Author Topic: energy storage (Read 8406 times)

pepa · « **on:** January 10, 2009, 07:17:45 PM »

Energy storage---

I need some help with a math problem.

I don't know how to figure this out and hope some of you would help. It is just an idea I playing with . Assume you have a storage of 500lbs. Of liquid or vapor storage and you filled the tank or tanks with 140psi of compressed air to pop off valve capacity, how long would it take to use it down to 70psi driving an air driven motor at 70psi. while Driving an induction motor a little over speed? Thanks pepa

tanner0441 · « **Reply #1 on:** January 10, 2009, 01:02:09 PM »

Hi

Your never going to find an answer to that, because it isn't a question. 500Lbs of one medium. 140Psi of air. What volume of air? Drawn off at what flow rate? what is the CFM of air required by your air motor? Is it all stored air or will it be replenished as you draw it off.

Perhaps if you try to explane what you are trying to do someone may be able to help, but if you ask a question you have to supply enough information to be able to work out an answer.

Try re wording your question.

Brian.

Tritium · « **Reply #2 on:** January 10, 2009, 02:53:20 PM »

Pepa, You need to supply the volume of the tank in cubic feet and the cubic feet per minute required by the air motor in addition as the information you posted. I suppose you are going to use a regulator in front of the motor set to 70 psi.

Thurmond

pepa · « **Reply #3 on:** January 10, 2009, 04:57:43 PM »

i have two 250 gallon propane tanks and one 60 gallon compressed air tank with with a twin cycle compressoe, another wheelborrow twin cycle compressor and three

one lung ten gal compressors. all five compressors will have their own windmill driving them with one compressor with a double lunger tied to grid for backup air. each compressor is capible of producing 140psi into the tanks. the motor that i have built works at 70psi through a regulator at the nozzle. what i need is a approx. time if no wind is avaible. i guess i can fill them up and bleed them back down at 70psi. i just figured that some of you might do the math because i never learned algebra. thanks anyway. pepa

pepa · « **Reply #4 on:** January 10, 2009, 05:12:28 PM »

thurmond...how do you get the cubic foot of 60 gal of lequid? i dont know, the tanks are marked in gal. thanks, pepa

kurt · « **Reply #5 on:** January 10, 2009, 05:23:31 PM »

60 gallon [US, liquid] = 8.020 833 333 3 cubic foot

http://www.onlineconversion.com/

pepa · « **Reply #6 on:** January 10, 2009, 06:38:23 PM »

think you kirt, i will work on it pepa

ghurd · « **Reply #7 on:** January 10, 2009, 06:53:48 PM »

I'd like to hear more about the whole plan.

Is the plan to direct drive the compressors, or run them from an inverter or DC?

I had it figured direct drive makes very little usable CFM in good wind with low tank pressure, or nothing in low wind with a higher pressure.

Best solution I could come up with is charge a battery with the windmill,

set up a dump controller with an inverter to run a fridge compressor.

Best math guesses I can find is the motor surge is lower than in a fridge, and the compressor will only pull about 75W running.

Should start and run from a cheap 800 ~ 1000W inverter?

Low CFM though, people tell me to expect more than 0.5, but probably less than 1.

About then is when my source of free NEW fridge compressors dried up.

G-

bob g · « **Reply #8 on:** January 10, 2009, 10:51:04 PM »

years ago i did research on windpowered compressed air systems, the concensus

at the time from various sources was an overall efficiency of about 60%

as opposed to flooded lead acid of between 80-85% efficiency.

having said that maybe the hit in efficiency is not a bad thing when you consider

that a good air resevoir if kept dry will likely outlast any battery technology, and

if left empty (uncharged) the tank won't be damaged from crystalline sulfation, overcharging is not an issue as long as you have a good redundant popoff valve system. in some ways air makes sense.

power density is the problem, air motors like a lot of air and at reasonable pressures

a common 1/2 inch air drill will consume about 4 cfm of air at 90 psi to produce

1/2 hp

i think it makes sense in certain circumstances, if you have a huge old propane tanker? maybe a large windpumping mill which makes a lot of torque but not much speed? and a large well made compressor that can be direct driven?

heat is the killer of efficiency with air systems, when you compress air you make heat, the power it equates to is generally useless.

now if you can fill the resevoir overnight when it is relatively cool, and then draw down the tank in the hot afternoon sun, and you paint the tank black

perhaps you might get closer to 70% efficiency?

very interesting, i would like to see such a system set up and running

bob g

Madscientist267 · « **Reply #9 on:** January 11, 2009, 12:15:57 AM »

No doubt - the physics alone paint a wild picture in my mind. This should be interesting.

