Neutral Buoyancy: How Does it Affect Subs?

[gloo]

I am curious about what it means to be neutral buoyant. If a submarine took in enough water so that it sinks down say 20 metres below the surface and stays floating at that level -- it is considered to be neutrally buoyant down at 20 meters below. If a force pulled the sub back up to 5 meters below the surface...would it stay at that level, or would gravity pull it back down to 20 meters below surface correct?

 

[DaveC426913]

A submarine is ideally a rigid structure. It will not change volume as it moves up and down. So, at 5 metres, it will be denser than the surrounding water, and will sink until its density equals that of the water around it.

 

[russ_watters]

You mean if this mysterious force that pulled it up to 5 meters was released? It would probably continue rising to the surface.

What you appear to be asking in general is whether or not neutral buoyancy is a stable or unstable equilibrium. The answer is that it depends on the object. Water is very slightly compressible. So buoyancy as a function of volume increases very slightly with depth. So an object with a fixed volume - a very rigid object - will get more buoyant as it sinks. Such an object might have a stable equilibrium.

An object that isn't very rigid (is very compressible), like a scuba diver's buoyancy vest, will change in volume substantially with changes in depth. So it will not achieve a stable equilibrium.

A submarine? Probably more compressible than water, but I'm not certain.

 

[SteamKing]

You mean if this mysterious force that pulled it up to 5 meters was released? It would probably continue rising to the surface.

What you appear to be asking in general is whether or not neutral buoyancy is a stable or unstable equilibrium. The answer is that it depends on the object. Water is very slightly compressible. So buoyancy as a function of volume increases very slightly with depth. So an object with a fixed volume - a very rigid object - will get more buoyant as it sinks. Such an object might have a stable equilibrium.

An object that isn't very rigid (is very compressible), like a scuba diver's buoyancy vest, will change in volume substantially with changes in depth. So it will not achieve a stable equilibrium.

The term "neutral buoyancy" is used to indicate that the average density of a submerged object is equal to the density of the surrounding fluid. Since the densities are the same, the submerged object shows no tendency to rise or sink, unless something changes to upset the balance between its average density and the density of the fluid. For submarines navigating underwater, this means filling and emptying various ballast tanks to maintain neutral buoyancy, since seawater is not completely homogeneous in terms of salinity, temperature, etc., all of which affect its density.

https://en.wikipedia.org/wiki/Neutral_buoyancy

The stability of a submerged object is a separate issue. In order to remain stable while submerged, the center of buoyancy of the vessel must always be lower than the center of gravity, otherwise any moments tending to rotate the vessel from the upright position could not be resisted easily.

http://www.southampton.ac.uk/~jps7/Aircraft%20Design%20Resources/Sydney%20aerodynamics%20for%20students/fprops/statics/node24.html

A submarine? Probably more compressible than water, but I'm not certain.

Definitely more compressible than water, because submarines are not made from solid hunks of metal. There are examples of this used in strength of materials courses to show how much the sphere on a deep-diving submarine compresses when it is at depth.

A solid block of steel will show very little compressibility versus an equal volume of water, due to the difference in the bulk moduli of elasticity of the two materials (160 GPa for steel v. 2.2 GPa for water)

https://en.wikipedia.org/wiki/Bulk_modulus

 

[russ_watters]

The term "neutral buoyancy" is used to indicate that the average density of a submerged object is equal to the density of the surrounding fluid.

Since the densities are the same, the submerged object shows no tendency to rise or sink, unless something changes to upset the balance between its average density and the density of the fluid. For submarines navigating underwater, this means filling and emptying various ballast tanks to maintain neutral buoyancy, since seawater is not completely homogeneous in terms of salinity, temperature, etc., all of which affect its density...

I'm not sure what your point is/why you quoted me. I'm aware of what "neutrally buoyant" means: The OP is referring to a situation where the submarine is not neutrally buoyant and asking if it finds a neutral buoyancy depth.

The stability of a submerged object is a separate issue. In order to remain stable while submerged, the center of buoyancy of the vessel must always be lower than the center of gravity, otherwise any moments tending to rotate the vessel from the upright position could not be resisted easily.

