I-d diagram humid air- a diagram widely used in calculations of ventilation, air conditioning, drying and other processes associated with changes in the state of moist air. It was first compiled in 1918 by the Soviet heating engineer Leonid Konstantinovich Ramzin.
Various I-d diagrams
I-d diagram of humid air (Ramzin diagram):
 Increase |
 Increase |
 Increase |
 Increase |
Description of the diagram
The I-d diagram of moist air graphically connects all the parameters that determine the thermal and moisture state of the air: enthalpy, moisture content, temperature, relative humidity, partial pressure of water vapor. The diagram is constructed in an oblique coordinate system, which allows you to expand the area of unsaturated moist air and makes the diagram convenient for graphic constructions. The ordinate axis of the diagram shows the values of enthalpy I, kJ/kg of the dry part of the air; the abscissa axis, directed at an angle of 135° to the I axis, shows the values of the moisture content d, g/kg of the dry part of the air.
The diagram field is divided by lines of constant values of enthalpy I = const and moisture content d = const. It also shows lines of constant temperature values t = const, which are not parallel to each other - the higher the temperature of the moist air, the more its isotherms deviate upward. In addition to the lines of constant values of I, d, t, lines of constant values are plotted on the diagram field relative humidity air φ = const. At the bottom of the I-d diagram there is a curve that has an independent ordinate. It connects the moisture content d, g/kg, with the water vapor pressure pp, kPa. The ordinate axis of this graph is the scale of partial pressure of water vapor pp.
It is very convenient to determine the parameters of moist air, as well as solve a number of practical issues related to the drying of various materials, graphically using using i-d diagrams, first proposed by the Soviet scientist L.K. Ramzin in 1918.
Built for 98 kPa barometric pressure. In practice, the diagram can be used in all cases of calculating dryers, since with normal fluctuations in atmospheric pressure the values i And d change little.
Diagram in i-d coordinates is a graphical interpretation of the enthalpy equation for moist air. It reflects the relationship between the main parameters of moist air. Each point on the diagram highlights a certain state with very specific parameters. To find any of the characteristics of moist air, it is enough to know only two parameters of its state.
The I-d diagram of moist air is constructed in an oblique coordinate system. On the ordinate axis up and down from the zero point (i = 0, d = 0), the enthalpy values are plotted and the lines i = const are drawn parallel to the abscissa axis, that is, at an angle of 135 0 to the vertical. In this case, the 0 o C isotherm in the unsaturated region is located almost horizontally. As for the scale for measuring moisture content d, for convenience it is transferred to a horizontal straight line passing through the origin of coordinates.
The water vapor partial pressure curve is also plotted on the i-d diagram. For this purpose, use the equation:
R p = B*d/(0.622 + d),
Solving which for variable values of d we obtain that, for example, for d=0 P p =0, for d=d 1 P p =P p1, for d=d 2 P p =P p2, etc. Specifying a certain scale for partial pressures, a curve Р p =f(d) is constructed at the bottom of the diagram in a rectangular system of coordinate axes at the indicated points. After this, curved lines of constant relative humidity (φ = const) are plotted on the i-d diagram. The lower curve φ = 100% characterizes the state of air saturated with water vapor ( saturation curve).
Also, on the i-d diagram of moist air, straight lines of isotherms (t = const) are drawn, characterizing the processes of moisture evaporation, taking into account the additional amount of heat introduced by water having a temperature of 0 o C.
During the process of evaporation of moisture, the enthalpy of the air remains constant, since the heat taken from the air to dry the materials returns back to it along with the evaporated moisture, that is, in the equation:
i = i in + d*i p
A decrease in the first term will be compensated by an increase in the second term. On the i-d diagram, this process runs along the line (i = const) and is called the process adiabatic evaporation. The limit of air cooling is the adiabatic temperature of the wet thermometer, which is found on the diagram as the temperature of the point at the intersection of the lines (i = const) with the saturation curve (φ = 100%).
