Dalton's laws

Dalton's laws

In the modern world, electronic technology is developing by leaps and bounds. Every day something new appears, and these are not only small improvements to existing models, but also the results of the use of innovative technologies that make it possible to significantly improve performance.

The instrument-making industry is not lagging behind the electronic technology - after all, in order to develop and launch new devices on the market, they must be thoroughly tested, both at the design and development stage, and at the production stage. New measuring equipment and new measurement methods are appearing, and, consequently, new terms and concepts.

This section is intended for those who often encounter incomprehensible abbreviations, abbreviations and terms and would like to better understand their meanings.

Dalton's laws- two physical laws that determine the total pressure and solubility of a mixture of gases. Formulated by John Dalton at the beginning of the 19th century.

Formulation of laws

Law on the total pressure of a mixture of gases

The pressure of a mixture of chemically non-interacting ideal gases is equal to the sum of the partial pressures.

Law on the solubility of gas mixture components

At a constant temperature, the solubility in a given liquid of each of the components of the gas mixture located above the liquid is proportional to their partial pressure.

Limits of applicability

Both Dalton's laws are strictly satisfied for ideal gases. For real gases, these laws are applicable provided that their solubility is low and their behavior is close to that of an ideal gas.

History of discovery

The law of addition of partial pressures was formulated in 1801. At the same time, the correct theoretical justification, based on the molecular kinetic theory, was made much later.

Statement of Dalton's laws

Law on the total pressure of a mixture of gases

Pressure mixtures chemically not interacting ideal gases equal to the sum partial pressures.

(\displaystyle P=\sum _(i=1)^(n)(p_(i))=p_(1)+p_(2)+\cdots +p_(n))

Law on the solubility of gas mixture components

At constant temperature solubility in a given liquid, each of the components of the gas mixture located above the liquid is proportional to their partial pressure.

(\displaystyle \m_(i)=(\frac (p_(i))(P)))

Limits of applicability

Both Dalton's laws are strictly satisfied for ideal gases. For real gases these laws are applicable provided that their solubility is low and their behavior is close to that of an ideal gas.

Explain the terms mass volume and mole fraction

The ratio of the mass of the solute to the total mass of the solution is called the mass fraction of the solute.

Volume fraction(Sometimes volumetric part) is a dimensionless quantity equal to the ratio of the volume of a substance in a mixture to the volume of the entire mixture. Denoted by the letter φ.

MOLAR FRACTION

dimensionless physical a value characterizing the concentration and equal to the ratio of the amount of a component to the total amount of the mixture. ppm is expressed in fractions of a unit, for example hundredths (percent), thousandths (ppm), millionths and is designated accordingly %, ooo, million -1 .

The principle of equivalence of energy conversion

The principle of equivalence can be formulated as follows. If different types of energy are taken in such quantities that they will cause the same change in the state of a given closed system, then they are equivalent.
Generalizations of the equivalence principle lead to the first law of thermodynamics (the law of conservation of energy). It states that in an isolated system the sum of all types of energy is constant, while different forms of energy can transform into each other. The law of conservation of energy covers all forms of energy that can be found in a given system. The sum of the different types of energy that a system possesses is called, according to Clausius's definition, internal energy and. Thus, the internal energy of a substance is the sum of various energies, for example, the kinetic energy of its atoms or molecules, potential energy, as well as the energy of electric and magnetic fields, etc.

5. Carnot's theorem

Let us briefly analyze the formula for the Terminian efficiency of a reversible direct Carnot cycle:

From this equality it follows:

1) thermal efficiency depends only on the temperatures of hot and cold sources;

2) h t(for the Carnot cycle) the higher the temperature of the hot spring (71) and the lower the temperature of the cold spring (72);

3) in the Carnot cycle, the thermal efficiency must be less than unity. Because h t= 1 can only be in the case of T 2 / T 1 = 0, when T 1 = 0, or T 2 = 0 (or T 2 = -273.15 o C). The cold source temperature 72 in real heat engines is usually a temperature T 2 = 260 – 300 K(environment). The heater temperature in the furnace of steam power plants is approximately 2000 K, and in internal combustion engines it is about 2500 K, since the walls of the piston cylinders of these engines are cooled, and combustion products become the working substance. This implies the same statement that all the heat supplied to the gas during the cycle cannot be completely converted into useful work; this transition must necessarily be accompanied by the loss of part of the heat (it is absorbed by the cold source);

4) in the Carnot cycle the thermal efficiency is zero in the case of T 1 = T 2 . It follows from this that if thermal equilibrium is maintained in the system, i.e., the temperature of all bodies in the system is the same, then the conversion of heat into useful work is impossible. For the Carnot cycle (direct) it is true: h t= 1 – T 2 / T 1 = 1 – 1 = 0 at T 1 = ? t = T 2 (in case of equal temperatures of both sources);

5) thermal efficiency? t characterizes a reversible Carnot cycle (circular process). All real processes are irreversible, this is explained by energy losses (due to heat transfer, friction, etc.). Therefore, the thermal efficiency of a real Carnot cycle (irreversible) is always less than 1 – T 2 /T 1 . The main feature of this cycle is that it is the same for both ideal and ordinary real gases, if the temperatures are given ( T 1 , T 2) sources. This statement is the essence Carnot's theorem, which reads: “In a heat engine for any reversible cycle, the thermal efficiency will not depend either on the nature of the cycle or on the type of substance (working fluid).” It will be determined only by the ratio of the temperatures of the heater (heat transmitter) and refrigerator (heat receiver). In other words, in a heat engine, for each reversible cycle, the thermal efficiency is calculated using the same formula as defined for the reversible Carnot cycle.

Single stage compressor

A single-stage compressor is unsuitable for obtaining high pressures and therefore it is used to obtain compressed air of no more than 10 - 12 sip. In addition, at high compression pressures, the temperature rises, which worsens lubrication conditions.

Single-stage compressors are used to compress gases to pressures that rarely exceed 6 - 7 atm. Higher pressures can only be achieved in multi-stage compressors for the following reasons.

