Properties of gases and gas mixtures used for diving descents. Elements of gas mechanics influence on gas density
Physicochemical properties of oil and parameters characterizing it: density, viscosity, compressibility, volumetric coefficient. Their dependence on temperature and pressure
The physical properties of reservoir oils are very different from those of surface degassed oils, which is due to the influence of temperature, pressure and dissolved gas. The change in the physical properties of reservoir oils associated with the thermodynamic conditions of their presence in the reservoirs is taken into account when calculating the reserves of oil and petroleum gas, when designing, developing and operating oil fields.
Density degassed oil varies over a wide range - from 600 to 1000 kg/m 3 and more, and depends mainly on the hydrocarbon composition and the content of asphalt-resinous substances.
The density of oil in reservoir conditions depends on the amount of dissolved gas, temperature and pressure. With an increase in pressure, the density slightly increases, and with an increase in the other two factors, it decreases. The influence of the latter factors is more pronounced. The density of oils saturated with nitrogen or carbon dioxide increases somewhat with increasing pressure.
The effect of the amount of dissolved gas and temperature is stronger. Therefore, the density of the gas as a result is always less than the density of degassed oil on the surface. With increasing pressure, the density of oil decreases significantly, which is associated with the saturation of oil with gas. An increase in pressure above the saturation pressure of oil with gas contributes to some increase in the density of oil.
The density of formation waters, in addition to pressure, temperature and dissolved gas, is strongly influenced by their salinity. When the concentration of salts in formation water is 643 kg/m 3 its density reaches 1450 kg/m 3 .
Volume ratio. When a gas dissolves in a liquid, its volume increases. The ratio of the volume of liquid with gas dissolved in it in reservoir conditions to the volume of the same liquid on the surface after its degassing is called the volumetric coefficient
b=V PL / V SOV
where V PL is the volume of oil in reservoir conditions; V POV - the volume of the same oil at atmospheric pressure and t=20°C after degassing.
Since a very large amount of hydrocarbon gas can be dissolved in oil (even 1000 or more m 3 in 1 m 3 of oil), depending on thermodynamic conditions, the volumetric coefficient of oil can reach 3.5 or more. Volumetric coefficients for formation water are 0.99-1.06.
The decrease in the volume of oil recovered compared to the volume of oil in the reservoir, expressed as a percentage, is called "shrinkage"
u=(b-1) / b *100%
When the pressure decreases from the initial reservoir p 0 to saturation pressure, the volumetric coefficient changes little, because oil with gas dissolved in it behaves in this area as an ordinary weakly compressible liquid, expanding slightly with decreasing pressure. As the pressure decreases, the gas is gradually released from the oil and the volume factor decreases. An increase in oil temperature worsens the solubility of gases, which leads to a decrease in the volumetric coefficient
Viscosity. Viscosity is one of the most important characteristics of oil. The viscosity of oil is taken into account in almost all hydrodynamic calculations associated with lifting fluid through tubing, flushing wells, transporting well products through infield pipes, processing bottomhole formation zones by various methods, as well as in calculations related to the movement of oil in the reservoir.
The viscosity of reservoir oil is very different from the viscosity of surface oil, since it contains dissolved gas in its composition and is under conditions of elevated pressures and temperatures. With an increase in the amount of dissolved gas and temperature, the viscosity of oils decreases.
An increase in pressure below the saturation pressure leads to an increase in the GOR and, as a result, to a decrease in viscosity. An increase in pressure above the saturation pressure for reservoir oil leads to an increase in the viscosity
With an increase in the molecular weight of oil, its viscosity increases. Also, the viscosity of oil is greatly influenced by the content of paraffins and asphalt-resinous substances in it, as a rule, in the direction of its increase.
Oil compressibility. Oil has elasticity, that is, the ability to change its volume under the influence of external pressure. The elasticity of a liquid is measured by the compressibility coefficient, which is defined as the ratio of the change in volume of a liquid to its original volume with a change in pressure:
β P =ΔV/(VΔP) , where
ΔV is the change in the volume of oil; V is the initial volume of oil; ΔP - pressure change
The compressibility coefficient of reservoir oil depends on the composition, content of dissolved gas in it, temperature and absolute pressure.
Degassed oils have a relatively low compressibility coefficient, of the order of (4-7) * 10 -10 1/Pa, and light oils containing a significant amount of dissolved gas in their composition - up to 140 * 10 -10 1 / Pa. The higher the temperature, the higher the compressibility factor.
Density.
Density is usually understood as the mass of a substance contained in a unit volume. Accordingly, the dimension of this quantity is kg / m 3 or g / cm 3.
ρ=m/V
The density of oil in reservoir conditions decreases due to the gas dissolved in it and due to an increase in temperature. However, when the pressure drops below the saturation pressure, the dependence of the oil density is nonmonotonic, and when the pressure increases above the saturation pressure, the oil contracts and the density slightly increases.
Viscosity of oil.
Viscosity characterizes the force of friction (internal resistance) that occurs between two adjacent layers inside a liquid or gas per unit surface during their mutual movement.
The viscosity of oil is determined experimentally on a special VVD-U viscometer. The principle of operation of the viscometer is based on measuring the time of falling of a metal ball in the investigated liquid.
The viscosity of oil is determined by the formula:
μ = t (ρ w - ρ l) k
t – ball fall time, s
ρ w and ρ w - the density of the ball and liquid, kg / m 3
k is the viscometer constant
An increase in temperature causes a decrease in oil viscosity (Fig. 2. a). An increase in pressure below the saturation pressure leads to an increase in the GOR and, as a result, to a decrease in viscosity. An increase in pressure above the saturation pressure for reservoir oil leads to an increase in viscosity (Fig. 2. b).
The minimum value of viscosity occurs when the pressure in the reservoir becomes equal to the reservoir saturation pressure.
Oil compressibility
Oil has elasticity. The elastic properties of oil are estimated by the oil compressibility factor. The compressibility of oil is understood as the ability of a liquid to change its volume under pressure:
β n = (1)
β n - oil compressibility coefficient, MPa -1-
V n - the initial volume of oil, m 3
∆V – oil volume measurement under pressure measurement ∆Р
The compressibility coefficient characterizes the relative change in the unit volume of oil with a change in pressure per unit. It depends on the composition of reservoir oil, temperature and absolute pressure. With increasing temperature, the compressibility coefficient increases.
Volume ratio
The volume factor is understood as a value showing how many times the volume of oil in reservoir conditions exceeds the volume of the same oil after gas release on the surface.
in \u003d V pl / V deg
c - volumetric coefficient
V pl and V deg - volumes of reservoir and degassed oil, m 3
With a decrease in pressure from the initial reservoir p 0 to saturation pressure (section ab), the volumetric coefficient changes little, because oil with gas dissolved in it behaves in this area as an ordinary weakly compressible liquid, expanding slightly with decreasing pressure.
As the pressure decreases, the gas is gradually released from the oil and the volume factor decreases. An increase in oil temperature worsens the solubility of gases, which leads to a decrease in the volumetric coefficient.
Copyrightã L.Kourenkov
Properties of gases
Gas pressure
A gas always fills a volume bounded by impenetrable walls. So, for example, a gas cylinder or a car tire chamber is almost evenly filled with gas.
