Enthalpy(from Greek. enthalpo- I heat) is a property of a substance that indicates the amount of energy that can be converted into heat.

Enthalpy- this is thermodynamic property a substance, which indicates the level of energy stored in its molecular structure. This means that while matter can have energy based on temperature and pressure, not all of it can be converted to heat. Part of the internal energy always remains in the substance and maintains it. molecular structure. Part of the kinetic energy of a substance is not available when its temperature approaches the temperature environment. Consequently, enthalpy is the amount of energy that is available for conversion into heat at a given temperature and pressure.

The units of enthalpy are BTU or Joule for energy and Btu/lbm or J/kg for specific energy.

Enthalpy amount

The amount of enthalpy of a substance is based on its given temperature.

Given temperature is the value chosen by scientists and engineers as the basis for calculations. This is the temperature at which the enthalpy of a substance is zero J. In other words, the substance has no available energy, which can be converted into heat. This temperature at various substances different. For example, a given water temperature is the triple point (0°C), nitrogen is −150°C, and methane and ethane refrigerants are −40°C.

If the temperature of a substance is above its given temperature, or changes state to gaseous at a given temperature, the enthalpy is expressed as positive number. Conversely, at a temperature below a given enthalpy of a substance is expressed as a negative number. Enthalpy is used in calculations to determine the difference in energy levels between two states. This is necessary to set up the equipment and determine the coefficient useful action process.

Enthalpy is often defined as the total energy of a substance, since it is equal to the sum of its internal energy (u) in given state along with his ability to get the job done ( pv ). But in reality, enthalpy does not indicate the total energy of a substance at a given temperature above absolute zero (-273°C). Therefore, instead of defining enthalpy as the total heat of a substance, it is more accurate to define it as total available energy of a substance that can be converted into heat.

H=U+pV ,

where V is the volume of the system. Full differential enthalpy has the form:

dH = TdS + Vdp

When working with any calculations, calculations and forecasting various phenomena related to heat engineering, everyone is faced with the concept of enthalpy. But for people whose specialty does not concern thermal power engineering or who only superficially encounter such terms, the word "enthalpy" will inspire fear and horror. So, let's see if everything is really so scary and incomprehensible?

If we try to say it quite simply, the term enthalpy refers to the energy that is available for conversion into heat at a certain constant pressure. The term enthalpy in Greek means "I heat". That is, a formula containing the elementary sum internal energy and the work done is called enthalpy. This value is denoted by the letter i.

If we write the above physical quantities, convert and derive the formula, you get i = u + pv (where u is the internal energy; p, u are the pressure and specific volume of the working fluid in the same state for which the internal energy value is taken). Enthalpy is an additive function, that is, the enthalpy of the entire system is equal to the sum of all its constituent parts.

The term "enthalpy" is complex and multifaceted.

But if you try to understand it, then everything will go very simply and clearly.

• First, in order to understand what enthalpy is, it is worth knowing general definition, which we did.
• Secondly, it is worth finding the mechanism for the appearance of this physical unit, to understand where it came from.
• Thirdly, you need to find a connection with others physical units that are inextricably linked to them.
• And finally, fourthly, you need to look at the examples and the formula.

Well, well, the mechanism of work is clear. You just need to carefully read and understand. We have already dealt with the term "Enthalpy", we have also given its formula. But another question immediately arises: where did this formula come from and why is entropy associated, for example, with internal energy and pressure?

Essence and meaning

To try and find out physical meaning the concept of "enthalpy" you need to know the first law of thermodynamics:

energy does not disappear into nowhere and does not arise from nothing, but only passes from one form to another in equal quantities. Such an example is the transition of heat (thermal energy) into mechanical energy, and vice versa.

We need to transform the equation of the first law of thermodynamics into the form dq = du + pdv = du + pdv + vdp - vdp = d(u + pv) - vdp. From here we see the expression (u + pv). It is this expression that is called enthalpy ( full formula given above).

Enthalpy is also a quantity of state, because the components u (stress) and p (pressure), v (specific volume) have for each quantity certain values. Knowing this, the first law of thermodynamics can be rewritten in the form: dq = di - vdp.

