From: CHRISTOPHER_ZAVATSON@udlp.com
Date: Tue, 12 Oct 1999 10:58:39 -0500
Message-ID: <0008EFB5.C21254@udlp.com>
Subject: Pitot tube (dynamic) pressure
To: Lancair.list@olsusa.com
Mime-Version: 1.0

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1.   H = V^2/(2*g)   

and

2.   P = 1/2 * rho * V^2

are equivalent once density is added to equ. 1 to make it fluid specific.  One
must be very careful with units in this case, especially with density (slug/ft3
vs lb/ft3)

P = H[ft] * dens[lb/ft^3] = v[ft/s]^2/(2*g[ft/s^2]) * dens[lb/ft^3] 
                          = 0.5 * v[ft/s]^2 * dens[lb/ft^3]/g[ft/s^2]
                          = 0.5 * v[ft/s]^2 * rho[slug/ft^3]

If density of the fluid can be considered a constant, as in incompressible pipe
flow, it is conventional to express pressures in terms of pressure heads. Then
density can be left out of the equation.  In aeronautical applications, density
of the fluid (air) varies enormously with changes in altitude (among other
things)and, therefore, dynamic pressure (1/2 * rho * V^2) is much easier to deal
with.  It also makes the equation independent of the gravitional 'constant'
which technically varies with altitude and latitude.

Chris Zavatson
L-360
N91CZ
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