How many times will the scuba diver’s air consumption change? Physical foundations of diving descents. Laws of hydrostatics and hydrodynamics
Tasks
Solution.
Solution.
Examples
A 20 liter oxygen cylinder is under pressure
10 MPa at 15 ºС. After some of the oxygen was consumed, the pressure dropped to 7.6 MPa and the temperature dropped to 10 ºС.
Determine the mass of oxygen consumed.
From the characteristic equation (2.5)
Consequently, before oxygen was consumed, its mass consisted
kg,
and after consumption
kg.
Thus, oxygen consumption
ΔМ = М 1 –М 2= 2.673 - 2.067 = 0.606 kg.
Determine the density and specific volume of carbon monoxide CO at a pressure of 0.1 MPa at a temperature of 27 ºС.
The specific volume is determined from the characteristic equation (2.6)
m 3 /kg .
Carbon monoxide density (1.2)
kg/m3.
A cylinder with a movable piston contains oxygen at
t= 80 ºС and vacuum (vacuum) equal to 427 hPa. At constant temperature, oxygen is compressed to excess pressure
p out= 1.2 MPa. Barometric pressure IN= 933 hPa.
How many times will the volume of oxygen decrease?
Answer:V 1 / V 2 = 22,96.
In a room with an area of 35 m2 and a height of 3.1 m, the air is at t= 23 ºС and barometric pressure IN= 973 hPa.
How much air will penetrate from the street into the room if barometric pressure increases to IN= 1013 hPa. The air temperature remains constant.
Answer:M = 5.1 kg .
A vessel with a volume of 5 m3 contains air at barometric pressure IN= 0.1 MPa and temperature 300 ºС. The air is then pumped out until a vacuum pressure of 80 kPa is formed in the vessel. The air temperature after pumping out remains the same.
How much air has been pumped out? What will be the pressure in the vessel after pumping out if the remaining air is cooled to a temperature t= 20 ºС?
Answer: 2.43 kg of air was pumped out. After cooling the air, the pressure will be 10.3 kPa.
The air heater of the steam boiler is supplied by a fan with 130,000 m 3 /h of air at a temperature of 30 ºС.
Determine the volumetric air flow rate at the outlet of the air heater if it is heated to 400 ºС at constant pressure.
Answer:V= 288700 m 3 /h.
How many times will the gas density in the vessel change if, at a constant temperature, the pressure gauge reading decreases from p 1= 1.8 MPa up to p 2= 0.3 MPa?
Take barometric pressure equal to 0.1 MPa.
Answer:
A vessel with a volume of 0.5 m3 contains air at a pressure of 0.2 MPa and a temperature of 20 ºC.
How much air must be pumped out of the vessel so that the vacuum in it is 56 kPa, provided that the temperature in the vessel does not change? Atmospheric pressure according to a mercury barometer is 102.4 kPa at a mercury temperature in it equal to 18 ºС. The vacuum in the vessel was measured with a mercury vacuum gauge at a mercury temperature of 20 ºС.
Answer: M= 1.527 kg.
Often we have to solve problems in which not individual gases are considered, but their mixtures. When mixing chemically non-reacting gases having different pressures and temperatures, it is usually necessary to determine the final state of the mixture. In this case, two cases are distinguished (Table 1).
Table 1
Gas mixing*
| Temperature, K | Pressure, Pa | Volume, m 3 (volume flow, m 3 / h) | |
| Mixing of gases at V=const |  | ||
| Mixing of gas flows** |  | ||
| * - all equations related to the mixing of gases are derived in the absence of heat exchange with the environment; ** - if mass flow rates ( M 1, M 2, …M n, kg/h) mixing flows are equal. |
Here k i– ratio of heat capacities of gases (see formula (4.2)).
Gas mixtures are understood as a mechanical mixture of several gases that do not chemically interact with each other. The composition of the gas mixture is determined by the amount of each gas included in the mixture and can be specified by mass m i or volumetric r i shares:
m i = M i / M; r i = V i / V, (3.1)
Where M i- weight i-th component
V i– partial or reduced volume i- th component;
M, V are the mass and volume of the entire mixture, respectively.