That's not the stability we're discussing. We're discussing whether the submarine returns to or moves away from its neutral buoyancy depth. In other words, whether it needs to constantly be adjusting its buoyancy (like a scuba diver) in order to remain at the same depth or if it just gets and stays there on its own.

 

[Chestermiller]

The density of water changes exceedingly little from a depth of 5 meters to a depth of 20 meters. The bulk modulus of water is 2.15x10⁹ Pa, the pressure at 5 meters depth is about 50000Pa gauge, and the pressure at 20 meters is about 200000 Pa gauge. So the pressure change between 5 meters and 20 meters is about 150000 Pa. So this would result in a density difference of < 0.01%. So there would be virtually no difference in what one would consider neutral buoyancy between these two depths. I doubt if one could accurately establish neutral buoyancy to within 0.01% at any depth.

Chet

 

[gloo]

A submarine is ideally a rigid structure. It will not change volume as it moves up and down. So, at 5 metres, it will be denser than the surrounding water, and will sink until its density equals that of the water around it.

Dave, I was thinking along the lines of what you concluded...but Russ, who is very sharp, concludes almost the complete opposite stating that it will continue to rise (maybe talking about momentum?). What is the bridge here between what Russ is saying and what you are saying??

 

[jbriggs444]

Dave, I was thinking along the lines of what you concluded...but Russ, who is very sharp, concludes almost the complete opposite stating that it will continue to rise (maybe talking about momentum?). What is the bridge here between what Russ is saying and what you are saying??

It seems to me that DaveC is assuming an ideally rigid submarine in water that is [slightly] compressible -- a stable situation. Russ expects a compressible submarine in water that is less so -- an unstable situation. Complete agreement about the physics, different idealizations/expectations about the situation.

 

[Chestermiller]

Dave, I was thinking along the lines of what you concluded...but Russ, who is very sharp, concludes almost the complete opposite stating that it will continue to rise (maybe talking about momentum?). What is the bridge here between what Russ is saying and what you are saying??

Did you not read what I wrote in post #6. In going from 5 m to 20 m, the change buoyant force will be too small to realistically observe. As I said, even if the submarine is considered incompressible, the change in the buoyant force resulting from water compressibility would be less than 0.01 %. That quantifies Dave's mechanistic assessment, namely, it would be negligible. I don't know how much a submarine's volume would decrease if we increased the outside pressure by 25 psi, but it can't be much either. Plus, it would act to offset the tiny effect of the water compressibility.

I'm beginning to feel like this thread has run its course.

Chet

 

[DaveC426913]

I'm beginning to feel like this thread has run its course.

It sounds like we all have some ideas about certain situations, but no one has yet stated an answer with conviction. I would suggest that it has not run its course until the OP has a well-defined answer. It may take time for an expert to step in.

 

[Chestermiller]

It sounds like we all have some ideas about certain situations, but no one has yet stated an answer with conviction. I would suggest that it has not run its course until the OP has a well-defined answer. It may take time for an expert to step in.

Was my quantitative calculation in post #6 not adequate for you? What part didn't you understand?

 

[anorlunda]

Chet's calculation in #6 does not consider changes in water density due to changing temperature or salinity. Vertical components of currents are also a factor (although not a neutral buoyancy issue, they can be countered with buoyancy.) Almost nothing is static in an ocean environment.https://en.m.wikipedia.org/wiki/Submarine#Submersion_and_trimming

For general submersion or surfacing, submarines use the forward and aft tanks, called Main Ballast Tanks (MBT), which are filled with water to submerge or with air to surface. Submerged, MBTs generally remain flooded, which simplifies their design, and on many submarines these tanks are a section of interhull space. For more precise and quick control of depth, submarines use smaller Depth Control Tanks (DCT) – also called hard tanks (due to their ability to withstand higher pressure), or trim tanks. The amount of water in depth control tanks can be controlled to change depth or to maintain a constant depth as outside conditions (chiefly water density) change. Depth control tanks may be located either near the submarine's center of gravity, or separated along the submarine body to prevent affecting trim.

 

[Chestermiller]

Chet's calculation in #6 does not consider changes in water density due to changing temperature or salinity.

Over a depth interval of 15 meters? Get real.

Chet

 

[anorlunda]

Over a depth interval of 15 meters? Get real.

Chet

No need to be testy.

Changes with respect to time, not just depth.