In an effort to expand, the gas exerts pressure on the walls of the cylinder, tire chamber or any other body, solid or liquid, with which it comes into contact. If we do not take into account the action of the Earth's gravitational field, which, with the usual dimensions of vessels, only negligibly changes the pressure, then at equilibrium, the pressure of the gas in the vessel seems to us to be completely uniform. This remark refers to the macrocosm. If we imagine what happens in the microcosm of the molecules that make up the gas in the vessel, then there can be no question of any uniform distribution of pressure. In some places on the surface of the wall, gas molecules hit the walls, while in other places there are no impacts. This picture changes all the time in a chaotic way. Gas molecules hit the walls of the vessels, and then fly off at a speed almost equal to the speed of the molecule before impact. Upon impact, the molecule transfers to the wall a momentum equal to mv, where m is the mass of the molecule and v is its velocity. Reflecting from the wall, the molecule gives it the same amount of motion mv. Thus, with each impact (perpendicular to the wall), the molecule transfers to it an amount of motion equal to 2mv. If in 1 second there are N impacts per 1 cm 2 of the wall, then the total amount of motion transferred to this section of the wall is 2Nmv. By virtue of Newton's second law, this amount of movement is equal to the product of the force F acting on this section of the wall by the time t during which it acts. In our case, t = 1 sec. So F=2Nmv, there is a force acting on 1 cm 2 of the wall, i.e. pressure, which is usually denoted p (moreover, p is numerically equal to F). So we have
p=2Nmv
No brainer that the number of impacts in 1 second depends on the speed of the molecules, and the number of molecules n per unit volume. For a not very compressed gas, we can assume that N is proportional to n and v, i.e. p is proportional to nmv 2 .
So, in order to calculate the pressure of a gas using molecular theory, we must know the following characteristics of the microcosm of molecules: mass m, velocity v, and the number of molecules n per unit volume. In order to find these micro-characteristics of molecules, we must establish on what characteristics of the macrocosm the gas pressure depends, i.e. establish by experience the laws of gas pressure. By comparing these experimental laws with the laws calculated using molecular theory, we will be able to determine the characteristics of the microcosm, for example, the speed of gas molecules.
So, let's establish what the pressure of a gas depends on?
Firstly, on the degree of gas compression, i.e. on how many gas molecules are in a certain volume. For example, by inflating a tire or compressing it, we force the gas to press harder on the chamber walls.
Secondly, on what is the temperature of the gas.
Usually, a change in pressure is caused by both causes at once: both a change in volume and a change in temperature. But it is possible to realize the phenomenon in such a way that when the volume changes, the temperature will change negligibly little, or when the temperature changes, the volume will practically remain unchanged. We will deal with these cases first, after making the following remark beforehand.
We will consider gas in a state of equilibrium. It means; that the gas is in both mechanical and thermal equilibrium.
Mechanical equilibrium means that there is no movement of individual parts of the gas. For this, it is necessary that the pressure of the gas be the same in all its parts, if we neglect the insignificant pressure difference in the upper and lower layers of the gas, which occurs under the action of gravity.
Thermal equilibrium means that there is no transfer of heat from one section of the gas to another. To do this, it is necessary that the temperature in the entire volume of the gas be the same.
Dependence of gas pressure on temperature
Let's start by finding out the dependence of gas pressure on temperature, subject to a constant volume of a certain mass of gas. These studies were first made in 1787 by Charles. It is possible to reproduce these experiments in a simplified form by heating the gas in a large flask connected to a mercury manometer in the form of a narrow curved tube.
Let us neglect the insignificant increase in the volume of the flask when heated and the insignificant change in volume when the mercury is displaced in a narrow manometric tube. Thus, the volume of gas can be considered unchanged. By heating the water in the vessel surrounding the flask, we will note the temperature of the gas with a thermometer , and the corresponding pressure - according to the pressure gauge . Having filled the vessel with melting ice, we measure the pressure corresponding to the temperature 0°C .
Experiments of this kind showed the following:
1. The increase in pressure of a certain mass of gas when heated by 1 ° is a certain part a of the pressure that this mass of gas had at a temperature of 0 ° C. If the pressure at 0 ° C is denoted by P, then the increment in gas pressure when heated by 1 ° C is aP.
When heated by t degrees, the pressure increment will be t times greater, i.e., the pressure increment proportional to the temperature increment.
2. The value a, showing by what part of the pressure at 0 ° C the gas pressure increases when heated by 1 °, has the same value (more precisely, almost the same) for all gases, namely . The quantity a is called thermal, pressure coefficient. Thus, the thermal pressure coefficient for all gases has the same value, equal to .
The pressure of a certain mass of gas when heated to 1° in unchanged volume increases by part of the pressure at 0°C (Charles law).
It should be borne in mind, however, that the temperature coefficient of gas pressure obtained by measuring the temperature with a mercury thermometer is not exactly the same for different temperatures: Charles's law is only approximately fulfilled, although with a very high degree of accuracy.
Formula expressing Charles' law.
Charles' law allows you to calculate the pressure of a gas at any temperature if its pressure at 0°C is known. Let the pressure at 0°C of a given mass of gas in a given volume be , and the pressure of the same gas at a temperature t there is p. There is a temperature increment t, therefore, the pressure increment is a t and the desired pressure is
P = + a t=(1+ a t )= (1+ ) (1)
This formula can also be used if the gas is cooled below 0°C; wherein t will have negative values. At very low temperatures, when the gas approaches the state of liquefaction, as well as in the case of highly compressed gases, Charles's law is inapplicable and formula (1) ceases to be valid.
Charles' law from the point of view of molecular theory
What happens in the microcosm of molecules when the temperature of a gas changes, for example, when the temperature of a gas rises and its pressure increases? From the point of view of molecular theory, there are two possible reasons for the increase in pressure of a given gas: firstly, the number of molecular impacts per 1 cm 2 could increase for 1 sec; secondly, the amount of motion transmitted when a single molecule hits the wall could increase. Both causes require an increase in the speed of molecules. From this it becomes clear that an increase in the temperature of a gas (in the macrocosm) is an increase in the average speed of the random movement of molecules (in the microcosm). Experiments to determine the velocities of gas molecules, which I will discuss a little later, confirm this conclusion.
When we are dealing not with a gas, but with a solid or liquid body, we do not have at our disposal such direct methods for determining the velocity of the body's molecules. However, in these cases, it is undoubted that with an increase in temperature, the speed of movement of molecules increases.
The change in temperature of a gas with a change in its volume. Adiabatic and isothermal processes.
We have established how the pressure of a gas depends on temperature if the volume remains unchanged. Now let's see how the pressure of a certain mass of gas changes depending on the volume it occupies, if the temperature remains unchanged. However, before moving on to this question, it is necessary to figure out how to keep the temperature of the gas constant. To do this, it is necessary to study what happens to the temperature of the gas, if its volume changes so quickly that there is practically no heat exchange between the gas and the surrounding bodies.
Let's do this experiment. In a thick-walled tube of transparent material closed at one end, we place a cotton wool slightly moistened with ether, and this will create a mixture of ether vapors with air inside the tube, which explodes when heated. Then quickly push the tightly fitting piston into the tube. We will see that a small explosion will occur inside the tube. This means that when the mixture of ether vapors with air was compressed, the temperature of the mixture increased sharply. This phenomenon is quite understandable. Compressing the gas with an external force, we produce work, as a result of which the internal energy of the gas should have increased; this happened - the gas heated up.