Wouldn't just getting that kind of torque transmitted down the tower to a compressor, in and of itself, prove to be quite the task for a DIY implement? I would think that at the volumes and pressures being mentioned that this would make for some seriously hefty hardware in the pole department.

Sounds stupid, but I am assuming we're not talking 'flying' a compressor, right?

Steve

SparWeb · « **Reply #10 on:** January 11, 2009, 12:40:36 AM »

Oh goody, a math question, and I'm not too late to jump in!

(dyed-in-the-wool nerd, here, you know)

I can start with the two 250 gallon tanks. Together that's 67 cubic feet. Ignore the other ones because they don't add up to much by the time we get to the end of this.

When the 67 cubic feet are full, the pressure is 140 psi. Removing air until the pressure is 70 psi... that is half of the air. Pressure and volume are always proportional to each other, so 70/140*67 = 33.5 cubic feet of air at 70 psi.

If you have 33 cubic feet of air available, then using it at different rates affects how long it lasts. If you use it at 33 cubic feet per minute, then obviously it will last 1 minute. At 3 CFM, it will last 10 minutes. To last for one hour, you can only use 0.5 CFM.

I don't know how fast you expect to use it, but here's one more thing to think about:

1/2 CFM multiplied by 70 psi equals 114 watts of mechanical power (100% efficiency).

It won't take long for a windmill to fill your tank, but it won't take you long to empty it, either.

Speaking for myself, it would really be nice to find my compressor tank full, not empty, every time I go to use it, so it does sound like a handy idea. I might be doing something wrong, but any tank that I own/use/borrow ALWAYS self-discharges.

ghurd · « **Reply #11 on:** January 11, 2009, 06:24:51 AM »

I was told how much hp it takes to compress a given amount of air is...

.2267 x cfm x ((((P2/P1)+1)^.283)-1) = hp

P1 and P2 would be the starting pressure and ending pressure.

Is that about what you figured?

At the time I figured Bob G's 60% was somehow in the formula.

G-

bob g · « **Reply #12 on:** January 11, 2009, 06:29:49 AM »

personally i was thinking of flying a compressor

yes they can be quite heavy, but lets face it so is a decent size generator

also my thinking is that an air gennie would be a better match to a farm style pumping rotor, low rpm and loads of torque, with the vast majority on 4 legged towers that certainly could hold the weight.

one air swivel on the down pressure line would be all that is needed to get the power down the tower.

another system i have always wondered about is the use of a refrigeration compressor

and building up a windpower freezer or refrigerator. maybe a dual system where wind does most of the work and a standard electric compressor system could be called on for backup.

now that would be a very useful piece of work in my opinion considering that refrigeration often times is the one item that draws the most power in an offgrid home.

that might be more easily done with an angle drive and shaft brought down the tower

where the compressor could be mounted in the base of the tower.

i think that would be very cool, no pun intended

bob g

jonas302 · « **Reply #13 on:** January 11, 2009, 08:17:48 AM »

From what I can remember my compressor pulled about 4500 watts at the meter

thats a 3phase 10 hp the pump is 30cfm at 650rpm 175psi 60%duty cycle weighs 400 pounds put a good sized air buffer on it and it will barely shut off it would take on big set of blades to keep that full running a moter

For the original question we need to figure how many cfm the moter uses under load probly easiest to fill one of the tanks and time it

Ungrounded Lightning Rod · « **Reply #14 on:** January 11, 2009, 08:38:10 AM »

... the concensus ... from various sources was an overall efficiency [of wind powered compressed air storage systems] of about 60% as opposed to flooded lead acid of between 80-85% efficiency.

Compressed air has the problem that compressors and air motors are also heat engines. Compress the air and it heats up - you have to dump the heat. If the compressor runs fast you deliver hot air at a higher pressure than you need and its pressure reduces as it cools to ambient - wasting energy pumping heat up a temperature gradient. Similarly, air cools as it expands, lowering its pressure further (just as you need it). If you can't feed it ambient heat while it expands you again waste power, refrigerating your exhaust air. These are the main sources of inefficiency with a compressed air energy storage system.