It's obvious that
M 1 + M 2 +…+M n = M; m 1 + m 2 +…+m n = 1, (3.2)
V 1 + V 2 +…+ V n = V ;r 1 + r 2 +…+r n = 1, (3.3)
Relationship between gas mixture pressure R and partial pressure of individual components p i included in the mixture is set Dalton's law
The fear of diving is one of the greatest human fears. It is inherent even in divers with good experience. What is the essence of this fear? Most often, this is not a fear of the fauna of the depths, nor a fear of decompression sickness. And even high deep pressure, as well as loss of consciousness as a result of hyperventilation, do not frighten us as much as the possibility of getting into a stupid situation frightens us.
Diving requires us to have many specific skills. And when we engage in this sport, we are more afraid of appearing flawed in the eyes of others. We are afraid to be under their gaze, afraid of their assessments.
Of course, diving is not a competition, but often we ourselves set the tone for it, especially when it comes to personal experience and skills.
The ability to properly use air underwater is one of the signs of experience. It is by this, as well as by the ability to relax and control the buoyancy of fins, that underwater skill is most often assessed. You cannot hide from your partners the lack of air and the need to float to the top, especially when the whole group is forced to interrupt the dive because of you. Nobody wants to be the first to give the thumbs up.
And these constant boastful comparisons of who has more air left are also depressing...
And your pressure gauge showed 15 bar. But you, of course, hoped against hope that this would escape the attention of your underwater guide. And your partner and wife in one person had a reserve of 90. And, to be completely honest, you were already tired of thinking with every dive that, most likely, in the end you would have to borrow her octopus.
But you shouldn’t hang your fins on the wall in despair or rush to buy a pair, because your lungs’ air consumption is not predisposed by your genes. Effective breathing is a skill. Moreover, it is the most important adaptive skill we acquire during scuba diving. But any skill can be worked on, and breathing is no exception.
Already on your next dive you have the opportunity to save air.
So, if our diver is a man from 30 to 45, of average physical fitness, who, diving in warm water with a standard 10 liter aluminum cylinder, can breathe normally at a depth of 22 meters.
Under such conditions, the cylinder lasts for an average of 20 minutes.
Our advice is to increase this time by another 5-17 minutes.
Of course, if you are already using some of these recommendations, then a little less time will be added.
1. The breathing cycle needs to be changed.
You need to change the order of holding your breath. If on land we pause while exhaling (inhale, then exhale and then a pause), then under water, in a relaxed scuba diver, the breathing itself changes in such a way that the pause is made immediately after inhalation: inhale, then pause, then exhale, inhale again and only then - a pause. The length of the pause when inhaling, as well as the degree of relaxation, distinguishes a beginner from an experienced diver.
A long pause during relaxed breathing reduces air consumption. Relaxation helps to avoid barotrauma during a pause, even when ascending to a shallower depth.
2. Try to breathe deeply.
Take slow, deep, relaxed breaths. You know this axiom from the first lesson, but what is the need for such breathing?
Under pressure, the air in our breathing system moves slightly differently. And in the air itself, in addition to oxygen, there are dense gases. Frequent breathing in such a situation does not allow oxygen to be absorbed. You must slow down your breathing rate so as not to simply push air through your respiratory organs, but to allow oxygen to penetrate well into your lungs. And the deeper you dive, the deeper and slower your breathing should become, this will ensure normal oxygen exchange.
3. Achieve slowness and relaxation in your movements.
Because water is 800 times more dense than air, you won't be able to move at your normal speed underwater without a lot of effort. This means you will use more air. Move very slowly, becoming relaxed and weightless, like a mime doing slow motion. Let your movements be smooth, easy, without the slightest effort.
Many divers benefit from the practice of yoga and various relaxation techniques - such practices allow you to slow down your breathing rate even more.
4. It is very important not to make unnecessary movements with your hands.
Do not use your arms when swimming, but use your fins to paddle slowly and deliberately. Don't be like a cyclist who pedals faster and harder as he climbs a steep hill. Cross your arms across your chest or down along your body, or tuck them behind your back under the tank, or under the weight belt in front. To achieve the state of weightless relaxation required in our case, you need to achieve neutral buoyancy - an important skill for saving air.
