To determine the pressure needed to blow air under 2 meters of water, you need to consider the pressure exerted by the water column as well as the atmospheric pressure at the surface.
Key Points:
- Water Pressure Calculation:
- The pressure exerted by a column of water can be calculated using the formula:
where:
- is the pressure (in pascals),
- is the density of water (approximately ),
- is the acceleration due to gravity (approximately ),
- is the height of the water column (in meters).
- Calculating Pressure at 2m Depth:
To determine the pressure needed to blow air under 2 meters of water, you need to consider the pressure exerted by the water column as well as the atmospheric pressure at the surface.
Key Points:
- Water Pressure Calculation:
- The pressure exerted by a column of water can be calculated using the formula:
where:
- is the pressure (in pascals),
- is the density of water (approximately ),
- is the acceleration due to gravity (approximately ),
- is the height of the water column (in meters).
- Calculating Pressure at 2m Depth:
- For 2 meters of water: - Total Pressure Required:
- In addition to the pressure needed to counteract the water, you must also account for atmospheric pressure (approximately 101.3 kPa at sea level). Therefore, the total pressure required from the compressor would be:
Conclusion:
To blow air under 2 meters of water, a compressor needs to generate a pressure of approximately 120.92 kPa (or about 1.21 atm).
The pressure exerted on a one square meter surface 2 meters under water effectively has the weight of two cubic meters of water pushing down on it. A cubic meter has a volume of 1000 liters and the original definition of one kilogram of mass was the mass of a liter of water, so we have the weight of 2000 liters of water with a mass of 2000 kilograms. But how much weight is this? Since weight is the product of mass and gravity, we need to multiply 2000 kilograms by earth’s acceleration of gravity which is 9.8 meters per second per second. This results in the water having a weight of 19,600 Newt
The pressure exerted on a one square meter surface 2 meters under water effectively has the weight of two cubic meters of water pushing down on it. A cubic meter has a volume of 1000 liters and the original definition of one kilogram of mass was the mass of a liter of water, so we have the weight of 2000 liters of water with a mass of 2000 kilograms. But how much weight is this? Since weight is the product of mass and gravity, we need to multiply 2000 kilograms by earth’s acceleration of gravity which is 9.8 meters per second per second. This results in the water having a weight of 19,600 Newtons of force on the square meter of area and a Newton per square meter is the definition of the S.I. pressure unit called a Pascal; thus, we have 19,600 Pascal's of pressure at that depth, or more practically, 19.6 kiloPascals usually abbreviated as “kPa” on pressure gauges.
19.6 Kpa = .193 atmospheres, 2.84 psi. Understand that these are all gauge, not absolute pressures.
To blow air under 2 meters of water with a compressor, you will need to apply a pressure of approximately 2 atmospheres, or about 30 pounds per square inch (psi). This is because the pressure at a depth of 2 meters under water is approximately 1 atmosphere (14.7 psi), and you will need to apply enough pressure to overcome that in order to force the air out of the compressor and into the water. It's important to note that the actual pressure required may vary depending on the specific compressor being used and the specific conditions under which it is being operated.
Pressure of water at a given depth is also called “head pressure”.
Head pressure is the same if you are lifting water up in a column or blowing air down a column. Either way you are displacing a distance of water in a vertical manner. Size of the column (diameter) makes no difference
So at the surface there is zero feet of water above or below you so the head pressure is zero. Since head pressure. For each foot of depth the head pressure is 0.433 Pounds per Square Inch (PSI), at 2 meters (6.56168 feet) the pressure is 2.84120744 PSI
If you want anything to come out of the pipe, you will need to provide a pressure greater than the ambient pressure that the pipe exit is seeing. That is the bad news.
The good news is, you only have to provide even 1/4 of one psi higher than water ambient pressure to achieve flow.
If you are at a depth where the water pressure is 2000 psi, and you push 2001 psi through the pipe, you will have flow
Do you need a pontoon that will withstand 2000 psi, in order to do deep-water salvage?
NOPE.
All you need are enough 2 liter soda bottles, empty, with a retaining collar on their necks to distribute the
If you want anything to come out of the pipe, you will need to provide a pressure greater than the ambient pressure that the pipe exit is seeing. That is the bad news.