AT technical thermodynamics enthalpy values ​​are used, which are calculated from the conventionally accepted zero. All absolute values These quantities are very difficult to determine, since for this it is necessary to take into account all the components of the internal energy of a substance when its state changes from O to K.

The formula and values ​​​​of enthalpy were given in 1909 by the scientist G. Kamerling-Onnes.

In the expression i is the specific enthalpy, for the entire mass of the body, the total enthalpy is denoted by the letter I, according to world system units enthalpy is measured in Joules per kilogram and is calculated as:

Functions

Enthalpy ("E") is one of the auxiliary functions, thanks to which the thermodynamic calculation can be greatly simplified. For example, great amount heat supply processes in thermal power engineering (in steam boilers or the combustion chamber of gas turbines and jet engines, as well as in heat exchangers) is carried out at constant pressure. For this reason, enthalpy values ​​are usually given in tables of thermodynamic properties.

The enthalpy conservation condition underlies, in particular, the Joule-Thomson theory. Or an effect that found important practical use when liquefying gases. Thus, enthalpy is the total energy of the expanded system, which is the sum of internal energy and external - potential energy pressure. Like any state parameter, enthalpy can be defined by any pair of independent state parameters.

Also, based on the above formulas, we can say: "E" chemical reaction is equal to the sum of the enthalpies of combustion starting materials minus the sum of the enthalpies of combustion of the reaction products.
AT general case energy change thermodynamic system is not necessary condition to change the entropy of this system.

So, here we have analyzed the concept of "enthalpy". It is worth noting that "E" is inextricably linked with entropy, which you can also read about later.

Enthalpy is a property of matter that indicates the amount of energy that can be converted into heat.

Enthalpy is a thermodynamic property of a substance that indicates energy level stored in its molecular structure. This means that although matter can have energy based on , not all of it can be converted into heat. Part of internal energy always remains in matter and maintains its molecular structure. Part of the substance is inaccessible when its temperature approaches the ambient temperature. Consequently, enthalpy is the amount of energy that is available for conversion into heat at a given temperature and pressure. Enthalpy units- British thermal unit or joule for energy and Btu/lbm or J/kg for specific energy.

Enthalpy amount

Quantity enthalpies of matter based on its given temperature. Given temperature is the value chosen by scientists and engineers as the basis for calculations. This is the temperature at which the enthalpy of a substance is zero J. In other words, the substance has no available energy that can be converted into heat. This temperature is different for different substances. For example, this temperature of water is the triple point (0°C), nitrogen is -150°C, and refrigerants based on methane and ethane are -40°C.

If the temperature of a substance is above its given temperature, or changes state to gaseous at a given temperature, the enthalpy is expressed as a positive number. Conversely, at a temperature below a given enthalpy of a substance is expressed as a negative number. Enthalpy is used in calculations to determine the difference in energy levels between two states. This is necessary to set up the equipment and determine the beneficial effect of the process.

enthalpy often defined as the total energy of matter, since it is equal to the sum of its internal energy (u) in a given state, along with its ability to do work (pv). But in reality, enthalpy does not indicate the total energy of a substance at a given temperature above absolute zero(-273°C). Therefore, instead of defining enthalpy as the total heat of a substance, more precisely define it as the total amount of available energy of a substance that can be converted into heat.
H=U+pV

Enthalpy. This elementI- dI dedicated the diagrams separate topic, because for me this element was the least understood among the others ( temperature, moisture content and relative humidity) and requiring analysis of other associated concepts.
Duplicate the picture from the previous article :

I will not go deep into the terminology, I will only say that I understand the enthalpy of air as the energy that a certain volume of air stores in itself. This energy is potential, that is, in the condition of equilibrium, the air does not spend this energy and does not absorb it from other sources.

I will not even give an example to clarify my definition ( although I wanted), because, in my opinion, it will confuse and lead astray.

Straight to the point - what is the most important thing we can take from enthalpy? - I answer - energy ( or the amount of heat), which must be transferred to the air to heat it or taken away to cool it ( or drain).

