Potential Energy stored in a SCUBA TANK?
Steve
My Question is how much potential energy is stored in a SCUBA tank.
Given the Parms. of
1 80cu ft. SCUBA Tank filled at 3000 pounds per sq inch of AIR.
Please also post if possible the formula used.
Uncle Al
The product of pressure and volume is energy, e.g., liter-atmospheres.
Look up the conversion (e.g, 1 l-atm is about 101 joules as I remember).
Multiply volume times pressure for your scuba tank, convert to energy, and then
be afraid, be very afraid.
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John Thornley
The answer depends on whether you really mean potential energy, or the
energy taken to compress the gas to adiabatically change its volume.
Do you really care, if not I'll knock the answer out when you reply.
What units? Ft.Lbs force, Joules or what?
On or about Wed, 23 Jul 1997 09:46:58 -0400, Steve <st...@nortel.com>
scribbled:
R.G. Lynn
> Steve wrote:
> >
> > My Question is how much potential energy is stored in a SCUBA tank.
> > Given the Parms. of
> >
> > 1 80cu ft. SCUBA Tank filled at 3000 pounds per sq inch of AIR.
> >
> > Please also post if possible the formula used.
If you use the assumption that the expansion process is adiabatic (ie no
heat transfer occurs out of or into the air as it expands) then if I've
worked my algebra right the equation you want is:
Total Energy = (Ptank*Vtank/(gamma-1))*(1-(Patmos/Ptank)^(1-1/gamma))
Using SI units as all sane people do:
Total Energy is in Joules [J] ie one Watt for one second or 1/3600000
kWh (a fairly standard unit of electrical power)
R = 287 [J/kg/K] for air
gamma = 1.4 [dimensionless] for air.
Patmos = absolute atmospheric pressure in Pascals [Pa] or [N/m^2] =
roughly 101 kPa at sea level.
Ptank = absolute pressure inside tank [Pa] = roughly 20.7 MPa at 3000
psi
Vtank = tank volume [m^3] = 0.0510 m^3 for 1.8 cubic feet
So leaving aside possible errors in my calculations total energy is
roughly 2060000 J or 0.57 of a kWh.
Robert
Ville Saari
> My Question is how much potential energy is stored in a SCUBA tank.
> Given the Parms. of
>
> 1 80cu ft. SCUBA Tank filled at 3000 pounds per sq inch of AIR.
>
> Please also post if possible the formula used.
1.22MJ or the energy of the explosion of 0.29kg of TNT.
It's calculated this way:
those stupid American units converted to SI:
V0 = 80ft³ = 2.2654m³
p1 = 3000lb/in² = 20.684MPa
atmospheric pressure:
p0 = 1atm = 101325Pa
potential energy:
deltaE = -p0V0log(p0/p1) = -101325*2.2654*log(101325/20684000) = 1220882
and the unit of the result is Pa*m³ = N/m²*m³ = Nm = J
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John Thornley
The answer depends on whether you really mean potential energy, or the
energy taken to compress the gas to adiabatically change its volume.
Do you really care, if not I'll knock the answer out when you reply.
What units? Ft.Lbs force, Joules or what?
If adiabatic then pV= k as compression begins. Imagine the cyliner is
ver very long and you are compressing it with a long piston.
I make it about 100 kJ for a 12.2 litre cyclinder at 232 BAR, but I
could be wrong.
On or about Wed, 23 Jul 1997 09:46:58 -0400, Steve <st...@nortel.com>
scribbled:
>My Question is how much potential energy is stored in a SCUBA tank.
John Thornley
Cock up sorry. Dropped a factor. Start again.
Working in SI Units
1 BAR= 15^5 Pa (approx)
p1=232Bar=232*10^5
v1=12.2 litres= 0.0122 m^3
p2=1Bar= 10^5 Pa
V2=232*0.012=2.83m^3 (the volume of the tank at 1BAR)
Allowing this to expand adiabatically (in this case so it doesn't
produce a change in temperature)
p1V1=p2V2=k (constant)
You can plot this function as
p=k/V
It turns out that the energy released in the process is the area UNDER
the curve between the two volumes, i.e.
