i have a sunk barrel (200 liter) full of water at a depth of 10 meters
of sea water and its fixed to the botton.

keeping the barrel at that depth, i want to pump the water our and
replace it with air. an air hose is attached to the barrel from the
surface and a sump pump drains the water out (assume proper valves are
used to let air in as the water is pumped out - no water gets into the
hose)

Question has 3 parts but basically the same Q, but using different methods 

i) how much work (in joules) is done to drain the barrel with the sump
pump and air hose?
ii)if the water were not replaced by air but rather a vacuum created
inside as it was drained by the sump pump, does it still take the same
amount of work to empty the barrel of water? what is the work effort
and energy for that?
iii) does it take the same amount of work and energy to replace the
water with a pressure hose from the surface to displace the water
instead of using the sump pump?

---

Clarification of Question by tankman-ga on 28 Oct 2005 23:49 PDT

for i) assume the tank does not distort due to pressure diffrential
between water pressure and sea level air pressure in the tank

---

Clarification of Question by tankman-ga on 30 Oct 2005 22:09 PST

please disregard points ii) and iii) - an answer for point i) only is required
thanks

The pressure under 10 meters of seawater is a bit over 100,000
Pascals, relative to atmospheric pressure.  Your volume is 0.2 m^3. 
Energy is pressure times volume, or about 20,000 Joules.  Another way
to do the calculation is to recognize that you are essentially lifting
slightly more than 200 kg of water a distance of 10 meters against
gravity.  Energy is mass times gravity times distance = 200kg *
9.8m/s^2 * 10 m = 20,000 Joules.  Of course, neither the sump pump nor
the air compressor is 100% efficient, so the amount of electrical
energy required to power those devices will be significantly more.

that seems to answers the Q thanks.

PS - are you not claiming the $10!?

If rracecarr-ga is willing to clarify, I'm puzzled that the first
calculation depends on the density of the seawater in which the barrel
is immersed (the 100,000 Pascals) while the second calculation depends
on the mass of the barrel itself (the 200 kg) and the depth.

Isn't it a matter of coincidence that both calculations return he same result?
If the pressure were 200,000 Pascals wouldn't the first calculation
yield 40,000 Joules and the second calculation still yield 20,000
Joules?

P.S.  rracecarr-ga is not a Google Answers Researcher, and so cannot
claim the $10 for the Comment.

Hi Richard-ga,

No, there is no coincidence, and both results depend on the density of
seawater.  200 kg is not the mass of the barrel, but the mass of 200 L
of seawater.  (Actually, it would be more like 205 kg).  If the
pressure were 200,000 Pascals, that would mean the depth was about 20
meters, not 10, so the 2nd calculation would be 200*9.8*20 = 40,000.

Pressure, density, and depth are not independent.  You can't just
change one and leave the others the same.  For an incompressible
fluid, pressure is density*gravity*depth.

Hi rracecarr-ga,

Can you clarify, if the pressure is 100k pascals at 10m - is the
pressure the same for two objects of same volume but different surface
area? what is the total pressure on a sphere with vol 200lt and the
total pressure on a barrel with a 200lt displacement?

Hi tankman-ga,

Pressure is a scalar field, like temperature, so it really doesn't
make sense to talk about the total pressure on an object.  The total
force is equal to the pressure times the surface area, and it is
greater for a 200 lt 'barrel' than for a 200 lt sphere. But I don't
think the total force is important.  The pressure under 10 m of
seawater is 100000 Pa greater than the pressure at the surface.  In
terms of the amount of work required to evacuate the tank, the shape
of the tank does not matter, as long as the center of the tank is kept
at the same depth.

Hi rracecarr-ga,

Much appreciate your feedback. Trust you have the patience for me to
grasp this! So, i'm trying to understand what is the relationship (if
any) between the energy needed to grow an object under 10m of water
and its surface area - which, according to what you mentioned
previously, it does not appear to have a relationship? but here's
where i'm confused. If i extend the tube of the barrel by 50% of its
original length, it adds 50% more volume to give me 300 lts and if i
grow the radius of a sphere from 36.25 cms to 41.5 cms, which produces
a sphere of 300 lts - does it still consume the same amount of energy
to expand both given the same volume or does it take less for the
spere given the smaller surface area and less total pressure?

thnx

Surface area doesn't matter, only volume.  

If you 'stretch' a cylindrical barrel lengthwise, the surface area of
the curved side part increases, but only the flat end does work
against pressure.  In inflating a sphere, the entire surface does work
against pressure.

Hi  rracecarr-ga,

So total presure doesn't come into the equation - i'll try to get my
little head around that! Mnay thanks.

PS - do you recommend any great site to continue my exploration? - and education!

Hi  rracecarr-ga,

hope you are still there - got an observation and a couple of Qs.

if it takes the equivalent of "essentially lifting slightly more than
200 kg of water a distance of 10 meters against gravity" expanding the
object requires the equivalent of 2000kgs (roughly) of force at a
depth of 10m

Q - in trying to figure out the barrel's KE as it floats freely to the surface 
what is its mass? the 200 kgs of water displaced?

if that's the case, and we assume an easy terminal velocity of 1 m/s
its KE is .5*200*1*1 = 100 J/m/s - is that correct?

look forward to hearing from you - many thanks
happy thanks giving