Now let's let the gas expand and do work against the forces of external pressure. It can be done. Let there be compressed air in a large bottle at room temperature. Having informed the bottle with external air, let the air in the bottle expand, leaving the not-large. holes outward, and place a thermometer or a flask with a tube in a stream of expanding air. The thermometer will show a temperature noticeably lower than room temperature, and a drop in the tube attached to the flask will run towards the flask, which will also indicate a decrease in the air temperature in the jet. So, when the gas expands and at the same time does work, it cools and its internal energy decreases. It is clear that the heating of a gas during compression and the cooling during expansion are expressions of the law of conservation of energy.
If we turn to the microworld, then the phenomena of gas heating during compression and cooling during expansion will become quite clear. When a molecule hits a stationary wall and bounces off it, the speed, and hence the kinetic energy of the molecule, is on average the same as before hitting the wall. But if the molecule hits and bounces off the piston advancing on it, its speed and kinetic energy are greater than before hitting the piston (just like the speed of a tennis ball increases if it is hit in the opposite direction with a racket). The advancing piston transfers additional energy to the molecule reflected from it. Therefore, the internal energy of the gas increases during compression. When rebounding from the receding piston, the speed of the molecule decreases, because the molecule does work by pushing the retracting piston. Therefore, the expansion of the gas, associated with the removal of the piston or layers of the surrounding gas, is accompanied by the performance of work and leads to a decrease in the internal energy of the gas.
So, the compression of a gas by an external force causes it to heat up, and the expansion of the gas is accompanied by its cooling. This phenomenon always takes place to some extent, but I notice it especially sharply when the exchange of heat with surrounding bodies is minimized, because such an exchange can more or less compensate for temperature changes.
Processes in which the transfer of heat is so negligible that it can be neglected are called adiabatic.
Let us return to the question posed at the beginning of the chapter. How to ensure the constancy of the temperature of the gas, despite changes in its volume? Obviously, for this it is necessary to continuously transfer heat from the outside to the gas if it expands, and continuously take heat from it, transferring it to the surrounding bodies, if the gas is compressed. In particular, the temperature of the gas remains fairly constant if the expansion or contraction of the gas is very slow, and the transfer of heat from the outside or outside can occur with sufficient speed. With slow expansion, heat from the surrounding bodies is transferred to the gas and its temperature decreases so little that this decrease can be neglected. With slow compression, on the contrary, heat is transferred from the gas to the surrounding bodies, and as a result, its temperature rises only negligibly.
Processes in which the temperature is kept constant are called isothermal.
Boyle's law - Mariotte
Let us now turn to a more detailed study of the question of how the pressure of a certain mass of gas changes if its temperature remains unchanged and only the volume of the gas changes. We have already found out what isothermal the process is carried out under the condition that the temperature of the bodies surrounding the gas is constant, and the volume of the gas changes so slowly that the temperature of the gas at any moment of the process does not differ from the temperature of the surrounding bodies.
Thus, we pose the question: how are volume and pressure related to each other during an isothermal change in the state of a gas? Daily experience teaches us that when the volume of a certain mass of gas decreases, its pressure increases. As an example, you can specify the increase in elasticity when inflating a soccer ball, bicycle or car tire. The question arises: how Does the pressure of a gas increase with a decrease in volume if the temperature of the gas remains the same?
The answer to this question was given by studies carried out in the 17th century by the English physicist and chemist Robert Boyle (1627-1691) and the French physicist Edem Mariotte (1620-1684).
Experiments that establish the relationship between the volume and pressure of a gas can be reproduced: on a vertical stand , equipped with divisions, there are glass tubes BUT and AT, connected by a rubber tube C. Mercury is poured into the tubes. Tube B is open at the top, tube A has a stopcock. Let us close this faucet, thus locking a certain mass of air in the tube BUT. As long as we don't move the tubes, the mercury level in both tubes is the same. This means that the pressure of the air trapped in the tube BUT, the same as the ambient air pressure.
Now let's slowly pick up the phone AT. We will see that the mercury in both tubes will rise, but not in the same way: in the tube AT the level of mercury will always be higher than in A. If, however, tube B is lowered, then the level of mercury in both knees decreases, but in the tube AT decrease more than BUT.
The volume of air trapped in the tube BUT, can be counted from the divisions of the tube BUT. The pressure of this air will differ from the atmospheric one by the pressure of the mercury column, the height of which is equal to the difference in the levels of mercury in tubes A and B. At. pick up the phone AT the pressure of the mercury column is added to the atmospheric pressure. The volume of air in A decreases. When dropping the tube AT the level of mercury in it is lower than in A, and the pressure of the mercury column is subtracted from atmospheric pressure; the volume of air in A increases accordingly.
Comparing the values of pressure and volume of air locked in tube A obtained in this way, we will make sure that when the volume of a certain mass of air increases a certain number of times, its pressure decreases by the same amount, and vice versa. The air temperature in the tube during our experiments can be considered unchanged.
Similar experiments can be made with other gases. The results are the same.
So, the pressure of a certain mass of gas at a constant temperature is inversely proportional to the volume of gas (Boyle-Mariotte law).
For rarefied gases, the Boyle-Mariotte law is fulfilled with a high degree of accuracy. For gases that are highly compressed or cooled, noticeable deviations from this law are found.
The formula expressing the Boyle-Mariotte law.
(2)
A graph expressing the Boyle-Mariotte law.
In physics and technology, graphs are often used to show the dependence of gas pressure on its volume. Draw such a schedule for an isothermal process. We will plot the volume of gas along the abscissa axis, and its pressure along the ordinate axis.
Let's take an example. Let the pressure of a given mass of gas with a volume of 1 m 3 be 3.6 kg/cm 2 . Based on the law, Boyle - Mariotte, we calculate that with a volume equal to 2 m 3 , pressure is 3.6 * 0.5 kg/cm 2 = 1,8kg/cm 2 . Continuing these calculations, we get the following table:
|
V (in m 3 ) |
||||||
|
P(in kg1cm 2 ) |
Putting this data on the drawing in the form of points, the abscissas of which are the values of V, and the ordinates are the corresponding values R, we get a curved line - a graph of an isothermal process in a gas (figure above).
The relationship between the density of a gas and its pressure
Recall that the density of a substance is the mass contained in a unit volume. If we somehow change the volume of a given mass of gas, then the density of the gas will also change. If, for example, we reduce the volume of a gas by a factor of five, then the density of the gas will increase by a factor of five. This will also increase the pressure of the gas; if the temperature has not changed, then, as the Boyle-Mariotte law shows, the pressure will also increase five times. From this example, it can be seen that in an isothermal process, the pressure of a gas changes in direct proportion to its density.
Denoting the gas density at pressures and letters and , we can write:
This important result can be considered another and more essential expression of the Boyle-Mariotte law. The fact is that instead of the volume of gas, which depends on a random circumstance - on what mass of gas is chosen, - the formula (3) includes the density of the gas, which, like pressure, characterizes the state of the gas and does not depend at all on random choice of its mass.
Molecular interpretation of the Boyle-Mariotte law.
In the previous chapter, we found out, on the basis of the Boyle-Mariotte law, that at a constant temperature, the pressure of a gas is proportional to its density. If the density of the gas changes, then the number of molecules in 1 cm 3 changes by the same amount. If the gas is not too compressed and the motion of gas molecules can be considered completely independent of each other, then the number of impacts per 1 sec per 1 cm 2 of the vessel wall is proportional to the number of molecules in 1 cm 3 . Therefore, if the average speed of molecules does not change over time (we have already seen that in the macrocosm this means a constant temperature), then the gas pressure should be proportional to the number of molecules in 1 cm 3 , i.e., the density of the gas. Thus, the Boyle-Mariotte law is an excellent confirmation of our ideas about the structure of a gas.