E=integ(pdV) between v1 and v2;
or
E=k ln (p1/p2) = p1.V1.ln(p1/p2)
= 232*10^5 . 0.0122 . ln (232) = 1.54 MJoule.
This is a theroretical maximum, where the energy is "released" very
slowly.
At 1 horse power, this would take about 35 minutes.
Hope this makes some sense.
Lawrence R. Mead
John Thornley (john.t...@hypno.demon.co.uk$) wrote:
: Cock up sorry. Dropped a factor. Start again.
:
: Working in SI Units
: 1 BAR= 15^5 Pa (approx)
: p1=232Bar=232*10^5
: v1=12.2 litres= 0.0122 m^3
:
: p2=1Bar= 10^5 Pa
: V2=232*0.012=2.83m^3 (the volume of the tank at 1BAR)
:
: Allowing this to expand adiabatically (in this case so it doesn't
: produce a change in temperature)
: p1V1=p2V2=k (constant) <---------- again, incorrect. For adiabatic
processes: p1 V1^gamma = p2 V2^ gamma, where gamma=C_p/C_v = 5/3 for
a monatomic gas. If you do not wish the temperature to change you must
mean *isothermally* .
: You can plot this function as
:
: p=k/V
:
: It turns out that the energy released in the process is the area UNDER
: the curve between the two volumes, i.e.
:
: E=integ(pdV) between v1 and v2;
:
: or
:
: E=k ln (p1/p2) = p1.V1.ln(p1/p2)
:
: = 232*10^5 . 0.0122 . ln (232) = 1.54 MJoule.
:
: This is a theroretical maximum, where the energy is "released" very
: slowly.
:
: At 1 horse power, this would take about 35 minutes.
:
: Hope this makes some sense.
:
--
Lawrence R. Mead (lrm...@whale.st.usm.edu)
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Two little stories. Oxygen bottle lying on side has neck snapped off. Takes
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wall.
Six foot long r=2ft, gas bottle standing up with top explosively released.
How
high does the two ton top rise and it is not 75 feet as originally
calculated.
(tank filled with freon at 6000psi).
Lawrence R. Mead <lrm...@whale.st.usm.edu> wrote in article
<5ra87l$6hg$2...@thorn.cc.usm.edu>...
> John Thornley (john.t...@hypno.demon.co.uk$) wrote:
> : The answer depends on whether you really mean potential energy, or the
> : energy taken to compress the gas to adiabatically change its volume.
> : Do you really care, if not I'll knock the answer out when you reply.
> : What units? Ft.Lbs force, Joules or what?
> :
> : If adiabatic then pV= k as compression begins. Imagine the cyliner is
>
> If adiabatic pV^gamma = K, if gas is sufficiently ideal.
Lawrence R. Mead
Phil Gibbs
In article <33D62C...@elec.canterbury.ac.nz>, "R.G. Lynn"
<lyn...@elec.canterbury.ac.nz> writes
>> Steve wrote:
>> >
>> > My Question is how much potential energy is stored in a SCUBA tank.
>> > Given the Parms. of
>> >
>> > 1 80cu ft. SCUBA Tank filled at 3000 pounds per sq inch of AIR.
>> >
>> > Please also post if possible the formula used.
>
>heat transfer occurs out of or into the air as it expands) then if I've
>worked my algebra right the equation you want is:
>
>Total Energy = (Ptank*Vtank/(gamma-1))*(1-(Patmos/Ptank)^(1-1/gamma))
>
>Using SI units as all sane people do:
>Total Energy is in Joules [J] ie one Watt for one second or 1/3600000
>kWh (a fairly standard unit of electrical power)
>R = 287 [J/kg/K] for air
>gamma = 1.4 [dimensionless] for air.
