Unless you change the laws of chemistry after the flood from what they are
today, you cannot quickly erode those caverns. The following is from a
previous note.
Russ Maatman, a chemist at Dordt Univ. writes: "But a cup of water dissolves
only two ten-thousands of an ounce of limestone." (The Impact of
Evolutionary Thought: A Christian Perspective, p. 55)
This is a baker's recipe for
calculating how long it would take to erode a cave 150 feet long,
10 feet in diameter. This is not a large cave. There are 2 cups
in a pint and two pints in a quart and 4 quarts in a gallon.
Thus 1 gallon of water would dissolve 0.0032 ounces (.09952 gm)
of limestone.
The math below shows that it would take several tens of thousands
of years before one can erode a cave starting from an initial
crack .05 inches in radius. A full solution will require an
interative program which I will write in the next couple of
weeks. But initial results are very discouraging for the young-
earth position. After 10000 years, the cave is only about 2-5 inches in
diameter! If one want to have men living in
caves in Europe and Asia only a couple of hundred years after the
flood, one needs to have mining equipment.
Christianity needs to seriously reconsider the young-earth
position.
**************Math.*************************************
All this is taken from Warren L. McCabe and Julian C. Smith,
_Unit Operations of Chemical Engineering_, McGraw-Hill Book Co.,
1956, p. 35-83. I will assume laminar flow which greatly
simplifies the problem and yet overestimates the speed with which
the erosion will occur.
Caves start from small cracks in rocks. Lets start with a
typical but large crack of .01 inch. We will assume that every
day it rains and that water drains off the land above within two
hours. I will let all the water that wants to, come down the
conduit for two hours. Initially the speed with which water can
travel the conduit limits the water flow and every day most of
the water goes down the creek rather than down the conduit.
ground above
I______ ________I -------
\\ I
\\ I cliff face
\\ I
\\ I
conduit=> \\ I Za
\\ I
\\ I
\\I
\ --------
Bernouli's equation with friction is
Pa g Va^2 Pb g Vb^2
-- + --Za+ --- - Wsh = --- + ---Zb + --- + Hf
rho gc 2@gc rho gc 2@gc
Where Pa=Pb= atmospheric pressure, Wsh=0 (it is for a pump and
there is none in this problem) rho is the density of the fluid,
Za is 300 feet in this case, zb=0 and is the datum, gc =32.174, g
is the gravitational acceleration g= 32 ft/s^2, @=1 (it is the
kinetic energy factor and Hf is the friction factor being
32 u L Vb
Hf= ---------
gc D^2 rho
where u is the viscosity (2.089 x 10^-5 lb-force-sec/ft^2), l is
the length of the conduit(150 ft), D is the diameter of the pipe
(yes, diameter--Engineers do things weird)which initially is
(.01"/(12"/')=.000833', rho=62.43 lbs/ft^3 (As a physicist, I
can't stand engineering units, but that is what they use and I
don't want to risk conversion errors)
Substituting these we have
Hf= 72.33 Vb
Since the Pa term and Pb term are equal, Zb=0, g~gc and assuming
that Va is negligible Bernoulli's equation reduces to
Vb^2
0 = ---- + 72.33 Vb - 150
64
This is the quadratic equation in Vb which is the velocity the
water travels through the conduit.
Vb = .0647 feet per second.
In two hours,(7200 seconds) the water will travel .0647*7200=466
ft. Thus the rainfall can seep through the entire 150 feet of
conduit in two hours. But the volume of water is only
466 x pi x (.0004165)^2 = 2.542 x 10^-4 cubic feet. THis is
2.542 x 10^-4/.133 = 1.909 x 10^-3 gallons of water. This would
erode only 1.9 x 10^-4 grams of limestone.
Now, how much water is needed to excavate this cave? The volume
of limestone we want to erode is 150 x 5^2 x pi= 11780 cubic feet
which, at 2.65 grams/cc this is 884,719,262 gm of limestone.
There are 31.1 grams per ounce so each gallon of water (from
text) dissolves 0.09952 grams. Dividing we have
Water needed = 884,719,262/.09952 = 8.88 x 10 ^9 gallons
or, since a gallon is .133 cubic feet, this is
1.18 x 10^9 cubic feet of water.
Here is the problem. The initial crack will only enlarge from
4.165 x 10^-4 feet to 4.171 x 10 -4 feet. Doing this for a year
at this rate, you have a channel 4.3 x 10-4 feet in radius. You
would not be much off to linearly extrapolate that in 10,000
years your cave would not be larger than 2-5 inches. I would dare
say that it would take several tens of thousand years for the
channel to enlarge enough for man to inhabit it.
glenn
Adam, Apes, and Anthropology: Finding the Soul of Fossil Man
and
Foundation, Fall and Flood
http://www.isource.net/~grmorton/dmd.htm