Cycle, diagram, term of efficiency of gas turbine unit

The working fluid of a gas turbine is the combustion products of liquid or gaseous fuel.

Figure 7.6 shows a diagram of the simplest gas turbine installation with fuel combustion at constant pressure. Fuel pump 5 and compressor 4 supply fuel and air through nozzles 6 and 7 into combustion chamber 1. From the chamber, combustion products are directed to combined nozzles 2, where they expand and enter the blades of gas turbine 3.

Figure 7.7 and Figure 7.8 show the ideal cycle of a gas turbine plant on PV and TS diagrams.

1-2 - adiabatic compression to pressure P 2;

2-3 – heat supply q 1 at constant pressure P 2 (fuel combustion);

3-4 – adiabatic expansion to the initial pressure P1;

4-1 – cooling of the working fluid at constant pressure P 1 (heat removal q 2);

The characteristics of the cycle are:

degree of increase pressure -  = P 2 / P 1;

degree of isobaric expansion-  =  3 / 2 .

Turbine operation:

l t = h 3 – h 4. (7.10)

Compressor operation:

l n = h 2 – h 1. (7.11)

The useful work of a gas turbine is equal to the difference between the work of the turbine and the compressor:

L GTU = l t – l c. (7.12)

Thermal efficiency GTU cycle has the form:

 t = 1 – 1/  (-1)/ . (7.13)

Theoretical power of the gas turbine, compressor and installation (GTU):

N t = l t D/3600 = (h 3 – h 4) D/3600, (7.14)

N k = l k D/3600 = (h 2 – h 1) D/3600 , (7.15)

N gas turbine unit = l gas turbine unit ·D/3600 = [(h 3 – h 4) (h 2 – h 1) ]·D/3600. (7.16)

The actual cycle of a gas turbine plant differs from the theoretical one in the presence of losses due to friction and vortex formation in the turbine and compressor. Effective methods for increasing the efficiency of gas turbine units are: heat recovery, stepwise compression and expansion of the working fluid, etc.

9. Scheme of steam turbine unit (STU)
N

and fig. 9.8, 9.9, 9.10 show the diagrams of a steam turbine unit (STU) and a reversible cycle in p-v- And T-s- diagrams (Rankine cycle).

Designations: PC – steam boiler; PP – steam superheater; ET – screen (evaporation) pipes of a steam boiler; VE – water economizer; T – steam turbine; K – water-cooled condenser; N – pump;
EG – electric current generator (consumer). Numbers on the diagram
correspond to the nodal points of the reversible cycle represented by
V p-v- And T-s- charts

Heat supplied to water and water steam in a steam boiler (in processes: 3-4 - heating water to boiling, 4-5 - evaporation of water, 5-1 - superheating of steam),

The work obtained in the turbine is the external work of the adiabatic expansion process 1-2:

The useful work of a reversible cycle (cycle area in p-v- And T-s- diagrams)

In practical calculations, it is often possible to neglect the work of the pump, which, due to the incompressibility of the fluid, is negligible compared to the work of the turbine. In this case, state 3 is not shown on the diagrams (Fig. 9.11), since point 3 coincides with point 2¢:

,
,	(9.11)
.	(9.12)

Analysis of formulas (9.9) – (9.12) shows that thermal efficiency depends on three parameters ( p 1 , t 1 , p 2), it increases with increasing pressure p 1 in a steam boiler, with increasing steam superheat temperature t 1 and with decreasing pressure p 2 in the capacitor.

In modern powerful steam turbine plants, steam parameters are used p 1 = 235...240 bar, t 1 = 535...565 o C, p 2 = 0.03...0.05 bar
(ts= 25...35 o C). Upgrading to higher settings p 1 and t 1 is determined by the level of development of metallurgy, since expensive high-alloy steels are required. Using lower pressures p 2 is limited by the temperature of the water cooling the condenser, which in summer is 18...20 o C.

IN steam turbine plant could implement a Carnot cycle a-4-5-b(Fig. 9.12): 4-5 – evaporation; 5- b– expansion of steam in the turbine; b-a – incomplete steam condensation; a-4 – compression of wet steam in the compressor.

In practice, this cycle is not carried out primarily because in a real cycle, due to losses in the compressor drive, most of the power generated by the turbine would be spent. It is more economical to condense the steam completely, and then use a pump to increase the water pressure from p 2 to p 1 in process 2¢-3. In addition, the expansion process of dry saturated steam in the turbine (5- b) is associated with large friction losses due to a significant decrease in the degree of dryness during the expansion process, i.e., an increase in the water content in the steam. Therefore, in steam turbine installations, steam superheating is used in the superheater pipes of the steam boiler. In this case, the expansion process 1-2 shifts to the area of ​​superheated steam, and friction losses during steam flow in the flow part of the turbine are reduced.

9 .3.2. Efficiency system
to evaluate the effectiveness of vocational training schools. Heat balance of PTU

In Fig. Figure 9.13 shows the actual Rankine cycle 1-2 d-2¢ (excluding pump work costs):

1-2d– irreversible adiabatic process of steam expansion in the turbine ( s 2d > s 1);

1-2 – reversible adiabatic expansion process ( s 2 = s 1).

Thermal efficiency characterizes the thermodynamic perfection of the reversible cycle 1-2-2¢:

Where N i = l i G– internal power (power of the actual cycle).

Heat losses in a steam boiler (from chemical and mechanical underburning of fuel, from heat exchange with the environment, with exhaust gases, etc.) are characterized by Steam boiler efficiency

Where N e = l e G– effective power (on the turbine shaft); l e- effective work.

All losses in the PTU (without taking into account the energy consumer) are characterized by efficient efficiency

Where l uh, N uh = l uh . G– electrical work and electrical power, respectively.

All losses in a steam turbine power plant generating electrical energy are taken into account electrical efficiency

,	(9.19)
.

