Dealing with air pressure differences inside a sealed hull

Discussion in 'Boat Design' started by W9GFO, Mar 14, 2019.

  1. philSweet
    Joined: May 2008
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    philSweet Senior Member

    If feel like this may be a solution looking for a problem, because all I ever did was run an SS sheetmetal screw into the chamber and call it good (same as I do with those stupid ventless plastic fuel jugs they sell nowadays), But if you are really concerned about where to drill the vent hole wrt flooding issues, drill it in the middle, then take a foot of rubber drip irrigation line and pull it through the hole and tack one side up and the other side down. This should create a (rather weak, but effective) seal in the event that a chamber floods regardless of orientation.

    You could also just run a single length of drip irrigation tubing from bow to stern and put a siphon loop in each compartment along with a dripper head. That ought to set you back about $2 for the complete system. I'd want to be able to blow the tube out with low pressure air.
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  2. Waterwitch
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    Waterwitch Senior Member

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  3. Barry
    Joined: Mar 2002
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    Barry Senior Member

    Boat is being built at 68 degrees F. The post cure is 145F or a 77 degree heat rise.
    P*V/T = PV/T

    P/T = P/T


    P/75 = P(2)/145

    P(2) = 145/75 * P

    Or the pressure inside each space is 1.933 the original pressure I believe.[/QUOTE]

    If you are using this formula, you need to use absolute temperatures not degrees above zero F as you have indicated.

    As the differential between 75 degrees and 145 degrees is small when compared to absolute zero -460 F, your result is close enough for what you are doing.

    But with larger temperature differences, to be accurate, you need to use absolute temps.
  4. Heimfried
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    Heimfried Senior Member

    Ideal gas law for both conditions:
    p(1) * V = n * R * T(1)
    p(2) * V = n * R * T(2)

    both equations devided by V and rearranged:

    p(1) = T(1) * n * R / V
    p(2) = T(2) * n * R / V

    both divided by T :

    p(1) / T(1) = n * R / V
    p(2) / T(2) = n * R / V

    as n * R / V is the same in both

    p(1) / T(1) = p(2) / T(2)

    multiplied by T(1):

    p(1) = p(2) * T(1) / T(2)

    68 deg. F = (68 + 460) * 5/9 K = 293 K (Kelvin)
    147 deg. F = 337 K

    p(1) = p(2) * 337 K / 293 K = p(2) * 1.15

    So the heated up pressure is 1.15 times the original pressure.
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  5. BlueBell
    Joined: May 2017
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    BlueBell WOT

    AKA 1.15 atmospheres, or 2 pounds per square inch at standard atmospheric pressure.

  6. fallguy
    Joined: Dec 2016
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    fallguy Senior Member

    Kelvin? Silly me. It has been about 30 years since my college chem (unapplied).

    And many thanks, I am less worried about having trouble now in post curing.

    Still have to vent all enclosures. Horrible things happen to fiberglass vessels umder pressure. Some area always seems to get the brunt of it.
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