>Patmos = absolute atmospheric pressure in Pascals [Pa] or [N/m^2] =
>roughly 101 kPa at sea level.
>Ptank = absolute pressure inside tank [Pa] = roughly 20.7 MPa at 3000
>psi
>Vtank = tank volume [m^3] = 0.0510 m^3 for 1.8 cubic feet
>
>So leaving aside possible errors in my calculations total energy is
>roughly 2060000 J or 0.57 of a kWh.
>
>Robert
You got the formula right but you have made an error in the volume.
Steve asked about an 80 cuft tank, this is the volume of the air when
expanded from the tanks working pressure to atmospheric pressure.
Actually 3000 psi is not quite the working pressure and the actual
volume of an 80 cuft tank is usually 10 litres.
So I make the answer about 405 KJ
If it was allowed to expand isothermally instead of adiabatically
the formula would be
Ptank*Vtank*log(Ptank/Patmos)
The answer is then about 1100 KJ
There are other interpretations of the question which
would give different answers. Try E=mc^2 for example :-)
Phil Gibbs
http://www.weburbia.com/ http://www.weburbia.demon.co.uk/
"When all you've got is a hammer, everything looks like a nail"
- Japanese proverb
Phil Gibbs
Actually I don't think this answer is really right because the
temperature would drop to below the boiling point of liquid
nitrogen if it was. The ideal gas laws probably cannot be
trusted for this calculation because of the high pressure
ratio. I think the true answer for adiabatic expansion must
be much less. I can't find the right book but I think the
amount of energy in there is only PV or 1/2PV. Perhaps
someone who knows what they are talking about should jump in :-)
If it was allowed to expand isothermally instead of adiabatically
the formula would be
Ptank*Vtank*log(Ptank/Patmos)
The answer is then about 1100 KJ
I think this is probably a more appropriate answer if
Steve was wondering about the energy which would be
released if the air is let out of the tank,
although most of this energy comes from ambient
heat rather than the energy in the tank itself.
John Thornley
I think you're right about pV^(gamma). However, the calc is near
enough for a rough estimate.
On or about 25 Jul 1997 16:23:45 GMT, lrm...@whale.st.usm.edu
(Lawrence R. Mead) 's pen grunted:
R.G. Lynn
Phil Gibbs wrote:
> If it was allowed to expand isothermally instead of adiabatically
> the formula would be
>
> Ptank*Vtank*log(Ptank/Patmos)
>
> The answer is then about 1100 KJ
>
> I think this is probably a more appropriate answer if
> Steve was wondering about the energy which would be
> released if the air is let out of the tank,
> although most of this energy comes from ambient
> heat rather than the energy in the tank itself.
Yeah, isothermal is probably a better approximation for more practical
situations, but an explosive release, while not strictly reversible
would be fairly close to adiabatic I think.
Robert
Duane C. Johnson
Hi all;
I have been following this thread for a while and wanted to add
my 2 cents.
I don't know what the original poster had in mind but I do know
there are real uses for the stored energy in the tank.
An inventor in the field of under water exploration has shown
that the stored energy can power under water swimming assist devices.
( Sorry but I know little about the terminology used by you SCUBA
guys. Can anyone tell me who I am talking about? )
The way his pneumatic motor/propeller worked was that as the air
for the breathing regulator was consumed the power in the expanding
air was used to propel the craft.
The breathing done by the diver was uninterrupted. Actually the
normal SCUBA apparatus wastes the stored energy by heating the air
before it arrives at the mouth piece. If the air was not heated
it would be very cold and not suitable for breathing.
I believe that the pneumatic air motor system performs this same
heating function as well as providing propulsion using the added
efficiency of the isothermal expansion.
What an elegant invention. It performs a useful task as well as the
original while providing an added benefit of motive power.
