Portable Power Best Solution Compare Discuss Evaluate

tahanson43206 · 2023-03-27 12:47:55

In this transcript, ChatGPT bravely attempted to find out how long my pneumatic tool would run if I invest in a 21 gallon system that charges to 175 PSI. The answer delivered is 24 minutes. The compressor in consideration is affordable, and it has received praise from over 400 verified purchasers. 24 minutes is less than my target time of 60 minutes, but since the tool would not be run continuously, 60 minutes at the job site is probably reasonable.

I'd appreciate someone who might be interested in the flow of computations to comment. I will point out that ChatGPT comes up with different answers depending upon the framing of the question.

hello again!
Hello! Welcome back! How can I assist you today?

Earlier today we were working on calculations to estimate the size needed for an air compressor. We decided to use the mass of air as a way to obtain measurements of flow rate. I have a new air compressor design to study. This one has a storage tank of 21 gallons, capable of holding air at 175 psi. The tool I have consumes 6.5 cfm at a pressure of 90 PSI. You calculated that the tool consumes 5.1 grams of air in one minute. I failed to ask you if that mass is for air at 1 Bar, or at 6 Bar. Please clarify the pressure at which 6.5 cubic feet weighs 5.1 grams at sea level on Earth.
I apologize for not specifying the pressure at which the mass flow rate of 5.1 grams per minute applies. The calculation was based on the assumption that the air pressure is at 1 bar or 14.7 psi.
To determine the pressure at which 6.5 cubic feet of air weighs 5.1 grams at sea level on Earth, we can use the Ideal Gas Law, which relates the pressure, volume, and temperature of a gas to its mass:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Assuming that the temperature is constant at 298 K (25°C) and that the air is an ideal gas, we can rewrite the equation as:
P = (nRT) / V
where n is the number of moles of air, which can be calculated from the mass and the molar mass of air (28.97 g/mol).
n = m / M
where m is the mass (5.1 g) and M is the molar mass of air (28.97 g/mol).
n = 5.1 g / 28.97 g/mol = 0.176 mol
Substituting the values into the Ideal Gas Law, we get:
P = (nRT) / V = (0.176 mol x 0.08206 L·atm/mol·K x 298 K) / (6.5 ft³ x 28.32 L/ft³)
P = 4.68 atm = 69.2 psi
Therefore, at sea level on Earth, 6.5 cubic feet of air weighs 5.1 grams at a pressure of 69.2 psi or 4.68 atm.

Just to confirm, please calculate the mass in grams of 6.5 cubic feet of air at 14.7 psi.
Sure! To calculate the mass of 6.5 cubic feet of air at 14.7 psi, we can use the same formula we used earlier:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Assuming that the temperature is constant at 298 K (25°C) and that the air is an ideal gas, we can rewrite the equation as:
n = (PV) / (RT)
where P is the pressure in atm (14.7 psi / 14.7 psi/atm = 1 atm), V is the volume in liters (6.5 ft³ x 28.32 L/ft³ = 184.08 L), R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
Substituting the values into the equation, we get:
n = (1 atm x 184.08 L) / (0.08206 L·atm/mol·K x 298.15 K) = 7.29 mol
The mass of air can be calculated using the molar mass of air (28.97 g/mol):
mass = n x M
where M is the molar mass of air (28.97 g/mol).
mass = 7.29 mol x 28.97 g/mol = 211.0 g
Therefore, 6.5 cubic feet of air at 14.7 psi and 25°C weighs approximately 211.0 grams.

What is the mass of 184.08 liters of air at sea level 14.7 psi in grams?
To calculate the mass of 184.08 liters of air at sea level and 14.7 psi in grams, we can use the Ideal Gas Law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Assuming that the temperature is constant at 298 K (25°C) and that the air is an ideal gas, we can rewrite the equation as:
n = (PV) / (RT)
where P is the pressure in atm (14.7 psi / 14.7 psi/atm = 1 atm), V is the volume in liters (184.08 L), R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
Substituting the values into the equation, we get:
n = (1 atm x 184.08 L) / (0.08206 L·atm/mol·K x 298.15 K) = 7.29 mol
The mass of air can be calculated using the molar mass of air (28.97 g/mol):
mass = n x M
where M is the molar mass of air (28.97 g/mol).
mass = 7.29 mol x 28.97 g/mol = 211.0 g
Therefore, the mass of 184.08 liters of air at sea level and 14.7 psi is approximately 211.0 grams.