However, Boyle's law - Mariotte ceases to be justified if we go to high pressures. And this circumstance can be clarified, as M. V. Lomonosov believed, on the basis of molecular concepts.
On the one hand, in highly compressed gases, the sizes of the molecules themselves are comparable with the distances between the molecules. Thus, the free space in which the molecules move is less than the total volume of the gas. This circumstance increases the number of molecular impacts on the wall, since it reduces the distance that a molecule must travel to reach the wall.
On the other hand, in a highly compressed and therefore denser gas, molecules are noticeably attracted to other molecules much more of the time than molecules in a rarefied gas. This, on the contrary, reduces the number of molecular impacts on the wall, since in the presence of attraction to other molecules, the gas molecules move towards the wall at a lower speed than in the absence of attraction. Not too high pressure. the second circumstance is more significant and the product PV decreases slightly. At very high pressures, the first circumstance plays an important role and the product PV increases.
So, the Boyle-Mariotte law itself and deviations from it confirm the molecular theory.
Change in the volume of gas with a change in temperature
We studied how the pressure of a certain mass of gas depends on temperature, if the volume remains unchanged, and on the volume , occupied by the gas if the temperature remains constant. Now we will establish how a gas behaves if its temperature and volume change, while the pressure remains constant.
If we knew how the temperature of the air in the vessel changed in our experiment, and if we accurately measured how the volume of the Gas changes, we could study this phenomenon from a quantitative side. Obviously, for this it is necessary to enclose the vessel in a shell, taking care that all parts of the device have the same temperature, accurately measure the volume of the locked mass of gas, then change this temperature and measure the increment in the volume of gas.
Gay-Lussac's law.
A quantitative study of the dependence of gas volume on temperature at constant pressure was carried out by the French physicist and chemist Gay-Lussac (1778-1850) in 1802.
Experiments have shown that the increase in gas volume is proportional to the increase in temperature. Therefore, the thermal expansion of a gas can, like for other bodies, be characterized by the volume expansion coefficient b. It turned out that for gases this law is observed much better than for solid and liquid bodies, so that the coefficient of volumetric expansion of gases is a value that is practically constant even with very significant increases in temperature, while for liquid and solid bodies it is; constancy is observed only approximately.
From here we find:
(4)
The experiments of Gay-Lussac and others revealed a remarkable result. It turned out that the volume expansion coefficient for all gases is the same (more precisely, almost the same) and equals = 0.00366 . Thus, at heating at a constant pressure by 1 °, the volume of a certain mass of gas increases by the volume that this mass of gas occupied at 0°С (Gay's law - Lussac ).
As can be seen, the expansion coefficient of gases coincides with their thermal pressure coefficient.
It should be noted that the thermal expansion of gases is very significant, so that the volume of gas at 0°C differs markedly from the volume at another, for example, at room temperature. Therefore, as already mentioned, in the case of gases, it is impossible without a noticeable error to replace in formula (4) the volume volume v. In accordance with this, it is convenient to give the expansion formula for gases the following form. For the initial volume, we take the volume at a temperature of 0°C. In this case, the increment in gas temperature t is equal to the temperature measured on the Celsius scale t . Therefore, the volume expansion coefficient
Where 
(5)
Formula (6) can be used to calculate the volume both at temperatures above O o C and at temperatures below 0°C. In this last case I negative. It should, however, be borne in mind that the Gay-Lussac law is not justified when the gas is highly compressed or so cooled that it approaches the state of liquefaction. In this case, formula (6) cannot be used.
Graphs expressing the laws of Char-la and Gay-Lussac
We will plot the temperature of the gas in a constant volume along the abscissa axis, and its pressure along the ordinate axis. Let the gas pressure be 1 at 0°С kg|cm 2 . Using Charles's law, we can calculate its pressure at 100 0 C, at 200°C, at 300°C, etc.
Let's plot this data on a graph. We get an inclined straight line. We can continue this graph in the direction of negative temperatures. However, as already mentioned, Charles's law is applicable only to temperatures that are not very low .. Therefore, the continuation of the graph until it intersects the abscissa axis, i.e., to the point where the pressure is zero, will not correspond to the behavior of a real gas.
Absolute temperature
It is easy to see that the pressure of a gas contained in a constant volume is not directly proportional to the temperature measured on the Celsius scale. This is clear, for example, from the table given in the previous chapter. If at 100 ° C the gas pressure is 1.37 kg1cm 2 , then at 200 ° C it is equal to 1.73 kg/cm 2 . The temperature measured by the Celsius thermometer doubled, and the gas pressure increased only 1.26 times. There is nothing surprising, of course, in this, because the Celsius thermometer scale is set conditionally, without any connection with the laws of gas expansion. It is possible, however, using gas laws, to establish such a scale of temperatures that gas pressure will directly proportional to temperature, measured on this new scale. Zero in this new scale is called absolute zero. This name is adopted because, as proved by the English physicist Kelvin (William Thomson) (1824-1907), no body can be cooled below this temperature. Accordingly, this new scale is called absolute temperature scale. Thus, absolute zero indicates a temperature equal to -273 ° Celsius, and represents a temperature below which no body can be cooled under any conditions. The temperature expressed by the figure 273 ° + is the absolute temperature of a body that, on the Celsius scale, has a temperature equal to. Usually absolute temperatures are denoted by the letter T. Thus, 273 o + = . The absolute temperature scale is often called the Kelvin scale and is written T° K. On the basis of what has been said
The result obtained can be expressed in words: The pressure of a given mass of gas enclosed in a constant volume is directly proportional to the absolute temperature. This is a new expression of Charles' law.
Formula (6) is also convenient to use when the pressure at 0°C is unknown.
Gas volume and absolute temperature
From formula (6), you can get the following formula:
- the volume of a certain mass of gas at constant pressure is directly proportional to the absolute temperature. This is a new expression of Gay-Lussac's law.
Dependence of gas density on temperature
What happens to the density of a certain mass of gas if the temperature rises and the pressure remains unchanged?
Recall that density is equal to the mass of the body divided by the volume. Since the mass of the gas is constant, when heated, the density of the gas decreases as many times as the volume increased.
As we know, the volume of a gas is directly proportional to the absolute temperature if the pressure remains constant. Hence, The density of a gas at constant pressure is inversely proportional to the absolute temperature. If and are the gas densities at temperatures and , then there is a ratio
Unified law of the gaseous state
We considered cases when one of the three quantities characterizing the state of the gas (pressure, temperature and volume) does not change. We have seen that if the temperature is constant, then pressure and volume are related to each other by the Boyle-Mariotte law; if the volume is constant, then pressure and temperature are related by Charles's law; if pressure is constant, then volume and temperature are related by the Gay-Lussac law. Let us establish a relationship between pressure, volume and temperature of a certain mass of gas if all three of these quantities change.
Let the initial volume, pressure and absolute temperature of a certain mass of gas be V 1 , P 1 and T 1 final - V 2 , P 2 and T 2 - One can imagine that the transition from the initial to the final state occurred in two stages. Let, for example, first change the volume of gas from V 1 to V 2 , and the temperature T 1 remained unchanged. The resulting gas pressure is denoted by P cf. . Then the temperature changed from T 1 to T 2 at a constant volume, and the pressure changed from P cf to P 2 . Let's make a table:
Boyle's law - Mariotte
Р 1 V 1 t 1
P cp V 2 T 1
Charles' Law
P cp V 2 T 1
Applying to the first transition the Boyle-Mariotte law, we write
Applying Charles' law to the second transition, one can write
Multiplying these equalities term by term and reducing by P cp we get:
(10)
So, the product of the volume of a certain mass, gas, and its pressure is proportional to the absolute temperature of the gas. This is the unified law of the gas state or the equation of state of the gas.
