Though Experiment - Density and Bouyancy
Positron
changes and have attempted to model the situation mathematically. The
main objective have been to investigate the relationship between
applied work and energy potentially harvested from the system by
changing the density of an object with with one mechanism and
harvesting the energy potential produced by this using another method.
Depending on the situation, my current mathematical model indicates a
theoretical net energy gain, depending on certain specifications such
as the relationship between size and weight of the object. Realizing
that this is widely considered to be impossible, I suspect that there
must be some mistake in my approach or model. I have been unable to
find it as yet, and would appreciate it if you could investigate the
situation and give your input.
I have made the spreadsheet I used to do my calculations as well as a
short explanatory slideshow available:
http://sites.google.com/site/thetied/discussions/thoughtexperiment-densityandbouyancy
I will appreciate it if you could send replies to:
http://groups.google.co.za/group/thetied/browse_thread/thread/376bc26acc14e5bf?hl=en
Thank you
Greg Neill
Many people will not be comfortable downloading Powerpoint or Excel
files from comlete strangers. Perhaps you could describe your model
in text here with maybe some links to diagrams placed on your web site?
Positron
> Positron wrote:
> > I have attempted a simple thought experiment with buoyancy and density
> > changes and have attempted to model the situation mathematically. The
> > main objective have been to investigate the relationship between
> > applied work and energy potentially harvested from the system by
> > changing the density of an object with with one mechanism and
> > harvesting the energy potential produced by this using another method.
>
> > Depending on the situation, my current mathematical model indicates a
> > theoretical net energy gain, depending on certain specifications such
> > as the relationship between size and weight of the object. Realizing
> > that this is widely considered to be impossible, I suspect that there
> > must be some mistake in my approach or model. I have been unable to
> > find it as yet, and would appreciate it if you could investigate the
> > situation and give your input.
>
> > I have made the spreadsheet I used to do my calculations as well as a
> > short explanatory slideshow available:
>
>
> > I will appreciate it if you could send replies to:
>
>
>
>
> > Thank you
>
> Many people will not be comfortable downloading Powerpoint or Excel
> files from comlete strangers. Perhaps you could describe your model
> in text here with maybe some links to diagrams placed on your web site?
I have saved the PowerPoint document as a series of .jpeg images and
uploaded them on the website:
http://sites.google.com/site/thetied/documents
Here is the explanatory text that appears in the slides:
The design constitutes of a hollow container separated into two
compartments by a movable diaphragm, one containing gas and the other
containing water. The water side of the container has an opening that
allows for free passage of water into and out of the container. The
gas side of the container is connected to an external gas storage
tank.
Step 1: Container at minimum depth with gas pressure equal to water
pressure and overall density of container equal or greater than that
of water.
(Zero net force, gas compartment relatively small)
Step 2: Container moved to maximum depth. This step requires no work
to be done on the system as the density is equal to or slightly
greater than that of water. The gas compartment size decreases further
due to increase in water pressure and loose diaphragm, resulting in an
increasing downwards force as the compartment lowers.
Step 3: Gas pumped into container to give net upwards force. Density
equal or slightly less than that of water. This step requires work to
be input into the system. Size of gas chamber increases.
Step 4: Container moves to minimum depth, performing work on attached
cable and generator system. The net force acting upwards increases
with decrease in height as water pressure decreases leading to greater
density difference. (Larger gas chamber)
Step 5: Gas pumped out of container to make overall density again
equal to that of water to allow container to move downwards.
Cycle is now complete and is repeated from Step 1.
If you require further explanation of the operation of the proposed
system, the accompanying spreadsheet that I used to model the
situation mathematically, want to know how I derived the equations I
used in the spreadsheet or want my explanation why I think it is valid
for such a system to operate within the normal laws of thermodynamics,
please let me know.
Greg Neill
>
> I have saved the PowerPoint document as a series of .jpeg images and
> uploaded them on the website:
> http://sites.google.com/site/thetied/documents
>
> Here is the explanatory text that appears in the slides:
I've mixed some comments in below:
>
> The design constitutes of a hollow container separated into two
> compartments by a movable diaphragm, one containing gas and the other
> containing water. The water side of the container has an opening that
> allows for free passage of water into and out of the container. The
> gas side of the container is connected to an external gas storage
> tank.
