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Old Jan 21st 2011, 12:45 PM   #1
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Basic Thermo Question

Hey guys, I'm looking for a hand on this problem I've got to solve. It's basic thermo stuff involving balancing some forces but my attempts to solve this have been futile thus far...

A 10000kg empty tank is put into the ocean with an open top.
Cross sectional area is 3m squared. It is 16m high.
the tank must ink 10m and float there.
To make the tank sink 10m, we will pour concrete into it.
How much concrete must we pour?

It was suggested to balance the tank so it floats on top of the water as state 1, then state 2 will be 10m under the ocean's surface. I feel like I'd need the density of water to solve this and as I go deeper with the tank, the upwards force of the water on the tank will increase exponentially. Problem with this is, we're not that advanced in this class and there MUST be a simpler solution.

Any tips?

Thanks in advance.

-Ry

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Old Jan 21st 2011, 01:35 PM   #2
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Progression on Problem

So i think I figured what I need to do.

roe(9.81)(height) to get pressure at 10 m deep.....I also have to factor in the 3m squared area..i think by multiplying.

i do something similar with the density of concrete and balance the equation at -10m.

Po - atmosphere pressure

Po + 10000(9.81) + roe(9.81)(height of concrete in tank)(3m) = roe(9.81)(10m)(3m)


I don't think that equation makes sense yet....I'll play with it more tonight.

-Ry
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Old Jan 21st 2011, 09:25 PM   #3
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My 2 cents:

At 10 m depth, the weight of the contained + concrete will balance with the upthrust on the container.

W = F

Mg = ρgV

rho (ρ) is the density of water displaced,
V is the volume of water displaced (which is also the volume of the container) and
M is the mass of concrete + container

The equation can be simplified:

M = ρV

I hope it helps
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Old Feb 5th 2011, 06:38 AM   #4
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Hi,

This is uniform motion with zero velocity. The 2nd law applies:

System is tank (T) and concrete (C).

[d(mV)/dt = sumF] dot K

0 = -(mT + mc)g + pbottom A -ptop A with

mT = 10000kg mc = ? pbottom = patm + rhog(10m)
Ptop= patm A = 3m2

If mc arrives negative, it means the tank won't float at all.

wwwTHERMO spoken here! is under revision - See Artic Shelf Ice there...
later

good luck Jim





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Old Feb 5th 2011, 06:39 AM   #5
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Hi,

This is uniiform motion with zero velocity. The 2nd law applies:

System is tank (T) and concrete (C).

[d(mV)/dt = sumF] dot K

0 = -(mT + mc)g + pbottom A -ptop A with mc = ?

Ptop= patm pbottom = patm + rhog(10m) A = 3m2

If mc arrives negative, it means the tank won't float at all.

THERMO spoken here! is under revision - See Artic Shelf Ice there...
later

good luck





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