Law Dalton
So far, we have been talking about the pressure of a single gas - oxygen, hydrogen, etc. But in nature and technology, we very often deal with a mixture of several gases. The most important example of this is air, which is a mixture of nitrogen, oxygen, argon, carbon dioxide and other gases. What does pressure depend on? mixture gases?
Let us place in the flask a piece of a substance that chemically binds oxygen from the air (for example, phosphorus), and quickly close the flask with a cork with a tube. attached to a mercury manometer. After some time, all the oxygen in the air will combine with phosphorus. We will see that the pressure gauge will show a lower pressure than before the removal of oxygen. This means that the presence of oxygen in the air increases its pressure.
An accurate study of the pressure of a mixture of gases was first made by the English chemist John Dalton (1766-1844) in 1809. The pressure that each of the gases that make up the mixture would have if the remaining gases were removed from the volume occupied by the mixture is called partial pressure this gas. Dalton found that the pressure of a mixture of gases is equal to the sum of their partial pressures(Dalton's law). Note that Dalton's law is inapplicable to highly compressed gases, as well as the Boyle-Mariotte law.
How to interpret Dalton's law from the point of view of molecular theory, I will say a little further.
Densities of gases
The density of a gas is one of the most important characteristics of its properties. Speaking of the density of a gas, they usually mean its density under normal conditions(i.e. at a temperature of 0 ° C and a pressure of 760 mm rt. Art.). In addition, they often use relative density gas, which means the ratio of the density of a given gas to the density of air under the same conditions. It is easy to see that the relative density of a gas does not depend on the conditions in which it is located, since, according to the laws of the gas state, the volumes of all gases change equally with changes in pressure and temperature.
Densities of some gases
|
Density under normal conditions in g/l or in kg/m 3 |
Relation to air density |
Relationship to hydrogen density |
Molecular or atomic weight |
|
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0,0899 1,25 1,43 1,977 0,179 |
0,0695 0,967 1.11 1,53 0,139 |
29 (medium) |
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Hydrogen (H 2) |
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Nitrogen (N 2 ) |
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Oxygen (O 2 ) |
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Carbon dioxide (CO 2 ) |
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Helium(Not) |
The density of a gas can be determined as follows. Let us weigh the flask with a cock twice: once by pumping out as much air as possible from it, the second time by filling the flask with the gas under study to a pressure that must be known. Dividing the difference in weights by the volume of the flask, which must be determined in advance, we find the density of the gas under these conditions. Then, using the equation of state of gases, we can easily find the density of the gas under normal conditions d n. Indeed, we put in the formula (10) P 2 \u003d\u003d R n, V 2 \u003d V n, T 2 \u003d T n and, multiplying the numerator and denominator
formulas for the mass of gas m, we get:
Hence, taking into account what we find:
The results of measurements of the density of some gases are given in the table above.
The last two columns indicate the proportionality between the density of the gas and its molecular weight (in the case of helium, the atomic weight).
Avogadro's Law
Comparing the numbers in the penultimate column of the table with the molecular weights of the gases under consideration, it is easy to see that the densities of gases under the same conditions are proportional to their molecular weights. A very important conclusion follows from this fact. Since molecular weights are related as masses of molecules, then
, where d is the density of gases, and m are the masses of their molecules.
the masses of their molecules. On the other hand, the masses of gases M 1 and M 2 , enclosed in equal volumes V, are related as their densities:
denoting the number of molecules of the first and second gases contained in the volume V, letters N 1 and N 2, we can write that the total mass of the gas is equal to the mass of one of its molecules, multiplied by the number of molecules: M 1 =t 1 N 1 and M 2 =t 2 N 2 That's why
Comparing this result with the formula , find,
that N 1 \u003d N 2. So , at the same pressure and temperature, equal volumes of different gases contain the same number of molecules.
This law was discovered by the Italian chemist Amedeo Avogadro (1776-1856) on the basis of chemical research. It refers to gases that are not very strongly compressed (for example, gases under atmospheric pressure). In the case of highly compressed gases, it cannot be considered valid.
Avogadro's law means that the pressure of a gas at a certain temperature depends only on the number of molecules per unit volume of the gas, but does not depend on whether these molecules are heavy or light. Having understood this, it is easy to understand the essence of Dalton's law. According to the Boyle-Mariotte law, if we increase the density of a gas, that is, we add a certain number of molecules of this gas to a certain volume, we increase the pressure of the gas. But according to Avogadro's law, the same increase in pressure should be obtained if, instead of adding molecules of the first gas, we add the same number of molecules of another gas. This is exactly what Dalton's law consists of, which states that it is possible to increase the pressure of a gas by adding molecules of another gas to the same volume, and if the number of molecules added is the same as in the first case, then the same increase in pressure will be obtained. It is clear that Dalton's law is a direct consequence of Avogadro's law.
Gram molecule. Avogadro's number.
The number giving the ratio of the masses of two molecules indicates at the same time the ratio of the masses of two portions of a substance containing the same number of molecules. Therefore, 2 g of hydrogen (the molecular weight of Na is 2), 32 G oxygen (molecular weight Od is 32) and 55.8 G iron (its molecular weight coincides with the atomic weight, equal to 55.8), etc. contain the same number of molecules.
The amount of a substance containing the number of grams equal to its molecular weight is called gram molecule or we pray.
From what has been said, it follows that moths of various substances contain the same number of molecules. Therefore, it often turns out to be convenient to use the mole as a special unit containing a different number of grams for different substances, but the same number of molecules.
The number of molecules in one mole of a substance that received the name Avogadro numbers, is an important physical quantity. Numerous and varied studies have been made to determine the Avogadro number. They relate to Brownian motion, to the phenomena of electrolysis, and to a number of others. These studies have produced fairly consistent results. It is currently assumed that Avogadro's number is
N= 6,02*10 23 mol -1 .
So, 2 g of hydrogen, 32 g of oxygen, etc. contain 6.02 * 10 23 molecules each. To imagine the enormity of this number, imagine a desert of 1 million square kilometers covered with a layer of sand 600 meters thick. m. Then, if each grain of sand has a volume of 1 mm 3 , then the total number of grains of sand in the desert will be equal to Avogadro's number.
It follows from Avogadro's law that moles of different gases have the same volume under the same conditions. The volume of one mole under normal conditions can be calculated by dividing the molecular weight of a gas by its density under normal conditions.
Thus, the volume of a mole of any gas under normal conditions is 22400 cm 3.
Speeds gas molecules
What are the speeds at which molecules, in particular gas molecules, move? This question naturally arose as soon as ideas about molecules were developed. For a long time, the velocities of molecules could be estimated only by indirect calculations, and only relatively recently were methods developed for directly determining the velocities of gas molecules.