Let's make things simple by making the container an upright
circular cylinder with a closed top and an open bottom:
. ~~~~~~~~~~~~~~~~~~~~~~~~ water surface
. **=======**
. || air ||
. || space || Container at initial depth
. ||~~~~~~~||
. || ||
. || ||
. || ||
. || ||
> Step 1: Container at minimum depth with gas pressure equal to water
> pressure and overall density of container equal or greater than that
> of water.
> (Zero net force, gas compartment relatively small)
With the open access for water to the container bottom, the
gas pressure will always be equal to that of the ambient water
at any given depth.
> Step 2: Container moved to maximum depth. This step requires no work
> to be done on the system as the density is equal to or slightly
> greater than that of water. The gas compartment size decreases further
> due to increase in water pressure and loose diaphragm, resulting in an
> increasing downwards force as the compartment lowers.
As the depth increases the pressure in the volume of gas
increases by compression. The net bouyant force created by
the void will increase as the pressure rises. Note that
the water in the container does not contribute to the
net weight of the container; it is always fully supported by
the water below the container, being free to move in and
out of the container at all times.
Suppose the initial weight of the container alone is
Wc, and the cross sectional area of the top of the air
space is A. The upward bouyant force due to to the air
space is in fact due to the air pressure exerted upon
the undersurface if the container top, which is given
by P*A, where P is the gas pressure.
But the pressure will depend upon the depth in the
water, which will be given by P = Po + rho*g*h,
where Po is the initial pressure at your minimum depth,
rho the liquid density, and h the depth below the
intitial depth. Thus the overall weight of the
ensemble with depth would be:
W(h) = Wc - Po*A - P(h)*A
= Wc - Po*A - rho*g*h*A
Now, if the ensemble is initially neutrally bouyant,
then:
Wc - Po*A = 0
and the weight with depth becomes
W(h) = -rho*g*A*h
which is increasingly negative with depth. In other words, it
requires more force to push the container deeper as it descends,
and you need to do work to force the container down. It's not
a free lunch!
> Step 3: Gas pumped into container to give net upwards force. Density
> equal or slightly less than that of water. This step requires work to
> be input into the system. Size of gas chamber increases.
Surprise surprise, this work should be about equal to that
required to submerge the vessel (plus frictional, radiative,
and convective losses as compressed gas temperature soars;
PV = nRT and all that).
It should be clear byu this point that you aren't going to be
able to extract more work out of the ascent of the ensemble
than you put into it to submerge it.
Positron
reasoning. I will give a short description why I think it to be
erroneous, followed by an explanation of how I evaluated the
situation. I could be mistaken, maybe you can explain to why you
evaluated the situation as you did after you have seen the problems I
have with it. Also, please let me know at what point I make an error
in my reasoning if you can see one.
You stated that:
"The upward buoyant force due to the air space is in fact due to the
air pressure exerted upon the under surface if the container top,
which is given by P*A, where P is the gas pressure."
How sure are you of this? The way I see it, assuming the cross section
area A stays the same, an equal downwards force will be applied to the
bottom of the container by the gas, causing a net resultant force of
zero to be applied to the system by the gas. Therefore, the pressure
the gas exerts on the sides of the container cannot cause the buoyancy
force. Even if it did, which is not possible, it would have been an
internal force on the system, causing strain on the rest of the
container to react the force. This would have no effect on the effect
of external forces acting on the system. If this premise is incorrect,
then the rest of your deduction is also incorrect.
This is how I reasoned:
Archimedes's principle states that "Any object, wholly or partially
immersed in a fluid, is buoyed up by a force equal to the weight of
the fluid displaced by the object."
Using this principle, the net force acting on an object can be
calculated by adding the buoyancy force obtained by V*ρ*g minus the
total weight of the container, obtained by m*g, where m is the mass of
the container plus the mass of the gas in the container at any
specific point.