Rankine cycle for vocational schools

Heat removal from wet steam in the condenser must be carried out until all the steam is completely condensed. In this case, it is not low-density wet steam that is subject to compression from pressure p2 to pressure p1, but water. To supply water to the boiler, a feed water pump is used, which has small dimensions and high efficiency. Such a cycle was proposed in the 50s by the Scottish physicist and engineer W. Rankine. In the Rankine cycle, it is possible to use superheated steam, which makes it possible to increase the average integral temperature of the heat supply and thereby increase the efficiency of the cycle.

In turbine 3, heat is converted into work. The exhaust steam gives off part of the heat to the cooling water in the refrigerator 4 and is supplied back to the boiler by pump 5. In steam boiler 1, due to the heat of the fuel burning in the furnace, the process of steam formation occurs; in steam superheater 2, the necessary steam parameters are achieved.

In the boiler at pressure p1 = const, process 4-5 - heating and 5-6 - evaporation of water occurs.

Process 6-1 – superheating of steam in the superheater to T1. Thus, at the outlet of the superheater the steam has parameters p1, T1, h1, we assume that from the boiler to the turbine p1 = const

1-2 – adiabatic expansion of steam in the turbine to pressure p2 Parameters after the turbine p2 T2 h2

2-3 – isobaric steam condensation. The result will be water with parameters h¢2 T¢2. The condensate, after adiabatic compression from p2 to p1 in the feed pump, enters the boiler.

3-4 – adiabatic compression of water by a pump.

Binary cycles

Based on the considered cycles, we can formulate requirements for the properties of the most convenient working fluid:

The working fluid must provide a higher cycle fill factor. For this, the working fluid must have a lower isobaric heat capacity in the liquid state, higher critical parameters,

The properties of the working fluid must be such that the upper temperature at a sufficiently high cycle fill factor is ensured at a not too high steam pressure, because high pressure makes installation more difficult,

The working fluid must be inexpensive, non-toxic, and aggressive towards structural materials.

Since there are currently no working fluids that satisfy these requirements, it is possible to carry out a cycle using a combination of two working fluids, using each of them in the temperature range where this working fluid has the greatest advantages. Cycles of this kind are called binary cycles.

Cycle of a combined cycle plant

The steam-gas cycle is a binary cycle that uses two working fluids - combustion products and water vapor. In gas

cycle, the gas temperature at the turbine inlet is 900...1000°C, and at the outlet 350°C or more. In steam power plants, the temperature of superheated steam reaches 650°C, but the temperature of the water in the condenser will be only 30°C. Thus, in a binary cycle it is possible to realize a temperature difference significantly greater than in each of the individual cycles, and thereby increase the thermal efficiency of the cycle.

Combined-cycle plant- an electricity generating station used to produce electricity. It differs from steam power and gas turbine plants in its increased efficiency.

Advantages of PSU:

Combined-cycle plants make it possible to achieve electrical efficiency of more than 50%. For comparison, for steam power plants operating separately, the efficiency is usually in the range of 33-45%, for gas turbine plants - in the range of 28-42%

Low cost per unit of installed capacity

Combined-cycle plants consume significantly less water per unit of generated electricity compared to steam power plants

1- Short construction time (9-12 months)

2- There is no need for constant supply of fuel by rail or sea transport

3- Compact dimensions allow construction directly at the consumer (factory or within the city), which reduces the cost of power lines and electricity transportation. Energy

4- More environmentally friendly compared to steam turbine plants.

Disadvantages of PSU:

5- Low unit power of equipment (160-972.1 MW per unit), while modern thermal power plants have a unit power of up to 1200 MW, and nuclear power plants have a unit capacity of up to 1200 MW, and nuclear power plants have a unit capacity of up to 1200-1600 MW.

The need to filter the air used for fuel combustion.

An ideal steam-gas cycle is a cycle where the temperatures of the working fluids during the supply and removal of heat are constant and equal to the temperatures of the corresponding heat sources.

The practical implementation of isothermal heat supply in the gas cycle 1-2-3-6-1 can be carried out as a result of a multi-stage

heat supply, and isothermal heat removal in the steam cycle 3-4-5-6-3 can be carried out in the process of condensation of water vapor (process 4-5). Heat is transferred from combustion products to water vapor in a heat exchanger. The gas cycle in this scheme is open (combustion products are removed into the atmosphere), and the steam cycle is closed.

The working fluid in a closed cycle can be not only water, but also carbon dioxide or other substances that utilize the heat of the gas cycle.

Gas is also carried out in the form of a gas turbine unit cycle with heat supply at p = const (on the Ts diagram, area 72367).

The gas and steam cycles can be combined in a gas-steam cycle (the working fluid of such a cycle is a vapor-gas mixture consisting of combustion products and water vapor). In combined cycle plants, injection of water in front of the turbine leads to a decrease in the temperature of the gases and at the same time to an increase in the enthalpy of the working fluid, since the specific enthalpy of water is greater than that of the combustion products. Such a cycle was proposed by academician S.A. Khristianovich.

There are two main schemes of combined cycle gas plants. In the first type of installation, combustion gases and water vapor are mixed with each other and then enter the turbine.

In the second type, the working fluids, each separately, are directed respectively to the gas and steam parts of the installation.

Let's consider a combined cycle plant with separate flows of combustion products and water vapor.

In this installation, the air, after being compressed in a compressor, is supplied to a high-pressure steam generator operating on gas or liquid fuel at constant pressure. The heat of combustion products is partially spent on steam formation and superheating of steam in the superheater. Combustion products at a reduced temperature are sent to a gas turbine and then to a gas-water heater to heat the feed water sent to the steam generator.

Specific amount of heat received by both working fluids in a theoretical cycle

qi = m(hy - h4") + (h - h).

The thermal efficiency of the steam-gas cycle will be

r]t = (h l - h 2 ) + m(h i - h 2") -(h 4 - h 3)-(h 4" -h 3") m(hy - h^)+ (hi - h$)

Efficient combined-cycle plants are those that discharge exhaust gases from gas turbines into steam generators. Efficiency can reach 0.45.

20) 4Cycles of refrigeration units.