= = = =

But pepa's scheme has an additional loss: The regulator. If he's storing air at 140 PSI and running his engine with 70 PSI regulated air pressure, he's throwing away half the energy in the regulator. It's exactly the same as taking, say, an amphour from a 24V battery, dropping the voltage to 12V with a resistor, and using the amphour at 12 volts rather than 24. (Except that the lost energy shows up as refrigeration of the air rather than heating of the resistor.) Assuming his compressors and air motor achieved a 60% overall efficiency (a big assumption, given that they probably weren't designed for high efficiency in the cycle where they're being used), this regulation step would cut it to 30%.

If you want to do this efficiently you need to design an air motor that can run at the "wild" range of pressures available from the storage tank rather than from regulated air pressure.

You'll also need to vary the air consumption - to match the load: An air motor wants to consume a particular amount of air per revolution and produce a torque proportional to the air pressure, while a genny wants to produce a voltage proportional to RPM and a load torque proportional to load current. This is a bad match. You can try to fix it by using an excited generator with controlled excitation (rather than a permanent magnet type or a synchronous induction generator pushed slightly above synchronous speed). But a better approach might be to design an air motor with a number of expansion chambers of varying sizes that can be switched in-and-out to let it run at a constant speed and vary its output torque (and air consumption) according to load (and air prssure, thus killing both birds with one stone).

Or you could build your wind farm 'WAY oversized and eat the inefficiencies. When the fuel is free the correct measure of "efficiency" is dollars per output watt. B-)

wooferhound · « **Reply #15 on:** January 11, 2009, 11:41:17 AM »

Great story about using the wind to compress air

http://www.fieldlines.com/story/2006/1/9/103933/9137

Talking about using different items as a compressor

http://www.fieldlines.com/story/2005/4/17/131823/172

More talk about compressing air with wind

http://www.fieldlines.com/story/2005/8/24/135518/795

http://www.fieldlines.com/story/2005/2/25/193846/326

http://www.fieldlines.com/story/2004/10/23/11533/778

pepa · « **Reply #16 on:** January 11, 2009, 01:50:17 PM »

i would like to think you woofer and everyone else for their input and help. i will type up a reply with an explation of what i am thinking about. pepa

dnix71 · « **Reply #17 on:** January 11, 2009, 08:30:22 PM »

If you can use the compressed air directly then you can win with a windmill air compressor. But using air to make electricity is not very efficient.

There are people trying to make a car run on 4500 psi air. It doesn't work and won't work. 4500 psi and 300 liters (80 gallons) only has about 16 kwh of potential energy, about the same as a third of a gallon of gasoline.

In the only real world test done with the car in Paris, it went about 4 miles (7 km).

A student with a degree in aeronautics made an air powered moped that would go 7 miles on twin tanks.

http://www.treehugger.com/files/2008/04/air-powered-motorcycle-diy-moped.php

Madscientist267 · « **Reply #18 on:** January 11, 2009, 09:00:58 PM »

I think theres a little bit of a misconception with the theory about the regulator:

If he's storing air at 140 PSI and running his engine with 70 PSI regulated air pressure, he's throwing away half the energy in the regulator. It's exactly the same as taking, say, an amphour from a 24V battery, dropping the voltage to 12V with a resistor, and using the amphour at 12 volts rather than 24.

Unlike a resistor or a 7812, there is no loss of energy (other than friction of the air passing through at the valve) in the regulator. An air regulator's valve closes in response to a balance of pressure opposing a spring, pushing from the other side, trying to keep the valve open.

Assuming everything is functioning as it should (ie no leaks etc), there is no loss of air (and therefore energy) to the atmosphere unless the regulator is tweaked toward a lower pressure than is currently in the load side of the regulator, causing the regulator to bleed off excess pressure in the line, until the air pressure and spring are once again back in balance. The only loss here is a function of the volume of air released and the difference in pressure.

It's not like a resistor where energy is dissipated in the act of restricting current, it just simply stops supplying air when the diaphragm reaches equilibrium with the spring. The analogy is therefore imperfect, as if compared directly to electronics, the effect is more like a switch mode power supply than a pass transistor.

Steve

SparWeb · « **Reply #19 on:** January 11, 2009, 11:14:32 PM »

Ghurd,

Your equation is good. The basic equation wants P2 to be the High pressure, and P1 to be the Low pressure.

0.2267 x (0.5cfm) x ((((140psi/14.7psi)+1)^.283)-1) / 60% = 0.178 hp = 133 watts. My estimation was much more off the cuff, so I low-balled it by using a much simpler + less accurate formula.

Any way you look at it, a small amount of energy. If there is some novel way of using it that has convenience that makes it worthwhile, then I'm still interesting in hearing about what Pepa has in mind.