The good news is, you only have to provide even 1/4 of one psi higher than water ambient pressure to achieve flow.
If you are at a depth where the water pressure is 2000 psi, and you push 2001 psi through the pipe, you will have flow
Do you need a pontoon that will withstand 2000 psi, in order to do deep-water salvage?
NOPE.
All you need are enough 2 liter soda bottles, empty, with a retaining collar on their necks to distribute the weight of the bottle’s potential lift. Once they are tied off to whatever it is that you want to salvage, you pump air into the open mouth of the soda bottle. You keep pumping air until the soda bottle is nearly full. After you have attached enough soda bottles, and inflated them with air, the item you are trying to salvage will float right off the sea bottom and head for the surface.
Don’t you have to regulate the pressure in the bottles as they lift, so they don’t explode?
NOPE.
As the bottles lift, the air inside will expand until it matches the outside ambient pressure. It will just bubble away as the object lifts. The total lift will remain constant at all times, and the pressure inside the soda bottles will remain exactly at ambient (whatever that might be at any particular depth) until the object reaches the surface.
All you have to do is wait on the surface, by the big bubble storm in the water, until the object you are trying to raise pops to the surface. Or rather, the plastic bottles that are supporting the object, because all you will see will be soda bottle butts bobbing in the water.
While working offshore on oil rigs in the 80’s they injected mud thru the drill pipe which was pretty thick as it was going down 300–500 ft of water before it hits ground with the big Rock drill head. The Derrick men took care of the mud pits onboard, they would run at high pressure and so did Halliburton's cement equipment to cap or plug the well after drilling and testing. The cement ran at much higher pressures due to its viscosity ,the mud was much lighter so took less but exactly how much I would guess 10,000 plus for cement and a few for the mud. It was used to make the well flow and coo
While working offshore on oil rigs in the 80’s they injected mud thru the drill pipe which was pretty thick as it was going down 300–500 ft of water before it hits ground with the big Rock drill head. The Derrick men took care of the mud pits onboard, they would run at high pressure and so did Halliburton's cement equipment to cap or plug the well after drilling and testing. The cement ran at much higher pressures due to its viscosity ,the mud was much lighter so took less but exactly how much I would guess 10,000 plus for cement and a few for the mud. It was used to make the well flow and cool the bit and pipe. If you were just pumping air it wouldn't require that much pressure without a material or substance included.
The first part William May’s answer is all you really need. In order for you to have any kind of movement you have to have in inbalance of force. The driection of movement willl allways be in favour of the higher force. Air is no differnt. If you have 2000 psa on both sides of the pipe, youo have no flow. Create an imbalance enen a minute amount and you have flow.
In your case, you want the air to come out into the water, so the higher force (PSI) has to be in favour of he air.
Now, the weight of the material (air, concrete etc) that goes down the pipe will contribute the the pressure at the exi
The first part William May’s answer is all you really need. In order for you to have any kind of movement you have to have in inbalance of force. The driection of movement willl allways be in favour of the higher force. Air is no differnt. If you have 2000 psa on both sides of the pipe, youo have no flow. Create an imbalance enen a minute amount and you have flow.
In your case, you want the air to come out into the water, so the higher force (PSI) has to be in favour of he air.
Now, the weight of the material (air, concrete etc) that goes down the pipe will contribute the the pressure at the exit of the pipe. In this case, the air pressure required the top of the pipe to create 2000 psi at the bottom will be proportionally less than the weight of the contents.
For example: if the pipe has a frontal internal area of 10 sq inches and the weight of the contents of the pipe is 10,000 lbs (10 000/10=1000), then I will only need 1000 psi air pressure at the inlet to result in 2000psi at the outlet.
Assuming (falsely, this is nowhere near true), you can get ALL of the energy going into the compressor to convert into energy in the tank, you’ll need to accrue 4.184 gigajoules (one ton of TNT). That translates by dividing by 3600 into 1.16 megaWatthours. Divide by 1490, and you get 780 hours, or 30-something days. But like I said, you will need to take into account how efficient (or rather inefficient) the compressor is. I’d hazard a guess at 30%. So make it about three months.