First of all, let us clarify what is meant by the speed of molecules. Recall that due to incessant collisions, the speed of each individual molecule changes all the time: the molecule moves either quickly or slowly, and for some time the velocity of the molecule takes on many different values. On the other hand, at any particular moment in the vast number of molecules that make up the volume of gas under consideration, there are molecules with very different velocities. Obviously, to characterize the state of a gas, one must speak of a certain average speed. It can be considered that this is the average velocity of one of the molecules over a sufficiently long period of time, or that it is the average velocity of all gas molecules in a given volume at some point in time.
Let us dwell on the arguments that make it possible to calculate the average velocity of gas molecules.
Gas pressure proportional Friv 2 , where t - mass of the molecule v- average speed and P - the number of molecules per unit volume. A more accurate calculation leads to the formula
A number of important consequences can be deduced from formula (12). Let us rewrite formula (12) in the following form:
where e is the average kinetic energy of one molecule. Let us denote the gas pressure at temperatures T 1 and T 2 with the letters p 1 and p 2 and the average kinetic energies of the molecules at these temperatures e 1 and e 2 . In this case
Comparing this ratio with Charles' law
So, the absolute temperature of the gas is proportional to the average kinetic energy of the gas molecules. Since the average kinetic energy of the molecules is proportional to the square of the average velocity of the molecules, our comparison leads to the conclusion that the absolute temperature of the gas is proportional to the square of the average velocity of the gas molecules and that the speed of molecules increases in proportion to the square root of the absolute temperature.
Average velocities of molecules of some gases
As you can see, the average velocities of the molecules are very significant. At room temperature, they typically reach hundreds of meters per second. In a gas, the average speed of movement of molecules is about one and a half times greater than the speed of sound in the same gas.
At first glance, this result seems very strange. It seems that molecules cannot move at such high speeds: after all, diffusion even in gases, and even more so in liquids, proceeds relatively very slowly, in any case, much more slowly than sound propagates. The point, however, is that, while moving, the molecules very often collide with each other and at the same time change the direction of their movement. As a result, they move in one direction or the other, mostly pushing in one place. As a result, despite the high speed of movement in the intervals between collisions, despite the fact that the molecules do not linger anywhere, they move in any particular direction rather slowly.
The table also shows that the difference in the speeds of different molecules is due to the difference in their masses. This circumstance is confirmed by a number of observations. For example, hydrogen penetrates through narrow holes (pores) at a higher rate than oxygen or nitrogen. It can be found in this experience.
The glass funnel is closed with a porous vessel or sealed with paper and lowered end into the water. If the funnel is covered with a glass, under which hydrogen (or luminous gas) is let in, we will see that the water level at the end of the funnel will drop and bubbles will begin to come out of it. How to explain it?
Through narrow pores in a vessel or paper, both air molecules (from inside the funnel under the glass) and hydrogen molecules (from under the glass into the funnel) can pass. But the speed of these processes is different. The difference in the size of the molecules does not play a significant role in this, because the difference is small, especially compared to the size of the pores: the hydrogen molecule has a “length” of about 2.3 * 10 -8 cm, and a molecule of oxygen or nitrogen is about 3 * 10 -8 cm, the diameter of the holes, which are pores, is thousands of times larger. The high rate of hydrogen penetration through the porous wall is explained by the higher speed of movement of its molecules. Therefore, hydrogen molecules quickly penetrate from the glass into the funnel. As a result, molecules accumulate in the funnel, the pressure increases and the mixture of gases in the form of bubbles comes out.
Such devices are used to detect the admixture of firedamp gases to the air, which can cause an explosion in mines.
Heat capacity of gases
Suppose we have 1 G gas. How much heat must be imparted to it in order for its temperature to increase by 1 ° C, in other words, what specific heat capacity of the gas? This question, as experience shows, cannot be answered unambiguously. The answer depends on the conditions under which the gas is heated. If its volume does not change, then a certain amount of heat is needed to heat the gas; this also increases the pressure of the gas. If the heating is carried out in such a way that its pressure remains unchanged, then a different, larger amount of heat will be required than in the first case; this will increase the volume of the gas. Finally, other cases are possible when both volume and pressure change during heating; in this case, an amount of heat will be required, depending on the extent to which these changes occur. According to what has been said, the gas can have a wide variety of specific heat capacities, depending on the heating conditions. There are usually two of all these specific heat capacities: specific heat capacity at constant volume (C v ) and specific heat at constant pressure (C p ).
To determine C v, it is necessary to heat the gas placed in a closed vessel. The expansion of the vessel itself during heating can be neglected. When determining C p, it is necessary to heat the gas placed in a cylinder closed by a piston, the load on which remains unchanged.
The heat capacity at constant pressure C p is greater than the heat capacity at constant volume C v . Indeed, when heated 1 G gas by 1 ° at a constant volume, the heat supplied is used only to increase the internal energy of the gas. To heat the same mass of gas at a constant pressure by 1 °, it is necessary to impart heat to it, due to which not only the internal energy of the gas will increase, but also the work associated with the expansion of the gas will be performed. To obtain C p to the value of C v, you must add another amount of heat equivalent to the work done during the expansion of the gas.
Abstract on the topic:
Air density
Plan:
- Introduction
- 1
Relationships within the ideal gas model
- 1.1 Temperature, pressure and density
- 1.2 Influence of air humidity
- 1.3 Influence of height above sea level in the troposphere
Notes
Introduction
Air density- the mass of gas of the Earth's atmosphere per unit volume or the specific mass of air under natural conditions. Value air density is a function of the height of the measurements taken, of its temperature and humidity. Usually the standard value is considered to be 1.225 kg ⁄ m 3 , which corresponds to the density of dry air at 15°C at sea level.
1. Relationships within the ideal gas model
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1.1. Temperature, pressure and density
The density of dry air can be calculated using the Clapeyron equation for an ideal gas at a given temperature. and pressure:
Here ρ - air density, p- absolute pressure, R- specific gas constant for dry air (287.058 J ⁄ (kg K)), T is the absolute temperature in Kelvin. So by substitution we get:
- under the standard atmosphere of the International Union of Pure and Applied Chemistry (temperature 0 ° C, pressure 100 kPa, zero humidity), air density is 1.2754 kg ⁄ m³;
- at 20 °C, 101.325 kPa and dry air, the density of the atmosphere is 1.2041 kg ⁄ m³.
The table below shows various air parameters calculated on the basis of the corresponding elementary formulas, depending on temperature (pressure is taken as 101.325 kPa)
1.2. Influence of air humidity
Humidity refers to the presence of gaseous water vapor in the air, the partial pressure of which does not exceed the saturated vapor pressure for given atmospheric conditions. The addition of water vapor to air leads to a decrease in its density, which is explained by the lower molar mass of water (18 g ⁄ mol) compared to the molar mass of dry air (29 g ⁄ mol). Humid air can be considered as a mixture of ideal gases, the combination of the densities of each of which makes it possible to obtain the required value for their mixture. This interpretation allows the determination of the density value with an error level of less than 0.2% in the temperature range from -10 °C to 50 °C and can be expressed as follows:
where is the density of moist air (kg ⁄ m³); p d- partial pressure of dry air (Pa); R d- universal gas constant for dry air (287.058 J ⁄ (kg K)); T- temperature (K); p v- water vapor pressure (Pa) and R v- universal constant for steam (461.495 J ⁄ (kg K)). Water vapor pressure can be determined from relative humidity:
where p v- water vapor pressure; φ - relative humidity and p sat is the partial pressure of saturated vapor, the latter can be represented as the following simplified expression:
which gives the result in millibars. Dry air pressure p d determined by a simple difference:
where p denotes the absolute pressure of the system under consideration.
