
#1
Feb309, 07:56 AM

P: 30

can compressed air be used to lift a load of 400lbs about 30 feet into the air
i came up with using an air compressor to route compressed air into 4 small rocket nozzles to lift it off the ground how powerful of an air compressor would i need if i wanted a constant thrust of 500lbs coming out of 4 small nozzles (125lbs out of each nozzle)? if that is actually possible can the compressor be driven by something like a 50hp engine? ie. a compressor attached directly onto the engine? also are there any alternatives to lift a 400lb load while having a small footprint on the ground the only criteria being that the power required to get the load airborne should be less than 100hp and the total weight should be less than 180lbs 



#2
Feb309, 08:16 AM

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In simple terms to lift something with a rocket nozzle  the pressure at the open end * the area of the open end is the force. So if you had four nozzles with 1inch diameter and a pressure of 100psi, you would have 400lbs of lifting force.
Now how much power would it take to compress air to 100psi if it was flowing out of 4sq.inches of nozzle  



#3
Feb409, 07:07 AM

P: 30

really?
i thought if you had 100psi air coming out of 4, 1inch diameter nozzles you would only get 25lbs of thrust from each nozzle and that you would need 400psi to get the required 400lbs of thrust there are many air compressors which can compress air at 100psi although i havent found any that can constantly produce air at that pressure but isnt the more important thing the flow rate of the air coming out of the compressor, because when you look at the specifications for compressors, many can do high pressures but the cheaper ones have lower CFM (cubic foot per minute) and you would need a compressor which has enough flow rate to produce 100psi at the 4 nozzles wouldnt you? if i'm wrong about anything can you explain to me on why i am wrong? 



#4
Feb409, 08:59 AM

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Using Compressed Air to Lift a 400lb Load
Pressure is force * area so if you have 100psi and a 1inch^2 hole then you have 100pounds of force. The point is that you are going to need an awfull lot of CFM to keep 100psi with a 4inch^2 hole in something.
This is why lifting something with a compressed air jet purely using the reaction force is a bit impractical. A hovercraft which traps the air in a skirt, gets round this by using the pressure between the skirt and the ground  it is really more like a piston that a jet. 



#5
Feb409, 09:39 AM

P: 30

how can you work out the amount of CFM that will be needed?
at first i thought of going the hovercraft route as well but i had to let go of the idea because the load will have to be lifted to at least 10 feet off the ground do you know of any alternative ways of lifting the load? (without using large rotors) while still having a small ground footprint i appreciate you taking the time to help me 



#6
Feb409, 10:20 AM

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I haven't worked out the CFM but imagine that a pinhole can empty the 100psi in a road bike tyre in under a second a 4inch^2 hole is going to need a pump of the size normally hanging underneath airliner wings to maintain 100psi.
You need to suspend a 400lb load in mid air without using ropes? Then you choice is either a very large helium balloon or a helicopter 



#7
Feb409, 11:17 AM

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It takes more than just pressure to determine the thrust from a jet of air coming out of a nozzle. You'll basically need to know the mass flow rate, and the velocity of the air leaving the nozzle.
It sounds to me like you're trying to invent a flying machine that can carry you at about 30 feet off the ground with a nozzle pushing out compressed air at each corner; let me assure you that it will take a lot more than just a compressor and some nozzles to get something like that off the ground. 



#9
Feb409, 11:37 AM

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If you assume the air coming out of the nozzles has a valocity of 300 mi/hr, getting 500 pounds of thrust would require somewhere in the vicinity of 36.5 lb/s of mass flow, or roughly equivalent to 30,000 cfm of air flow. And just for a very rough guess, if you take the mass flow rate of the air and the velocity its travelling at, the air flowing through the nozzle has approximately 200hp of power hehind it. Depending on the efficiency of the device being used to propel the air (propeller, nozzle, whatever) the engine would have to be much more powerful than the minimum required power. 



#10
Feb409, 03:33 PM

P: 30

wow thats a lot of air, i never realised it would be so much
the device isn't specifically for lifting people but i see how you could swap out the load for a person what about 5ft diameter coxial rotors powered by a 100hp engine? or each rotor being powered by a 50hp engine? would that produce the 400500lbs of thrust needed? is there an equation where i can work this out? 



#11
Feb409, 05:24 PM

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P: 22,001

For low pressures, Bernoulli's equation and the fan laws will work. Velocity pressure times the area of the rotor gives you lift. The shorthand is (V/4005)^2=VP in inches of water and there are 2.31 feet of water per psi
Fan power (horsepower) = cfm * VP [in inches] / (fan efficiency) / 6365 For example, for a 5' rotor to get 500 lb of lift requires .177 psi of velocity pressure (force/area). That's 4.9" of water or 8,867 fpm or 174,000 CFM (using the first equation). Using the second equation and a 75% efficient fan, that's 178 horsepower. 



#12
Feb509, 10:08 AM

P: 30

this is probably a silly question but i'm curious:
if i had a propellor spinning in a duct and the thrust was forced to come out of a small diameter nozzle would it produce more thrust than if the nozzle was not there? i.e. if i had the 5 foot diameter rotor producing 500lbs thrust which had to come out of a 1 foot diamter nozzle would i get a larger amount of thrust or would it be of no use? on another note if you can't lift 400lbs with compressed air is it possible to lift it with pressurised water due to water being much more dense than air? where you have an engine pumping water via a pipe to lift the load 



#13
Feb509, 05:24 PM

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http://en.wikipedia.org/wiki/Water_rocket 



#14
Feb609, 04:15 AM

P: 30

well if thrust is being provided by water you would only be able to lift the load on a body of water, where the engine and the equipment needed will just be floating on the water while the load can be lifted up and be connected to the ground via a long pipe pumping the water to lift the load also i was looking of other ways of getting thrust and i ended up looking at paramotors which generate static thrust of over 200lbs, so if you attached 3 of them to the 400lb load wouldn't there be enough lift generated? 



#15
Feb609, 12:13 PM

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Just curious; is there any reason you can't use pneumatic cylinders?




#16
Feb709, 06:38 AM

P: 30

the load has to be lifted with thrust, not through pneumatic or hydraulic systems which are attached to the ground
what about my paramotor idea from my last post, even if you take away the weight of the paramotors you still have enough thrust left to lift up the load 



#17
Feb909, 08:36 AM

P: 30

i have a question:
say i have a prop which produces x amount of thrust with a 30hp engine at 3000rpm, would you get the same x amount of thrust if you swapped the engine for a 75hp engine running the prop at the same 3000rpm? also the thrust from a paramotor seems too good to be true because they are getting 230lbs of static thrust with 35hp engines and relatively small props (60 inches) am i missing something? 



#18
Feb909, 11:26 AM

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P: 5,095

The engine is there to provide the power the propeller needs to turn at the given speed. If you could turn the prop with a 30 HP motor, the only thing adding a 75 HP motor would do is give you 45 HP of excess power you're not using. The power requirement comes from the mass flow of the prop. If you don't change the prop or the speed at which it turns, excess power gets you nowhere except with more weight to carry.



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