BigBreaker · « **Reply #20 on:** January 12, 2009, 08:36:32 AM »

Compressed air is bad for energy storage but great for power storage. Weird statement huh? Well compressed air delivers energy very fast (high power) but not for very long. Numbers below from Wikipedia along with an example you can extrapolate from:

...it is useful to describe the maximum energy storable using the isothermal case, which works out to about 100 ln{Pressure_Tank/Pressure_Ambient) kJ/m3. Thus if 1.0 m3 of ambient air is very slowly compressed into a 5-liter bottle at 200 bar, the potential energy stored is 530 kJ (or 0.15 kWh).

A standard 200 bar 5 liter steel bottle has a mass of 7.5 kg, a superior one, 5 kg. Bottles reinforced with, or built from, high-tensile fibers such as carbon-fiber or kevlar can be below 2 kg in this size, consistent with the legal safety codes. 1 m3 of air contained inside such a full bottle has a mass of 1.225 kg (at 0°C). Thus, theoretical energy densities are from roughly 70 kJ/kg at the motor shaft for a plain steel bottle to 180 kJ/kg at the motor shaft for an advanced fiber-wound one, whereas practical achievable energy densities for the same containers would be from 40 kJ/kg to 100 kJ/kg.

pepa · « **Reply #21 on:** January 12, 2009, 12:16:31 PM »

Hello friends, the air motor was an accidental discovery while trying to build an inline motor for my in ground salt water swimming pool. I have two large high pressure pumps running about twelve hours per day in the winter and about sixteen hours in the summer and I was hoping to use some of that high pressure to turn a generator for a little power. I built a 14" disk Tulsa motor with six disk and a wooden housing split in half and carved to fit the disk, mounted disk on a piece of ¾" all thread for a shaft and two pillow blocks.

I had it set up and was fitting the top piece in place by sanding where the disk was rubbing the side and blowing out the dust with my blowgun. When I hit the disk with the air pressure at an angle, the thing took off to full speed in a heart beat without the top part being on, and a new idea was born.

I mounted a large compressor flywheel to each side of the motor shaft and it still spun up pretty quick and I grabbed the all thread to test power and the threads pulled my hand into the flywheel and I turned loose with only a small burn on my wrist, lots of power To it. I have seen working on a large vault for a long time and will still finish it one day but now for something to power the compressor. I have a good supply of long fiberglass poles and two of them are twenty inches in dia. I am cutting them down to thirty ft. and splitting them down the middle to create two thirty foot vaults that will be mounted on my shop roof ridge.

I am very lucky that my building is running at ninety degrees to my prevailing which is from the south east (ocean) and some big winds from the north west the half roof to vaults is 72'x32' for a total of 72'x64' of area to channel the air. It may not work but I wont know until I try. Thanks pepa. any comments wilcome and looked for.

you can see the shop ridge behind the first pole.

spinningmagnets · « **Reply #22 on:** January 12, 2009, 01:06:47 PM »

Dear Pepa, I have heard of engine-driven Tesla turbines being used as a pump for concrete because it allows lumpy fluid to flow in spite of its inefficiencies. Are you using a Tesla turbine because of salt-water corrosion in common-style impellor pumps?

To use it as an air-motor, as soon as any load is applied, efficiency will drop completely off the meter IMHO. A while back I wanted to learn whatever I could about turbochargers, they had to be spinning thousands of RPM's to do any work, and their impellors had many blades on the disc (as opposed to a smooth Tesla disc).

If an enthusiast had access to lots of free junk, and could store a large volume of air pressure from a wind-gen, then using an theoretically inefficient system to generate a couple of Watts couldn't hurt. I would still try a small salvaged 2-cylinder engine conversion as a PMA-motor if it was me.

Whatever you try, best of luck, sounds like fun!

Bruce S · « **Reply #23 on:** January 12, 2009, 02:58:22 PM »

Pepa;

I see the pup dog is still guarding the cement pond :-)

Since I was last out there , I knew of what you were talking about, but since work demands more time, I see you've updated your post.

Glad to read that you decided to got with multiple tanks inline. I knew that as smooth as the Tesla type was turning on our last visit you would get this going.

Cheers

Bruce S

Bruce S · « **Reply #24 on:** January 12, 2009, 03:24:28 PM »

All ;-)

If Pepa doesn't mind too much , let me see if I can help explain what he has built/ devised.

The motor is of the Tesla style.

