Question:
72F tank air. 100psi
72F outside air.
72F engine block.
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Air into spark plug hole at top center.
Air decompresses and chills.
Help needed- driving in the UK?
15psi at 30F (just example)
What car should i but thats not to much like less then $50,000?
Stop the engine at bottom center.
All valves closed.
Air trapped in engine.
Come back a week later.
S type wiring on car for caravan?
Will the air be recompressed to 100psi and 72F?
Answers:
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Mythbusters did this, with a vaporizing wheel in a cold river. It turned itself but extremely slowly.What is the "safest" and most efficient car?
But what he is saying, the atmosphere is a source of heat which can do work on cold decompressed air.
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Now if he uses a('air to air' atmospheric) heat exchanger for the cold air leaving the engine and used a compressor to also add HEAT needed to reach full recompression, it may be practical.
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Remember the energy source of compressed air is the latent heat of the atmosphere(which is of course, the sun). 100% of the energy to access that energy is wasted (electric, chemical, whatever) as heat. Every air compressor is 200% ENERGY efficient like a heat pump.
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What would happen if that 100% wasted heat energy was used to recompress air? Is it possible?
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how about a car powered by the tides..forever people have tried to get something for nothing..didnt happen 2000 yrs ago,,isnt gona happen nowWhat are the options for a hard top, four seater sports car?
what turns the engine to draw the air in?what do you do with the compressed air ?
you are at the bottom of your stroke.
this motor is to slow to be practical
Air tight assumes an absolute that cannot exist. There would be bleeding within the system due to cooling and heating, variations in barometric and atmospheric pressures. This also assumes that all systems would remain the same. If this were absolute then the wait would be irrelevant whether a week or 100 years. Setting all systems at 72 degrees (I'm unsure how you arrived at that number.) would make it less stable, not more. Variation is the key, and all systems should be calibrated to within 10 tolerances of the preceding variable. In answer to your question, no the psi would not be 100 probably closer to 86 and the temp would probably be lower by 6 to 8 degrees. Of course this answer is also subject to variations dependent upon the nature of actual changes within the engine.
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Huh? How do you maintain the stable air temperature? The pressure is possible, but the temp would be a challenge. Why 72F? Is this some sort of magic temperature?What is the web address for the road charge petition?
You will only ever get as much energy out of the system as you put in it. There will always be a loss, by which I mean there will be energy out in a form you cannot use.What you are trying to do is create perpetual motion. This breaks the 1st law of thermodynamics, which basically states that you can't cook a potato by putting it in the freezer.
I think, from your question, that the pressure and temperature of the air in the tank don't matter. You have defined the cylinder as being filled at top center with 30F air at 15 PSI. Work is then done to move the cynlinder to the bottom position, with will reduce the pressure by the "compression ratio". For this example, lets assume 10::1.
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The gas formula is PV=nrT. For our purposes, the Pressure times the Volule divided by Temperature is constant. For this formula, all temperatures are absolute temperatures. We'll use the Rankine scale, which adds 459.67º to the Fahrenheit scale to give Rankine degrees.
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What is the volume of the cylinder? Lets assume a big engine, like a 409 cubic inch displacement V8. That gives us a volume of 51.125 cubic inches for the cylinder.
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We know the temperature, volume, and pressure at the beginning. We know the temperature and volume at the end, and want to know the pressure.
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Pi = Pressure initial
Vi = Volume initial
Ti = Temperature initial
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Pf = Pressure final
Vf = Volume final
Tf = Temperature final
Pi = 15 psi (given)
Vi = 5.1125 (from compression ratio)
Ti = 489.67ºR
Pf = unknown
Vf = 51.125 (from assumption)
Tf = 531.67ºR
(Pi * Vi)/Ti = (Pf * Vf)/Tf
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(15 * 5.1125) / 489.67 = (Pf * 51.125) / 531.67
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Solving for Pf gives
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Pf = 1.63 psi
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The temperature difference between 30 degrees and 72 degrees does not come close to making up for the volume difference between the top and bottom of the stroke.
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