Water intrusion formula

[Stumble]

As part of the process of resizing the dewatering capability of my boat I ran across a snag. Trying to estimate the 'worst case' scenario for water intrusion into the hull.

Basically I am assuming that the worst case would be defined as a complete failure of the through hull fitting that will allow the largest volume of water into the hull. At this point that seems to be a failure of the drive shaft such that the shaft is ripped out of the hull, and nothing can be found to slow down the leak.

The problem is in determining the expected water flow from such a hole. The best formula I have been able to find is from a flow rate calculator for a spout in a holding tank at http://www.efunda.com/formulae/fluid..._tank.cfm#calc

My question though is is this computational model reasonable for a hole in a boat, and if not is someone familure with a formula that can be used to determin the flow rate of a hull breach. Assuming a normalized size hole (Tring to model a complex breach shape) seems to be outside of this exercise.

[gonzo]

That is a reasonable formula to use. Because the boat is moving, there would be pressure differences to consider. However, you don't need to be so accurate. One thing to take into account, is that once water covers the hole on the inside, there will be a restriction to the flow. The formula does not take that into account. If you are trying to estimate the worst case scenario, it is OK to be on the pessimistic side anyway.

[Eric Sponberg]

Stumble,

The formula written out is:

Q = A x (2gh)^0.5

In words, the rate of flow, Q, in cubic feet per minute, equals the Area, A, of the opening, in square feet, times the square root of the product 2 x g x h, where g is the acceleration of gravity, 32.174 ft.sec^2, and h is the height difference between the water level outside and the level of the opening inside, in feet.

Once you calculate the cubic feet per minute, you can convert that to gallons per minute; there are 7.48 gallons in a cubic foot. So, multiply Q by 7.48 to get gallons per minute. You want a pump that will pump quite a bit faster than that.

Good luck, stay afloat!

Eric

[SheetWise]

Originally Posted by gonzo

That is a reasonable formula to use. Because the boat is moving, there would be pressure differences to consider.

Not to pick nits, but he has mentioned a failure in the drive shaft. In what sense is the boat "moving" that would cause pressure differences to be considered?

[gonzo]

If a boat is pitching, the pressure will be higher when it slams down and lower or even negative when it rises. Fore and aft movement also changes the pressure.

[Stumble]

Thanks for the advice. I am planning on sizing off of the largest through hull instead of the shaft because my drive shaft is only at 12" under water and while it is large, I assumed the deeper hole would allow more water in. Though I plan to calculate it both ways to be sure.

My thought was to provide a maximum installed pump capacity of 125% of the worst case single point failure. Plus whatever capacity the engines and generator pick ups use. I don't really think it is reasonable to include their flow rate into the pumping capacity since you never know what if they will be operational. I have a standard 25% bilge pump running off of 12V power for standard dewatering and accidents. Then a 100% dewatering pump operating from its own engine. Plus the engines that can pull down pretty significantly as well (total installed hp of 1150hp).

I just wanted to make sure I was using a reasonable formula to predict the water flow into the boat since all this work would be worthless if I miscalculated significantly.