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Re: [PSUBS-MAILIST] Precision prop shaft



Hello Alec,
Have you considered the reduction in the through hull diameter due to the pressure loading on the hull at depth? If this reduction is in the order of the clearance you mention the shaft might get stuck. I have done some calculations in the past to make sure for instance the weight release system would still work at depth. i am not sure this would apply to your geometry,
the way I did it is a bit simplistic but should give a good order of magnitude:
 
1-estimate the stress in the hull
(for a spherical shell it should be close to sea_pressure*Diameter/(4*hull_thickness) )
so for instance for 500psi sea pressure, diameter of 30", thickness of 0.25" you get 15,000psi stress in the hull

2-estimate the radial displacement inside the through hull
for this one the type of formula above does not apply because the through hull is too thick to be considered as a shell.
So instead I used a stress/strain solution known for a hollow cylinder subject to external pressure, where the hollow cylinder is the through hull, and the external pressure is chosen to be a fraction of the stress in the hull calculated above.
 ('a fraction of' because the through hull sticks up on either side of the hull to reduce the stress level in the through hull)
 the displacement on the internal face of the through hull reads:
ri* ((1-nu)*A - nu *longitudinalstress) / E + B / ( 2*ri * mu)
 
where ri is the internal radius of the through hull, re the external radius of the through hull
nu the poisson ratio, E the Young's modulus,
mu = E/(2*(1+nu))
A= (internalpressure*ri^2-externalpressure*re^2)/(re^2-ri^2)
B= (internalpressure-externalpressure)*ri^2*re^2/(re^2-ri^2)
 
Taking an internal pressure of 1atm, and a longitudinal stress equal to sea pressure (not much influence though), E=28446000psi nu=0.3, ri=0.25" re=1", and let's take a fraction of 2/3 of the hull stress as the outer pressure boundary condition.  
If the calculation is correct I get a displacement of .0002", could be close to the clearance..
 
 
cheers,
 
Antoine
 
 
On Mon, May 25, 2009 at 10:26 AM, Smyth, Alec <Alec.Smyth@compuware.com> wrote:
Hi all,

You may recall some months ago there was an interesting thread in which
many of you helped with calculations for the leak rate expected of a
prop shaft with a failed seal. In a nutshell, what I was proposing was
to machine my prop shaft and it's associated components very closely so
that, in the event of a seal failure, the leakage rate would be small
enough that I'd have time to surface before getting too heavy.

Well, it's built and tested, and it worked. The leak rate as tested
turned out to be 1 liter per minute at 2800 psi. Ambient pressure will
be 500 psi, and I had planned to test at that pressure. I used a
pressure washer for the test, and should say I know nothing about them.
My intention was to dial in the pressure by adjusting the throttle on
the washer. Wrong! It seems pressure washer throttle just adjusts the
water volume, as the pressure was a constant 2800 psi. So I got a lot
more pressure than I'd bargained for. No worries, it was an error on the
side of increased safety.

Based on this test, the amount of weight taken on during an emergency
ascent would not get anywhere near reserve buoyancy. I could make a
further modification to the shaft tunnel that would shut off most of the
leakage that did occur. However, since I'm within acceptable leak rates
I've decided not to pursue that change, because it involves adding weld
heat that could distort close-tolerance parts.

You're probably wondering what tolerance I got to. I don't know. My
digital caliper, as best I can measure, says I've got 0.0005" diametric
clearance. But the instrument reads anywhere from 0 to 0.001 depending
perhaps on the blood pressure in my thumb at the particular moment. In
short, the instrument's measurement is not reliable to the precision it
displays. In any case I don't rely on measurement when working to close
tolerances. Instead, I rely on fit. Once the parts are close, it comes
down to a series of fit tests alternating with polishing.

Photos are up on the project page.


Cheers,

Alec



-----Original Message-----
From: owner-personal_submersibles@psubs.org
[mailto:owner-personal_submersibles@psubs.org] On Behalf Of Sean T.
Stevenson
Sent: Thursday, February 05, 2009 22:20
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
>
> The contents of this e-mail are intended for the named addressee only.

> It contains information that may be confidential. Unless you are the
> named addressee or an authorized designee, you may not copy or use it,

> or disclose it to anyone else. If you received it in error please
> notify us immediately and then destroy it.
>




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