Real life Is not like a laboratory. The atmosphere and the oceans are dynamic with respect to time. The ocean also has significant (and dynamic) temperature and salinity layering with sharp boundaries.

 

[Chestermiller]

No need to be testy.

Changes with respect to time, not just depth.

Real life Is not like a laboratory. The atmosphere and the oceans are dynamic with respect to time. The ocean also has significant (and dynamic) temperature and salinity layering with sharp boundaries.

Not significant at depths down to 20 meters. Would you like me to do the calculations for the effect of temperature and salinity variations from 5 to 20 meters also?

 

[mfig]

If some object is neutrally buoyant, this means that the average density of the submerged object is equal to the density of the surrounding fluid.

So let's start simple. Imagine a system made of a bunch of particular water molecules at rest in still water at the same temperature, and at a depth of $x_1$. The system will have a certain density, which depends on the pressure, which in turn depends on $x_1$. Since the system experiences no net force, we know that its density is equal to that of the surrounding water at depth $x_1$. Now let's move this system to a new depth within the still water, $x_2$. Will the system experience net vertical force? No. The reason why is that when it moved to $x_2$ it experienced a new pressure and thus attained a new density - the same density as the surrounding water at $x_2$. Both the system and the water change their density at the same rate, as a function of depth (or pressure).

If that makes sense so far, then we can ask the question,

"What would have to be true for a system to experience a net force at $x_2$ when it experienced no net force at $x_1$?"

The answer is that the system would have to change density as a function of depth (pressure) at a rate different from the surrounding water.

If my logic is on target so far, this means that in order to answer your question about a particular physical system (a submarine, for instance) we would have to know how that particular system's density changes as a function of pressure. The direction of the net force at $x_2$, if the system's $\frac{d\rho}{dp}$ is different from water's, will depend on whether the system compresses "faster" as a function of pressure, or slower, than water.

Note that nothing in this analysis suggests that we cannot include the effects of temperature or other variables. We would simply have to take into account how the water and the system change density with those other variables when answering the question posed above.

 

[anorlunda]

http://courses.washington.edu/uwtoce05/webg1/results.html

Are you talking about the turbulent mixed layer near the surface? That is average behavior. I don't believe that back of envelope calculations can capture the complexity of real life like in the link and picture above.

Those submarine designers put the DC tanks in for a reason.

I also know from swimming in the same places day after day, that the depth of the thermocline rises to less than a foot on some days and that its depth can change in less than an hour. Can your calculations show that?

 

[Chestermiller]

http://courses.washington.edu/uwtoce05/webg1/results.html

Are you talking about the turbulent mixed layer near the surface? That is average behavior. I don't believe that back of envelope calculations can capture the complexity of real life like in the link and picture above.

Those submarine designers put the DC tanks in for a reason.

I also know from swimming in the same places day after day, that the depth of the thermocline rises to less than a foot on some days and that its depth can change in less than an hour. Can your calculations show that?

I don't think so. I was thinking more of the average density profile as shown in this figure: http://www.windows2universe.org/earth/Water/density.html

For any given density profile, we can, of course, perform the calculation to quantify the effect of the variation in buoyancy between 5 and 20 meters in depth. What is your best estimate of the range in the difference in water density between 5 m and 20 meters, taking into account the possible range of salinity and temperature that might be encountered in practice?

 

[russ_watters]

It sounds like we all have some ideas about certain situations, but no one has yet stated an answer with conviction. I would suggest that it has not run its course until the OP has a well-defined answer. It may take time for an expert to step in.

The wikipedia article on submarines has the same conclusion I reached:

When submerged, the water pressure on a submarine's hull can reach 4 MPa (580 psi) for steel submarines and up to 10 MPa (1,500 psi) for titanium submarines like K-278 Komsomolets, while interior pressure remains relatively unchanged. This difference results in hull compression, which decreases displacement. Water density also marginally increases with depth, as the salinity and pressure are higher.[38] This change in density incompletely compensates for hull compression, so buoyancy decreases as depth increases. A submerged submarine is in an unstable equilibrium, having a tendency to either sink or float to the surface. Keeping a constant depth requires continual operation of either the depth control tanks or control surfaces.[39][40] [emphasis added]

https://en.wikipedia.org/wiki/Submarine

BTW, I don't see Chet's post as disagreeing with mine. He's going a step further to say that even if you could make a submarine rigid enough for a stable equilibrium to exist, water is so incompressible that that equilibrium wouldn't be very stable anyway. So it would still be hard to call it "stable". I agree.

