I Will an AUV Ascend and Descend at the same rate?

[Zachb34]

Assuming the AUV only relies on a ballast system to ascend and descend, and that the ballast is completely empty of water at the surface and completely full of water at the seafloor, will they ascend and descend at the same rate? Assume the AUV is neutrally buoyant at the surface for simplicity's sake. Using the terminal velocity equation Vt=sqrt(2mg/pACd), I use the density of saltwater for p, but unsure if m should be the mass of the whole system in water or just the mass of the ballast. Any help would be great, a friend and I are having a disagreeing over it.

 

[Dale]

What is an AUV?

 

[kuruman]

What is an AUV?

https://oceanexplorer.noaa.gov/facts/auv.html

 

[hutchphd]

The terminal velocity fluidics equation $Vt=sqrt(2mg/pACd)$ is better written as $Vt=sqrt(2F/pACd)$ where F is whatever is pushing the vehicle. Is the sinking force equal to the buoyant force?

See post below!

 

[berkeman]

the ballast is completely empty at the surface and completely full at the seafloor

That sounds backwards, no? You need the ballast at the surface in order to sink, and you need to drop the ballast at the sea floor in order to ascend.

Or is the "ballast" water that is blown out to add buoyancy, like in a submarine?

 

[hutchphd]

Sorry I missed that it is neutrally buoyant sans ballast. The ballast will need to be replaced by air (or perhaps water) and the answer will depend. If it is displaced by water then it is symmetric.

 

[Dale]

If it is neutrally buoyant when empty then I would think it could only sink

 

[anorlunda]

The question is confusing because full could mean full of air or full of water. Ballast could mean something neither air nor water.

I think you want the ballast tank to be half full when at the surface. You fill it with water to sink and fill it with air to rise. If it is completely full of water and you have neutral buoyancy, it can never sink. If it is completely full of air and you have neutral buoyancy, it can never rise.

If you have neutral buoyancy, then the net force is zero. Vt=sqrt(2mg/pACd) can never be zero unless m is zero. So can m mean the difference in mass compared to neutral buoyancy. Where did you get that equation from? What do they say about m?

 

[Dale]

The question is confusing because full could mean full of air or full of water. Ballast could mean something neither air nor water.

Yes, good point. I am not a sailor or a submariner, so I was thinking of tractor ballast which is always dense and adding more always makes you sink (in the mud) more.

 

[hutchphd]

Difficult to add negative ballast to the tractor (big Helium balloons I guess). But I think I may have confused myself!

 

[berkeman]

Looks like we will need to wait a day for @Zachb34 to return and clarify...

 

[hutchphd]

If it is neutrally buoyant when empty then I would think it could only sink

Yes I don't know what logical detour I wandered down!

 

[kuruman]

I think OP's question could be answered thusly. As pointed out by @hutchphd, the terminal velocity for a descending AUV is better written as $$v_{t}=\sqrt{ \frac{mg-BF}{\rho_0 AC}}$$ where the buoyant force is $BF = \rho_0 V g$ ($\rho_0$= water density, $V$= external volume) and $m$ is the total mass of the AUV, i.e. what it would measure if you placed it on a scale as is. You can write $m$ in terms of the average density of the AUV, $m=\rho_{\text{avg}}Vg$ in which case $$v_{t}=\sqrt{ \frac{(\rho_{\text{avg}}-\rho_0)Vg}{\rho_0 AC}}.$$ The AUV must be designed so that when the ballast tank is more than half-empty, $\rho_{\text{avg}}<\rho_0$ (don't worry about the square root of a negative number) so the AUV can rise and when it's full, $\rho_{\text{avg}}>\rho_0$ so that it can sink. The magnitude of the difference between the two decides the terminal velocity and hence the rate of ascent or descent.

That's my guess without the benefit of solving a differential equation. My answer to the original question is that the mass and the terminal velocity vary together if the amount of water in the ballast tank changes, nevertheless "$m$" is the total mass of whatever is enclosed by volume $V$.

 

[jbriggs444]

If we are being realistic, I believe (without actual data) that one effect of external pressure on the submersible will be to squeeze it into a smaller volume. The designers will need to worry about whether the vessel is more or less compressible than the fluid within which it operates.

 

[Zachb34]

Thank you all for the replies, I realize my question was not very clear and that shows my lack of knowledge on the topic. I initially assumed an AUV would be able to remain neutral on the surface, but as others have pointed out, it needs to be buoyant from the start to become buoyant again if the whole ballast wants to be used. I guess to salvage by initial intentions to the question, I will reword it to be more practical.
Say there's an AUV at the surface with a ballast tank full of air. For convenience, let's say it starts at the surface with 1N of buoyancy force. The ballast then takes on 2 liters of water, taking away about 2N of buoyancy force and reaching the seafloor with -1N of buoyancy force. The AUV then uses a compressed air tank (ignore the change in volume) to expel the water and generate 1N of positive buoyancy force and return to the surface. Will the ascent and decent rate be similar? Also for simplicity assume this is in somewhat shallow water and pressure differences can be ignored.

 

[hutchphd]

First: one liter of water masses 1kg but weighs 10 N (g=10 in SI uniits) ! But your intent is clear.
Because water is nearly incompressible there are no asymmetric density effects from the increasing pressure. But the craft might be slightly compressible (as per @jbriggs444 ).

Having screwed this up once I will fearlessly say that going down and coming up are pretty much the same. This assumes that the vehicle is basicly symmetric and not streamlined to produce a preference... which would require different values for C up and down.