How much water escapes from the vessel?

Question

A thin-walled cylindrical steel pressure vessel with internal volume 1.5 m³ is to be tested. The vessel is entirely filled with water. Then two pistons at the ends of the cylinder are pushed inwardly until the pressure inside the vessel has increased by 3 MPa. Suddenly, a safety plug at the top bursts. How many liters of water escape from the vessel?

Numerical values:

Young’s modulus for steel, E = 197.4 GPa

Internal radius to thickness ratio of unstressed cylinder, R/t = 10

Bulk modulus of water, β = 2.2 GPa

Answer 1

1.5 m^3 -- The pistons keep pushing and all of the water escapes. lol.

Nice easy one.  We don't have to worry about (or at least I will ignore) longitudinal stress because we have the pistons.  So that means hoop stress

R/t = 10 (pretty thick sounds like a pipe)

P = 3 MPa

E = 197.4 GPa = Stress/Strain

Length vessel = 1.5 m^3 / pi R^2

Stress = F/Area

= 3 MPa *2R / 2t = 3MPa R/t

Strain = Stress/E

= 3 MPa R/t /197.4 GPa

= 1.519e-5 * R/t

= 1.519e-4

Since the distance around is 2pi R, the length increase of the hoop is

= 1.519e-4 * 2pi R

and the increase in volume is

d V = L * [pi (R + 1.519e-4 * 2pi R)^2 - piR^2]

= (1.5 m^3 / pi R^2)* pi *[ 2R* (1.519e-4 * 2pi R) + (1.519e-4 * 2pi R)^2]

We can, under most circumstance, ignore the (1.519e-4 * 2pi R)^2 term, so we have:

dV = (1.5 m^3 / pi R^2)* pi *[ 2R* (1.519e-4 * 2pi R)]

= 1.5 m^3* (4pi*1.519e-4)

= 2.8647e-3 m^3

= 2.8647 liters

Ok, lets add in the compression of the water -- should be small.

1.5 m^3 * 3 MPa/ 2.2 GPa = .002045 m^3 = 2.045 liters

(a bit bigger than I anticipated.  Probably because your R/T ratio was so small)

Total = 2.865 l + 2.045 l = 4.910 l

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Sure, I can decouple length and volume -- I just did!!  lol.  The question is whether or not I am correct in doing so.  I'll get back to you. ;-)

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The coupling with length should be weak.  I neglected it.  Just assumed that it was as if the piston stayed in place and you added 4.90 liters of water.

First, lets write an expression for dL

dL/L = compression water + expansion hoop

dL/L = (2.045 l / 1500 l) * dP/3 MPa + (2.8647l/1500 l) dP/3MPa

dL/L = 4.910 l/1500 l * dP/3MPa

dL/L = 0.03273% dP/3MPa

assuming 3 MPa

dL/L = 0.03273%

Ok, lets figure the volume:

dV =  dL/L * V  -  dL/L * V * (4pi*1.519e-4)

= .03273% * 1500L + - 0.3273% 1500 L* (4pi*1.519e-4)

= 1500 L * 0.0327% (1- (4pi*1.519e-4))

= 4.9006 L

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I mispoke on the R/t ratio -- look where I put it, when talking about the compressibilty of water.  

What I was thinking was -- that a lot of compressibility -- it is almost as much as the gain in volume caused by the hoop stress.  Oh, it's because the R/t is small therefore the vessel didn't expand much.

I think we are thinking alike even if I talk about small R/t and you talk about pression.  ;-)

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Revised and Restated:

You are correct.

I did everything correctly except I threw in an extra factor of 2pi into the cross-sectional area.

The correct answer for volume from hoop increase should be:

2.8647 liters /2pi = .4559 liters

You made it look simple.