Message-ID: <001d01bf1311$f2ada020$5550aecf@djpryan.ruralnet.net>
From: "DON J RYAN" <djpryan@ria.net>
To: "LANCAIR LIST" <lancair.list@olsusa.com>
Subject: PRESSURE VS SPEED
Date: Sun, 10 Oct 1999 06:23:58 -0500
Mime-Version: 1.0

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The relationship between air velocity and pitot pressure is as follows:

        H = V^2/(2*g)   ft of air head is equal to the velocity in ft/sec
squared divided by twice the gravity constant 32.2.  To get inches of water
correct for the density ratio and the multiply by twelve to convert feet  to
inches.

At standard conditions 15 deg C and 1 atm
                                 200mph  =  19.7 inches of water
                                  120mph  =   7.1 inches of water
                                    90mph  =   4.0 inches of water
                                     60mph =   1.8 inches of water

Note that 200 mph is also = to about 19.7/13.6 = 1.45" of mercury, and thus
you could get that much manifold boost with a good induction system and
neglecting propellor influence.

I may be a little rusty, and don't have the texts handy, but pretty sure
this is about right.  The relationship is out of the fluid flow book.

Hope this helps.


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