Now, if we look at compressed air as stored energy. I found that 5 liter at 200 bar holds 530 kJ. 4 GJ would be 4000
Assuming (falsely, this is nowhere near true), you can get ALL of the energy going into the compressor to convert into energy in the tank, you’ll need to accrue 4.184 gigajoules (one ton of TNT). That translates by dividing by 3600 into 1.16 megaWatthours. Divide by 1490, and you get 780 hours, or 30-something days. But like I said, you will need to take into account how efficient (or rather inefficient) the compressor is. I’d hazard a guess at 30%. So make it about three months.
Now, if we look at compressed air as stored energy. I found that 5 liter at 200 bar holds 530 kJ. 4 GJ would be 4000000/530, multiplied by 5 makes just short of 40000 liter - roughly the size of a large US tanker truck. 160 psi is about 11 bar, so a lot less than 200 bar, by a factor of approximately 18. Somewhere around 700000 liter, 700 cubic meter. Or about 18 large tanker trucks… If you make one round cylinder shape, it would be around 4 meter in diameter and 14 meter tall. Hopefully it holds pressure nicely, or you will be running more than 3 months for your destructive power.
I’m pretty sure that’s not entirely correct, because I’m sure whatever text-book you have will have the correct formulae and stuff in it, but I thought it would be fun to just rough-guess it with simple information that I could find and guess from experience.
Water squirts out the other side. It’s how a pressure washer works.
You thought water was incompressible, eh? Well it pretty much is, but that doesn’t stop the compressor raising the static pressure of the water so that it comes out of a nozzle faster.
Back pressure from the water is very low….use 0.433 psig per foot of depth (you can work out the conversions to your units). For comparison, the average person can blow at least 7 psig.
If you wish to get maximum effective stripping and minimize the time needed to accomplish it, then what you want is lots of very fine bubbles, not a few big ones. Fine bubbles maximize the air/water contact area.
It sounds like a pretty small scale operation. I haven’t had an aquarium for years, but I recall using a porous bubble stone to create those fine bubbles. Your tank is shallow, so I think an aquarium co
Back pressure from the water is very low….use 0.433 psig per foot of depth (you can work out the conversions to your units). For comparison, the average person can blow at least 7 psig.
If you wish to get maximum effective stripping and minimize the time needed to accomplish it, then what you want is lots of very fine bubbles, not a few big ones. Fine bubbles maximize the air/water contact area.
It sounds like a pretty small scale operation. I haven’t had an aquarium for years, but I recall using a porous bubble stone to create those fine bubbles. Your tank is shallow, so I think an aquarium compressor would do just fine. If you want to go upscale, look for a used compressor like the little ones made hobby air-brushing. In a pinch, you could use a compressed air tank, fill it up at a vehicle service station (free air), and meter it slowly to your air bubbler. I made one out of an empty propane tank.
Only if their thirsty. No they do not need water. They generate water as a byproduct of compressing air that has moisture in it.
The year was 1647, the location was France, and the guy's name is Blaise Pascal.
His law simply states that a pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. Yes, it says an incompressible fluid. While it is possible to apply pressure to a liquid, it really does not compress. As others have already pointed out, huge amounts of pressure do not result in significant compression. Some analysis suggests that in compressing a liquid, you are not so much compressing the liquid as the gases which ar
The year was 1647, the location was France, and the guy's name is Blaise Pascal.
His law simply states that a pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. Yes, it says an incompressible fluid. While it is possible to apply pressure to a liquid, it really does not compress. As others have already pointed out, huge amounts of pressure do not result in significant compression. Some analysis suggests that in compressing a liquid, you are not so much compressing the liquid as the gases which are dissolved in the liquid. In any event, it does highlight a few things:
You can indeed fill a rigid glass bottle with water (with as little dissolved gas in it as possible), cap it tightly with absolutely no air gaps/bubbles in the bottle (and with a cap that cannot leak or expand or deform due to pressure from within the bottle), and drive a nail with it. I *had* to try it and it worked. (I used a champagne bottle.) (Oh and eye, hand, and body protection.)
The compressor itself, very little until resistance is introduced into the equation - remember, pressure is resistance to flow. You take a mobile air compressor like used for a blower…
They’re not going to give a PSI/kpa rating because there’s no reservoir… only a CFM or LPM rating. Too much resistance to air flow, and it continues to run in an unloaded state, the way the same way the compressor for an air brake system does when the system reaches cut out pressure.
