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Re: Math ahead (was Re: Oil Cooler Hose Connections

Subject: Re: Math ahead (was Re: Oil Cooler Hose Connections
Date: Tue, 03 Sep 2002 23:26:07 -0400
Cc: Spridgets@autox.team.net
References: <20020903173900.96209.qmail@web11702.mail.yahoo.com> pop018.verizon.net from [151.201.239.51] using ID <vze2z49v@verizon.net> at Tue, 3 Sep 2002 22:26:03 -0500
User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:0.9.4.1) Gecko/20020508 Netscape6/6.2.3
OWWWWWWWWW, my head hurts after that.

Dave

Ron Soave wrote:

> --- "Wm. Severin Thompson" <wsthompson@thicko.com>
> wrote:
> 
>>Ron (Grasshopper...the Unlearned One) is no doubt
>>sitting in his newly
>>purchased racecar in his garage, making vroom vroom
>>noises, with the wrong
>>stick shift in his hand...
>>
> 
> Grasshoppah back from walking on rice paper without
> leaving mark...
> 
> This is all from memory, so no one bitch if I leave
> out a few things.  Pressure drop - the resistances
> need to be considered in series.  Glen is correct that
> small fittings will control the resistance, but small
> lines + small fittings is worse than large lines +
> small fittings.  To figure the difference, take
> diameter 1 and raise it to the 4th power.  Take
> diameter 2, raise it to the 4th power it.  Divide one
> by the other - that's your difference.
> 
> Basically, the pressure drop is a defined by:
> 
> DeltaP = K * q
> 
> Looks a lot like V=I*R, right?  and it is...
> 
> q is the dynamic pressure, aka velocity pressure, aka
> "dynamic head" (easy, Flounder...).  It is also .5 *
> density * velocity squared/gravity.  So the velocity
> squared term dominates.  What determines velocity?  
> 
> V=w/(density * flow area).  So as the area gets
> smaller, the velocity gets higher at a given flow
> rate.  The area is a function of diameter squared, and
> the q is a function of the velocity squared, hence the
> 4th power relationship above.
> 
> So what is "k"?  It's a resistance factor, also a
> function of geometry.  For a hose, K=4fL/d, where 4f
> is the friction factor, L is the length of
> tubing/duct/etc, and d is the effective diameter of
> the duct.  So as d decreases, K increases, delta P
> goes up.  Again, small diameter is not good.  
> 
> What about 4F?  f is equal to (.046)/(Re^.2).  And Re
> is the Reynolds number, which is density X velocity X
> diameter divided by viscosity.  Again, small diameter
> = high velocity, equals high Re, equals high f, equals
> more pressure drop.
> 
> Enough of that, keep the restrictions small, the
> pressure DROP will be low.  

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