5. Learn neutral buoyancy.
When you have succeeded, you are absolutely motionless and feel as if you are completely suspended in water. And this water around your body holds you up. This is one of the most wonderful sensations, and this is what makes our movements underwater effective.
The standard for checking for ideal buoyancy is as follows: you take with you the minimum weight with which a safety stop is possible at a depth of 3-5 meters with a remaining 30 bar in the cylinder, without air or with its minimum in the compensator. The goal is to maintain neutral buoyancy, regardless of depth, correcting it only with breathing.
6. Try to keep your body horizontal.
Now that you know how to weigh yourself correctly, using a buoyancy compensator, while being neutrally weightless, you will be able to move horizontally in the water. This is the most effective way. If your body is as parallel as possible to the direction of movement, this will save you air. Most often, beginners, moving at an angle to the movement vector and, in addition, making many unnecessary movements, waste air and energy unproductively.
7. It is necessary to tidy up the equipment and try to make it more streamlined.
In order to reduce the level of resistance to the water elements, you need to keep all the hoses as close to you as possible. Use a small cylinder with the volume of breathing gas you need for a given dive. The streamlining of the compensator is of great importance; its lifting force must correspond to the conditions in which you are diving.
It is better to place various items you need during the dive in the pockets of the compensator.
There is no need to take ballast weight, the exception will be the load that you will need during a safety stop, at a depth of 3-5 meters. It is also possible to reduce the number of hoses by using an alternative type of air source or inflator, as well as a computer with the ability to connect without using hoses. Take only the equipment you need for your dive.
8. The importance of the breathing regulator.
Despite the apparent ease, breathing under water is a rather difficult and time-consuming task.
It requires certain physical expenditures and skills. In order to reduce the load, it is necessary to use a high-power regulator, highest performance.
Be sure to thoroughly rinse the regulator before diving. It is important to take it to technical experts once every twelve months. inspection, as well as every time before using the regulator, if before that you have not used it for a long time. Try setting the breathing ease controls to the maximum position, but make sure that air does not escape from the cylinder in an arbitrary manner.
9. Techniques for saving air by being on the surface of the water.
Stay on the surface as much as possible, breathing either into a tube, or slightly inflating the compensator, and float on your back. The efficiency of movements on the surface of the water is reduced, but you will have enough air to breathe. Diving to shallow depths requires less air. You won't need to surf frequently to determine where you are, allowing you to stay underwater longer.
10. Suppression of arbitrary air loss.
There are cases of inevitable consumption of air, for example, to equalize pressure, blowing a mask, adjusting buoyancy, creating a layer of air in dry suits. When removing the regulator, turn on the air flow suppression function, if available. Control the position of the mouthpiece; it needs to be turned down. O-rings on scuba gear can also sometimes leak, but usually only a minimal amount of air escapes through them. The illusion that you can use air more economically by blowing up the compensator under water with your mouth is just an illusion. A power inflator is more preferable and effective in this case. While on the surface it makes sense to do this, while observing the necessary safety measures.
11. Less load, more savings.
The less you use your fins underwater, the less air you will waste. Use the force of the current, when diving and ascending, use buoyancy control, when moving along the bottom, use your fingertips, provided that this does not harm the environment.
12. Stay warm.
The warmer you are underwater, the less air you will use. Even in the tropics, where water temperatures reach thirty degrees, you lose a lot of heat when diving without a wetsuit. Consequently, you get tired faster, begin to breathe more often and thereby increase your air consumption. Based on this, choose a wetsuit that provides you with the best protection from the cold. The best option is a dry wetsuit complete with thermal underwear.
13. The importance of physical fitness.
Being in good physical shape allows you to make better use of the oxygen in the air. Proper nutrition, relaxation without various stresses, regular sports activities, quitting smoking and alcohol, all this will give you the opportunity to endure diving easier and save air.
14. Experience and level of training.
The more often you dive underwater, the more you improve your deep-diving skills. Various diving courses supervised by experienced instructors will increase your level and understanding of diving tactics. Training in water and underwater rescue operations will provide you with good physical fitness. All this will undoubtedly help you understand the underwater world, as well as learn to feel calm and free under water.
