How Much Water Escapes from a Pressurized Steel Vessel?

[2020vision]

Homework Statement

A cylindrical steel pressure vessel with volume 1.31 m^3 is to be tested. The vessel is entirely filled with water, then a piston at one end of the cylinder is pushed in until the pressure inside the vessel has increased by 1000 kPa. Suddenly, a safety plug on the top bursts. How many liters of water come out?

Homework Equations

B=0.2x10^10N/m^2
P1=P0+(rho)gh
P1=-B((Delta V)/V)

The Attempt at a Solution

I am assuming that I have to look for delta V as that would be the water that comes out causing the change in volume.

Delta V=-V(Delta P)/B=-1.31(1000)/(0.2x10^10)
Delta V= 6.55*10^-7

But this is not the right answer. I am confused as to where I'm going wrong.
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A second problem

Homework Statement

A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over.

a) As it hits the table, what is the angular velocity of the tip of the rod?
b) What is the speed of the tip of the rod?

Homework Equations

v = wr
a = v^2/r=(w^2)r
v= (2(pi)(r))/T

The Attempt at a Solution

I am having trouble picturing what the question is asking.

a) w = v/r = (2(pi)r)/T/r = 2pi/T...don't know where to go from here

b) I think after I find w in a i just use v = wr

 

[LowlyPion]

For 1) don't you have 1000 kPa and not 1000 Pa?

 

[LowlyPion]

For 2) consider the moment of inertia of the bar. Consider too the energy in the system as it starts to fall. The total potential energy of the bar should be rotational kinetic energy when it's at the horizontal shouldn't it?

 

[2020vision]

For 1) don't you have 1000 kPa and not 1000 Pa?

Yeah its kpa, so in the formula I should use 1 000 000pa and I get 6.55*10^-4 m^3 which will be 0.655L...is that right?

 

[LowlyPion]

Yeah its kpa, so in the formula I should use 1 000 000pa and I get 6.55*10^-4 m^3 which will be 0.655L...is that right?

If your bulk modulus for water is good.

I might use 4.58 * 10^-10 for compressibility k, where k = 1/B, which yields a slightly smaller result.

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html#c1

 

[2020vision]

For 2) consider the moment of inertia of the bar. Consider too the energy in the system as it starts to fall. The total potential energy of the bar should be rotational kinetic energy when it's at the horizontal shouldn't it?

Ok I made potential energy = rotational energy

so i get:

Ug = Krot
MgL = 1/2 (1/3ML^2)w^2
GL = 1/6 (L^2)(w^2)
sqrt(6g/L) = w

so I put the answer as sqrt (6g/L) on the online system and its wrong, but I get this message:
"Your answer either contains an incorrect numerical multiplier or is missing one."

Where did I go wrong?

 

[2020vision]

If your bulk modulus for water is good.

I might use 4.58 * 10^-10 for compressibility k, where k = 1/B, which yields a slightly smaller result.

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html#c1

We are supposed to use the bulk modulus from our textbook, and that one is 0.2*10^10. Anything else would give a wrong answer in the system. So with this bulk modulus, is 0.655L right?

 

[LowlyPion]

For 2) your potential energy will be the height of the center of mass won't it?

 

[LowlyPion]

We are supposed to use the bulk modulus from our textbook, and that one is 0.2*10^10. Anything else would give a wrong answer in the system. So with this bulk modulus, is 0.655L right?

That what it looks like to me then.

 

[2020vision]

For 2) your potential energy will be the height of the center of mass won't it?

oh so it should be sqrt (3g/L) right?

 

[LowlyPion]

oh so it should be sqrt (3g/L) right?

Yes.