Cooling of bodies to a temperature below the ambient temperature is carried out using refrigeration units operating on a reverse thermal cycle.

A reverse cycle is a cycle in which the work of compression exceeds the work of expansion and, due to the supplied work, heat is transferred from the lower source to the upper one.

Let q2 be the heat taken from the cold source; q\ - heat given off to the hot spring;

L c =q 1 -q 2 - work supplied in a cycle.

A refrigeration unit includes a device for compressing the working fluid (compressor or pump) and a device in which expansion of the working fluid occurs (the working fluid is called refrigerant); expansion of the working fluid can occur with the performance of useful work (in a piston machine or turbomachine) and without the performance of useful work, i.e. fundamentally irreversible (by throttling).

Air refrigeration cycle (Lorentz cycle)

The air refrigeration unit was one of the first types of refrigeration units used in practice.

with a piston compressor were common in the second half of the 19th century. Currently, installations with turbocompressors and regeneration are widely used, due to which the efficiency of air refrigeration units increases and the scope of their application expands. [ 1 ]

Air refrigeration units have a number of inconveniences and have recently fallen out of use. Instead, refrigeration units are becoming widespread, in which low-boiling liquids are used as working fluids: ammonia, carbon dioxide, sulfur dioxide, freons.

The main elements of the installation for producing cold (Figure 14.1) are the compressor 1 and expander 3 . In addition to them, there are two heat exchangers, one of them is a refrigerator 4 the air receives heat from the cooled container, and in the second - the refrigerator 2 releases heat to the environment or the refrigerator water.

The refrigerant (air) expands in expander 1 from pressure p1 up to pressure P2, performing the work given by the expander to the external consumer. Air cooled as a result of the process of adiabatic expansion in an expander from temperature

T1 to temperature T2, enters cooled volume 2, from which it takes heat. The process of heat transfer from the cooled volume to the air occurs at constant air pressure (p = const). Heat removal from

cooled volume is possible only if the air temperature during the entire isobaric heat removal process is less than the temperature of the cooled volume. In principle, the air temperature at the outlet of the cooled volume T3 can be equal to the temperature of the cooled bodies, but in practice it is always slightly lower than this temperature. After leaving the cooled volume, the air is directed to compressor 3, where its pressure increases from the pressure R 2 to pressure R 1 (at the same time the air temperature increases from T3 to T4). The air compressed by the compressor enters the cooler

6- The cooler is a surface type heat exchanger in which the air temperature is reduced due to the transfer of heat to the cooling water circulating through the cooler. In principle, the temperature of the air leaving the cooler T 1 can be made arbitrarily close to the cooling water temperature, but in practice the air temperature is always slightly higher than the cooling water temperature. The process in the cooler occurs at constant air pressure ( pi= const).

Nuclear power plant cycle

Thermal power plants using fossil fuels use superheated steam cycle, and at nuclear power plants with thermal neutron reactors (RTN) – saturated steam cycle. This is due to the fact that RTN uses zirconium-based alloys as the main structural material of the core. They make it possible to reduce the harmful absorption of neutrons compared to various grades of steel, but can withstand temperatures of no more than 340-350 o C. This is less than the critical temperature of water vapor, equal to approximately 374 o C, and at subcritical parameters the efficiency of the saturated steam cycle is greater than that of the superheated steam cycle pair (Fig. 7).

T, TO

Rice. 7. Rankine cycle on saturated (solid lines) and superheated (dashed lines) steam at subcritical initial steam parameters

Initial steam parameters - this is his pressure P o and temperature T o at the turbine inlet. Note that for the saturated steam cycle, only the initial pressure can be selected, since the saturation pressure uniquely determines the temperature of the working fluid.

Increasing the initial steam parameters is one of the main ways to increase the thermal efficiency of the cycle.

The higher the initial temperature assumed, the lower the pressure should be - according to the reliability of the metal. Paired values P o and T o that provide the same strength of power equipment are called equal strength initial parameters working fluid.q (π *) =1, and at π *<π≤1 убывает до q(1) = 0.

Let us consider the dependence of the reduced flow rate of gas flowing out of the nozzle on the backpressure ratio p n to tank pressure p 0 – π n = p n/ p 0 .

As…………the outflow velocity…increases from……It follows that the decrease in pressure…., which can be considered as a weak disturbance propagating relative to the gas flow at the local speed of sound A, will propagate towards the jet flowing from the nozzle and reach the exit section of this nozzle. Thus, for ……….and the dependence of …..on ….will be described by formula (10.2) for …Note that for…………..

With a further decrease in backpressure, this disturbance will no longer be able to reach the exit section of the nozzle, since the component of the absolute velocity of its propagation in the direction opposite to the jet will turn to 0, i.e. this pressure disturbance will be, as it were, carried away by a counter flow of gas. This leads to a peculiar phenomenon called “blocking” of the flow or flow crisis. A change in pressure in the specified range will not affect the outflow parameters, so in this case, i.e. part of the graph of the dependence ...at ....will be represented by a segment of a horizontal straight line ...., and not by a dashed curve corresponding to formula (10.2).

The maximum possible mass flow rate of the flowing gas for the given parameters of the gas state in the boiler ... and ... is determined by formula (9.4) at

(10.3)

and is implemented in the back pressure range 0<p n ≤ p * (0<π н ≤π *).

When the equality holds

, (10.4)

in which q in = q(π n) is determined by formula (10.2) with π in =π n.

Critical speed is called the gas velocity in the outlet section of the channel, at a pressure equal to or less than critical - P K.

w K = Ö 2(g/(g + 1)) P 1 x 1

During the outflow of an ideal gas, the critical speed depends only on the initial parameters and its nature and is equal to the speed of sound of the gas (a) at critical parameters.

w K = a = Ö g P K x K

The Laval combination nozzle is designed to utilize large pressure drops and to produce flow velocities in excess of the critical or sound speed. The Laval nozzle consists of a short tapering section and an expanding conical nozzle (Fig. 5.1). Experiments show that the cone angle of the expanding part should be equal to  = 8-12 o. At large angles, separation of the jet from the channel walls is observed.