Systems with storage reservoirs, it will vary by application. An FMVSS 121 air brake system, the requirements are for it to be gover
The compressor itself, very little until resistance is introduced into the equation - remember, pressure is resistance to flow. You take a mobile air compressor like used for a blower…
They’re not going to give a PSI/kpa rating because there’s no reservoir… only a CFM or LPM rating. Too much resistance to air flow, and it continues to run in an unloaded state, the way the same way the compressor for an air brake system does when the system reaches cut out pressure.
Systems with storage reservoirs, it will vary by application. An FMVSS 121 air brake system, the requirements are for it to be governed between 115 - 135 PSI, and the governor is typically set at 120. Shop air compressors will be set in the same range.
Compressors for industrial machines, they could be set considerably higher.
If we’re talking about an air compressor, and speaking of the compressor itself and not the air tank, then it probably rusts inside. If not emptied out very soon, it probably rusts very badly inside. If it’s a piston-type air compressor and an attempt is made to start it while still full of water, that attempt may destroy the head, the valves, the piston(s), and/or the crankshaft.
If we’re only speaking of the compressor’s air tank, the working volume of the tank is diminished by the water inside it. The tank will probably rust somewhat inside. With luck, a conscientious maintenance person will
If we’re talking about an air compressor, and speaking of the compressor itself and not the air tank, then it probably rusts inside. If not emptied out very soon, it probably rusts very badly inside. If it’s a piston-type air compressor and an attempt is made to start it while still full of water, that attempt may destroy the head, the valves, the piston(s), and/or the crankshaft.
If we’re only speaking of the compressor’s air tank, the working volume of the tank is diminished by the water inside it. The tank will probably rust somewhat inside. With luck, a conscientious maintenance person will (cautiously) bleed off the water through the bleed plug or valve at the bottom of the tank, taking great care to not permit the drain plug (if so equipped) to escape and ricochet around the room, finally coming to rest in the least-accessible possible location in the shop after having grazed the maintenance person’s left eyebrow hard enough to draw blood. Hint: always blow down the pressure from an air tank before bleeding the water from that tank.
Just to give this thread a little substance, here is a video of the modern industrial blown glass process:
And another one showing the old way:
And to answer the question : one to three psi :
http://dallasglassart.com/our-process/#:~:text=Inflating%20hot%20glass%20does%20not,and%20lip%20of%20the%20piece.My first job was in the technical office of a bottle plant, so I clearly identify with the subject. The “finish” (mouth of the bottle) when hand blown, is nearly the last step, whereas in mass production it is formed first.
I derived the formula for hydrostatic pressure here:
Assuming the depth of water is 1m:
or
I noticed that the only good other answer here was by Andrew Chee and it was collapsed as needing improvement. Just saying . . .
Ideal gas law:
P (pressure) = 100psi = 689475.728 pascals
V (volume) = 5 gal = 0.0189271 cubic meters
n (amount of air) in moles
R (gas constant) = 8.3144621
T (temperature) = 80F = 310.928K
molecular weight of air is 28.97 g/mol
n = ( 689475.728 x 0.0189271)
/ (8.3144621 x 310.928)
n = 5.047880526352362 mols = 146.2370988484279 g = 0.32239761627 lbs
I changed the topic from physics to chemistry
Ideal gas law:
P (pressure) = 100psi = 689475.728 pascals
V (volume) = 5 gal = 0.0189271 cubic meters
n (amount of air) in moles
R (gas constant) = 8.3144621
T (temperature) = 80F = 310.928K
molecular weight of air is 28.97 g/mol
n = ( 689475.728 x 0.0189271)
/ (8.3144621 x 310.928)
n = 5.047880526352362 mols = 146.2370988484279 g = 0.32239761627 lbs
I changed the topic from physics to chemistry
The latent heat of fusion for water is 334 kj/kg which equals 143.5941531 Btu/lb. Since water has a mass of 8 lb/gallon that makes it 1148.7532248 Btu/gallon. Dividing 12000Btu by 1148.7532249 Btu/gallon makes it 10.446107781 gallons or 39.54281949 liters.
That's assuming I didn't miss calculate somewhere.
