rjnoe@ihlts.UUCP (Roger Noe) (09/22/84)
Thanks to the several people who came up with partial solutions to the
differential equation I posted earlier to net.math. It turned out that
there was a good "trick" one could use to make some progress with it and
what's more interesting is that this wasn't even a textbook problem
meant to demonstrate such a technique. (I probably should have spotted
it myself, and I'm somewhat embarrassed I didn't. But my joy at having
the problem solved this far exceeds my embarrassment several times over.)
The equation generated considerable curiousity among respondents, not all
of whom had partial solutions to the differential equation. To answer your
questions (and the reason I am also posting this to net.physics), the equation
did indeed come from a (somewhat idealized) real situation. I outline this
and partial solution of the differential equation below.
GIVEN:
1. A bubble of some ideal gas, molar mass M (gm/mol).
2. The bubble is submerged at some depth D in a liquid of constant density p.
3. At time t = 0 the bubble has no net instantaneous motion.
4. There is some gas over the liquid at pressure P0.
5. The entire system is in thermal equilibrium at temperature T.
Assume the liquid imparts no resistance to the bubble. Characterize the upward
speed of the bubble as a function of its depth d, i.e. v(d). If possible, find
either d(t) or v(t).
SOLUTION:
By Archimedes' principle, the net upward acceleration of the bubble is
[1] a = g(Vp/m - 1)
where m is the mass of the bubble and V is its volume. Since the gas is ideal
[2] V = mRT/MP
where P = P0 + gpd is the pressure of the fully submerged bubble at depth d.
Substituting [2] into [1]:
[3] a = g[ RTp/M(P0 + gpd) - 1 ]
Let f(t) = -d. Then v = f'(t) is the bubble's upward speed and a = f"(t):
gpRT
[4] f"(t) = -------------- - g
M[P0 - gpf(t)]
Multiply both sides of [4] by f'(t) (which is not uniformly zero):
gpRT f'(t)
[5] f'(t)*f"(t) = ---- * ----------- - g*f'(t)
M P0 - gpf(t)
Integrate [5] with respect to t using the initial conditions f(0) = -D
and f'(0) = 0:
2RT P0 + gpD
[6] f'(t)^2 = --- * ln ----------- - 2g[D + f(t)]
M P0 - gpf(t)
Substituting f(t) = -d and v = f'(t):
2RT P0 + gpD
[7] v^2 = --- * ln -------- - 2g(D - d)
M P0 + gpd
This is what was initially sought, v(d). Of course, it's not valid for d < 0
because the net acceleration changes when the bubble is no longer submerged.
If it's possible to integrate df/sqrt{-K*ln[1 - A(f-D)] - L*(f+D)} for some
constants K, A, and L then one would have the inverse of f(t), i.e. time
elapsed as a function of depth. This looks difficult, to say the least.
Inverting the function to get f(t) could very well be impossible.
The condition for the bubble remaining motionless, i.e. f(t) = -D uniformly,
is that the initial depth D equal:
RT P0
[8] D0 = -- - --
gM gp
When D < D0 the density of the bubble at depth D is less than the density of
the liquid and so the bubble rises and [7] is valid for all d in [0,D]. If
D > D0 the density of the bubble is greater than that of the surrounding liquid
and the bubble should sink. In this event, [7] is valid only for d in [D,X]
for some maximum depth X.
Some typical values for these parameters are: RT/M = 86030 m^2/sec^2 for
T = 298 K and M = 28.8 gm/mol (somewhat moist air); of course g = 9.807 m/sec^2
and p = 1025 kg/m^3 for sea water at 298 K (typically) so that gp = 0.099204
atm/m; P0 = 1 atm is a good assumption.
Substituting these values into [8] we find that D0 = 8762.5 m = 28748 feet.
Even taking T = 275 K only reduces D0 to 8085.5 m = 26527 feet. In any event
humans have been taken deeper, to 35800 feet = 10912 m. The foregoing
analysis would indicate that air bubbles at these great depths would sink!
I do not know if this is what really happens.
To answer this, we need to know if at these great pressures (over 800 atm)
is air still a gas, is sea water still about the same density, does the
humidity (and therefore molar mass) of the air change significantly, and
are there any thermodynamic properties which need to be taken into
consideration? There are others better qualified than I to tackle these
questions. I do plan to start investigating the frictional properties
of water and see if there is such a thing as a terminal velocity for rising
air bubbles.
--
Roger Noe ihnp4!ihlts!rjnoe