If this answer is positive, the net force is upwards, and if it is
negative, the net force is downwards.
You are correct in stating that the water can be eliminated from the
equation without any effect. Keeping the water in the equation makes
no difference though, as the addition of volume and mass to the
equation in the relationship of ρ makes no difference to the answer
when the whole system is submerged in water.
As the container moves to the bottom, the pressure exerted on the gas
increases causing it to compress and the gas chamber becomes
increasingly smaller, causing either V in F = Vρg - mg to decrease if
water is not considered or m to increase if water is considered.
Either of these two situations will cause the net force to become more
negative, and since it was zero at the start, the net force acting on
the system becomes increasingly negative (downwards) with increase in
depth.
Thus, if you start with a net external force of zero acting on the
system, the moment that the container is moved down a bit its overall
density will increase beyond that of water and it will start
experiencing an increasing net downwards force with increase in depth,
simply because V decreases while m stays the same (without water).
Greg Neill
> Thank you for your comments, however I fail to agree with your
> reasoning. I will give a short description why I think it to be
> erroneous, followed by an explanation of how I evaluated the
> situation. I could be mistaken, maybe you can explain to why you
> evaluated the situation as you did after you have seen the problems I
> have with it. Also, please let me know at what point I make an error
> in my reasoning if you can see one.
>
> You stated that:
> "The upward buoyant force due to the air space is in fact due to the
> which is given by P*A, where P is the gas pressure."
> How sure are you of this? The way I see it, assuming the cross section
> area A stays the same, an equal downwards force will be applied to the
> bottom of the container by the gas, causing a net resultant force of
> zero to be applied to the system by the gas.
There is no bottom of the container; it is open to the liquid; There is no
surface
there upon which the gas pressure can act that will transfer a force to the
container. Pressure exerted laterally upon the sides of the container will
sum to zero as expected.
Positron
Thanks, I think I now understand how you visualised the setup, almost
like an inverted cup.
However, that will not change much since the gas is at the same
pressure as the water. The way I see it, any upwards force created by
the gas pressing against the top surface of the container from below
will be countered by a downwards force from the water pressing against
the top surface of the container from above. Therefore there will be
no net force acting on the system that could account for buoyancy.
The Wikipedia article explains the real cause of buoyancy as caused by
the difference of pressure applied to the bottom areas and top areas
of an object submerged with nonzero vertical length. The deduction
leads to the equation of F = Vgρ. This is the only why how buoyancy is
explained. You may also find that the container in my thought
experiment resembles a compressible object with a compressibility
greater than that of water, according to the “Compressible Objects”
subsection, such an is in unstable equilibrium and will continue to
descend if disturbed, as predicted by my approach.
Did you see anything wrong with my approach?
Greg Neill
The pressure on the under surface of the container top is equal to the
pressure of the water at the bottom of the gas pocket, which is lower
in the liquid than is the top of the container, so it is at a higher
pressure
than the water over the container.
Thus, there is a pressure differential between the top surface and
under surface of the container top. The bouyancy force due to the
gas pocket is thus this pressure differential multiplied by the
area of the container top.
>
> The Wikipedia article explains the real cause of buoyancy as caused by
> the difference of pressure applied to the bottom areas and top areas
> of an object submerged with nonzero vertical length.
Right. The volume of gas represents an object with nonzero vertical
length.
> The deduction
> leads to the equation of F = Vg?. This is the only why how buoyancy is
> explained. You may also find that the container in my thought
> experiment resembles a compressible object with a compressibility
> greater than that of water, according to the "Compressible Objects"
> subsection, such an is in unstable equilibrium and will continue to
> descend if disturbed, as predicted by my approach.
Okay, let's say the tank is neutrally bouyant with a small volume of gas V.
If the tank descends to a depth d such that the volume of gas becomes
negligible due to the increase in pressure with depth, then to regain
neutral bouyancy you'll have to pump down an additional volume of gas
V at that pressure.