His is built using 14 inch diameter carbide discs and is encased in wood.

These are mounted on an ALL thread that is very well balanced, so much so a quick twist will have it turning several turns afterwards.

While on a way too short visit, he had at the time only one flywheel on it and a pulley on the other side.

I know, these quick twist and several turns, comments aren't easy numbers to put to paper, but...

During our visit, we talked about having the blades turn the shaft of the 2-piston compressor. The blades at the time were to be attached directly to the shaft ( this may have changed). The "load as it were" will be how much air the system can compress back into the system causing something similar to a fully charged battery to load a 'mill down.

The compressor will have the down-link air, coming to the tanks, these could be considered as his batteries, he has two of the large tanks and several smaller tanks to be used as air storage ALL setup with 140psi pop-off valves.

He had/has in place one-way valves that will let the incoming air into the tanks, the dump control will be pop-off valves rated at 140psi.

The Tesla motor has as air pressure regulator set at 70 psi with a nozzle directed to the 14" discs for it to turn this is used to turn the PMA ( I think, he was going to first try an Amtek) using either direct wheel to wheel or pulley to get it up to needed RPMs.

PEPA: Hope you don't mind me putting this in.

Hope this helps other to see in their mind's eye what he is attempting.

Bruce S

Bruce S · « **Reply #25 on:** January 12, 2009, 03:27:38 PM »

PEPA:

Send a couple measurements and we can figure the volume a little better. Need to know the diameter of each tank , plus how high they are.

Cheers

Bruce S

pepa · « **Reply #26 on:** January 12, 2009, 04:34:59 PM »

welcome back my good friend bruce, i am a lot better at doing than i am at talking, i do thank you for helping me explain what i am up to and i will make a diary intry with pictures when i get time. to many projects to give each one the time it deserves but my mind builds these things at night and all day long. i believe i have been addected and i dont know how find a cure. i do think that i have come up with a in line pump idea that mite work for the pool system, thanks pepa

Ungrounded Lightning Rod · « **Reply #27 on:** January 12, 2009, 07:48:47 PM »

There's the loss of air from a high pressure to a lower pressure without any useful work being done. That's the same as the loss of electrons from a high voltage to a low one without any useful work being done.

If you had to pump that air from the 70 PSI pipe to the 140 PSI pipe you'd have to put in a bunch of energy to do it. You could get this energy back by running an air motor on it.

It's easier to see this way:

Build a SECOND 70psi pressure-difference airmotor - designed for higher-density air. Remove the regulator and put the second air motor in its place, with the input at 140 PSI and the output at 70 PSI.

Do you think you'd get no energy from this second airmotor? The energy you get from it is the energy that would be wasted by the regulator.

pepa · « **Reply #28 on:** January 12, 2009, 08:19:48 PM »

thanks ungrounded, i will try that. pepa

Bruce S · « **Reply #29 on:** January 13, 2009, 07:09:23 AM »

Hello My Friend :-)

Sure wish Amy and I could've stayed longer. Robin is looking to come along this year and maybe become squatters ;-) could do a little misquito fishing as well.

I will help all I can, send your grand-daughter out to get the measurements of each one of the tanks and we'll see what we can do. Get how high and how wide plus how many of each one, I don't have a memory of how many you had ready when we had to leave.

Did you decide to attach the blades directly to the shaft? we both knew it is certainly thick enough (3/4" or full 1" thick) or put the 10 foot "S" backup in the air, it'll turn that pulley system easy enough.

I also see you have the other 'mill backup in the air, good to see it flying again.

That pool sure look good right about now with ST.Louis at 18F :-/

Stay well;

Bruce S

pepa · « **Reply #30 on:** January 13, 2009, 09:15:43 AM »

hi Bruce, it's 65 and partly cloudy, good short sleeve working weather. we are looking forward to your next longer visit and meeting Robin. tell everyone hi for me, pepa.

ghurd · « **Reply #31 on:** January 13, 2009, 09:25:18 AM »

Someone is missing something. Probably me.

The air going from 140 PSI to 70 PSI increases in volume.

One CF at 140PSI increases to more than 1 CF at 70 PSI.

Almost sounds like a buck converter.

G-

Tritium · « **Reply #32 on:** January 13, 2009, 10:40:51 AM »

If half of the energy is lost in the regulator, what form does the loss take, heat?

I know this isn't true. The regulator is more like a PWM scenario than a resistor.

Thurmond

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Author Topic: energy storage (Read 8406 times)