The flow rate and second flow rate of an ideal gas are determined by formulas (5.7) and (5.9).
The length of the expanding part of the nozzle can be determined by the equation:

l = (D – d) / 2 tan(/2) , (5.13)

where:  - nozzle taper angle;
D - outlet diameter;
d is the diameter of the nozzle in the minimum section.

Question

Laval nozzle- a gas channel of a special profile that accelerates the gas flow passing through it to supersonic speeds. Widely used on some types of steam turbines and is an important part of modern rocket engines and supersonic jet aircraft engines.

The nozzle is a channel narrowed in the middle. In the simplest case, such a nozzle may consist of a pair of truncated cones connected by narrow ends.

The use of a Laval nozzle is advisable in the case when the total gas pressure in front of the nozzle is sufficient to obtain a critical pressure in the narrow section of the nozzle p, which is greater than the pressure of the medium into which the gas flows, i.e. when p rnar. The mode in which p rka ( is called supercritical; at p r ri and p r we have, respectively, critical and subcritical regimes

The use of a Laval nozzle allows you to obtain an increased steam speed, thereby improving the quality of atomization and reducing steam consumption. In the Danilin nozzle, a certain amount of air is introduced into the fuel oil supply channels, which is sucked in along with the fuel oil due to the injection action of the steam jet. Some believe that this air significantly improves the combustion process.

Give a classification of thermal power plants by the type of primary natural energy used and by the type released. Decoding of the State District Power Plant. Examples of power plant types in the region.

By type of primary natural energy used There are the following types of thermal power plants:

Thermal power plants fossil fuel(coal, fuel oil, natural gas, oil shale, etc.); such power plants are called thermal power plants (in the narrow sense of the word); the main types of thermal power plants using fossil fuels are pulverized coal and gas and oil thermal power plants; for pulverized coal stations, gas can be a backup fuel;

Thermal power plants nuclear fueled, i.e. nuclear power plants (NPP);

Thermal power plants using non-traditional and renewable energy sources(NRES), in particular, the energy of direct solar radiation. Let us note that the primary source of almost all types of primary natural energy is the Sun. For example, coal was formed in the earth's crust from products of organic origin, primarily vegetation, and its growth occurs due to solar energy. The cause of ocean tides is the rotation of the Moon around the Earth, and the latter around the Sun. The flow of rivers is caused by the evaporation of water from the surface of large reservoirs due to solar energy and subsequent precipitation in the form of rain and snow.

Partial pressure is that part of the total pressure of a gas mixture that is due to a given gas or steam. The partial gas in the mixture is equal to the pressure of the gas in the mixture that it would have alone, occupying the same volume as the mixture occupies at the same temperature.

Dalton's law.In the absence of chemical reactions, the total pressure of the gas mixture P total is equal to the sum of the partial pressures of all gases included in it p 1, p 2, p 3 ..., p n:

P total = p 1 + p 2 + ... + p n. (62)

The partial pressure of a given gas is proportional to the proportion of its molecules from the total number of molecules of the mixture (mole fraction):

p i = P total ·X i = P total · . (63)

Mole fraction X i is the ratio of the number of moles of a given substance - n i (or a certain type of particle) to the total number of moles of the substance (or particles) located in the system n i.

The mole fraction can be attributed either to the entire system or to some phase. In the latter case, the ratio of the number of moles of a given substance in this phase to the total number of moles of the substance forming this phase is taken. The sum of the mole fractions of all substances forming a system (or phase) is equal to unity.

The composition of gas mixtures can also be expressed using weight and volumetric parts. The weight fraction of a given gas in a mixture is the ratio of the mass of this gas to the mass of the gas mixture. If we denote the weight fractions of gases by G 1, G 2, G 3, ..., G i; and the masses of gases in the mixture - through m 1, m 2, m 3, ..., m i and the total mass of the gas mixture - through m, then we get:

G 1 = G 2 = G 3 = … G n = (64)

G 1 + G 2 + G 3 + … + G n =1

m 1 + m 2 + m 3 + … + m n = m.

To express the composition of a gas mixture in volumetric units, it is necessary to bring the volumes of the gases that make up the mixture to the same pressure and temperature. The volume of an individual gas included in a mixture reduced to the pressure of the mixture is called reduced volume. In order to find the reduced volume of gas at the pressure of the gas mixture Ptot and temperature T, it is necessary to use the Boyle-Mariotte law:

p 1 V total = v 1 P total; p 2 V total = v 2 P total; p 3 V total = v 3 P total; ... ; p n V total = v n P total,

where v 1, v 2, v 3, ..., v n are the reduced volumes of the individual gases that make up the mixture; р 1, р 2, р 3, …, р n – partial pressures of individual gases;

v 1 = v 2 = v 3 = …; v n = (65)

The sum of the reduced volumes of individual gases equals the total volume of the mixture:

v 1 + v 2 + v 3 + … + v n = V total.

The ratio of the reduced volumes of individual gases to the total volume of the mixture is called the volume fraction and is expressed through r:

r 1 = r 2 = r 3 = ...; r n = (66)

For gas mixtures, the composition expressed by volume and mole fractions is the same, i.e.:

…; (67)

The average molecular weight of a gas mixture, if the volume fractions of gases in the mixture are known, is calculated using the formula:

M av = M 1 r 1 + M 2 r 2 + M 3 r 3 + … + M n r n. (68)

Example 7. Bring the gas to normal conditions (calculate the volume V that a given amount of gas occupies at 273 K and 1.0133 10 5 Pa), if at 373 K and 1.333 10 3 Pa its volume is 3 10 -2 m 3.

Solution. Using equation (59), we determine the volume of gas:

Example 8. Gas under a pressure of 1.2 10 5 N/m 2 occupies a volume of 4.5 liters. What will the pressure be if, without changing the temperature, the volume is increased to 5.5 liters?