The pressure required at depth d is P = rho*g*d. If the pressure of the
gas at the surface is P0 then the work you need to do to compress the
gas to this pressure and volume is at least
W = (rho*g*d - P0)*V
As an example, suppose that the liquid is water, the volume is 1 m^3,
the intiial pressure is 1 atmosphere, and the depth is 1000 m. The
work required to compress that volume of gas is about 9.7 x 10^6 J,
and doesn't include the usual losses do to the hot gas cooling to
operating temperature, compressor efficiency, etc.
Can you show that the work you can extract from the rising container
is greater than this? And no, I won't load your Excel spreadsheet; you'll
have to post your equations here.
Positron
Assume the following conditions:
Pressure in storage tank is 1 atmosphere.
Gas used is Helium.
Container mass is 1000kg.
Temperature is 15°C throughout descent (not practical, but it appeared
as if temperature difference makes very little difference)
The minimum depth is 1m and the maximum depth is 1000m.
I will be using the following equations:
w = PΔV w in Joules (J); P in Pa; V in m^3
PV = nRT P = nRT/V
w = Fd w in Joules (J); F in Newton (N); d in meters (m)
P = 101325 + ρgh P in Pa; ρ in kg/m^3; g in m/s^2; h in m
The following constants will be needed:
Gravity acceleration 9.81m/s^2
Water density 988kg/m^3
Gas constant (R) 8.3145J / (mol*K)
0.000082057(m^3*atm) / (mol*K)
Pascal (Pa) 9.86923E-06 atmosphere
Atmosphere 101325 Pa
Mr of He 4.0026 g/mol
The following values can be calculated:
Temperature 288.15K
Initial water pressure 111017.28 Pa
1.095655403 atm
Final water pressure 9793605 Pa
96.65536965 atm
Buoyancy force 96922.8N
Storage tank pressure 101325 Pa
The net force acting on the container at any point is:
F = (nRT/P)gρ -mg -n(Mr/1000)g
Where n is the amount of moles of gas, R is the gas constant, T is the
temperature, P is the pressure acting on the gas (the water pressure
at a specific depth), m is the mass of the container, Mr is the molar
mass of the gas and g is acceleration due to gravity.
This formula was derived using Archimedes’ principle, where the (nRT/P)
gρ term is used to calculate the upwards buoyant force due to the
displacement of water, and the mg and n(Mr/1000)g terms are used to
calculate the total weight of the system (total downwards force).
Step 1: Container is at minium depth with gas pressure equal to water
pressure and overall density equal to that of water.
Rearranging this equation and setting F = 0 for zero net force allows
the amount of moles required to keep the container stationary at a
specific dept to be calculated:
n = (mg)/(RTgp/P - (Mr/1000)g)
Setting the pressure P equal to the initial water pressure in atm
(1.095655403 atm) gives the required amount of moles of gas at the
start of the cycle to be 46.9098795 moles. The corresponding volume
(obtained using V = nRT/P) is 1.012335791m^3.
Step 2: Container sinks to maximum depth.
Step 3: Gas pumped into container to give buoyancy.
After the container has sunk to the bottom, the gas compartment is now
only 0.011475526m^3 big due to the increased external pressure.
According to the previously derived equation, n = (mg)/(RTgp/P - (Mr/
1000)g), the amount of moles og gas required to get the container
buoyant is 4207.142227 moles. At a depth of 1000m (a pressure of
96.65536965 atm) this will take up 1.029189785m^3 of space.
The work that must therefore be done at the maximum depth to get the
container buoyant again so it can start its upwards ascent is given by
w = PΔV, where P is the pressure of the water at the maximum depth (in
Pascal) minus the pressure in the storage tank, and ΔV is the change
in volume caused by the addittion of the extra amount of moles of gas
required to get the container buoyant. This requires 9863971.56
Joules (9.864 x 10^6 J, very close to your calculated value) of work
to be input into the system. (This is the first of two occassions
where work needs to be input into the system in one cycle. Additionnal
work needs to be input when pumping gas out of the container at the
minimum depth to the storage tank so it can sink again, but it is much
less. This will be calculated in Step 5)
Step 4: Container drifts to minimum depth, performing work on attached
cable and generator system.
Now for the part you questioned me about.