15. Selection and operation of fins.
According to various tests, there is no universal fin suitable for all underwater enthusiasts. When choosing, you need to rely on your experience, physical fitness, as well as skills in working with fins.
The principles of working with fins are as follows: in the water you need to move in a horizontal position, strokes are carried out with a straight leg from the hip, you should not strain too much, be nervous and make various jerks, and so on.
Fins that are large in size and have high rigidity are not the most effective, as they create unnecessary stress on the legs. When choosing, pay the main importance and attention to the convenience of the fins.
16. Relax.
This is the main secret to saving respiratory resources. Don't try to keep up with someone.
People have different parameters: physical, psychological, metabolism, and so on and so forth. A large, physically strong, trained man will not be able to compete with a miniature, fragile woman in the matter of saving air. A woman will spend much less air when breathing than a man, and there is no escape from this.
Understanding these simple rules can greatly reduce the risk of diving and diving.
Compression of air in a vessel immersed in water
Consider the following situation. An empty, open glass bottle is lowered into water to a depth h.
1. Explain why when a bottle is immersed bottom down, the air comes out of it in bubbles and the bottle fills with water (Fig. 46.1).
2. Why does the bottle immediately sink?
3. Explain why when a bottle is immersed upside down, air does not come out of it (Fig. 46.2).
4. Explain why when a bottle is immersed upside down, the volume of air in it decreases with increasing depth.
Let us denote the density of water ρ in, the internal volume of the bottle V 0, the volume of air contained in it V air, and the atmospheric pressure p a. Let us assume that the temperature of the air in the bottle remains constant.
5. Explain why when a bottle is immersed to a depth h the following equation is valid:
V air (p a + ρ in gh) = V 0 p a. (1)
6. How many times will the volume of air in the bottle decrease when it is immersed to a depth of 10 m?
7. How does the Archimedes force acting on a bottle of air change with increasing depth?
8. Explain why in this case, when finding the Archimedes force, the volume of a body immersed in water should be considered equal to the total volume of glass and air in the bottle.
At a certain depth of immersion, the force of Archimedes will become equal to the force of gravity. When diving to an even greater depth, the force of Archimedes will be less than the force of gravity, so the bottle with air will begin to sink.
Let us pose the question: can the force of gravity acting on the air be neglected in comparison with the force of gravity acting on the bottle?
9. How many times is the mass of air contained in a half-liter bottle less than the mass of the bottle? Take the mass of the bottle to be 0.5 kg; The air density at 20 ºС is approximately 1.2 kg/m3.
So, we see that the mass of air in the bottle can be neglected with good accuracy compared to the mass of the bottle.
Let us denote the density of glass ρ с, and the volume of glass V с.
10. Explain why, when a bottle of air completely submerged in water is in equilibrium, the following equation holds:
ρ with V with g = ρ in g(V air + V c). (2)
Equations (1) and (2) can be considered as a system of two equations with two unknowns. For example, if the values of all quantities included in these equations are known, except for Vair and h, they can be found using these equations.
11. An open bottle containing air at atmospheric pressure is lowered into water, bottom up. Bottle capacity 0.5 l, glass volume 0.2 l, glass density 2.5 times the density of water, atmospheric pressure 100 kPa.
a) What is the volume of air in the bottle when the bottle immersed in water is in equilibrium?
b) At what depth will the bottle be?
In the situation considered, the mass of air can be neglected, because at pressure close to atmospheric pressure, the density of air is much less than the density of water and solids.
But in cases where we are talking about lifting loads from great depths using compressed air, the mass of compressed air can be significant.
Let's look at an example.
12. Explorers of the ocean depths discovered a sunken treasure chest at a depth of 1 km. The weight of the chest is 2.5 tons, the volume is 1 m 3. They tied the chest with a cable to a strong, empty waterproof bag and began to pump air into the bag until it began to float up along with the chest. To simplify the calculations, we will assume that the density of sea water is equal to the density of fresh water. We will assume that the water is incompressible, and the volume of the bag shell is negligible. The water temperature at great depths can be considered close to 0 ºС.
a) Is it necessary to take into account atmospheric pressure to determine the air pressure in the bag?
b) Let us denote by ρ the density of water, m c and m in the mass of the chest and the mass of air in the bag, V c and V in the volume of the chest and the volume of air at the beginning of ascent, M in – the molar mass of air, T – the absolute temperature of the water. Write down a system of two equations with two unknowns (m in and V in), assuming that atmospheric pressure can be neglected.
c) What is the volume of air in the bag at the moment when the bag with the chest began to float?
d) What is the mass of air in the bag when the bag with the chest begins to float?
e) Is it possible not to let the air out of the bag until the bag and the chest float to the surface?