Isothermal pneumatic methods have the added efficiency over
adiabatic pneumatic systems. Some of the energy used to compress
the air is stored in the surrounding environment to be recaptured
when the air is released from the tank as it passes through the
expander. When the air is expanding it will tend to become cold.
This cooled air can pick up heat, and thus energy, from the surrounding
water which adds to the output power.
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Steve
Steve wrote:
>
> My Question is how much potential energy is stored in a SCUBA tank.
> Given the Parms. of
>
> 1 80cu ft. SCUBA Tank filled at 3000 pounds per sq inch of AIR.
>
> Please also post if possible the formula used.
Thank-you to everyone working this problem out.
you've basically answered my question. My original conception was
a battle of stored energy underwater. Most ROVs (Remote operated Vehic)
use electricity for propulsion, but I was wondering if it would be more
practical to use compressed air. (ala WWII Torpedoes pre electric)
Discounting the Economics of producing the storage energy
i.e. cost of compressing a SCUBA Tank vs. Charging a Battery.
And also discounting the weight differences of a energy storage
unit (Battery vs Tank),
For a given Size (About 1-2 cu ft. physical size) which means of energy
storage would be most efficient to produce approx. 700-1000watts for
approx 30-40 minutes.
The Debate Continues....
Paul Skoczylas
Steve wrote:
>>
> Thank-you to everyone working this problem out.
> you've basically answered my question. My original conception was
> a battle of stored energy underwater. Most ROVs (Remote operated Vehic)
> use electricity for propulsion, but I was wondering if it would be more
> practical to use compressed air. (ala WWII Torpedoes pre electric)
Ah, a new constraint. As ambient pressure increases, the amount of
energy you can get out of the tank decreases. Presuming that you want
to use the compressed air to drive a turbine, the amount of energy you
can get will decrease substantially as your ROV goes deeper. An
electric battery does not have this limitation.
-Paul
Ian Woollard
There's one more problem. I can't remember my thermodynamics that
well, but technically compressed air cylinders aren't stores of
energy- they store entropy.
The difference is there because as you release the air from
the cylinder- the cylinder cools. Because the cylinder is
cooler the pressure is reduced and you aren't getting so much
energy out of it and the efficiency goes way down.
In fact the energy that you can obtain from the cylinder comes
from the heat energy in and around the cylinder.
So if you are diving in cold water, the water around the
compressed air cylinder may have a tendency to freeze,
and form an insulating layer.
It will probably limit the maximum power that the cylinder can
give you.
Sounds like its time for you to crack those thermodynamics books.
Good luck...
> -Paul
--
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Ed Price
Paul Skoczylas wrote:
>
> Steve wrote:
> >>
> > Thank-you to everyone working this problem out.
> > you've basically answered my question. My original conception was
> > a battle of stored energy underwater. Most ROVs (Remote operated Vehic)
> > use electricity for propulsion, but I was wondering if it would be more
> > practical to use compressed air. (ala WWII Torpedoes pre electric)
>
> Ah, a new constraint. As ambient pressure increases, the amount of
> energy you can get out of the tank decreases. Presuming that you want
> to use the compressed air to drive a turbine, the amount of energy you
> can get will decrease substantially as your ROV goes deeper. An
> electric battery does not have this limitation.
>
You mechanical guys are into some scary stuff, what with energy
estimates up to megajoules. Maybe we might be interested in finding out
how much energy is required to charge a tank. We might find the electric
power or gasoline needed to do the compressing, and then make a decent
guess at efficiency?
Another interesting point is that I think a turbine would be very
wasteful of the stored energy. Most ROV's will run slow in the water,
making a bellows or diaphram type motor more attractive. Anyway, one
nice feature would be that it's easier to slip in a new tank, while
underwater, than replacing the battery.
Paul Skoczylas
Ed Price wrote:
>
> You mechanical guys are into some scary stuff, what with energy
>estimates up to megajoules.