Or in other words, if from point A (with coordinates i = 72 kJ/kg, d = 12.5 g/kg dry air, t = 40 °C, V = 0.905 m 3 /kg dry air. φ = 27%), releasing a certain state of moist air, draw down a vertical beam d = const, then it will represent a process of cooling the air without changing its moisture content; the value of relative humidity φ gradually increases. When this ray is continued until it intersects with the curve φ = 100% (point “B” with coordinates i = 49 kJ/kg, d = 12.5 g/kg dry air, t = 17.5 °C, V = 0 .84 m 3 /kg dry air j = 100%), we get the lowest temperature t p (it is called dew point temperature), at which air with a given moisture content d is still able to retain vapors in non-condensed form; a further decrease in temperature leads to the precipitation of moisture either in a suspended state (fog), or in the form of dew on the surfaces of fences (car walls, products), or frost and snow (pipes of the evaporator of a refrigeration machine).
If air is able to be humidified in state A without supplying or removing heat (for example, from an open water surface), then the process characterized by the AC line will occur without a change in enthalpy (i = const). Temperature t m at the intersection of this line with the saturation curve (point “C” with coordinates i = 72 kJ/kg, d = 19 g/kg dry air, t = 24 °C, V = 0.87 m 3 /kg dry air φ = 100%) is wet bulb temperature.
Using i-d, it is convenient to analyze the processes that occur when mixing moist air flows.
Also, the i-d diagram of humid air is widely used to calculate air conditioning parameters, which is understood as a set of means and methods of influencing the temperature and humidity of the air.
For practical purposes, it is most important to calculate the cooling time of the cargo using the equipment available on board the ship. Since the capabilities of a ship's gas liquefaction plant largely determine the time a ship stays in a port, knowledge of these capabilities will allow you to plan your stay time in advance and avoid unnecessary downtime, and therefore claims against the ship.
Mollier diagram. which is given below (Fig. 62), is calculated only for propane, but the method of its use is the same for all gases (Fig. 63).
The Mollier chart uses a logarithmic absolute pressure scale (R log) - on the vertical axis, on the horizontal axis h - natural scale of specific enthalpy (see Fig. 62, 63). Pressure is in MPa, 0.1 MPa = 1 bar, so in the future we will use bars. Specific enthalpy is measured in p kJ/kg. In the future, when solving practical problems, we will constantly use the Mollier diagram (but only its schematic representation in order to understand the physics of thermal processes occurring with the load).
In the diagram you can easily notice a kind of “net” formed by the curves. The boundaries of this “net” are outlined by the boundary curves of changes in the aggregate states of liquefied gas, which reflect the transition of LIQUID TO saturated vapor. Everything that is to the left of the “net” refers to supercooled liquid, and everything that is to the right of the “net” refers to superheated steam (see Fig. 63).
The space between these curves represents different states of the mixture of saturated propane vapor and liquid, reflecting the process of phase transition. Using a number of examples, we will consider the practical use* of the Mollier diagram.
Example 1: Draw a line corresponding to a pressure of 2 bar (0.2 MPa) through the section of the diagram reflecting the phase change (Fig. 64).
To do this, we determine the enthalpy for 1 kg of boiling propane at an absolute pressure of 2 bar.
As noted above, boiling liquid propane is characterized by the left curve of the diagram. In our case this will be a point A, Swiping from a point A vertical line to scale A, we determine the enthalpy value, which will be 460 kJ/kg. This means that each kilogram of propane in this state (at its boiling point at 2 bar pressure) has an energy of 460 kJ. Therefore, 10 kg of propane will have an enthalpy of 4600 kJ.
Next, we determine the enthalpy value for dry saturated propane steam at the same pressure (2 bar). To do this, draw a vertical line from the point IN until it intersects with the enthalpy scale. As a result, we find that the maximum enthalpy value for 1 kg of propane in the saturated vapor phase will be 870 kJ. Inside the chart
* For calculations, data from thermodynamic tables of propane are used (see Appendices).
