Fluid Mechanics

BbyBlue24

A thin spherical shell of mass 4 kg and diameter 0.2 m is filled with helium (density 0.18 kg/m3). It is then released from rest on the bottom of a pool of water that is 4 m deep. Determine the value of the upward acceleration of the shell.

Pyrrhus

what have you done???

Apply newton's 2nd Law to the forces acting on the shell, buoyant force and the weight.

BbyBlue24

I determined the weight to be 39.2 N. I used the formula B=ρgV = (.18)(9.8)(.004) = .007 and V=(4πr3)/3 = .004.

Acceleration = Force/Mass = (.004-39.2)/4 = -9.799

And this answer doesn't make sense to me, I think my problem is with the buoyant force. Thanks!

Pyrrhus

You should have out for the buoyant force the density of the water, remember the buoyant force is equal to the weight of fluid displaced.

BbyBlue24

Doing that I get (1)(9.8)(.004) = .0392...sorry I am not getting this at all

Pyrrhus

You should have

Applying Newton's 2nd Law

[tex] B - m_{shell}g -m_{He}g = (m_{He} + m_{shell}) a [/tex]

BbyBlue24

So that makes the upward acceleration -9.79?

Pyrrhus

Check well, i get a = 0.46 m/s^2

[tex] (\rho_{water}V - \rho_{He}V - m_{shell})g = (m_{shell} + \rho_{He}V)a [/tex]

BbyBlue24

These are the values that I entered into the equation, I think the V (volume) might be incorrect, I am using .004 (V=(4Πr^3)/3)


((1∙.004)-(.18∙.004)-4)9.8=(4+.18∙.004)a = -9.79

Thank you!

Pyrrhus

I still get the same answer, i will write it to the simpler form

[tex] a = g \frac{(\rho_{water} - \rho_{He}) \frac{\pi d^3}{6} - m_{shell}}{m_{shell} + \rho_{He} \frac{\pi d^3}{6}} [/tex]

The density of water is 1000 kg/m^3!

BbyBlue24

Yes, thank you. Also thank you very much for helping me out with this problem and the rest of the problems I have had tonight!!

pack_rat2

The upward force will be equal to the difference between the weight of the He plus the material of the shell, and that of the water the sphere displaces. But there will be a considerable drag force exerted against the sphere's upward travel by the water. The D.E. for this system will be: -m*x'' - k*x' + F = 0, where m is the mass of the sphere, m*x'' is the inertial force on the sphere (downward, hence negative), k is the coefficient of drag (the force is downward), and F is the buoyant force (upward). x' and x'' are the sphere's velocity and acceleration, respectively.

pack_rat2

Hey, check this out: http://www.ma.iup.edu/projects/CalcD...ag.html#drag15
It's got a complete explanation!

Pyrrhus

This problem probably was meant for a non-viscous, stable, incompresible and irrotational fluid.

pack_rat2

The problem says it's hydrogen hydroxide, aka hydrogen monoxide...HOH...H2O...