Solution. Using the Boyle–Mariotte law (52):

where do we get it from

Example 9. Calculate the partial volumes of water vapor, nitrogen and oxygen and the partial pressures of nitrogen and oxygen in humid air. The total volume of the mixture is 2·10 -3 m3, the total pressure is 1.0133·10 5 Pa, the partial pressure of water vapor is 1.233·10 4 Pa. The volumetric composition of air is 21% O 2 and 79% N 2,

Solution. We calculate the partial volume of water vapor V using equation (65):

V=

We calculate the partial volumes of O 2 and N 2:

V + V = V - V = 0.002 – 0.00024 = 1.76·10 -3 m3.

V/V = 0.21/0.79.

V = 1.76·10 -3 m 3 · 0.21 = 0.37·10 -3 m 3 ;

V = 1.76·10 -3 m3 · 0.79 = 1.39·10 -3 m3.

We calculate the partial pressure of O 2 using equation (63):

P = P x ,

x = V / V = ​​0.37 10 -3 m 3 / 2 10 -3 m 3 = 0.186;

P = 1.0133 10 5 Pa 0.186 = 1.866 10 4 Pa;

and since P = P + P + P, then

P = 1.0133 10 5 Pa - 1.866 10 4 Pa ​​- 1.233 10 4 Pa ​​= 7.033 10 4 Pa.

Option 1.

5. Under normal conditions, the density of carbon dioxide is 1.977 kg/m 3. What pressure must be used to compress the gas so that its density at 0ºC reaches 10 kg/m3?

6. The gas mixture consists of 3 m 3 CO 2 taken at a pressure of 95,940 N/m 2, 4 m 3 O 2 at a pressure of 106,600 N/m 2, 6 m 3 N 2 at a pressure of 93,280 N/m 2. The volume of the mixture is 10 m3. Determine the partial pressures of gases in the mixture and the total pressure of the mixture (temperature is constant).

Option 2.

5. The mass of 1 m 3 of nitrogen at 10ºC and a pressure of 9.86·10 4 N/m 2 is equal to 1.175 kg. What is the mass of the same volume of nitrogen under a pressure of 1.092·10 5 N/m 2 at the same temperature?

6. Dry air has approximately the following composition (vol.%): N 2 78.09; O 2 20.95; Ar 0.93; CO 2 0.03%. Determine the mass of 40 m 3 of dry air at 22ºС and normal pressure.

Option 3.

5. At 37ºС, the volume of gas is 0.50 m 3. What volume will the gas occupy at 100ºC if the pressure remains constant?

6. The gas mixture was prepared from 3 liters of CH 4 at a pressure of 95,940 N/m 2 ; 4 l H 2 at a pressure of 83,950 N/m 2 and 1 l of CO at a pressure of 108,700 N/m 2. The volume of the mixture is 8 liters. determine the partial pressures of gases in the mixture and the total pressure of the mixture.

Option 4.

5. At 18ºC, the pressure in the nitrogen cylinder is 1.621·10 6 N/m 2 . At what temperature will the pressure double?

Option 5.

5. Calculate the volume of flue gases at normal pressure, if their volume at a pressure of 9.888·10 4 N/m 2 and constant temperature is equal to 10 m 3?

6. The blast furnace gas has the approximate composition (vol.%): CO 28; N 2 3; CO 2 10; N 2 59. calculate the partial pressures of the gases that make up the mixture if the total pressure of the gas mixture is 106,400 N/m 2.

Option 6.

5. What pressure is oxygen under if its density at 0ºC is 6.242 kg/m3? Oxygen density at no. 1.429 kg/m3.

6. The underground gasification gas has approximately the following composition (vol.%): CO 12; N 2 14; N 2 62.2; CO 2 10 and CH 4 1.8. Determine the composition of this mixture of gases in weight percent.

Option 7.

5. The oxygen pressure in the cylinder at 15ºC is 1.255·10 7 N/m 2. How much will the gas pressure decrease if the cylinder is cooled to -33ºС?

6. Producer gas has approximately the following composition (wt.%): CO 2 12; N 2 14; CO 20; N 2 54. Calculate the content of each component of the generator gas by volume.

Option 8.

5. How many cubic meters of carbon dioxide at 22ºC and 99289 N/m 2 can be obtained by firing 1000 kg of limestone containing 90% CaCO 3?

6. A cylinder with a capacity of 20 liters at 18ºC contains a mixture of 28 g of oxygen and 24 g of ammonia. Determine the partial pressures of each gas and the total pressure of the mixture.

Option 9.

5. Determine the pressure under which 13.5 g of carbon monoxide will be in a vessel with a capacity of 8 liters at 150ºC?

Option 10.

5. The highest temperature in the gas tank in summer is 40ºС, the lowest in winter is -30ºС. How much more methane (by mass) can a gas tank with a capacity of 2000 m3 hold in winter than in summer at normal pressure?

6. A vessel with a capacity of 2 liters contains 5.23 g of nitrogen and 7.10 g of hydrogen. calculate the total pressure of the gas mixture at 25ºС.

Ideal gas mixtures are mixtures that obey the laws of ideal gases. In the absence of chemical reactions, the total pressure of an ideal gas mixture R total is equal to the sum of the proportional pressures of all gases entering it R 1 , R 2 , R 3 , …, р n(Dalton's law). The partial pressure of the gas of the mixture is equal to the pressure of the gas that it would have alone, occupying the volume of the mixture at the same temperature:

R total = R 1 + R 2 + R 3 + …+ р n.

The composition of gas mixtures can be expressed by mass, volume fractions, number of moles or mole fractions. The mass fraction of a given gas in a mixture is the ratio of the mass of this gas to the mass of the gas mixture. Designating the mass fractions of gases through G 1 ,G 2 ,G 3 , …,G, masses of gases in the mixture – through m 1 ,m 2 ,m 3 , …,m and the total mass of the gas mixture – through m, we get

G 1 = m 1 /m; G 2 = m 2 /m; G 3 = m 3 /m; …; G = m/m,

G 1 +G 2 +G 3 + … + G= 1, a m 1 + m 2 + m 3 + … + m .

To express the composition of a gas mixture in volume fractions, it is necessary to bring the volumes of gases making up the mixture to the same pressure and temperature. The volume of an individual gas included in a mixture reduced to the pressure of the mixture is called reduced volume. To find the reduced volume of gas at the pressure of the gas mixture R total and temperature T it is necessary to use the Boyle-Mariotte law:

R 1 V total = υ 1 R general; R 2 V total = υ 2 R general;

R 3 V total = υ 3 R general; pV total = υ Р total

Where υ 1 , υ 2 ,υ 3 , …, υ - given volumes of individual gases making up the mixture; R 1 , R 2 , R 3 , …, R– partial pressures of individual gases;

υ 1 = υ 2 =

υ 3 = υ =

The sum of the reduced volumes of individual gases equals the total volume of the mixture

υ 1 + υ 2 + υ 3 + …+υ= V total

The ratio of the reduced volumes of individual gases to the total volume of the mixture is called the fractional volume and is denoted by r:

r 1 = υ 1 / V general; r 2 = υ 2 / V general; r 3 = υ 3 / V general; ...; r= υ / V total

The kilomolar (molar) fraction of a gas in a gas mixture is the ratio of the number of kmoles (moles) of a given gas n 1 ,n 2 , n 3 , …, n to the total number of kmoles (moles) of gases making up this mixture:

∑n = n 1 + n 2 + n 3 + … + n

n 1 / ∑n;n 2 / ∑n;n 3 / ∑n;…;n/ ∑n.

For ideal gas mixtures, the composition, expressed in volume and mole fractions, is the same, i.e.

n 1 / ∑n = υ 1 / V total = r 1; n 2 / ∑n = υ 2 / V total = r 2;

n 3 / ∑n = υ 3 / V total = r 3; n/ ∑n = υ / V total = r.

The number of kmoles can be determined by dividing the masses m 1 ,m 2 ,m 3 , …,m(kg) for the molecular weights of gases in the mixture:

n 1 = m 1 /M 1 ;n 2 = m 2 /M 2 ; n 3 = m 3 /M 3 ;…; n= m/M.

It is convenient to calculate the partial pressure of each gas based on the total pressure of the mixture of gases (determined experimentally) and the molar content of gases in the mixture according to the formula

R =(n/ ∑n)R total

If the masses of gases and the temperature of the mixture are known, then for an individual gas the equation of state of an ideal gas is used:

R =nRT/V total

The equation of state for a mixture of gases is written as follows:

R generally V total = ∑ nRT,

R generally V total =( m cm / M Wed) RT,

Where m cm – mass of gas mixture, kg; ∑ n– the sum of gases making up the mixture, kmol; M cf is the average molecular weight of the gas mixture, which is calculated using the mixing rule formula, taking into account the molecular weights of the gases that make up the mixture and their volume fractions:

M avg = M 1 r 1 + M 2 r 2 + M 3 r 3 + … + Mr.

Volume, molar and mass fractions of gases in a gas mixture can be expressed as percentages. To change from volume fractions expressed as percentages r(%), to mass fractions in percent m(%) uses the formula

m(%) =r(%) (MM Wed),

Where M– molecular weight of a given gas; M cf is the average molecular weight of a mixture of gases.

If the composition of the gas mixture is expressed in mass fractions of individual gases, then the average molecular weight is calculated using the formula

M cf = .

Example 1. The gas mixture consists of 3 m 3 of carbon dioxide taken at a pressure of 95,940 Pa, 4 m 3 of oxygen at a pressure of 106,600 Pa, 6 m 3 of nitrogen at a pressure of 93,280 Pa. The volume of the mixture is 10 m3. Determine the partial pressure of gases in the mixture and the total pressure of the mixture. The temperature is constant.

Solution. We calculate the partial pressures of each gas using the formula of the Boyle-Mariotte law:

Pa;
Pa;

Pa.

R total = 28,782 + 42,640 + 55,968 = 127,390 Pa.

Example 2. Dry air has approximately the following composition (%): N 2 78.09; O 2 20.95;Ar 0.93; CO 2 0.03. Determine the mass of 40 m 3 of dry air at 22º C and normal pressure.

Solution. Using the formula, we calculate the average molecular weight of air:

M av = ,

M avg = 28.02 ∙ 0.7809 + 32.00 ∙ 0.2095 + 39.94 ∙ 0.0093 + 44.01 ∙ 0.0003 = 28.97.

For determining m air we use the equation

m air =
kg.

Example 3. In a vessel with a volume of 2000 m3, 1 kg of nitrogen, 2 kg of oxygen and 3 kg of hydrogen are mixed. Calculate the partial volumes and pressures of the gas mixture components, as well as the total pressure of the gas mixture at 17º C.

Solution. We calculate the number of kmoles of gases using the equation:

; ; ;

∑n = 0,03569 + 0,0625 + 1,485 = 1,583;T= 273 + 17 = 290 K.

Determine the total pressure of the gas mixture R general:

R total =
Pa.

We calculate the partial pressure of gases in the mixture:

Pa,

Pa;
Pa.

We determine the partial volumes of gases:

m 3;

m 3;
m 3.

21. A cylinder with a capacity of 20 liters at 18º C contains a mixture of 28 g of oxygen and 24 g of ammonia. Determine the partial pressures of each gas and the total pressure of the mixture

22. A vessel with a volume of 7 liters contains 0.4 g of hydrogen and 3.15 g of nitrogen at 0º C. Determine the partial pressures of the gases and the total pressure of the gas mixture.

23. In a vessel with a volume of 6 liters, 1 g of water and hexane C 6 H 14, heated to 250º C, were introduced under vacuum. Calculate the partial volumes of gases in the mixture.

24. 5 liters of nitrogen, 2 liters of oxygen and 3 liters of carbon dioxide are taken under pressure of 2.3 ∙ 10 5, respectively; 2.7 ∙ 10 5 and 5.6 ∙ 10 5 Pa and mixed, and the volume of the mixture is 15 liters. Calculate the partial pressures, partial volumes of gases in the mixture and the total pressure of the gas mixture.

25. A gas mixture is prepared from 3 liters of methane at a pressure of 95,940 Pa, 4 liters of hydrogen at a pressure of 83,950 Pa and 1 liter of carbon monoxide at a pressure of 108,700 Pa. The volume of the mixture is 8 liters. Determine the partial pressures, partial volumes of individual gases in the mixture and the total pressure of the mixture of gases.

26. Two oxygen cylinders with a capacity of 3 and 4 liters are connected to each other by a tube with a tap. With the tap closed, the oxygen pressure in the first cylinder is 55,970 Pa, in the second – 103,500 Pa. The gas temperature is the same. What will be the pressure in the cylinders at the same temperature if you open the tap? Neglect the volume of the tube.

27. Three cylinders with a capacity of 3, 7 and 5 liters are filled with oxygen respectively ( Pa), nitrogen ( Pa) and carbon dioxide ( Pa) at the same temperature. The cylinders are connected to each other, and a mixture of the same temperature is formed. What is the total pressure of the gas mixture?

28. A mixture of nitrogen and hydrogen is in a gasometer with a capacity of 8 liters at 20º C. The partial pressure of hydrogen is 50,660 Pa, the amount of nitrogen is 0.85 mol. Determine the pressure of the gas mixture in the gasometer.

29. A mixture containing nitrogen and 0.854 mol of hydrogen, at a pressure of 3.55 ∙ 10 5 Pa and 20º C, occupies a volume of 25 liters. Determine the number of moles of nitrogen and the mass of nitrogen.

30. A mixture of gases has the composition (volume fraction, %): H 2 3.0; CO 2 11.0; CO 26.0; N 2 60.0. Determine the mass of 80 m 3 of this mixture at 15º C and normal atmospheric pressure.

At the end of the 18th and in the first half of the 19th century, scientists from different countries actively studied the behavior of gaseous, liquid and solid matter under various external conditions, basing their research on ideas about the atomic and molecular structure of matter. One of these scientists was the British Law for a mixture of gases, which currently bears his name, is discussed in this article.

Special conditions

Before formulating Dalton's law for a mixture of gases, one of the concepts should be understood. This is very important, since this law is valid only for such a substance. We are talking about an ideal gas. What is it?

An ideal gas is one for which the following requirements apply:

• the sizes of molecules and atoms in it are so small that they can be considered material points with zero volume;
• molecules and atoms do not interact with each other.

Thus, an ideal gas is a collection of material points moving randomly. The speed of their movement and mass uniquely determine the temperature of the entire mixture. The pressure that the test substance exerts on the walls of the vessel depends on such macroscopic parameters as temperature, volume of the vessel and the number of molecules.

For such a gas model the following equality holds:

It is called and combines pressure (P), temperature (T), volume (V) and the amount of substance in moles (n). The value of R is the proportionality coefficient, which is equal to 8.314 J/(K*mol).

The surprising thing about this formula is that it does not include a single parameter that would depend on the chemical nature of molecules and atoms.

Partial pressure

Dalton's law for a mixture of ideal gases presupposes knowledge of one more macroscopic parameter - partial pressure.

Let's assume that there is some mixture consisting of 2 components, for example, H 2 and He. This mixture is in a vessel of a specific volume and creates a certain pressure on its walls. Since hydrogen molecules and helium atoms do not interact with each other, then for any calculations of macroscopic characteristics both components can be considered independently of each other.

The partial pressure of a component is the pressure that it creates independently of the other components of the mixture, occupying the volume provided to it. In the example under consideration, we can talk about the partial pressure of H 2 and the same characteristics for He. This value is expressed in pascals and is denoted for the i-th component as Pi.

Gas mixtures and Dalton's law

John Dalton, studying various volatiles, including water vapor, at different temperatures and pressures, came to the following conclusion: the pressure of a mixture of absolutely any similar substances in any proportions is equal to the sum of the partial pressures of all its components. This formulation is called Dalton’s law for the pressure of a mixture of gases and is written as follows:

Here P tot is the total pressure of the mixture.

This fairly simple law is true only for ideal gas mixtures, the components of which do not react chemically with each other.

Another formulation of Dalton's law

Dalton's law for a mixture of gases can be expressed not only in terms of partial pressures, but also in terms of the mole fractions of each component. We obtain the corresponding formula.

Since each component behaves independently of the others in the gas mixture, then the equation of state can be written for it:

This equation is valid for each i-th component, since for all of them the temperature T and volume V are the same. The value n i is the number of moles of component i in the mixture.

Let us now express the partial pressure and divide it by the total pressure of the entire mixture, then we get:

P i /P tot = n i *R*T / V / (n *R*T/V) = n i /n

Here n is the total amount of substance in the entire mixture. It can be obtained by summing all n i . The ratio n i /n is called the mole fraction of component i in the mixture. It is usually denoted by the symbol x i. In terms of mole fractions, Dalton's law is written as follows:

Often represented as atomic percentages of the components in a mixture. For example, 21% O 2 in the air means that its mole fraction is 0.21, that is, every fifth air molecule is oxygen.

Application of the considered law to solve the problem

It is known that a gas mixture of oxygen and nitrogen is under a pressure of 5 atmospheres in a cylinder. Knowing that it contains 10 moles of nitrogen and 3 moles of oxygen, it is necessary to determine the partial pressure of each substance.

To answer the question of the problem, let’s first find the total amount of substance:

n = n N2 + n O2 = 10 + 3 = 13 mol

x N2 = n N2 /n = 10/13 = 0.7692

x O2 = n O2 /n = 3/13 = 0.2308

Using the formula of Dalton’s law through the mole fraction of a component, we calculate the partial pressure of each gas in the cylinder:

P N2 = 5*0.7692 = 3.846 atm.

P O2 = 5*0.2308 = 1.154 atm.

As can be seen from the obtained figures, the sum of these pressures will give 5 atmospheres. The partial pressure of each gas is directly proportional to its mole fraction in the mixture.