For example, we have a task - to calculate what power we need a heater in order to supply 1200 m3 / h of outdoor air heated to a temperature of plus 20 degrees in autumn or spring. Design temperature outdoor air during the transition period - plus 10 degrees at an enthalpy of 26.5 kJ / kg ( according to SP 60.13330.2012).

The task is easily solved. To solve such a simple problem using i-d chart, we need to introduce into the level of understanding the units of measurement of some physical quantities:
1) Enthalpy - kilojoule/kilogram. That is the amount of potential energy in one kilogram of air. Everything is simple here - if the enthalpy is 20, then this means that in one kilogram of this air there are 20 kilojoules of potential heat or 20,000 joules.
2) The power of the heater is Watts, but at the same time, watts can be decomposed into Joule / second. That is, how much energy can the heater give out in one second. How more energy we can give out a heater in a second, the more powerful it is. And here everything is simple.

So we takeI- dchart and put on it a point of outdoor air. After, we draw a straight line up ( air is heated without changing the moisture content).

We get a point onj- ddiagram with a temperature of plus 20 degrees and an enthalpy of 36.5 kJ / kg. The question arises - what the hell are we supposed to do with this fucking information?! :)

First, let's pay attention to the fact that we performed all operations with one kilogram of air ( this is indirectly seen from the unit of enthalpy kJ/ kg ).

Secondly, we had a kilogram of air with 26.5 kJ, and now with 36.5 kJ of potential energy. That is, 10 kJ was reported to a kilogram of air in order for its temperature to rise from plus 10 degrees to plus twenty.

Next, we will convert 1200 m 3 / h to kg / s ( kilograms/second, because on the I - d chart uses these units ), multiplying 1200 by 1.25 kg / m 3 ( one cubic meter of ten-degree air weighs 1.25 kilograms) which will give us 1500 kg/h and then dividing by 3600 ( pay attention to the logic of translation between systems - we divide by 3600 not because we memorized or memorized it, but because in a second we will have less than an hour of air, less than 3600 times) we get a total of 0.417 kg/s.

Move on. We got that 0.417 kg of air passes in one second. And we know that every kilogram needs to be transferred ( to report) 10 kJ in order to heat it to a temperature of plus 20 degrees. We report by multiplying 0.417 kg / s by 10 kJ / kg, and getting 4.17 kJ / s ( kilograms decreased) or 4170 J/s, which is equal to 4170 W ( defined by us earlier in the text). So we got the power of our heater.

Conditioning

Cooling follows the same principle, but is only slightly more complicated due to the release of moisture from the air.

Moisture release ( condensate) from air occurs when the temperature of the air during cooling reaches the dew point on the 100% relative humidity line. In a previous article, I described this process:

It seems to be nothing complicated - we cool the air with a temperature of plus 20 degrees and a relative humidity of 50% to plus 12 degrees ( as it usually happens in split systems), drawing a straight line vertically downward from a point of 20 degrees air to a point of 12 degrees air.

And what we see is no moisture release. The moisture content remained the same level- 8 g/kg. But we know that during the operation of the air conditioner there is an abundant moisture release ( condensate is actively dripping from the drainage pipe, brought to the facade of the building) - this fact is confirmed by the repeated observation of a person walking along the summer streets.

The question arises - where does the moisture come from? Answer: the fact is that copper tubes pass through the indoor unit of the air conditioner, which are cooled by the refrigerant to temperatures below plus 12 degrees, and in connection with this, the cooled air is divided into layers with different temperature, approximately as in the figure below ( suppose that the tubes are cooled to plus 5 degrees). I must say right away that this is a drawing far from reality, but showing the general meaning of the words I have said above ( please don't scold me for it)

Therefore, from the air that comes into contact with the tubes ( and fins) and moisture is released. And the air that did not have time to cool to the dew point, or managed, but avoided contact with the chilled surface, bypasses the process of moisture release and carries the same amount of moisture as it carried before cooling ( in fact).