Now please calculate the volume of the 184.08 liters of gas if we apply a pressure of 90 psi.
Sure! To calculate the volume of 184.08 liters of gas at a pressure of 90 psi, we can again use the Ideal Gas Law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Assuming that the temperature is constant at 298 K (25°C) and that the air is an ideal gas, we can rewrite the equation as:
V = (nRT) / P
where P is the pressure in atm (90 psi / 14.7 psi/atm = 6.12 atm), n is the number of moles of gas (which we calculated earlier to be 7.29 mol), R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
Substituting the values into the equation, we get:
V = (7.29 mol x 0.08206 L·atm/mol·K x 298.15 K) / 6.12 atm = 138.5 L
Therefore, the volume of 184.08 liters of air at sea level and 14.7 psi, if compressed to 90 psi, would be approximately 138.5 liters (or 3.97 cubic feet).

The tool I am using consumes 6.5 cubic feet per minute of air at a pressure of 90 psi. You have just shown that approximately 211 grams of air are contained in a volume of air of 138.5 liters at 90 psi. How long will the tool run if I supply 422 grams of air at 90 psi?
To answer this question, we can use the information that the tool consumes 6.5 cubic feet per minute of air at a pressure of 90 psi, and that you have just shown that approximately 211 grams of air are contained in a volume of air of 138.5 liters at 90 psi.
First, we can calculate how many grams of air are contained in 6.5 cubic feet at 90 psi:
6.5 cubic feet x 28.32 liters/cubic foot = 184.08 liters
Using the calculation from before, we know that 184.08 liters of air at 90 psi contains approximately 211 grams of air. Therefore, we can say that 6.5 cubic feet of air at 90 psi contains approximately 211 grams.
Next, we can use a proportion to calculate how long the tool will run if supplied with 422 grams of air at 90 psi:
6.5 cubic feet / 211 grams = X cubic feet / 422 grams
Solving for X, we get:
X = (6.5 cubic feet / 211 grams) x 422 grams = 12.97 cubic feet
Therefore, if you supply the tool with 422 grams of air at 90 psi, it will run for approximately 12.97 cubic feet / 6.5 cubic feet per minute = 1.99 minutes (or about 2 minutes).

we seem to be making progress! Thank you and your developers for your assistance. Here is the final question: Given a tank of 21 gallons, and a run time of 1.99 minutes for 422 grams of air at 90 PSI, what will be the run time if I fill the tank to 175 PSI and let it run down to 90 PSI?
To calculate the run time for the compressor given a 21-gallon tank filled to 175 PSI and allowed to run down to 90 PSI, we need to know the volume of air contained in the tank and the volume of air that the tool requires to operate.
At 175 PSI, the volume of air in the tank can be calculated as follows:
V = (nRT) / P
where V is the volume, n is the number of moles of gas, R is the gas constant, T is the temperature, and P is the pressure.
Assuming the temperature is constant at 298 K (25°C) and the air is an ideal gas, we can rearrange the Ideal Gas Law to solve for the volume:
V = (nRT) / P
= (21 gallons * 3.785 liters/gallon) * (175 psi / 14.7 psi/atm) * (1 atm / 14.69595 psi) / (0.08206 L·atm/mol·K * 298 K)
= 4.42 cubic meters
Now, we can use the information that the tool requires 6.5 cubic feet of air at 90 PSI, which we calculated earlier to be approximately 0.184 cubic meters, to determine the run time of the compressor.
The volume of air contained in the tank (4.42 cubic meters) divided by the volume of air consumed by the tool per minute (0.184 cubic meters) gives us the total run time of the compressor:
4.42 cubic meters / 0.184 cubic meters per minute = 24 minutes
Therefore, the compressor with a 21-gallon tank filled to 175 PSI and allowed to run down to 90 PSI should run for approximately 24 minutes before the tank pressure reaches 90 PSI.