I have deduced that the net force acting on the system at any depth is
given by:
F = (nRT/P)gρ -mg -n(Mr/1000)g
Since this force varies with depth as the container moves from the
maximum to the minimum depth, increasing in magnitude (just as the
force that caused the system to sink became increasingly negative) the
equation w = FΔd cannot be directly applied, but integration is
necessary. Taking d to be the depth the container is submerged and
integrating results in the following equation:
w = (mg + n(Mr/1000)g)*(df - di) - (nRT/9.86923E-06)*ln|df/di|
Where df is the minimum depth and di is the maximum depth, the
9.86923E-06 constant term is the Pascal to atmosphere conversion
factor and n is the amount of moles after pumping. This equation
assumes that the resistance of the generator is always exactly equal
to the net force on the container.
Solving gives a value of 59661348.53 Joules (5.966135 x 10^7 J) as
output.
Step 5: Gas pumped out of container to make overall density again
equal to that of water.
More work needs to be applied to complete the cycle. The amount of
moles of gas must be reduced so that the container experiences no net
force and can start sinking down again.
Again using w = PΔV, where P is the pressure of the water at the
minimum depth (in Pascal) minus the pressure in the storage tank, and
ΔV is the change in volume caused by the removal of the extra amount
of moles of gas required to get the container buoyant.
This requires 870169.4114 Joules (8.7017 x 10^5 J) of work to be input
into the system. The relatively small value is a result of the
pressure difference of only 9692.28 Pascal between the minimum water
pressure level and the storage tank.
The cycle is now complete. Work was required in steps 3 and 5, and
gained in step 4.
Total work required (Step 3 + Step 5): 10734141 J
Total work gained (Step 4): 59661349 J
Net work: 48927208 J
Work out/in ration 5.558
Percentage gain 455.81%
The main reason I believe it is possible for this system to give a
positive output of energy is because of the use of two totally
independent mechanisms: one to harvest energy from potential energy
stored in the system and another to reload that potential energy.
Since the buoyancy laws and the gas laws are not directly linked, I
would find it very strange if the ratio works out to be exactly 1:1.
Positron
and rho is ρ.
Greg Neill
[snip]
> Step 3: Gas pumped into container to give buoyancy.
>
> After the container has sunk to the bottom, the gas compartment is now
> only 0.011475526m^3 big due to the increased external pressure.
> According to the previously derived equation, n = (mg)/(RTgp/P - (Mr/
> 1000)g), the amount of moles og gas required to get the container
> buoyant is 4207.142227 moles. At a depth of 1000m (a pressure of
> 96.65536965 atm) this will take up 1.029189785m^3 of space.
> The work that must therefore be done at the maximum depth to get the
> container buoyant again so it can start its upwards ascent is given by
> w = P�V, where P is the pressure of the water at the maximum depth (in
> Pascal) minus the pressure in the storage tank, and �V is the change
> in volume caused by the addittion of the extra amount of moles of gas
> required to get the container buoyant. This requires 9863971.56
> Joules (9.864 x 10^6 J, very close to your calculated value) of work
> to be input into the system.
This takes care of the work required to force the gas at a pressure of
96.66 atm into the container up to a volume of 1.03 m^3. But you
haven't calculated the work required to first create the gas at that
pressure. Remember, the storage tank is at 1 atm.
Here's a rough estimate of the energy required to provide the gas at
working pressure.
If you very slowly compress a gas from volume v0 to volume v1 so that
the temperature remains constant throughout, the work required is
W_c = n*R*T*ln(v0/v1)
In your case you wish to produce 1.03 m^3 of gas at 96.66 atm
starting with some volume of gas at about 1 atm. So you are
compressing about (99.66/1)*1.03 m^3 of gas down to 1.03 m^3.
