Verified answer

This is just an estimate.

P = 400 * 10 ^ 6 Pa ==> ultimate strength

r = ((30 / 4) / π) ^ (1 / 3)

T = 2 * r / 10

tensile stress of this shell using

P = ultimate strength: rim force = area force

∑ σ 2 π r T - P π r^2 = 0

σ= P r / (2 T)

bulk = 160 * 10 ^ 9 Pa

ΔV = (4 / 3) π r ^ 3 σ / bulk

ΔV = 6.25E-02 m^3

V = Vi + ΔV m^3

I know the formula for ΔV is not formally correct.

Further, If I can find more time I'll work water

compressibility into the mix.

*****************************************

Edit: Another Estimate:

V = (4 / 3) * pi * r ^ 3 = 10 m^3

r = ((30 / 4) / pi) ^ (1 / 3)

T = 2 * r / 10

Maximum tensile stress allowed:

∑F = ∑ σ 2 π r T - P π r^2 = 0

σ= P r / (2 T)

σ = 400 * 10 ^ 6 N/m^2

Maximum allowable pressure in the sphere

P = σ * 2 * T / r

P = 160000000 Pa

Let elastic modulus of steel:

Y = 200 * 10 ^ 9 N/m^2

u = 0.3 ==> Poisson's ratio for steel

ΔV = (P 2 π r ^ 4 / (Y*T)) (1 - u)

Change in volume of steel sphere:

ΔV = 4.2E-02 m^3

Volume change of water at P:

Bw = 2.2 * 10 ^ 9 N/m^2

ΔVw /V = P / Bw

v = 10 m^3

ΔVw = v * P / Bw

ΔVw = 0.727 m^3

Total volume opened up:

V = ΔV + ΔVw= 0.769 m^3

Filling this with uncompressed water

which I shouldn't do, I get:

Total volume of water = 10.769 m^3

Thanks for the boost First Grade, but I'm

still not to sure of my answer!

I think you haven't asked the question very clearly. Are you asking about how much it can hold before it bursts, under normal earth gravity? Or maybe how much it can hold before you hit some safety factor before it bursts?

Realize that you're asking this in the physics section, not in an engineering forum!

Anyway, there's surely a standard formula in a structural mechanics text for the maximum pressure a spherical vessel could take, given the yield strength of the steel, the radius, and the thickness. Then you'd figure out how deep the water is that gives you that pressure. If that depth is less than the inner diameter, then you do the integral to get the volume. If it's more than the inner diameter, then you get to keep adding pressurized water to the thing . . . and now the result depends on the bulk modulus of water and the stiffness of the shell!

So . . . what question are you asking, exactly?

Is this a trick question?

If the vessel has a 10 m^3 capacity, then it can hold 10 m^3 of water.

Assuming that the rest of the information is not extraneous, there must be an error in the question. If the intended question is:

A steel spherical vessel which occupies 10 m^3 has a diameter to wall thickness ratio or 10. What volume of water can be pumped into the vessel?

then the solution is the following:

Volume of a sphere is 4/3 * pi * r^3; solve for r

---> r = 1.337 m

---> diameter = 2.673 m

thickness = dia/10 = .2673 m

internal radius = external radius = thickness

internal radius = 1.069 m

internal volume = 4/3 * pi * (internal radius)^3

internal volume = 5.12 m^3

if the vessel is spherical it would contain the same volume as a cubic vessel having a side dimension of approx 80 % of the diameter of the sphere.

It would take less thickness on a spherical container to contain the same volume as a cubic container.

If the internal capacity of the vessel in question is ten cubic meters the container has to be strong enough to hold ten tons of mass and withstand a pressure of 10000 pascals.

If the wall thickness of the vessel is about 26 centimetres and the tensile stress of steel is about 18000 lbs per square inch it should not have a problem holding the water.

Are you looking for how much water you can pump in before it bursts? Or is there some other problem you are asking.

If it is the burst point, the variables are:

a. Compressiblity of water,

b. Elasticity of steel

c. Failure strength to steel.

d. Temperature.

I'll solve it when I have a little more time. (Want to do it without looking up formulas -- more fun). Unless someone beats me to it.

.......

If you are at atmospheric pressure, then it is slightly more complicated because you would be asking a question related to the differential pressure and exapansion cause by gravity as you fill up the tank

*********

Adding to Al P.'s answer, (he has the correct formula).

The bulk modulus of water is 2.2 x 10^9 Pa.

I don't know. The way you present your question, I'm not sure if you get the air out of it! And how much you could get to compress it? And I am assuming that the walls won't collapse...

Ana

As some others said, I also agree with 10.000 L. It seems to me like the only logical, possible way ... Good luck in finding your answer!

1m^3 of water = 1000L

1m^3 of water = 35.314 cubic feet

1 cubic foot of water = 7.4805 gallons

So

10m^3 of water = 10,000L = 353.14 cubic feet = 2641.66 gallons