 

[Chestermiller]

I'm going to try to summarize where we stand as a result of our deliberations.

Over the depth interval from 5 to 20 meters:

• Variations in density as a result of pressure variations do not have a material effect on the buoyant force on an object
• Variations in density as a result of salinity and temperature variations may have a material effect on the buoyant force on an object (we are awaiting anorlunda's estimate on the magnitude of the maximum density variation from these)
• Variations in the volume of a typical object (e.g., a submarine) as a result of pressure variations probably don't have a material effect on the buoyant force on an object (over this small depth interval, where the pressure changes by < 2 bars).

Over much larger depth intervals like hundreds of meters: Variations in water density and of volume of typical objects as a result of pressure, temperature, and salinity variations can have a significant effect on the buoyant force on an object.

Please correct this summary if my perception is incorrect.

Chet

 

[anorlunda]

So the pressure change between 5 meters and 20 meters is about 150000 Pa. So this would result in a density difference of < 0.01%.

Over the depth interval from 5 to 20 meters:
Variations in density as a result of pressure variations do not have a material effect on the buoyant force on an object

I have given this more thought. If we have a sub at depth $y=5$ meters that is neutrally buoyant, then fluid density $\rho_f$ equals sub density $\rho_s$ by definition. Then we push the sub down to $y=20$ meters where the fluid density is 0.01% higher, and let it go. What will happen?

Just using Newton's Second Law, I calculate that the sub will return to $y=5$ with a time constant of 9800 seconds (about 2.7 hours). That is very material.

However the above calculation applies if the sub is incompressible. If the sub's density also increases 0.01% it will remain at $y=20$. If the sub's density increases more than 0.01%, it will sink.

The error was introducing into the discussion, the numerical value for $\frac{\partial{\rho_f}}{\partial y}$ considered in isolation.

I believe the correct governing expression to be
$\left(\frac{\partial{\rho_f}}{\partial y}-\frac{\partial{\rho_s}}{\partial y}\right)$ evaluated at the initial neutral buoyancy depth. Let's call that term "differential compressibility" for discussion purposes.

The sign of differential compressibility determines whether the sub will rise or sink or remain stable. The magnitude of the term determines how fast.

I also believe that "differential compressibility" is the answer sought in the OP.

 

[Chestermiller]

Hi. I don't match your result of 9800 seconds. Can you please show how you got your result.

Chet

 

[Rx7man]

I agree it's the difference in buoyancy at different depth that will govern whether the sub will sink or rise.

I don't see any simple way to calculate how fast it will sink or rise without VERY involved calculations.. viscosity and shape of the vessel will be very large factors, as well as inertia.

 

[Chestermiller]

I agree it's the difference in buoyancy at different depth that will govern whether the sub will sink or rise.

I don't see any simple way to calculate how fast it will sink or rise without VERY involved calculations.. viscosity and shape of the vessel will be very large factors, as well as inertia.

If you neglect drag and only include inertia, anorlunda did the calculation. Including drag shouldn't be too hard to approximate (roughly), modeling the sub as a cylinder. I ran a calculation on this, but am not yet ready to share.

Chet

 

[mfig]

I believe the correct governing expression to be
(∂ρf∂y−∂ρs∂y)(∂ρf∂y−∂ρs∂y)\left(\frac{\partial{\rho_f}}{\partial y}-\frac{\partial{\rho_s}}{\partial y}\right) evaluated at the initial neutral buoyancy depth. Let's call that term "differential compressibility" for discussion purposes.

The sign of differential compressibility determines whether the sub will rise or sink or remain stable. The magnitude of the term determines how fast.

I also believe that "differential compressibility" is the answer sought in the OP.

Post #16.

 

[anorlunda]

Hi. I don't match your result of 9800 seconds. Can you please show how you got your result.

Chet

I'll post the calculation tomorrow.

Post #16.

Pardon me. I missed #16. Your did indeed seem to hit the nail on the head.

 

[mfig]

Pardon me. I missed #1

No pardon necessary. I am happy someone else went down the same road and saw similar stuff.

