RE: [PSUBS-MAILIST] formula for leak volume?

Wow, what outstanding responses guys, I love PSUBS. Thanks Brian, Sean and Frank!

In the meantime I found a downloadable program that yielded 0.09 lpm, so at least we're in the same order of magnitude. That program was intended for piping, and I was putting in all sorts of outlier parameters so had no confidence at all in the result.

Where I was going with this is that I'm wondering whether to build a last-ditch safety device onto the shaft. There would be an o-ring on the shaft, and a moveable threaded piece to be engaged in an emergency. With the shaft stopped, this moveable bit would push the o-ring into a conical groove in the shaft housing, and form a seal between itself, the housing, and the shaft. What I have to weigh is the complication of such a device, potential for vibration, etc. against the simpler alternative of a built-in restricor. Actually it's two restrictors. Each end of my shaft tunnel has a conical bearing. The shaft openings in the bearing seats could be bored to close tolerance because being right next to the bearing, I wouldn't expect the shaft vibrations to have any great amplitude there. I should be able to make it quite close, I learned my machining making plastic injection molds.

I'm hoping two such restrictors in series would produce significantly less flow. But let's go with 0.168 lps. That means I'd be shipping roughly 10 kg per minute. My reserve buoyancy is 515 kg, so I'd have 51 minutes before the sub became terminally heavy.

I'm leaning toward building it with the restrictors, doing a 500 psi test to see what the flow rate turns out to be, and then only if necessary adding the shaft blocker. If the approach with no moving parts turns out to be sufficient, I much prefer it for it's simplicity. To define "sufficient" I have to dig out the Captain's book, where I remember the emergency ascent rate for a K-250 was quantified in a story involving the recently restored Trilobite. I'd expect a comparable ascent rate for this boat, it's quite similar to the Ks.

As always thank you for your help.

Alec

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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