Surely you've heard of people knocking the tops off of gas
cylinders--the cylinder will literally fly through concrete walls.
Also, don't overestimate the value of a megajoule. 1 kWh (the energy
used by ten 100 W bulbs on for one hour) is a whopping 3.6 MJ. A 1/4 hp
motor running at 90% efficiency can run for only 20 minutes on 1 MJ of
energy.
> Another interesting point is that I think a turbine would be very
>wasteful of the stored energy. Most ROV's will run slow in the water,
>making a bellows or diaphram type motor more attractive.
By turbine I meant a device that converts pressure into mechanical
work. You're right that a bellows/diaphragm would probably be more
efficient than a rotary turbine.
>Anyway, one
>nice feature would be that it's easier to slip in a new tank, while
>underwater, than replacing the battery.
I question that.
Anyway, someone earlier in this thread said that pre-electric WWII
torpedoes ran on compressed air. I don't think that's true. I believe
they ran on steam--they would have had to carry compressed air and fuel
to burn together to create steam (obviously they wouldn't need to carry
water). This might be a more efficient way to carry energy. It's
almost certainly better than straight compressed air. It may or may not
be better than electricity--they didn't switch from steam to electricity
in torpedoes for efficiency, they did it because electric torps don't
leave a bubble trail on the surface.
-Paul
Paul F. Dietz
>Anyway, someone earlier in this thread said that pre-electric WWII
>torpedoes ran on compressed air. I don't think that's true. I believe
>they ran on steam--they would have had to carry compressed air and fuel
>to burn together to create steam (obviously they wouldn't need to carry
>water).
I believe they actually had steam loaded before launch, stored
in one or more internal pressure vessels.
Steam can also be produced by decomposition of hydrogen peroxide
(along with oxygen).
A modern gasless chemical heat source is the reaction of lithium
with sulfur hexafluoride.
Paul
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Gary Coffman
On Sat, 02 Aug 1997 01:49:23 GMT, di...@interaccess.com (Paul F. Dietz) wrote:
>On Fri, 01 Aug 1997 08:13:37 -0600, Paul Skoczylas
><P.Sko...@cfer.ualberta.ca> wrote:
>>Anyway, someone earlier in this thread said that pre-electric WWII
>>torpedoes ran on compressed air. I don't think that's true. I believe
>>they ran on steam--they would have had to carry compressed air and fuel
>>to burn together to create steam (obviously they wouldn't need to carry
>>water).
>
>I believe they actually had steam loaded before launch, stored
>in one or more internal pressure vessels.
Never heard of that. Subs in the steam torpedo era didn't have steam plants.
Maybe some torpedoes launched from surface ships may have worked this way,
though I don't see any advantage to using stored steam over using compressed
air.
>Steam can also be produced by decomposition of hydrogen peroxide
>(along with oxygen).
Right. And that was the normal power source of steam torpedoes.
They had two big drawbacks. They were very noisy, and they
left a bubble trail. Modern electric torpedoes have a chance of
sneaking in close to the target undetected before they make their
final attack run. Firing a steam torpedo gave away the location of
the sub and clearly indicated an attack was underway, often early
enough to allow the target to evade the attack.
Gary
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Grover Larkins
This might be a more efficient way to carry energy. It's
> almost certainly better than straight compressed air. It may or may not
> be better than electricity--they didn't switch from steam to electricity
> in torpedoes for efficiency, they did it because electric torps don't
> leave a bubble trail on the surface.
>
> -Paul
Huh?
Most modern torpedos use an internal combustion engine -- NOT BATTERIES!
How in the He&& are you going to catch a 50knot sub with a 25 Knot
electric Torpedo in a tail chase? Duhhh?
Grover Larkins
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Gregory Loren Hansen
In article <33E5F4...@imap.eng.fiu.edu>,
Grover Larkins <lar...@imap.eng.fiu.edu> wrote:
>Huh?