Air in a tube with a mercury column
There is air in a glass tube sealed at one end. This air is separated from atmospheric air by a column of mercury with a length of l Hg (Fig. 46.3). 
Let's consider how the length of the air-filled part of the tube depends on the position of the tube and the air temperature in it. We will assume that the length of the tube is large enough so that mercury does not pour out of the tube in any position.
Let us denote the atmospheric pressure p a, the density of mercury ρ rt, and the length of the air-filled part of the tube, when it is located horizontally, we denote l 0.
Let us first assume that the air temperature in the tube is constant.
13. Write down an equation that relates the quantities l rt, l 0 and the length l of the air-filled part of the tube when it is located:
a) vertically with the open end up;
b) vertically with the open end down.
14. At the initial moment, the tube is positioned with the open end down. When it was turned upside down, the length of the air-filled part of the tube decreased by 10%. What is the length of a column of mercury if the atmospheric pressure is 760 mm Hg? Art.?
Let us now consider the case when the air temperature in the cabin changes.
15. At the initial moment, the tube with air and a column of mercury is located horizontally. When it was lowered into boiling water with the open end up, the length of the air-filled part of the tube increased by 20%. What is the initial temperature of the air in the tube if the length of the mercury column is 5 cm? Atmospheric pressure is 760 mmHg. Art.
2. Two gases in a cylinder with a piston or baffle
The cylinder is located horizontally
Let us first consider the case where a cylinder with different gases is located horizontally (in Figure 46.4, different gases are schematically indicated by different colors). In this case, you can ignore the weight of the piston. 
A piston may have various properties that must be taken into account when solving problems.
16. What can be said about the pressure and temperature of two gases separated by a piston if it:
a) heat conductive and can move without friction?
b) does not conduct heat, but can move without friction?
c) heat-conducting, but it is necessary to take into account the friction between the piston and the walls of the vessel?
17. In a horizontally located cylinder with a piston, hydrogen and oxygen are located on opposite sides of the piston.
a) What is the relationship between the volumes of gases and the amount of substance in them if the piston is movable and heat-conducting?
b) What is the relationship between the volumes and masses of gases in this case?
c) How are the volumes, masses and temperatures of gases related if the piston is movable but does not conduct heat?
If it is said that the vessel is divided not by a piston, but by a partition, then it is implied that the volumes of the parts of the vessel remain constant. The partition can also have different properties.
18. What can be said about the temperature and partial pressure of two gases separated by a partition if it:
a) heat-conducting?
b) porous (this usually means that molecules of one gas can penetrate the partition, but molecules of another gas cannot)?
19. A thermally insulated vessel is divided into two equal parts by a porous partition. At the initial moment, there are 2 moles of helium on the left side of the vessel, and 1 mole of argon on the right side. The initial temperature of helium is 300 K, and the initial temperature of argon is 600 K. Helium atoms can freely penetrate through the pores in the partition, but argon atoms cannot.
a) Does it matter whether the partition conducts heat or not?
b) Which gas atoms initially have a greater average kinetic energy? How many times greater?
c) The internal energy of which gas is greater at the initial moment? How many times more?
d) Explain why the average kinetic energies of atoms of different gases are equal after reaching thermal equilibrium.
e) What temperature will be in the vessel at thermal equilibrium?
f) How many times will the average kinetic energy of helium atoms at thermal equilibrium be greater than their average kinetic energy in the initial state?
g) How will the helium pressure in the left side of the vessel change compared to the initial one after equilibrium is established?
h) How will the argon pressure change compared to the initial one after equilibrium is established?
i) The pressure in which part of the vessel will be greater after equilibrium is established? How many times more?
