BACK IN THE GARAGE!!! Obviously you haven't shifted your stick enough.
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Date: Tue, 3 Sep 2002 10:39:00 -0700 (PDT)
From Ron Soave <soavero at yahoo.com>
Subject: Math ahead (was Re: Oil Cooler Hose Connections
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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