>
>Most modern torpedos use an internal combustion engine -- NOT BATTERIES!
An internal combustion engine?
BWAHAHAHAHAHA!
Sorry, I don't doubt that you're right. It just seems incredibly funny.
I suppose they have a tank of compressed air or compressed oxygen to keep
the engine running?
--
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Lift the seat before you piss.
Ed Price
Grover Larkins wrote:
>
> This might be a more efficient way to carry energy. It's
> > almost certainly better than straight compressed air. It may or may not
> > be better than electricity--they didn't switch from steam to electricity
> > in torpedoes for efficiency, they did it because electric torps don't
> > leave a bubble trail on the surface.
> > -Paul
> Most modern torpedos use an internal combustion engine -- NOT BATTERIES!
> electric Torpedo in a tail chase? Duhhh?
>
> Grover Larkins
Well, Grover, you don't need more than a few minutes run time. I think a
thermal battery could work pretty well in this application, especially
since you could probably afford a couple of hundred pounds of weight for
the battery. A torp must weigh a ton or so. Also, I would expect the
torp to have an exotic skin to minimize hydraulic drag.
Steve Spence
have seen a few web sites from military archives and manufacturers, that
reference both piston engined and electric (39mph) torpedoes.
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Gregory Loren Hansen wrote in article
<5s517n$s28$1...@dismay.ucs.indiana.edu>...
>In article <33E5F4...@imap.eng.fiu.edu>,
>Grover Larkins <lar...@imap.eng.fiu.edu> wrote:
>
>>Huh?
>>
>>Most modern torpedos use an internal combustion engine -- NOT BATTERIES!
>
Paul Skoczylas
Grover Larkins wrote:
>
> Most modern torpedos use an internal combustion engine -- NOT BATTERIES!
>
> electric Torpedo in a tail chase? Duhhh?
Perhaps you could try reading the posts that you respond to before
responding. I was talking about World War II torpedoes (and that was
pretty clear from the post I made). It's a simple matter of history
that the USN switched from steam powered torpedoes to electric powered
torpedoes at some point during the war.
I don't know nor care what they use now.
-Paul
Steve
> Steam can also be produced by decomposition of hydrogen peroxide
> (along with oxygen).
>
> with sulfur hexafluoride.
>
> Paul
Please Explain more about the "decomposition of hydrogen peroxide
(along with oxygen)." Reaction.
Do you just mix the two together and light it?
I remember some rocket engines were powered by hydrogen peroxide
but I never knew how exactly that worked.
Also as for a internal combustion engine, an engine is basically a air
pump so, take a typical RC Model Marine Boat engine
2 cycle say about .45 cu in displacement.
It puts out roughly 1-2hp@~~20,000rpm depending on fuel and make/model.
So if it displaces .45 cu in of air each revolution
that would be .45 * 20000 = 9000 cu in of air per minute
9000/1728= 5.2 cu ft of air per minute.
1 cu ft of air = 1728 cu inches of air. =
So this engine would basically consume a 80cuft SCUBA tank's air supply
at suface level in about (80/5.2=15 minutes.)
And that is a very small engine.
Larry Charlot
>Anyway, someone earlier in this thread said that pre-electric WWII
>torpedoes ran on compressed air. I don't think that's true. I believe
The torpedo referred to in this post is the US Navy Mark 14. It's fuel
was ethanol (grain alcohol), and oxidizer was pure O2, stored in a high
pressure steel cylinder. I don't know exactly what size the cylinder
was, approximately 42 cubic feet at 2400 psig, I think. Maximum range
at "slow" speed setting was about 12,000 yards.
Larry Charlot
To reply, remove "nospam" from address
<lcha...@nospam.jps.net>
Wim Schuerman
You could of course use the energy lost in the first stage to power a
dynamo, this might reduce the size of your batteries or increase the radius
or time of yur ROV.
--