The cylinder is located vertically
If the cylinder is located vertically (Fig. 46.5), then it is necessary to take into account the weight of the piston, which presses on the gas located at the bottom of the cylinder. Because of this, the pressure at the bottom of the cylinder is greater than at the top. Let's look at an example.
20. A vertically located cylindrical vessel of height l is divided into two parts by a movable piston. In the upper part with height l in there are ν moles of helium, and in the lower part with height l n - the same number of moles of hydrogen. The temperature of the gases remains equal to T all the time. The mass of the piston m, the area S, and the thickness of the piston can be neglected in comparison with the height of the vessel.
a) Express the pressure in each part of the vessel in terms of other quantities. Does the type of gas in the parts of the vessel matter?
b) Write an equation relating the pressure of gases in each part of the vessel with the mass of the piston and its area.
c) What is the mass of the piston if l = 50 cm, ν = 0.22 mol, T = 361 K, l in = 30 cm?
Clue. Use the ideal gas equation of state.
Balloon lift
A balloon (Fig. 46.6) can be in equilibrium in the air only if the Archimedes force acting on it from the air is equal in magnitude to the total force of gravity acting on the ball and the load suspended from it:
F A = F t.sh + F t.gr. (3)
In the case of a balloon, the Archimedes sludge is equal to the weight of the surrounding air in the volume occupied by the balloon and its weight. We have highlighted the word “surrounding” in italics because the density of atmospheric air changes during rise for two reasons: firstly, its pressure decreases, and secondly, its temperature decreases.
Let's denote the volume of the ball V. The volume of the load and shell of the ball is usually neglected compared to the volume of the ball itself, but the masses of the load and shell of the ball are of great importance! We denote the mass of the load as m g, and the mass of the shell as m vol. Then
F t.sh = (m int + m about)g,
where m int is the mass of gas with which the ball is filled.
Let us denote the density of the air surrounding the ball as ρ ext, and the density of the gas located inside the ball as ρ int.
21. Explain why the following equations are true:
F A = ρ ext gV,
m internal = ρ internal V,
V(ρ ext – ρ int) = m gr + m vol. (4)
Clue. Use equation (3) and the relationship between mass, volume and density.
The ground force of a balloon is the weight of the load that the balloon can lift.
22. Explain why the lift modulus of a balloon is expressed by the formula
F under = Vg(ρ external – ρ internal) – m about g. (5)
From formulas (4) and (5) it follows that a balloon can lift a load only if the density of the gas with which the balloon is filled is less than the density of the surrounding air.
If the ball were rigid, this could be achieved by partially pumping the air out of it: the rigid shell could withstand the difference in air pressure inside and outside the ball. However, the shell of a hard ball would be too heavy. The soft shell, which is always used for balloons, cannot withstand any significant pressure difference. Therefore, the gas pressure inside the ball is equal to the pressure of the surrounding air.
23. Explain why if the pressure inside the ball is equal to the pressure of the surrounding air, then the equality is true
ρ internal /ρ external = (M internal * T external) / (M external * T internal). (6)
Clue. Use the ideal gas equation of state.
From formula (6) it is clear that the density of the gas with which the ball is filled can be made less than the density of the surrounding air in two ways:
– use heated air as an “internal” gas;
– use a gas with a lower molar mass.
The first method is used for pleasure balloons (Fig. 46.6), and the second - for weather balloons (Fig. 46.7), which rise to a great height (in this case, the balloon is usually filled with helium). 
24. Explain why from formulas (5) and (6) it follows that the module of the balloon’s lifting force is expressed by the formula
? 25. A balloon with a volume of 3000 m3 has a hole in the lower part through which the air inside the balloon is heated by a burner to a temperature of 77 ºC. The ball is in equilibrium at a height where the ambient temperature is 7 ºC and its density is 1.2 kg/m3. The mass of the ball shell is 300 kg. What is the mass of the load?
Additional questions and tasks
26. Air is pumped from above into a pontoon lying on the bottom of a lake at a depth of 90 m (Fig. 46.8). When atom water is forced out of the pontoon through a hole located in its lower part. What volume of atmospheric air must be supplied to the pontoon so that it can lift the load, if the total mass of the pontoon with the load is 20 tons, and the total volume of the cargo and the walls of the pontoon is 5 m 3? Assume that the water temperature is close to 0 ºС and the atmospheric pressure is 10 5 Pa. 
