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Re: Air Flow and Pressure



At 07:59  01/02/2000 -0700, Gary Weaver wrote:

>If I blow air threw a short smooth pipe with a constant diameter the air
>velocity and air pressure will say the same though out the inside of the pipe.
>
>If I take a second short smooth pipe whose area is exactly half that of the
>first pipe and blow air threw it as I did before in the first experement the
>air velocity will double.  What will happen to the air pressure inside of
>the pipe???

Your approaching this from the wrong perspective.

The Hagen-Poiseuille Equation states, for laminar flow in straight hollow 
tubes of constant radius, flow (Q) is determined by the driving pressure:

         Q  =  ( Pi . r^4 . dP ) / ( 8 . n . l ) (n = eta, or viscosity)

That is you don't "apply a flow" and "get a pressure gradient".  You apply 
a pressure gradient and there is a subsequent flow of fluid.

Subtle difference but relevant ;-)

Now, I seem to recall someone saying R = V/I, so we get relative 
resistances of tubes being:

         R  =  ( 8 . n . l ) / (Pi . r^4)

So, if we take your 2 tubes in series, we can work out the relative 
resistances of each.  This will divide the driving pressure as for any 
series resistance circuit, so we can then work out the division of driving 
pressure for any given flow rate.

Then ..... you then get into .....

>If a smooth pipe makes a 90 degree turn and air is blown threw it as before
>in the other 2 experiments will the air on the outside of the curve be
>higher density air because of centrifical force???

All bets are off here, because it depends upon the nature of flow, laminar 
vs turbulent.
With turbulence flow becomes directly proportional to the square root of 
the driving pressure;
therefore, as pressure flow is not linear, resistance is not constant, and 
the flow at which the resistance is measured must be specified.

Other factors in turbulent flow may be summarized,

         Q  =  k  x  ( r^2 . Sqrt(dP) ) / ( p . l )

         where,  k =  some constant
                 p =  rho, the density of the fluid in kg.m-3

         thus, radius has less of an effect on turbulent flow.

The likelihood of the onset of turbulent flow is predicted by,

         Reynold's number  (Re)  =    pvd / n

         where,  d =  the diameter of the tube
                 v =  the velocity of flow
                 p =  rho, the density of the fluid in kg.m-3
                 n =  eta, the viscosity of the fluid in pascal seconds

Empirical studies show that for cylindrical tubes, if Re > 2000 turbulent 
flow becomes more likely
for a given set of conditions there is a critical velocity at which Re = 2000

  ......... or something like this ;-)

Cheers

Mark


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