Not accounting for surface tension of the bubble, the pressure required would be approximately 1/2 PSI.
Not accounting for surface tension of the bubble, the pressure required would be approximately 1/2 PSI.
Given: Water Tank
Dimensions 3 m. x 2 m. x 2 m.(Assume Length,Width Height)
Half Filled with Water
Find: Pressure Exerted on Bottom (Pressure = Weight/Area)
Plan: Find water Volume, convert it to gallons of water and then to weight and divide by the bottom area.
Half Tank Volume = (3 x 2 x 2)/2 = 6 m^3
1 m^3 = 264.172 gallons Just look it up.
Therefore: 6 m^3 x 264.172 gallons/1 m^3 = 6 x 264.172 gals. ≈ 1585.03 gals (US)
1 gal = 3.785 kg => 1585.03 gals. x 3.785 kg/1 gal. ≈ 5999.34 kg. ✅
To get Pressure divide by tank base area: 3 m. x 2 m. = 6 m^2 ✅
Pressure(P) = Weight/Base Area = 5999.34 kg/6 m^2 =
Given: Water Tank
Dimensions 3 m. x 2 m. x 2 m.(Assume Length,Width Height)
Half Filled with Water
Find: Pressure Exerted on Bottom (Pressure = Weight/Area)
Plan: Find water Volume, convert it to gallons of water and then to weight and divide by the bottom area.
Half Tank Volume = (3 x 2 x 2)/2 = 6 m^3
1 m^3 = 264.172 gallons Just look it up.
Therefore: 6 m^3 x 264.172 gallons/1 m^3 = 6 x 264.172 gals. ≈ 1585.03 gals (US)
1 gal = 3.785 kg => 1585.03 gals. x 3.785 kg/1 gal. ≈ 5999.34 kg. ✅
To get Pressure divide by tank base area: 3 m. x 2 m. = 6 m^2 ✅
Pressure(P) = Weight/Base Area = 5999.34 kg/6 m^2 = 999.89 kg/m^2. Because of roundoff, let’s just say 1000 kg/m^2 ✅
Double Check: Reasonable/Recalculated ✅ ✅
Answer: 1000 kg/m^2 Approximately
Air compressors, for the most part, are powered by either gas or electric motors — it varies by model. At one end of the cylinder are the inlet and discharge valves. Shaped like metal flaps, the two valves appear at opposite sides of the cylinder’s top end. The inlet sucks air in for the piston to compress. The compressed air is then released through the discharge valve.
In certain air compressor models, the pressure is produced with rotating impellers. However, the models that are typically used by mechanics, construction workers and crafts people tend to run on positive displacement, in which
Air compressors, for the most part, are powered by either gas or electric motors — it varies by model. At one end of the cylinder are the inlet and discharge valves. Shaped like metal flaps, the two valves appear at opposite sides of the cylinder’s top end. The inlet sucks air in for the piston to compress. The compressed air is then released through the discharge valve.
In certain air compressor models, the pressure is produced with rotating impellers. However, the models that are typically used by mechanics, construction workers and crafts people tend to run on positive displacement, in which air is compressed within compartments that reduce its space. Even though some of the smallest air compressors consist of merely a motor and pump, the vast majority have air tanks. The purpose of the air tank is to store amounts of air within specified ranges of pressure until it’s needed to perform work. In turn, the compressed air is used to power the pneumatic tools connected to the unit supply lines. While all of this is going on, the motor repeatedly starts and stops to keep the pressure at a desired consistency.
What the piston effectively does with its back and forth movements is create a vacuum. As the piston retracts, the space in front gets filled with air, which is sucked through the inlets from the outside. When the piston extends, that same air is compressed and therefore given the strength to push through the discharge valve — simultaneously holding the inlet shut — and into the tank. As more air is sent into the tank, the pressure gains intensity.
Using the ideal gas law, PV=nRT, and assuming temperature is permitted to equalize, pressure P is inversely proportional to volume V. Since the pressure is raised by a factor of 200, volume is reduced by a factor of 200. The beginning volume of the air must therefore be 200 times the ending volume, or 200*0.45L = 90L.
Note the ideal gas law doesn’t provide an exact solution because it assumes the limit of low density. A more precise result would require empirical tables or a more sophisticated equation.