We have n*R*T = P*V = 96.66atm * 1.03m^3
n*R*T = 1.01 x 10^7 J
and the work required to compress the gas is
W_c = 1.01 x 10^7 J * ln(96.66)
= 4.61 x 10^7 J
Now your total work (ideal -- does not include losses due to inefficiencies)
to load the container will be approximately
W = 9.864 x 10^6 J + W_c
= 5.59 x 10^7 J
> (This is the first of two occassions
> where work needs to be input into the system in one cycle. Additionnal
> work needs to be input when pumping gas out of the container at the
> minimum depth to the storage tank so it can sink again, but it is much
> less. This will be calculated in Step 5)
>
> Step 4: Container drifts to minimum depth, performing work on attached
> cable and generator system.
>
> Now for the part you questioned me about.
> I have deduced that the net force acting on the system at any depth is
> given by:
> F = (nRT/P)g� -mg -n(Mr/1000)g
> Since this force varies with depth as the container moves from the
> maximum to the minimum depth, increasing in magnitude (just as the
> force that caused the system to sink became increasingly negative) the
> equation w = F�d cannot be directly applied, but integration is
> necessary. Taking d to be the depth the container is submerged and
> integrating results in the following equation:
> w = (mg + n(Mr/1000)g)*(df - di) - (nRT/9.86923E-06)*ln|df/di|
> Where df is the minimum depth and di is the maximum depth, the
> 9.86923E-06 constant term is the Pascal to atmosphere conversion
> factor and n is the amount of moles after pumping. This equation
> assumes that the resistance of the generator is always exactly equal
> to the net force on the container.
> Solving gives a value of 59661348.53 Joules (5.966135 x 10^7 J) as
> output.
>
> Step 5: Gas pumped out of container to make overall density again
> equal to that of water.
> More work needs to be applied to complete the cycle. The amount of
> moles of gas must be reduced so that the container experiences no net
> force and can start sinking down again.
> Again using w = P�V, where P is the pressure of the water at the
> minimum depth (in Pascal) minus the pressure in the storage tank, and
> �V is the change in volume caused by the removal of the extra amount
> of moles of gas required to get the container buoyant.
> This requires 870169.4114 Joules (8.7017 x 10^5 J) of work to be input
> into the system. The relatively small value is a result of the
> pressure difference of only 9692.28 Pascal between the minimum water
> pressure level and the storage tank.
>
> The cycle is now complete. Work was required in steps 3 and 5, and
> gained in step 4.
> Total work required (Step 3 + Step 5): 10734141 J
> Total work gained (Step 4): 59661349 J
> Net work: 48927208 J
> Work out/in ration 5.558
> Percentage gain 455.81%
>
> The main reason I believe it is possible for this system to give a
> positive output of energy is because of the use of two totally
> independent mechanisms: one to harvest energy from potential energy
> stored in the system and another to reload that potential energy.
> Since the buoyancy laws and the gas laws are not directly linked, I
> would find it very strange if the ratio works out to be exactly 1:1.
Energy is conserved. Always.
If you think you are getting a net positive output of energy from a
closed system, you'd better check to see what your model is missing.
Here's an idea for you. Instead of pumping compressed air down to
your container, start it off neutrally bouyant with some fixed amount
of gas. When the container is at the bottom, add heat to the gas
with an electric heating element to bring the gas volume up to that
required to make the tank bouyant. You are allowed to take advantage
of the heat rise in the gas as it compresses while descending by
assuming that the gas is enclosed in a perfectly insulating chamber.
Calculate the energy required to make the container bouyant when
at the bottom.
Positron
> Positron wrote:
>
> [snip]
>
>
>
>
>
> > Step 3: Gas pumped into container to give buoyancy.
>
> > After the container has sunk to the bottom, the gas compartment is now
> > only 0.011475526m^3 big due to the increased external pressure.
> > According to the previously derived equation, n = (mg)/(RTgp/P - (Mr/
> > 1000)g), the amount of moles og gas required to get the container
> > buoyant is 4207.142227 moles. At a depth of 1000m (a pressure of
> > 96.65536965 atm) this will take up 1.029189785m^3 of space.