 

[haruspex]

The stability of a submerged object is a separate issue. In order to remain stable while submerged, the center of buoyancy of the vessel must always be lower than the center

Off topic I know, but I think you have that backwards. The centre of buoyancy needs to be above the centre of gravity for stability.
(I've heard this also quoted in reference to buoyancy of surface vessels, but it is not correct there. The cross-sectional profile also matters in that case.)

 

[anorlunda]

Sorry for the delay, this week was hectic. I promised to post how I calculated the 9800 second time constant in #21. This time around I used a different approach and came up with a different number. However, the conclusion stands that even a 0.01% change in water density is material.

• The starting point is Newton’s 2nd law $f=ma$, plus Chet’s calculation that the water density changes 0.01% from $y= 5\ meters$ depth to $y=20\ meters$depth. Viscosity and friction are not specifically addressed because the velocities are so slow.
• First I note that the ratio $\frac{f}{m}$ for buoyant forces is the same for small objects as large ones. So with scale independence, I simply choose m=1 kg.
• The problem is, given a 1kg submarine neutrally buoyant at 5 meters depth, push it down to 20 meters depth and release. What happens then? (An aside: we could also be analyzing debris which floats below the surface, but does not sink to the bottom, nor rise to the surface. I live on the water and see such objects from time to time and I know how they behave.)
• If we integrate acceleration twice to get position, then put a negative feedback loop around it, we have described a harmonic oscillator. Real life experience contradicts that, so I am arbitrarily asserting that the object will return to 5 m with a transfer function $\frac{1}{1+\tau s}$. The step response is $1-e^{-\frac{t}{\tau}}$. Given the specifics of this problem, depth y as a function of time becomes $20-(1-e^{-\frac{t}{\tau}})(20-5)=y$.
• Assume that the initial acceleration is constant $a=\frac{f}{m}=\frac{10^{-4}}{9.8}$ where $10^{-4}$ is 0.01% and 9.8 is required to convert $kg_f$ to Newtons. The time to rise the first meter (from 20 to 19) is $t=\sqrt{2\cdot9.8\cdot10^4}=443\ seconds$.
• Here’s the approximation. Combine the y versus time behavior assumed with the calculated time to rise the first meter with constant acceleration. Thus we evaluate the depth using t=443 and y=19 which becomes $\left[20-(1-e^{-\frac{443}{\tau}})(20-5)=19\right]$ , and solve for $\tau$. We get $\tau=1985\ seconds$.

So 1985 seconds is my new estimate for tau.(Unless I made calculation errors which is something I do frequently. It has been 60 years since I was a student working problems.) It depends on three assumptions that are admittedly a bit of hand waving but good enough for rough estimates.

1. That the behavior of the object is to respond to step changes like a simple time constant, rather than like a harmonic oscillator.

2. That the time constant value can be approximated by the behavior for the first 1 meter rise out of 15, and that rise is characterized by constant acceleration.

3. That the submarine object is incompressible.

Of course, the numerical value of $\tau$ is not the point. As I said in #21, the only material point in this thread's question about neutral buoyancy is the differential compressiblity.

 

[Chestermiller]

I don't know how you got an acceleration of $10^{-4}/9.8$, but I get an initial acceleration of $9.8\times 10^{-4}$ from the equation:$$ma=0.0001mg$$. If x is the depth (in meters) below 5 meters, the acceleration at any time will be (0.0001)(9.8)x/15. So, we have:$$\frac{d^2x}{dt^2}=-(0.0001)(9.8)\frac{x}{15}$$. The solution to this equation under the initial condition that x = 15 at t = 0 and dx/dt = 0 at t =0 is: $$x = 15\cos(0.008t)$$According to this result, x will reach a value of 0 (sub will reach 5 meters) at 0.008t=π/2. This comes out to t =196 seconds. If we neglect the fact that the acceleration is decreasing as the sub rises, then we get $t=\sqrt{\frac{(2)(15)}{(0.0001)(9.8)}}=175$ seconds. My numbers for t are about 10x smaller than yours because of the location of the 9.8 in the numerator vs the denominator of the acceleration. This is why I was questioning your results in that earlier post.

The time will be longer if we include drag. I've done some calculations with drag, and may present them in a later post.

Chet