Thank you once again! Best wishes for success as you serve your thousands of eager customers, world wide!
You're welcome! It was my pleasure to assist you. Don't hesitate to reach out if you have any more questions in the future. Have a great day!

(th)

kbd512 · 2023-03-27 16:08:18

tahanson43206,

The challenge is how to provide 2.4kW / 3.2hp, for one hour / "an hour", using a portable energy storage system, not "20 amps for an hour". Get a Lithium-ion battery. If all you need is 1 hour of power and 20 Amps of current, then the battery will do the job and it can be carried in a backpack for quite some distance. It'll weigh about 20kg with the power inverter to be operable over most temperatures found on Earth. Alternatively, get a chain saw motor. That will also produce 3.2hp for an hour, and it should weigh about 10kg if it needs to have an electric generator connected to it. If you can use the power directly and care more abou weight than specific voltages and amperages, then a 3.2hp gasoline motor can fit in the palm of your hand, but it won't last very long, so how many uses before we received our money's worth?

A Nitto Manufacturing Wankel model aircraft engine can produce 3hp and it weighs 1.493kg with an integrated cooling fan. It runs on Alcohol fuel. It will consume 100cc/min at WOT, so 1.585gph, or 10.769lbs of Ethanol and 3.291lbs for the Wankel engine, or 14.06lbs / 6.38kg in total. No load / idle rpm consumes 38.5cc/min. If you want it to be really quiet, then you'll need a muffler about the size of the engine. A nearly identical engine that's spark-ignited / gasoline fueled, but only 1.8hp, consumes 50cc/min. Both models are all-Aluminum construction and small enough for you to close your hand around the engine. There are similarly powerful and light Nitro-Methane single piston model aircraft engines available, but I only used Alcohol and gasoline, never Nitro. I would expect to pay about $250 to $400 for one of these engines. They were less expensive when I was a kid, but still not cheap. They're precision machined and can be reasonably durable if maintained well, but will rattle themselves apart unless you re-torque screws / bolts regularly and apply loctite. I would expect them to last for a couple hundred hours at most before taking them apart for cleaning and replacement of bearings. Clean fuel and clean air is a must. Dust and grit will ruin them. You need a little lubricant mixed into the fuel as well.

If the generator has to be operable in a mildly cryogenic environment with severe dust storms and no O2 in the atmosphere, like Mars, then a RTG can supply that kind of power while remaining "portable" using a hand cart.

The challenge is to get across a small river with a 20 foot board.

In my head that made perfect sense, but someone not reading through every prior post might be a bit puzzled by it.

By using it as a flotation aid?
As a pole vault over a very small river?
As a plank of wood to walk over the very small river?
How functional does the 20 foot board need to be (how many river crossings)?
Is the board thick enough for someone to walk on without snapping it in half (20 foot only tells me 1 dimension)?

I could expect every person to remember every detail of what I wrote or to go back to the first post in every topic I write about, or I could just reiterate salient details of what I wrote and be succinct in what I'm asking for if I'm going to both with reiterating part of what I wrote about in the OP.

tahanson43206 · 2023-03-27 18:59:21

For kbd512 re neat additions to the topic!

The lithium battery with inverter solution is excellent, and there are several complete systems available on the market. Apparently there are real universe customers for equipment at that scale, because I found several vendors.

The lithium battery solution has the distinct advantage of working quietly, if the work site calls for that.

The small engine/generator combination is also quite practical and there are an amazing number of vendors offering products in this category.

Your particular suggestions have the advantage of appealing to folks who need (or want) to build custom equipment.

In any case, if someone reads the topic from the top, they would find your suggestions to enliven their thought process.

(th)

kbd512 · 2023-03-27 23:26:18

tahanson43206,

What kind of power tools will you be using with this portable power generator, which will make less noise than the generator?

The engine would have to be pretty loud to be louder than a circular saw or a nail gun. A drill could be quieter.

In general, for tasks that require intermittent power and true portability, Wankel engines do pretty well. An all-Aluminum variant of the Mazda 13b Wankel engine block that can power a car is light enough to pick up with both hands, weighing in at about 110 pounds and providing up to 200hp in naturally aspirated form. Turbocharging a Wankel engine significantly quiets the exhaust note. A gas turbine engine would weigh about half of that, but would be louder still and consume even more fuel. A 40hp to 50hp Wankel engine variant made in Germany for go-karts and ultralight aircraft weighs about 35 to 50lbs, dependent upon optional extras added to it, such as alternators, the radiator assembly (it's liquid-cooled), and various reduction drive units for light aircraft and helicopters. The larger engines are better on fuel economy for the power they produce, because less fuel is used to cool the engine, as with air-cooled engines, where a lot of fuel is never combusted because it's consumed in the process of cooling the engine. Another German-based company makes 100hp Wankel engines for experimental light aircraft, although their engine may have recently achieved certification.

All Wankel engines consume a little bit of oil to lubricate their apex seals, so emissions are not the greatest. The apex seals and springs are the parts that normally fail first. There is no camshaft or valve train parts, which makes them much simpler than conventional piston engines. Their eccentric crankshafts almost never fail because they're not subjected to fully reversing stresses the way piston engines are, which is another way of stating that there's no reciprocating motion involved. They still experience torque pulses like any other piston-type engine. All small engines are a bit temperamental by nature. That's why I sold off all of my gas powered tools except for my lawn mower, for which no acceptable substitute exists (and it's a four-stroke rather than a two-stroke). I grew tired of messing with the carbs and tune. It's irritating. I replaced them with corded electric models because I don't want the irritation of batteries, which have proven equally troublesome. If some manner of extension cord handling apparatus existed for use with an electric lawn mower, I'd have one of those as well. As small engines go, Wankels are known to be finnicky, not that similarly powerful / small / light two-strokes are any better. Wankels require very precise machining, clean oil, clean fuel, clean air, and precise metering of fuel. If you do all those things and run the tool up to full operating temperature at least once per week, then the engine can last multiple years before disassembly and refurbishment is required. Fail to start it for 6 months during the winter and you could be rebuilding it before its usable again. Batteries are just very expensive bricks if you forget to charge them. Air-cooled piston aircraft engines work the same way. They need to be run hard on a routine basis for longevity.

One of the most harmful things you can do to engines or batteries is to let them sit unused. Air cooled engines need to be pickled and batteries need to be trickle-charged if you're going to let them sit unused for long periods. Corded electric tools and air tools don't have that problem. The air tools require a healthy coat of lubricant before you store them, but not the extensive preservation of a piston engine.

If you can't afford the weight and bulk of an air compressor or large battery, or have no practical way to recharge a battery, then gas powered tools make a lot more sense. Most construction sites make use of a mix of air-powered tools and corded electric tools, with a smattering of gas powered tools. The battery tools are less of a hassle to use, especially in tight spaces, but only if you can recharge them fast enough, and batteries are expensive. I would not want to try to use a gas or air powered tool in an attic or basement, for example. Drills for drilling through wood and impact drivers tend to be battery powered, except in automotive shops with shop air. Concrete saws / drills / jackhammers tend to be gas powered or pneumatic, as appropriate electric power cables would be quite large and heavy. Saws tend to be corded electric, with chainsaws being the only notable exceptions, but a good corded electric chainsaw will chew through just about anything. A gas or diesel powered generator on a pickup truck is used to power all of the tools. Here in America it's mostly gas powered generators until you get into heavy duty construction for large buildings or roadways, and then towed diesel generators are much more common.

tahanson43206 · 2023-03-28 06:33:16

For kbd512 re #54

Thank you for your substantial contribution to the topic!

I just scanned it because I am here to post a new update, and I will return later to read it more slowly.

At first impression, I ** think ** this is a contribution that would be worth a reader's valuable time.

This forum is competing with an effectively infinite number of other ways for folks to use their time on Earth.

(th)

tahanson43206 · 2023-03-28 06:53:34

After considerable back and forth, ChatGPT and I appear to have arrived at figures for a pneumatic tool set that look reasonable.

Given a reciprocating saw that consumes 6.5 CFM, and an air compressor that refills at 3 CFM, to a capacity of 21 gallons of air at 175 PSI, the expected run time is just over 12 minutes, and the expected refill time is 30 minutes.

The cost of the air compressor from a local vendor is reasonable, and ratings for the compressor are favorable from over 400 purchasers.

While the run time of 12 minutes might seem short, that is for operation of the saw making cuts. Moving work pieces into place would take much more time than the cuts, so 12 minutes of actual saw time looks reasonable to me at this point.

The tank has to be filled at the base site, and transported to the work site, along with tools to hold work pieces in place and miscellaneous other tools and supplies.

(th)

SpaceNut · 2023-03-28 19:34:49

A tank that is 1,000 gallon is 133.68 cubic feet. https://www.unitconverters.net/volume-converter.html

The PSI relate to how many of that volume are compiled in the same space.

When a tool uses the air, it is telling you how much has passed that is in a period of time that is at 1 bar or 14.7 psi.

The 175 psi to the regulated psi for the tool is all you get to use and then performance speed will drop. That also mean the useable cfm are also reduced.

SpaceNut · 2024-02-19 08:48:53

tahanson43206 wrote:

In a new topic Terraformer showed us a report on maritime application of thermal energy storage.
https://newmars.com/forums/viewtopic.php?id=10702
What is different about the report (from my perspective) is that for the first time, I have seen the idea of combining the two sides of the thermal energy storage concept in the same (hypothetical) system.
In the case of the maritime application, liquefied air is carried in an insulated tank, and a thermal energy supply is carried in another tank. The two are matched so that a supply of pressurized air is available to power an air motor or other pneumatic devices.
I would like to see this concept developed for the consumer market. The challenge I am offering our members is to design a practical energy storage system able to deliver 20 amps at 120 VAC for an hour in a mobile cart that can be transported to a job site.
A battery system able to provide that amount of stored energy is available in 2024 for ...
KILOVAULT
2.4 kWh KiloVault HLX+ Lithium LFP Solar Battery 12V
(No reviews yet)Write a Review
SKU:KLV2400HLXPLUSMPN:KLV2400HLXPLUSAVAILABILITY:SHIPS 1 - 2 WEEKSSHIPPING:$250.00 (Fixed Shipping Cost)
$1,995.00
The battery above would need to be combined with a converter ....
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3000 WATT PURE SINE POWER INVERTER 12 VDC to 120 VAC ETL LISTED
$567.00 $528.00
13 in stock (can be backordered)
Subtotal $528.00
The AIMS Power pure sine inverter is capable of producing 3000 watts of clean power ideal for running your sensitive electronics. Great for use in vehicles, boats, camping and emergencies where back up power is needed. The inverter features dual outlets, along with a USB outlet and a direct connect terminal block for hard wiring the inverter’s full capacity. An optional wired remote is available for this unit, AIMS Power part number REMOTEHF.
The package cost is about $2500 at this point. A complete system would include a mobile platform for the energy storage system, suitable to move the system on a reasonably flat solid surface with just one worker.
The web site offering the Inverter also carries miscellaneous items such as cables.
I think a reasonable outlay for a complete system would be $3000.
I'd like to see a thermal storage energy supply system come in under that level.
The energy supply system will need a LOX tank, a hot supply tank, an air motor and control electronics, and the same mobile platform.
Hopefully one or more NewMars members are able to show a design for a thermal energy storage system able to function at this level for under $3000 USD.
(th)

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#51 2023-03-27 12:47:55

Re: Portable Power Best Solution Compare Discuss Evaluate

#52 2023-03-27 16:08:18

Re: Portable Power Best Solution Compare Discuss Evaluate

#53 2023-03-27 18:59:21

Re: Portable Power Best Solution Compare Discuss Evaluate

#54 2023-03-27 23:26:18

Re: Portable Power Best Solution Compare Discuss Evaluate

#55 2023-03-28 06:33:16

Re: Portable Power Best Solution Compare Discuss Evaluate

#56 2023-03-28 06:53:34

Re: Portable Power Best Solution Compare Discuss Evaluate

#57 2023-03-28 19:34:49

Re: Portable Power Best Solution Compare Discuss Evaluate

#58 2024-02-19 08:48:53

Re: Portable Power Best Solution Compare Discuss Evaluate

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