Question

how large do ballast tanks need to be to travel a depth of 1000 ft?

given:
weight = 33800 tons
volume = 11.5 million gallons

Answer 1

You expressed volume in terms of weight, do you mean the volume of water that weighs 11.5 tons?

Answer 2

Sorry, wrote the wrong units. correction made.

Answer 3

The bathysphere weighs 67.6 E6 lbs. Since its volume is 11.5 E6 gallons and 1 gal of sea water weighs 8.75 lbs the buoyant force is 100.6 E6 lbs. So this vessel is floating. So to sink it you need to add more than 33.0 E6 lbs of internal ballast ( the difference between the weight and buoyant force). If I take into account the change in density with depth it needs to be 33.07 E6 lbs. If the ballast is water then you need to use 3.79 E6 gal. of the 11.5 E6 gal of volume for the water ballast. This is one big bathysphere.
143 ft diameter sphere!!!!!!!!!!!!!!!!

Answer 4

how did you calculate the difference between the weight and the boyant force?
also how did you calculate your density with depth?

Answer 5

hey man sorry to keep bothering but i really need help on how you found change in density with depth to be 33.07 E6 lbs?...

Answer 6

Back again.


I used British units because you gave the data in those units.

Archimedes Principle state that a submerged object is buoyed up by a force equal to the weight of the volume of liquid that it displaces. You submersible displaces 11.5 Mill gal of (sea) water which weighs 8.6 lbs/gal. (this is a more correct water density). Therefore the total buoyant force is 11.5 E6 x 8.6 = 98.9 E6 lbs. Since the craft "Only" weighs 33,800 tons or 67.6 E6 lbs. it is floating. the net buoyant force is 98.9 E6 - 67.6 E6 = 31.3 E6 lbs. So in order to sink we need to add 31.3 E6 lbs of ballast which is place inside the 11.5 E6 gals of volume. This amount of ballast will make the craft neutrally buoyant and it will stay at whatever depth it is placed. However the density of water changes with depth getting denser as you get deeper because of the increase in water pressure as you get deeper. This not a big effect but since the volume you displace is so large this small increase in density results is a significant weight of water or ballast. The equation for change in density with change in pressure is \[\rho = \rho _{0}/\left( 1 - \Delta P/E \right)\]

where Rho is the density at depth
Rho0 is the density at the surface
delta P is the change in pressure at depth
E is what is called the Bulk Modulus a measure of the compressibility of the liquid
and for water is 4.5 E 7 lbs/ft^2

To find the pressure at depth in lbs/ft^2 find the weight of a column of water of 1 sq ft 1000 ft high Seas water weights 8.6 lbs/gal and there are 7.48 gal/ft ^3. So a column of water 1000 ft high weighs 1000 x 8.6 x 7.48 lbs = 64328 lbs which means the pressure 64328 lbs/ft^2. This is pretty high. Putting in this info into the equation above the new density is 8.612 lbs/gal. Not a big increase only a 0.143% increase but because of volume is so big the buoyant force is increased by that amount at depth. So we need to add this much more ballast to keep the craft at that depth. This added amount is .00143 x 31.3 e6 = 44759 lbs. I should note that the temperature of the water has been neglected. The colder temp will additionally increase the density but It can be variable depending on where the vessel dives.