Water is essentially incompressible. Trying to put an incompressible fluid through a positive displacement machine, as most compressors are, will lead to the destruction of the machine. Sure, you can increase the pressure but so what? If you are thinking of a hydraulic accumulator, it is not the fluid that is compressed, but air that exerts a force on the fluid.
Is the answer .38 lbs? (114.7psia)(.66cuft) /14.7psia= 5.15cuft * .0731lbm/ft3 = .38lbs (really not sure if I did it right)
Industrial air compressors are robust machines engineered for heavy-duty applications in industrial settings, providing compressed air for powering pneumatic tools, assembly lines, and manufacturing processes. These compressors are characterized by their large size, high capacity, and pressure ratings, making them suitable for continuous and demanding operations. Designed to withstand harsh conditions, industrial air compressors are essential for maintaining productivity and efficiency in various industrial sectors.
The pressure generated by an air compressor varies, depending on what it is used for. On trucks, equipped with air brakes around 100 psi. Stationary compressor used to supply pressure to handheld or fixed tools or equipment may be considerable higher, like 125 or more. The design of the compressor and the pressure relief valve setting, determines what the outlet pressure will be.
Adrian D.
first of all pressure less than 1 atm is termed as vacuum formation at the surface of the earth,which are the reasons for the destructive windy calamity that occurs around us, the temperature differences of air causes them.
A compressor is a device that take the air from atmosphere and collects it inside it self in the pressurised form which is used for a lot of purposes, a compressor just takes air from the atmosphere which is free it cannot create it's own air, and the compressor tank is considered empty when it reach 1 atm or the surrounding air pressure as it cannot deliver that air as it w
first of all pressure less than 1 atm is termed as vacuum formation at the surface of the earth,which are the reasons for the destructive windy calamity that occurs around us, the temperature differences of air causes them.
A compressor is a device that take the air from atmosphere and collects it inside it self in the pressurised form which is used for a lot of purposes, a compressor just takes air from the atmosphere which is free it cannot create it's own air, and the compressor tank is considered empty when it reach 1 atm or the surrounding air pressure as it cannot deliver that air as it will create vacuum inside it ,to do it external device should be used to create the vacuum in the tank.
The idea of the question is not clear. If the suction depth is high then compressed air assisted pumping is commonly used. In this case, the compressed air will force the water to go up at the bottom and this is also called jet pump. In this case, in addition to water pipe compressed air piping also laid in parallel till the water suction point. In this case, we need electric power to run normal pump and as well as the compressor.
In another case, in place of electric motor as a prime mover to run the impeller of the pump, air motor can be used as a prime mover to run the impeller of the pump.
The idea of the question is not clear. If the suction depth is high then compressed air assisted pumping is commonly used. In this case, the compressed air will force the water to go up at the bottom and this is also called jet pump. In this case, in addition to water pipe compressed air piping also laid in parallel till the water suction point. In this case, we need electric power to run normal pump and as well as the compressor.
In another case, in place of electric motor as a prime mover to run the impeller of the pump, air motor can be used as a prime mover to run the impeller of the pump. This is very very rarely used.
This question cannot be answered as written. The question does not specify the time, or pressure. Is the water to be pumped to a height of a kilometer, or down 5 meters? is it to be pumped in an hour, or a day? Does it have to be pumped out of the ground first? Define the parameters of your problem before you ask for it to be solved.
Which one of those dimensions is the depth? The formula is then rho(density of water) x g(acceleration due to gravity) x h(depth or height of the water). The other dimensions have no effect. Assuming the tank walls are upright then the centre of pressure acts 2/3 down from the surface or 1/3 up from the tank bottom.
It's all on the web…
Pressure increases by roughly 1 bar per 10m.
Given that (and the ability to convert between archaic to modern units of measurement) you can calculate the pressure at any depth.
If you are trying to control the dissolved gases (e.g. remove CO2, add O2) then the issue is not the depth as such, but the surface area and the circulation. Chemical engineers commonly use a ring with holes in it to get small bubbles going upwards. On a small scale, porous stones or plastic can create bubbles.
If the tank is very wide you may need to provide some circulation.
Yes there will be water, which could be taken out from tank after air is released.
Usually compressors have auto drain valves which drain water regularly due to pressure in tank.