?
>
> Draw a circle of unit radius. The circle depicts the speed and  
> direction of any air particle in the middle of a pipe.
> Draw two parallel horizontal lines touching the circle, one on each  
> side. The two straight lines represent a pipe.
>
> From the center of the circle, draw an arrow perpendicular to the  
> top horizontal.
> The perpendicular arrow represents the speed of one particle hitting  
> the pipe, contributing some pressure, proportional to the arrow's  
> length.
>
> Now draw a third short horizontal line with a arrow between the  
> other two horizontal lines - representing the speed with which all  
> the air is now moved.
>
> This 'wind' arrow will move any particle moving towards a wall.  The  
> vertically moving particle now moves instead at a forward slant. So  
> a particle with the SAME speed as before now has a forward  
> component. The outward component of its speed is reduced for this  
> reason....
>
> This argument cries out for a diagram, in the Newtonian manner!
>
> Brian W