1.3. Influence of height above sea level in the troposphere
The dependence of pressure, temperature and air density on altitude compared to the standard atmosphere ( p 0 \u003d 101325 Pa, T0\u003d 288.15 K, ρ 0 \u003d 1.225 kg / m³).
The following parameters can be used to calculate the air density at a certain height in the troposphere (the value for the standard atmosphere is indicated in the atmospheric parameters):
- standard atmospheric pressure at sea level - p 0 = 101325 Pa;
- standard temperature at sea level - T0= 288.15K;
- acceleration of free fall over the surface of the Earth - g\u003d 9.80665 m ⁄ sec 2 (for these calculations it is considered a value independent of height);
- rate of temperature drop (eng.) rus. with height, within the troposphere - L= 0.0065 K ⁄ m;
- universal gas constant - R\u003d 8.31447 J ⁄ (Mol K) ;
- molar mass of dry air - M= 0.0289644 kg ⁄ Mol.
For the troposphere (i.e., the region of linear temperature decrease - this is the only property of the troposphere used here), the temperature at altitude h above sea level can be given by the formula:
pressure at altitude h:
Then the density can be calculated by substituting the temperature T and pressure P corresponding to a given height h into the formula:
These three formulas (dependence of temperature, pressure and density on height) are used to construct the graphs shown on the right. Graphs are normalized - they show the general behavior of the parameters. "Zero" values for correct calculations must each time be substituted in accordance with the readings of the relevant instruments (thermometer and barometer) at the moment at sea level.
The derived differential equations (1.2, 1.4) contain parameters that characterize a liquid or gas: density r , viscosity m , as well as the parameters of the porous medium - the porosity coefficients m and permeability k . For further calculations, it is necessary to know the dependence of these coefficients on pressure.
Dropping Liquid Density. With steady filtration of a dropping liquid, its density can be considered independent of pressure, i.e., the liquid can be considered as incompressible: r = const .
In transient processes, it is necessary to take into account the compressibility of the liquid, which is characterized by liquid volumetric compression ratio b . This coefficient is usually considered constant:
Integrating the last equality from the initial pressure values p 0 and density r0 to the current values, we get:
In this case, we obtain a linear dependence of density on pressure.
Density of gases. Compressible liquids (gases) with small changes in pressure and temperature can also be characterized by volumetric compression and thermal expansion coefficients. But with large changes in pressure and temperature, these coefficients change within wide limits, so the dependence of the density of an ideal gas on pressure and temperature is based on Claiperon–Mendeleev equations of state:
where R' = R/M m is the gas constant, which depends on the composition of the gas.
The gas constant for air and methane, respectively, are equal, R΄ of air = 287 J/kg K˚; R΄ methane = 520 J/kg K˚.
The last equation is sometimes written as:
 | (1.50) |
From the last equation it can be seen that the density of a gas depends on pressure and temperature, so if the density of a gas is known, then it is necessary to indicate the pressure, temperature and composition of the gas, which is inconvenient. Therefore, the concepts of normal and standard physical conditions are introduced.
Normal conditions correspond to temperature t = 0°C and pressure p at = 0.1013°MPa. The density of air under normal conditions is equal to ρ v.n.us = 1.29 kg / m 3.
Standard Conditions correspond to temperature t = 20°C and pressure p at = 0.1013°MPa. The density of air under standard conditions is ρ w.st.us = 1.22 kg / m 3.
Therefore, from the known density under given conditions, it is possible to calculate the gas density at other values of pressure and temperature:
Excluding reservoir temperature, we obtain the ideal gas equation of state, which we will use in the future:
where z - coefficient characterizing the degree of deviation of the state of a real gas from the law of ideal gases (supercompressibility coefficient) and depending for a given gas on pressure and temperature z = z(p, T) . Values of the coefficient of supercompressibility z are determined by the graphs of D. Brown.
Oil viscosity. Experiments show that the viscosity coefficients of oil (at pressures above saturation pressure) and gas increase with increasing pressure. With significant pressure changes (up to 100 MPa), the dependence of the viscosity of reservoir oils and natural gases on pressure can be taken exponential:
 | (1.56) |
For small changes in pressure, this dependence is linear.
Here m0 – viscosity at fixed pressure p0 ; β m - coefficient determined experimentally and depending on the composition of oil or gas.
Formation porosity. To find out how the porosity coefficient depends on pressure, consider the question of stresses acting in a porous medium filled with liquid. When the pressure in the liquid decreases, the forces on the skeleton of the porous medium increase, so the porosity decreases.
Due to the small deformation of the solid phase, it is usually considered that the change in porosity depends linearly on the change in pressure. The rock compressibility law is written as follows, introducing formation volumetric elasticity coefficient b c:
where m0 – coefficient of porosity at pressure p0 .
Laboratory experiments for various granular rocks and field studies show that the coefficient of volumetric elasticity of the formation is (0.3 - 2) 10 -10 Pa -1 .
With significant changes in pressure, the change in porosity is described by the equation:
and for large - exponential:
 | (1.61) |
In fractured reservoirs, the permeability changes more intensively depending on pressure than in porous ones; therefore, in fractured reservoirs, taking into account the dependence k(p) more necessary than in granular.
The equations of state of the liquid or gas saturating the formation and the porous medium complete the system of differential equations.
As a rule, as the temperature decreases, the density increases, although there are substances whose density behaves differently, such as water, bronze, and cast iron. Thus, the density of water has a maximum value at 4 ° C and decreases both with an increase and a decrease in temperature relative to this number.
When the state of aggregation changes, the density of a substance changes abruptly: the density increases during the transition from a gaseous state to a liquid state and when a liquid solidifies. True, water is an exception to this rule, its density decreases during solidification.
The ratio of P. of two substances under certain standard physical conditions is called relative P.: for liquid and solid substances, it is usually determined in relation to the P. of distilled water at 4 ° C, for gases, in relation to the P. of dry air or hydrogen under normal conditions.
The unit of P. in SI is kg/m 3 , in the CGS system of units g / cm 3. In practice, non-systemic units of P. are also used: g/l, t/m 3 and etc.
Densitometers, pycnometers, hydrometers, and hydrostatic weighing are used to measure P. of substances (see Mora scales) . Dr. methods for determining P. are based on the connection of P. with the parameters of the state of a substance or with the dependence of the processes occurring in a substance on its P. Thus, the density ideal gas can be calculated from equation of state r= pm/RT where p - gas pressure, m - its molecular mass (molar mass), R - gas constant , T - absolute temperature, or determined, for example, by the speed of propagation of ultrasound (here b is the adiabatic compressibility gas).
The range of values of P. of natural bodies and environments is exceptionally wide. For example, the density of the interstellar medium does not exceed 10 -21 kg/m 3 , the average P. of the Sun is 1410 kg/m 3 , Lands - 5520 kg/m 3 , the largest P. of metals - 22,500 kg/m 3 (osmium), P. substances of atomic nuclei - 10 17 kg/m 3 , finally, the neutron stars can apparently reach 10 20 kg/m 3 .
pressure gauge- This is a mechanical measuring device, structurally representing a steel or plastic dial with a spring in the form of a tube, designed to measure the pressure of liquid and gaseous substances.
In mechanical pressure gauges, the measured pressure with the help of a sensitive element is converted into mechanical movement, causing mechanical deviation of the arrows or other parts of the reference mechanisms, recording the measurement result, as well as signaling and pressure stabilization devices in the systems of the controlled object. As sensitive elements of mechanical pressure gauges, tubular springs, harmonic (bellows) and flat membranes and other measuring mechanisms are used, in which, under the action of pressure, elastic deformations or elasticity of special springs are caused.