In order to carry out the correct straight line of the air cooling process in such a cooler ( where the refrigerant temperature is below the dew point temperature), we need to take into account each air flow with different thermal and humidity parameters of the air and find on the graph the mixing points of all these flows - which, in my opinion, is not realistic ( I just don't have the brains for it)! But…

I came up with this solution I'm probably not the only one) - we have the temperature of the incoming air, there is the temperature of the refrigerant and there is the temperature of the air received, and I believe that it is enough for us to draw a line for the process of cooling part of the air to plus 5 degrees and find the mixing point of 5-degree air and 20-degree air . That is, I assume that passing through the indoor unit of the air conditioner, the air is divided into two streams - the one that is cooled to plus five degrees and gives us the largest number moisture, and one that is not cooled at all, and at the exit these two streams mix and form an air stream with a temperature of plus 12 degrees and a certain moisture content.

I believe that in order to achieve the goals that I pursue, the result obtained with such a simplification is quite sufficient. And what are my goals?

The first goal is to determine the maximum dehumidification in order to design the condensate drainage system ( This is especially true for air conditioning systems, which include two or more cooling units.)

The second goal is to take into account the amount of cold used to transfer water from gaseous state into liquid ( for moisture condensation; so-called hidden cooling capacity ). This is especially true when cooling heat dissipation) in wet areas. For example, we need to remove 2 kW of heat from a certain pump, which it releases into the room. If we do not take into account that the room is humid ( wet, for whatever reason) and install a 2.5 kW split system in the room, then we can get ( under certain conditions), that the split system spends 1 kW only to convert steam into moisture, and spends the remaining 1.5 kW to remove excess heat, which is 500 W less than necessary, and which can lead to overheating of the pump and its early failure .

So, we divide the flow into two flows, one of which is cooled to plus five - segment 1-2, and the other is left untouched - point 1.

We mix these two streams, uniting the resulting points of the line 1-3-2, and find our 12-degree point on the resulting line.

We leave straight line 1-3 as a line of the air cooling process in a dry cooler from a temperature of plus 20 degrees to plus 12 degrees with the release of condensate.

To find out the amount of condensate that has fallen on the fins and tubes of the cooler we need to subtract the moisture content of the resulting air from the moisture content of the raw air 7.3 g/kg - 6.3 g/kg. As a result, we get that 1 gram of condensate will be released from each kilogram of air passed through the cooler. To find out the condensate flow rate, we need to know how many kilograms of air passes through the heat exchanger in certain time. For example, if we need to cool 1400 m 3 / h of air from a temperature of plus 20 degrees with a relative humidity of 50% to a temperature of plus 12 degrees, then we will translate 1400 m 3 / h into 1680 kg / h and get 1680 grams of condensate ( one gram for every kilogram of air), which is equal to 0.47 g/s ( gram/second) and 0.47 * 10 -3 kg/s.

Total cooling capacity is located in the same way as we were looking for the heat output of the heater earlier. We take enthalpy starting point 28 kJ/kg, subtract from it the end point enthalpy of 38.5 kJ/kg, getting a negative number 10.5 kJ/kg ( minus indicates that energy is being given to the refrigerant). We convert 1680 kg / h to kilogram / second, which will equal 0.47 kg / s. As a result, we get 4.935 kJ / s, which is equal to 4.935 kW of power.

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If necessary determine the latent cooling capacity, you can find it, starting from the amount of condensate released, using the specific heat of vaporization:
The heat required to condense moisture is found by the formula:Q = L * m,
whereL - specific heat vaporization;m - mass of moisture.
L water is equal to: 2260 kJ / kg.

In order to convert 0.47 grams of water from a gaseous state to liquid state per second we need 2260 J * 10 3 * 0.47 kg / s * 10 -3 \u003d 1063 J / s, which is equal to 1063 watts.

So the latent cooling capacity this process equal to 1063 watts.

It's all

Actually, this is all that I wanted to consider in this article. Please do not scold me for the naive simplification of what I have described - I tried to explain to myself first of all - what enthalpy is and how to use it. I hope you find it interesting and useful. Thank you for your attention.

P.S. This article is by no means study guide. It is only my subjective vision of the issue. I would even say - every word written in this article is erroneous. Information worthy of the title " scientific truth"Look in textbooks.

