didier
Junior Partner
Posts: 29

Post by didier on Sept 22, 2012 1:44:09 GMT 8



Post by Uncle Buddy on Sept 22, 2012 5:54:47 GMT 8
Thanks Didier. I corresponded with him several years ago when he was getting a patent. He is a smart inventor. He had been working with a university in Texas and then went back to Uruguay. Will read with pleasure.
His engine has constant torque output, like the torquerack engine on my website, but uses chains and sprockets instead of gears to do the same thing. No crankshaft in other words, but instead constant maximum torque.


greatballoffire
Deleted Member
Posts: 0

Post by greatballoffire on Sept 26, 2012 12:48:23 GMT 8
Thanks for posting another great interview. It definitely was more reality based than the interview with MDI that you posted. MDI's calculated numbers just seem out of place where as the numbers here are much closer to the real world. Thanks for sharing. Mark



Post by tommy on Sept 26, 2012 17:29:52 GMT 8
Many Thanks Didier, for posting this interview. I am very grateful, this information is very helpful. Very good, thank you again!


didier
Junior Partner
Posts: 29

Post by didier on Oct 11, 2012 12:36:32 GMT 8
You're wellcome. I wish I could do the same now with Angelo di Pietro, but it looks like he's too busy to bother. Also I have now added new information on the news page of www.aircars.tk about the Peter Dearman engine running on liquid air.



Post by tommy on Oct 12, 2012 12:58:35 GMT 8
Excellent website, Thank you for the link!



Post by tommy on Feb 22, 2013 17:14:06 GMT 8
Hello Didier, Have spent some time studying Armando Regusci's website, but have an unanswered question. I was wondering if you might have some avenue to obtain an answer? His calculations are based on the energy content of the compressed air in the tank, and that energy content per liter is based on the formula E=10(xLxx+1). I think I figured out that the 10 multiplier is to correct the units since the pressure is in Kg/cm^2 and the energy is in Kgm and it is per liter. However, I do not understand where the L value is obtained. This would be a most helpful answer to help us to understand his most interesting work. Any information on the derivation of this formula would also be most helpful. Thanks much!



Post by Uncle Buddy on Feb 23, 2013 3:02:32 GMT 8
What's the rest of the formula mean? What's x refer to? Is that x times L with subnotation x? Is L referring to the number of liters?
My guess is that E should equal PV or pressure times volume, so all that other stuff must equal PV?
PV = force times distance
P times area = force
Is L referring to length? Or is it a Spanish initial?



Post by Uncle Buddy on Feb 23, 2013 3:05:37 GMT 8
Subscribing to this thread. I don't like the way proboards makes us remember to subscribe to each thread individually, and you have to post in order to do it. That is gonna keep lurkers from lurking loyally because they don't post so they're not subscribed to anything. Hmmm....



Post by tommy on Feb 23, 2013 11:21:11 GMT 8
What's the rest of the formula mean? What's x refer to? Is that x times L with subnotation x? Is L referring to the number of liters? My guess is that E should equal PV or pressure times volume, so all that other stuff must equal PV? PV = force times distance P times area = force Is L referring to length? Or is it a Spanish initial? Correct, I would think that E is Pressure x Volume in this case. The x is the absolute pressure in Kg/cm^2 which is same as atm. The 1 is subtracting the 1 atm to get guage pressure. But the formula still does not make sense to me. The value for L changes depending on what the beginning pressure is in the tank. Also this formula is based on 1 liter, so the volume may not appear since it is =1. Armando states this formula is based on Boyles Law so I would assume he is taking the PxV in the tank and subtracting the PxV when fully expanded to the environment, and ignoring any losses. But, the formula still does not make sense, to me anyway.



Post by Uncle Buddy on Feb 23, 2013 16:11:38 GMT 8
Is this the formula?
E=10(xL_{x}  x + 1)
I don't see the " 1" you refer to.
Are you saying that x = atmospheric pressure, measured in kilograms per square centimeter or Kg/cm^{2}?
I'll look for an equivalent statement somewhere of the energy contained in compressed air, and compare the two formulas.



Post by Uncle Buddy on Feb 23, 2013 16:41:22 GMT 8
...look for an equivalent statement somewhere of the energy contained in compressed air, and compare the two formulas... Let me briefly step up on my soapbox. Googling "energy in a compressed air tank" should not get 5 results. It should get thousands. This is the quandary we are in. While the world has gone crazy on high tech, nanotech, biotech, genetically modified zombiebrained tenureseeking everupward "growth is everything" progresshounds, we have forgotten to take an interest in basic, simple solutions, especially compressed air. Finding usable info on air is sometimes very difficult. It's like, Mr. Scientist gets a new hightech chainsaw for Christmas so he calls up all his friends and they're gonna go mow down a field of daisies with it while shooting paintballs at each other. That is our exact quandary. But I did find a version of the formula, maybe. There are bound to be others in my notes or books somewhere so I'll keep looking. This is what I do when I don't understand a formula. Since it's no good guessing, I find at least two different versions of the same formula and work out the differences among them, why they purport to say the same thing, but look like different formulas. By the time I figure that out, I understand them both and they turn out to be the same formula. This one might not be useful but I'll quote a whole section so it's not out of context too much, maybe it will answer the questions it raises. This was written by air_car_dude aka Charlie at the airpoweredvehicles forum on yahoo, files section. The whole document is 5 or 6 pages long. I haven't digested his paper yet because it's all in SI units and I'm wired for footpounds and horsepower and stuff like that. tech.groups.yahoo.com/group/airpoweredvehicles/files/POWER STORED IN A COMPRESSED AIR TANK Power (in kilojoules) = 100 * Volume1 * ln(P2) Where Volume1 = the volume at 1 bar P2 = the final pressure ln is the natural log Example: 300 liter tank at 300 bar. Assume starting pressure is 1 bar. 300 liters is 0.3 cubic meters internal volume. For an ideal gas, the starting volume at 1 bar would simply be 0.3 m^3 * 300 bar = 90 cubic meters. (Air is fairly close to an ideal gas up to 200 bar, and then deviates from ideal by 3 or 4 % by 300 bar, in the direction such that the stored air volume is a few percent less than predicted for an ideal gas. I choose to ignore this few percent of error.) The ln ( 300) is 5.7 So power in kJ in 300 liters at 300 bar = 100 * 90 * 5.7 = 51, 300 kJ or 51.3 MJ since 1 kWhr = 3.6 MJ, 51.3 MJ = 14.25 kWhr. Assuming that the air in the tank is at room temperature, this is the maximum amount of energy than can be extracted by an air engine operating at room temperature with 100% theoretical efficiency. Since the energy stored at a given pressure is directly proportional to the internal volume of the tank, the following table can be used to calculate the energy stored in tanks various pressure. The weight of the air (just the air, excluding the tank) is also included. Figure are rounded to 0.1 MJ and 0.1 kg. 100 liters @ 200 bar = 10.6 MJ and weighs 24 kg 100 liters @ 300 bar = 17.1 MJ and weighs 36 kg 100 liters @ 350 bar = 20.5 MJ and weighs 42 kg 100 liters @ 400 bar = 24.0 MJ and weighs 48 kg 100 liters @ 450 bar = 27.5 MJ and weighs 54 kg * 100 liters @ 700 bar = 45.9 MJ and weighs 84 kg * * a more accurate calculation at these pressures would take into account the increasing deviation of air from the ideal gas laws at very high pressures. This table can be used to very simply calculate the energy stored in other sizes of air banks. For example 175 liters at 450 bar will have 1.75 times the energy as a 100 liter bank, or 1.75 * 27.5 MJ or 48.1 MJ. If energy storage is desired in kilwatt hours, the conversion is 1 kWhr = 3.6 MJ.


newt
Air Enthusiast
Posts: 87

Post by newt on Feb 24, 2013 14:16:46 GMT 8
Inspired by the Unc's maths textbook, and his explanation that cfm/hp is a measure of efficiency, I've had a go at comparing claimed ranges of depetro, MDI and regusci to calculate km/m3:
Vehicle Depetro Regussi MDI Tank volume litres 30 56 260 Tank pressure bar 207 229.5 248 Range km 100 52 120 Free air equiv m3 6.2 12.9 64.7 Efficiency km/m3 16.0 4.0 1.9
Sorry for not showing my working out, will try to next time.
Also aircaraccess has some really useful and interesting stuff on heating compressed air and that it is 10 times more cost effective to heat compressed air than to compress more air but Unc joked about heating with peanut oil or whatever the greenies are recommending. Surely some kind of biomass heating is a good idea? Better than MDI's petrol burner? although the petrol consumption is very low for the range it can get. Also does anyone know about any more recent attempts to design or build or use internal combustion reheaters as outlined in "reheating compressed air" from aircaraccess?
Thanks



Post by Uncle Buddy on Feb 24, 2013 17:32:56 GMT 8
Newt, thanks for your contribution, that's great. I used to do the math for Terry Miller's air car all the time, and found that his engine's maximum theoretical horsepower was 5 hp or less. The engineer who later tested his engine on a dynamometer got 1.5 hp out of it, but I think he could have gotten better if he'd wanted to. Terry claimed a range of 42 miles, and my calculations said slightly less. So this kind of math is worth the effort, it gives you an idea of what to expect from a machine. It is also always bound to be flawed and/or less than complete in what it can predict.
The next step is to become as smart as someone like Bill Truitt who claims to have "learned how air works" by doing it. That took him more than 50 years of experimentation.
A reheating device is always external combustion, I believe, since the combustion is not taking place in the piston cylinder. Depending on design, this kind of combustor can burn anything from sawdust to liquid fuel. Part of what's wrong with the internal combustion engine is the attempt to burn stuff inside a chamber of changing volume.
Newt, there is an interesting study in my online books somewhere by the Indian Institute of Science in Bangalore. It came out in the 1950s and was published in two or three detailed papers. It is about compressed air engines combined with lots of added heat. You might want to check it out. It's the only thing I know of that's been done since the heyday of the air loco, regarding using lots of added heat. The results were good, if I remember.
Let me know if you can't find it.
One of the first engineers I consulted had the task of informing me that compressed air wasn't about pressure alone. He really busted my bubble when he told me that the amount of air I wanted to store in my car to make it go 200 miles between fillups would weigh a ton. I had always assumed that the weight of air is negligible. Not so!
He also informed me that compressed air is all about heat. It only took me a couple more years after that to figure out the the heat of our atmosphere is hundreds of degrees warmer than absolute zero, essentially making it a huge air tank, a tank of air stored at 14.7 psia or less, depending on elevation. I guess that would make gravity the walls of the tank.


newt
Air Enthusiast
Posts: 87

Post by newt on Feb 25, 2013 10:42:09 GMT 8
Unc, thanks for your tip off for more reading on heat. I'll look for it and put in some time to read it.
I thought the difference between internal and external combustion is where the exhuast goes. In external combust it goes up a chimney and its heat is lost, with internal combustion it goes through the expansion motor. Yes standard ICE s are piston based with combustion in the cylinder but I've re read your heating compressed air document and there are three examples of internal combustion there!
The original oily rag heater for rock drills was internal combustion, the Edison reheater is int comb and a "type III internal combustion reheater" on p27. There are also external combustion reheaters there.