By accuracy, all mechanical pressure gauges are divided into: technical, control and exemplary. Technical pressure gauges have accuracy classes 1.5; 2.5; 4; control 0.5; 1.0; exemplary 0.16; 0.45.
Manometric tubular springs are hollow tubes of oval or other section, bent along an arc of a circle, along helical or spiral lines and having one or more turns. The conventional design, which is most commonly used in practice, uses single coil springs. Principal and structural diagrams of a pressure gauge with a single-coil tubular spring are shown in Fig.2.
Fig.2. Mechanical pressure gauge and its characteristics
The end of the manometric spring 5 is soldered to the fitting 1. The second soldered end K is pivotally connected by a rod 3 to the lever of the gear sector 4. The teeth of the sector are engaged with the driven gear wheel 6, which is mounted on the axis 7 of the arrow 9. To eliminate fluctuations of the arrow due to gaps between the teeth gears use a spiral spring 2, the ends of which are connected to the body and the axis 7. Under the arrow is a fixed scale.
Under the influence of the pressure difference inside and outside, the tubular spring changes the shape of its section, as a result of which its sealed end K moves in proportion to the existing pressure difference.
The structural diagram of a mechanical pressure gauge (Fig. 2b) consists of three linear links I, II, III, the static characteristics of which are represented by graphs , and, where is the displacement of the free end of the tubular spring, is the initial central angle of the tubular spring. Due to the linearity of all links, the overall static characteristic of the pressure gauge is linear and the scale is uniform. The input value of link I is the measured pressure, and the output value is the displacement of the free (soldered) end of the manometric spring5. Link 3 with the lever of the gear sector 4 forms the second link. The input value of link II is , and the output value is the angular deviation of the end of the manometric spring. The input value of link III (link III is a gear sector engaged with the driven gear wheel 6) is the angular deviation, and the output value is the angular deviation of the arrow 9 from the zero mark of the scale 8.
Mechanical pressure gauges are used for measurements in the low vacuum region. In strain gauges, the elastic element associated with the indicator sags under the influence of the difference between the measured and reference pressures (atmosphere or high vacuum). In industrial bellows pressure gauges of the BC-7 series, the measured pressure causes the bellows to move, which is transmitted to the recorder. These devices have a linear scale up to 760 torr and an accuracy of 1.6%.
Influence of Temperature and Pressure on Gas Density Gases, in contrast to dropping liquids, are characterized by significant compressibility and high values of thermal expansion coefficient. The dependence of gas density on pressure and temperature is established by the equation of state. The simplest properties are possessed by a gas rarefied to such an extent that the interaction between its molecules can be ignored. This is an ideal (perfect) gas, for which the Mendeleev-Clapeyron equation is valid:
Influence of temperature and pressure on gas density р - absolute pressure; R - specific gas constant, different for different gases, but independent of temperature and pressure (for air R = 287 J / (kg K); T - absolute temperature. The behavior of real gases in conditions far from liquefaction differs only slightly on the behavior of perfect gases, and for them in a wide range it is possible to use the equations of state of perfect gases.
Influence of temperature and pressure on gas density In technical calculations, gas density is usually brought to normal physical conditions: T=20°C; p = 101325 Pa. For air under these conditions, ρ = 1.2 kg / m 3. The density of air under other conditions is determined by the formula:
Influence of temperature and pressure on gas density According to this formula for an isothermal process (T = const): An adiabatic process is a process that proceeds without external heat transfer. For an adiabatic process, k=cp /cv is the adiabatic constant of the gas; cp - heat capacity, gas at constant pressure; cv - the same, at a constant volume.
Influence of temperature and pressure on gas density An important characteristic that determines the dependence of the change in density with a change in pressure in a moving stream is the speed of sound propagation a. In a homogeneous medium, the speed of sound propagation is determined from the expression: For air a = 330 m/s; for carbon dioxide 261 m/s.
Influence of temperature and pressure on gas density Since the volume of a gas depends to a large extent on temperature and pressure, the conclusions obtained in the study of dropping liquids can be extended to gases only if pressure and temperature changes within the phenomenon under consideration are insignificant. 3 Significant pressure differences, causing a significant change in the density of gases, can occur when they move at high speeds. The ratio between the speed of movement and the speed of sound in it makes it possible to judge the need to take into account the compressibility in each specific case.
Influence of temperature and pressure on gas density If a liquid or gas is moving, then to assess the compressibility, they use not the absolute value of the speed of sound, but the Mach number, which is equal to the ratio of the flow velocity to the speed of sound. М = ν/а If the Mach number is much less than unity, then the dropping liquid or gas can be considered practically incompressible
Equilibrium of gas At a small height of the gas column, its density can be considered the same along the height of the column: then the pressure created by this column is determined by the basic equation of hydrostatics. When the air column is high, its density at different points is no longer the same, so the hydrostatic equation does not apply in this case.
Gas equilibrium Considering the differential pressure equation for the case of absolute rest and substituting the density value into it, we have In order to integrate this equation, it is necessary to know the law of change in air temperature with respect to the height of the air column. It is not possible to express the change in temperature with a simple function of height or pressure, so the solution of the equation can only be approximate.
Gas equilibrium For individual layers of the atmosphere, with sufficient accuracy, it can be assumed that the change in temperature depending on the height (and for the mine - on the depth) occurs according to a linear law: T \u003d T 0 + αz, where T and T 0 are the absolute air temperature, respectively, at height (depth) z and on the surface of the earth α-temperature gradient characterizing the change in air temperature with an increase in height (-α) or depth (+α) by 1 m, K / m.
Gas equilibrium The values of the coefficient α are different in different areas along the height in the atmosphere or along the depth in the mine. In addition, they also depend on meteorological conditions, time of year, and other factors. When determining the temperature within the troposphere (i.e., up to 11000 m), they usually take α = 0.0065 K/m; for deep mines, the average value of α is taken equal to 0.004÷ 0.006 K/m wet - 0.01.
Gas equilibrium Substituting the formula for temperature change into the differential pressure equation and integrating it, we obtain The equation is solved with respect to H, replacing the natural logarithms with decimal ones, α - its value from the equation through temperature, R - the value for air, equal to 287 J / (kg K) ; and substitute g = 9.81 m/s2.
Gas equilibrium As a result of these actions, the barometric formula H \u003d 29, 3 (T-T 0) (lg p / p 0) / (lg. T 0 / T), as well as the formula for determining pressure, where n is determined by the formula
STEADY GAS MOVEMENT IN PIPES The law of conservation of energy in mechanical form for an element of length dx of a round pipe with a diameter d, provided that the change in geodesic height is small compared to the change in piezometric pressure, has the form process with a constant polytropic exponent n = const and assuming that λ= const after integration, the pressure distribution law along the gas pipeline is obtained
STEADY GAS MOVEMENT IN PIPES
STEADY GAS MOVEMENT IN PIPES M ω At n = 1, the formulas are valid for a steady isothermal gas flow. The coefficient of hydraulic resistance λ for gas, depending on the Reynolds number, can be calculated from the formulas used in fluid flow.
When moving real hydrocarbon gases for an isothermal process, the equation of state is used where the compressibility factor z of natural hydrocarbon gases is determined from experimental curves or analytically from approximate equations of state.
ω
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