27. In the sealed elbow of a U-shaped tube there is a column of air 30 cm high. The mercury in both elbows is at the same level. What will be the height of the air column if you slowly add mercury to the top? The pressure is equal to normal atmospheric pressure.
28. A helium-filled balloon is in equilibrium in the air. The mass of one square meter of the balloon shell is 50 g, the temperature of air and helium is 27 ºС, the pressure is equal to normal atmospheric pressure. What is the radius of the ball?
Accurate calculation of diving air is the second most important factor after the impeccable technical condition of the equipment. Since this task has been around since the invention of scuba gear, special methods for calculating the required volume of air have long been developed. The basis is the volume of air required by one diver per minute and then the resulting value is divided by the volume of gas in the cylinder.
These calculations are complicated by the fact that air consumption depends on physical activity. During quiet swimming it is much less than during intensive use of fins. Another factor that is also always taken into account is the depth of immersion. The greater the depth, the higher pressure the air must be supplied. All factors taken into account can be represented as a list:
- Cylinder volume.
- Cylinder pressure.
- Air consumption per minute (denoted as RMV)
- Immersion depth.
The first two parameters can be very accurate. Their accuracy depends only on how well they match the indicated volume, as well as how accurately the valve on the pump that was used for filling is adjusted. The compressor is switched off at the end of filling using a pressure sensor. It is responsible for ensuring that the volume of air in the cylinder exactly corresponds to the declared one.
The hardest part is calculating the RMV. Accurate data can only be obtained experimentally. This is exactly what they do when training divers. The student memorizes the pressure gauge readings during various dive modes, drifting with the current, ascent or standing still. Next, based on the data obtained, an individual RMV indicator is derived. The data is recorded in a table with three columns: dive time and depth, and tank pressure using a pressure gauge. By recalculating the pressure in the cylinder by the volume (you just need to multiply the indicators), we will get the exact value of air consumption per minute and make corrections for load and depth.
If there is no time for such measurements, which require trial dives with an instructor, then general indicators are taken. They are calculated with a certain margin, which is necessary to cover all individual features. So the air consumption on the surface by a diver weighing 80 kg is 20 - 25 l/min. (In reality, somewhat less - 16 - 22 l). Women have even less air consumption. Next, a correction is made for depth. As the diving depth increases, the volume of air required increases very quickly. At 50 meters (the maximum depth for amateur diving), you need almost twice as much (about 40 l/min.).
The maximum inhalation pressure differs for different mixtures. For oxygen it is only 1.3 - 1.4 atm. For this reason, special mixtures are required for deep sea diving. When compiling, they try to ensure that the oxygen content in them is slightly different from the natural one in ordinary air. The nitrogen content in the deep-sea mixture is also reduced, since if you use ordinary air, nitrogen narcosis begins already at 30 meters. For the deepest dives, a helium-oxygen mixture is optimal. It is almost never used in amateur diving. Filling cylinders with helium is difficult because it has ultra-high permeability, but when mixed with oxygen this disadvantage is almost eliminated.
When using clean air, it also matters where the cylinder was filled. There is only one main requirement here. Air purity is necessary. Therefore, it is better with an electric drive. Then the risk of carbon monoxide and excess carbon dioxide is minimal. It is optimal that cylinders are refilled in an environmentally friendly place, for example on the seashore or in the countryside.
Atmospheric air and its properties. The layer of air surrounding the globe is called the atmosphere. The higher you are from the earth's surface, the less air density.
Atmospheric air is a mixture of gases. One liter of it weighs 1.29 g at atmospheric pressure and temperature 15°C.
The composition of the air includes (by volume) nitrogen - 78.13%, oxygen - 20.90%, carbon dioxide - 0.03%, argon - 0.94%. In addition, the air contains small amounts of helium, hydrogen and other inert gases.
In addition to the listed gases, the air contains water vapor, the amount of which is not constant.