Enthalpy, also a thermal function and heat content, is a thermodynamic potential that characterizes the state of a system in thermodynamic equilibrium when pressure, entropy, and the number of particles are chosen as independent variables.

Simply put, enthalpy is that energy that is available for conversion into heat at a certain temperature and pressure.

The definition of this quantity is the identity: H=U+PV

The unit of enthalpy is J/mol.

In chemistry, the most common isobaric processes (P= const), and the thermal effect in this case is called the change in the enthalpy of the system or process enthalpy :

In a thermodynamic system, the released heat of a chemical process is considered to be negative (exothermic process, Δ H < 0), а поглощение системой теплоты соответствует эндотермическому процессу, ΔH > 0.

Entropy

and for spontaneous

The dependence of the change in entropy on temperature is expressed by the Kirchhoff law:

For an isolated system, the change in entropy is a criterion for the possibility of a spontaneous process. If , then the process is possible; if, then the process is impossible in the forward direction; if, then the system is in equilibrium.

Thermodynamic potentials. Free energy of Gibbs and Helmholtz.

To characterize the processes occurring in closed systems, we introduce new thermodynamic state functions: isobaric-isothermal potential (Gibbs free energy G) and isochoric-isothermal potential (Helmholtz free energy F).

For a closed system in which an equilibrium process is carried out at constant temperature and volume, we express the work of this process. Which we denote by A max (since the work of the process carried out in equilibrium is maximum):

A max =T∆S-∆U

We introduce the function F=U-TS-isochoric-isothermal potential, which determines the direction and limit of the spontaneous flow of the process in a closed system under isochoric-isothermal conditions and obtain:

The change in the Helmholtz energy is determined only by the initial and final state of the system and does not depend on the nature of the process, since it is determined by two state functions: U and S. Recall that the amount of work received or expended may depend on the method of carrying out the process during the transition of the system from the initial to the final state , but not changing the function.

A closed system under isobaric-isothermal conditions is characterized by the isobaric-isothermal potential G:

Gibbs differential energy for a system with a constant number of particles, expressed in eigenvariables - through pressure p and temperature T:

For a system with a variable number of particles, this differential is written as follows:

Here, is the chemical potential, which can be defined as the energy that must be expended to add one more particle to the system.

Analysis of the equation ∆G=∆H-T∆S allows you to determine which of the factors that make up the Gibbs energy is responsible for the direction of the chemical reaction, enthalpy (ΔH) or entropy (ΔS · T).

If ΔH< 0 и ΔS >0, then always ΔG< 0 и реакция возможна при любой температуре.

If ∆H > 0 and ∆S< 0, то всегда ΔG >0, and a reaction with the absorption of heat and a decrease in entropy is impossible under any conditions.

In other cases (ΔH< 0, ΔS < 0 и ΔH >0, ΔS > 0), the sign of ΔG depends on the relation between ΔH and TΔS. The reaction is possible if it is accompanied by a decrease in the isobaric potential; at room temperature, when the T value is small, the TΔS value is also small, and usually the enthalpy change is larger than TΔS. Therefore, most reactions occurring at room temperature are exothermic. The higher the temperature, the greater the TΔS, and even endothermic reactions become feasible.

Under the standard Gibbs energy of formation ΔG°, we understand the change in the Gibbs energy during the reaction of formation of 1 mol of a substance in the standard state. This definition implies that the standard Gibbs energy of formation of a simple substance that is stable under standard conditions is zero.

The change in the Gibbs energy does not depend on the path of the process, therefore, it is possible to obtain different unknown values ​​of the Gibbs energies of formation from equations in which, on the one hand, the sums of the energies of the reaction products are written, and on the other, the sums of the energies of the starting substances.

When using the values ​​of the standard Gibbs energy, the condition ΔG°< 0, а критерием принципиальной невозможности - условие ΔG° >0. At the same time, if the standard Gibbs energy is equal to zero, this does not mean that under real conditions (different from standard ones) the system will be in equilibrium.

Conditions for spontaneous processes in closed systems: