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high pressure tank for doors

To: Bricklin@autox.team.net
Subject: high pressure tank for doors
From: Jon Garruba <jgarru01@longisland.poly.edu>
Date: Wed, 05 Jan 2000 05:23:30 -0500
>Jon,
>   Maybe I am wrong but wouldn't the following line be different.
>Patomoshpere * Vatmoshpere = Psystem * Vsystem
> 1 atm    *  23 in^3   =   7.85 atm  *  Vsystem
> Vsystem = approx 3 in^3    : this is the volume of air the cylinder
>volume
>Vsystem
>would be 23 in^3 and Vatmosphere would be 180.55????  (Not 3).  That might
>throw
>a wrench into the plan.
>   Anyone??
your 100% right. I did the calculations backward and the long way its
much simpler than I made it out to be. But I believe the original was
still correct. I realized this shortly after I posted it.

1 atm = 14.7 (thanks for the correction Kim)
diameter of cyl = 1.5"   (are these correct? anybody)
throw of cyl  = 13" 
                                                      :still neglecting
temp variation
Patomoshpere * Vatmoshpere = Psystem         *   Vsystem
14.7 psi     * Vatm        = (110 psi+ 14.7) * 22.97 in^3 :110 is gauge
pressure
                                                          :we need
absolute pressure      Vatmosphere = 172.88 in^3  

20 ft^3 = 34560 in^3 / 172.88 in^3 = precisely 199.9 door openings 
(weird after making corrections an not rounding its virtually the same
result)

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