Nitrogen- under normal conditions, a gas neutral for the body. It is colorless, odorless and tasteless, does not burn and does not support combustion. One liter of nitrogen weighs 1.25 g, its density is 0.967. About one liter of nitrogen is dissolved in the human body at normal atmospheric pressure.
Oxygen- the most important gas for humans. Without it, life on Earth is impossible. Oxygen does not burn, but supports combustion. In its pure form it is flammable. One liter of oxygen weighs 1.43 g. Pure medical oxygen (98.99%) is used for breathing.
Carbon dioxide- the heaviest of all gases. One liter of it weighs 1.96 g. The density is 1.529 g. At a partial pressure of 0.03 ata, which corresponds to 3% in the air, carbon dioxide has a poisonous effect on the body.
Atmospheric pressure measurement. The weight of air presses on the ground and objects on it. The first to determine the value of atmospheric pressure was the Italian scientist Toricelli (in the 17th century). To do this, he used a long glass tube with a cross-sectional area of 1 cm2, sealed at one end and filled with mercury.
Having lowered the unsealed end of the tube into an open vessel with mercury, he noticed that the latter in the tube only dropped to a certain level. It did not go any lower, since this was prevented by the air pressure on the mercury in the vessel. When measuring, it turned out that the height of the mercury column in the tube was 760 mm, and its weight was 1.033 kg (Fig. 2). Thus, it was determined that the atmospheric pressure at the surface of the earth at sea level is 760 mm Hg. Art., which corresponds to a pressure with a force of 1.033 kg per 1 cm 2 or 10.33 m of water. Art. This pressure is called atmospheric, normal or barometric and is denoted atm. This is a physical atmosphere.
Rice. 2. Atmospheric air pressure
In practice, for the convenience of calculations, the technical atmosphere is taken as a unit of pressure, which is equal to a pressure of 1 kg per 1 cm 2 area. It is designated at.
Water pressure on the diver. We have already said above that when diving under water, a person experiences not only the pressure of atmospheric air, but also water. When diving, for every 10 m the pressure increases by 1 atm. This pressure is called excess and is designated ati.
The total (absolute) pressure of water and air on the diver. Underwater, the diver is exposed to both atmospheric and excess pressure from the water column.
Their total pressure is called absolute pressure and is designated ata. For example, at a depth of 10 m a diver is under pressure of 2 ati (1 ati + 1 ata), at a depth of 50 m - 6 ata, etc.
Compressibility and elasticity of gases. Gases consist of particles in continuous motion. Gas molecules are tiny in size, but occupy a large volume. The force of attraction between individual gas molecules is much less than in liquids or solids. Gases do not have a constant volume and take the shape and volume of the vessel in which they are located.
In contrast to liquids, gases are capable of expanding and compressing under pressure, thereby reducing their volume and increasing elasticity.
The relationship between the volume and pressure of gases is established by the Boyle-Mariotte law, which states that the volume occupied by a gas varies in inverse proportion to the pressure acting on it at a constant temperature. The product of gas volume (V) and the corresponding pressure (P) at a constant temperature does not change PхV=const.
For example, if you take 2 liters of gas under a pressure of 2 ata and change this pressure, the volume will change as follows:
In other words, by how many times the pressure increases, the volume of gas decreases by the same amount, and vice versa.
The significance of this law is (practical). It explains why the consumption of air for breathing increases with increasing depth (diving. If on the surface a diver consumes 30 liters of atmospheric air per minute, then at a depth of 20 m this air is compressed to 3 ata, which already corresponds 90 liters of air.Consumption actually triples.
Using this law, you can make the necessary calculations related to diving descents.
Calculation example:
Determine how many liters of compressed air a diver under pressure of 4 atm on a pressure gauge receives if he is supplied with 150 liters of free air per minute?
According to the Boyle-Mariotte law, P1 V1 = P2 V2.
In the example
These calculations are only valid for constant temperature. In practice, it is necessary to take into account changes in volume and pressure at different temperatures. The dependence of air volume and pressure on its temperature is determined by Gay-Lussac's law, which states that the change in gas volume at constant pressure is directly proportional to the heating temperature. The change in gas pressure at a constant volume is also directly proportional to the heating temperature.