Calculate Water Volume From Pressure Vessel Burst

[fredrick08]

Elasticity! Pressure!

Homework Statement

A cylindrical steel pressure vessel with volume 1.3m^3 is to be tested. The vessel is filled with water, then a piston at one end pushes until the pressure inside increases by 2000kPa, then suddenly a safety plug on the top bursts.
how many litres of water come out?
B=.02x10^10Pa

Homework Equations

P1=P0+\rhogh
P1=-B(\DeltaV/V)

The Attempt at a Solution

P1=P0+2000kPa
\DeltaV=-V(P1/B)=-1.3(P0+2000/.2x10^10)m^3
please i need help... i got no idea where to go from now... i don't have density, or a height, or youngs modulus of steel?? please can someone help?

 

[fredrick08]

am i missing any formulas? coz I am really stuck with this one

 

[alphysicist]

Hi fredrick08,

The container originally held 1.3 m^3 of water; once the pressure has increased (by the piston being pushed inwards) how much has that volume decreased? (This would be related to how far were they able to push in the piston, but you find it using the equation you have with \Delta V.) What number do you get?

Once the safety plug burst, the water goes back to its original volume (since it's open to the atmosphere), but the container is still at its new volume. So how much water escapes?

 

[fredrick08]

ok yes, but how can i find the distance?? coz i don't know the initial pressure??

 

[fredrick08]

i mean how can i find the distance the piston goes in... coz don't i need the initial pressure, coz the pressure increases by 2000kPa??

 

[alphysicist]

I just mentioned the distance to show what was happening in the experiment; you cannot find the distance here.

What you want to find first is the change in the volume, using the equation you had in your first post. What number do you get for the change in volume \Delta V?

 

[fredrick08]

i don't know, coz the \DeltaV=-1.3(P0+2000kPa/.2x10^10) i don't understand how i can find this because i don't know P0?

 

[alphysicist]

P0 is the pressure before the piston began pushing, when the water was just poured into the container. So it would be atmospheric pressure.

 

[fredrick08]

oh you ok ty lol ok well \DeltaV=-1.3x10^-3m^3

 

[fredrick08]

which is 1.3L? is this right??

 

[alphysicist]

It looks like your formula is a bit off. You just need the change in pressure, which is the 2000kPa. The formula is

<br /> \Delta P = -B \frac{\Delta V}{V}<br />

so

<br /> \Delta V = -V (\Delta P)/B<br />

 

[fredrick08]

ok thankyou, this book I am using is hopeless... btw that other tank question i got rite, ty for all help = )

 

[alphysicist]

Sure, glad to help!