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Re: [PSUBS-MAILIST] formula for leak volume?
Wait! My bad! I should have use 0.501 and not 0.5001.
Cheers,
Ian.
-----Original Message-----
>From: irox <irox@ix.netcom.com>
>Sent: Feb 5, 2009 11:45 PM
>To: personal_submersibles@psubs.org
>Subject: Re: [PSUBS-MAILIST] formula for leak volume?
>
>
>I just spend a while trying to figure out why your calculation
>give x10 mine.
>
>I believe:
>0.5001 inches = 0.01270254 meters
>and not 0.0127254 meters
>
>Cheers,
> Ian.
>
>-----Original Message-----
>>From: "Sean T. Stevenson" <cast55@telus.net>
>>Sent: Feb 5, 2009 10:20 PM
>>To: personal_submersibles@psubs.org
>>Subject: Re: [PSUBS-MAILIST] formula for leak volume?
>>
>>Alec, the basic (i.e. not calculus) formula is:
>>
>>Q = K * A* sqrt(2*g*h)
>>
>>where
>>
>>Q = flow rate (m^3 / s)
>>K = discharge coefficient
>>A = area of orifice (m^2)
>>g = gravitational constant (m / s^2)
>>h = head above orifice (m)
>>
>>Now, the K value above is where this becomes a difficult problem, as it
>>is dependent on the size and shape of the orifice, length ratio to
>>diameter, constricting or opening transition, surface roughness,
>>pressure differential, and whether the orifice boundaries are static or
>>moving with respect to one another. Determining the correct K value
>>requires an involved fluid mechanics analysis, and for your example in
>>which you have an annular orifice of extremely narrow aperture,
>>turbulent fluid flow, moving boundaries which may have non-constant
>>relative velocity, varying pressure differential across the orifice (due
>>to both vehicle movement in the water column, and increasing pressure
>>inside the hull as it takes on water), it is of questionable value to
>>calculate it that accurately.
>>
>>Approximating your problem as the simplest case, where you have a single
>>perfectly round hole, through a thin frictionless wall, into a fluid on
>>the other side of negligible density, completely incompressible fluid,
>>zero pressure gradient across the inlet area (i.e. orifice parallel to
>>sea surface), constant differential pressure across the orifice, etc.,
>>you have:
>>
>>A = pi * (0.0127254^2 - 0.0127^2) = 2.028856746E-6 m^2
>>
>>h = P / rho * g = 3447378.645 / 1000 * 9.80665 = 351.5347897 m
>>
>>Q = (1) * 2.0288566746E-6 * sqrt(2 * 9.80665 * 351.5347897) =
>>1.68465455E-4 m^3/s = 0.168 liters / second
>>
>>This is about 10 liters / minute, coming in at a velocity of about 82
>>meters per second (V = sqrt (2*g*h)). (186 miles per hour - a good
>>reason to have some sort of separation between your operator and likely
>>through-hull failure points).
>>
>>So, you only have about 10 liters per minute to deal with at the bottom
>>depth, provided your tolerance is as tight as you propose, which I would
>>guess is unlikely (0.001" gap on a prop shaft?). Recalculate for a more
>>realistic installation tolerance, and then make sure that your bilge
>>pumping arrangements can handle enough flood water to permit you to
>>initiate an ascent and reach the surface without accumulating enough
>>water to either prevent surfacing because you're too heavy, or to
>>disable critical system components. Once surfaced, as long as the
>>pumping arrangement is able to remove more water than comes in at
>>whatever depth the partially flooded hull exposes the breach to, you
>>will eventually pump it all out.
>>
>>-Sean
>>
>>
>>Smyth, Alec wrote
>>> I'm trying to calculate something but haven't found the right formula.
>>> Let's say the seal on my 1" prop shaft disintegrates. Just inboard of
>>> the seal, the shaft goes through an opening that is 1.002". This
>>> opening is not of any significant length The ambient pressure is 500
>>> psi. How many gallons or liters per minute would come in?
>>>
>>> I'd really appreciate any pointers!
>>>
>>>
>>> thanks,
>>>
>>> Alec
>>>
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>>
>>
>>
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