> > The work that must therefore be done at the maximum depth to get the
> > container buoyant again so it can start its upwards ascent is given by
> > Pascal) minus the pressure in the storage tank, and ÄV is the change
> > Since this force varies with depth as the container moves from the
> > maximum to the minimum depth, increasing in magnitude (just as the
> > force that caused the system to sink became increasingly negative) the
> > necessary. Taking d to be the depth the container is submerged and
> > integrating results in the following equation:
> > w = (mg + n(Mr/1000)g)*(df - di) - (nRT/9.86923E-06)*ln|df/di|
> > Where df is the minimum depth and di is the maximum depth, the
> > 9.86923E-06 constant term is the Pascal to atmosphere conversion
> > factor and n is the amount of moles after pumping. This equation
> > assumes that the resistance of the generator is always exactly equal
> > to the net force on the container.
> > Solving gives a value of 59661348.53 Joules (5.966135 x 10^7 J) as
> > output.
>
> > Step 5: Gas pumped out of container to make overall density again
> > equal to that of water.
> > More work needs to be applied to complete the cycle. The amount of
> > moles of gas must be reduced so that the container experiences no net
> > force and can start sinking down again.
> > minimum depth (in Pascal) minus the pressure in the storage tank, and
Thanks for that valuable input! As you will see at the start of this
tread I was really searching for the error as I doubted it could
really work, thats why I posted here. I included compression into my
model and now got a result of 6.25% gain, which could easily be
swallowed up by real life factors and inefficiencies. I still have
another configuration using a fixed amount of gas and just compressing
it to alter density that gives 25% gain under specific criteria, maybe
you can have a look at that one too? I will investigate the heating
idea.
Positron
current model is accurrate. My first model without the compression
part may not have been accurrate, but including the compression and
pumping parts do not seem to be correct either.
This is why I doubt:
As far as I know, in the equation work = pressure x change in volume,
the pressure term (in Pascal) is equal to the pressure difference
between the two sides of the diagram used to accomplish the pumping.
If the gas is first compressed to the exact same value as the gas
pressure at the maximum depth, then P = 0 and the w = PV term falls
away, leaving only the work required to compress the gas as the work
needed to give the container buoyancy at the maximum depth.
Now, if gas that is to be pumped always first needed to be compressed
to the pressure at which it will be if pumping has finished, can the
equation w = PV ever be applied to pumping a gas?
Greg Neill
Sure, but it's not always the best model.
Consider the case where you want to pump a gas at pressure
P2 into a liquid also at pressure P2. "Behind" the pump
the pressure is P1.
For a simple model consider a cylinder of cross section A
with a piston at one end and a one-way valve at the other.
The one-way valve leads to the environment at pressure P2,
and behind the piston is an ambient pressure at P1.
Initially the piston is full of gas at pressure P1, the
valve is closed, and the force on the piston is zero
since the pressure on either side of it is the same.
Now the piston is pushed into the cylinder to compress the
gas up to the pressure P2 at volume V where the valve will
be able to open. This takes some initial amount of work to
accomplish, reducing the initial volume of gas at pressure
P1 to a smaller volume V at higher pressure P2. How much
work it takes to accomplish this initial compression depends
upon the specifics of how it is done; the heat conduction
of the materials, specific heat of the gas, time period,
etc. The theoretical lowest amount of work, where the
compression is done at constant temperature, is
W1 = n*R*T*ln(P2/P1)
At this point the gas is at the "working" pressure of the
liquid environment and you're ready to push it out of the
cylinder and into the liquid. How much work does this take?
The pressure on the back side of the piston remains at P1,
while the pressure on the valve side will hold constant at
P2. Thus the force that must be applied to hold the piston
in place is
F = (P2 - P1)*A
that is, the pressure difference multiplied by the cross
sectional area of the piston. In order to move the volume
of gas V out of the cylinder and into the liquid the piston
must move forward a distance
d = V/A
So the work done to pump the gas is
W2 = F*d = (P2 - P1)*A*V/A
= (P2 - P1)*V
Positron
Thank you for this explanation, I appreciate it. I have thought of a
way of playing two of these systems off against each other and
harvesting some of the energy potentially produced when the container
is lowered to accomplish the compression step in the other system
through a type of hydraulic system. Will post here when I have
finished it and it seems to give a higher output (another mistake).