Rice. 64. For example 1 Fig. 65. For example 2
U 
unit enthalpy, kJ/kg (kcal/kg)
Rice. 63. Basic curves of the Mollier diagram
(Fig. 65) lines directed downwards from the point of critical state of the gas display the number of parts of gas and liquid in the transition phase. In other words, 0.1 means that the mixture contains 1 part gas vapor and 9 parts liquid. At the point of intersection of the saturated vapor pressure and these curves, we determine the composition of the mixture (its dryness or humidity). The transition temperature is constant throughout the entire condensation or vaporization process. If propane is in a closed system (cargo tank), it contains both liquid and gaseous phases of the cargo. You can determine the temperature of a liquid by knowing the vapor pressure, and the vapor pressure by knowing the temperature of the liquid. Pressure and temperature are related if liquid and vapor are in equilibrium in a closed system. Note that the temperature curves located on the left side of the diagram go down almost vertically, cross the vaporization phase in the horizontal direction, and on the right side of the diagram go down again almost vertically.
Example 2: Suppose that there is 1 kg of propane in the phase change stage (part of the propane is liquid, and part is steam). The saturated vapor pressure is 7.5 bar and the enthalpy of the mixture (vapor-liquid) is 635 kJ/kg.
It is necessary to determine what part of the propane is in the liquid phase and what part is in the gaseous phase. Let us first display the known values on the diagram: vapor pressure (7.5 bar) and enthalpy (635 kJ/kg). Next, we determine the intersection point of pressure and enthalpy - it lies on the curve, which is designated 0.2. And this, in turn, means that we have propane in the boiling stage, with 2 (20%) parts of propane being in a gaseous state, and 8 (80%) being in a liquid state.
You can also determine the gauge pressure of a liquid in a tank whose temperature is 60° F, or 15.5° C (to convert the temperature we will use the table of the thermodynamic characteristics of propane from the Appendix).
It must be remembered that this pressure is less than the saturated vapor pressure (absolute pressure) by the amount of atmospheric pressure equal to 1.013 mbar. In the future, to simplify calculations, we will use an atmospheric pressure value of 1 bar. In our case, the vapor pressure, or absolute pressure, is 7.5 bar, so the gauge pressure in the tank will be 6.5 bar.
Rice. 66. For example 3
It was already mentioned earlier that liquid and vapor are in equilibrium in a closed system at the same temperature. This is true, but in practice you can notice that the vapors located in the upper part of the tank (in the dome) have a temperature much higher than the temperature of the liquid. This is due to the heating of the tank. However, such heating does not affect the pressure in the tank, which corresponds to the temperature of the liquid (more precisely, the temperature at the surface of the liquid). The vapors directly above the surface of the liquid have the same temperature as the liquid itself on the surface, where the phase change of the substance occurs.As can be seen from Fig. 62-65, on the Mollier diagram the density curves are directed from the lower left corner of the net diagram to the upper right corner. The density value on the diagram can be given in Ib/ft 3 . For conversion to SI, a conversion factor of 16.02 is used (1.0 Ib/ft 3 = 16.02 kg/m 3).
Example 3: In this example we will use density curves. It is required to determine the density of superheated propane vapor at an absolute pressure of 0.95 bar and a temperature of 49°C (120°F). 
We will also determine the specific enthalpy of these vapors.
The solution to the example can be seen in Figure 66.
Our examples use the thermodynamic characteristics of one gas - propane.
In such calculations for any gas, only absolute values will change thermodynamic parameters, the principle remains the same for all gases. In the future, to simplify, increase the accuracy of calculations and reduce time, we will use tables of the thermodynamic properties of gases.
Almost all the information included in the Mollier diagram is presented in tabular form.
WITH 
Using tables you can find the values of the cargo parameters, but it is difficult. Rice. 67. For example 4 imagine how the process goes. . cooling, if you do not use at least a schematic diagram display p-
h.
Example 4: There is propane in a cargo tank at a temperature of -20" C. It is necessary to determine as accurately as possible the gas pressure in the tank at a given temperature. Next, it is necessary to determine the density and enthalpy of vapor and liquid, as well as the difference in enthalpy between liquid and vapor. Vapors above the surface of the liquid are in a state of saturation at the same temperature as the liquid itself. Atmospheric pressure is 980 mlbar. It is necessary to construct a simplified Mollier diagram and display all the parameters on it.
Using the table (see Appendix 1), we determine the saturated vapor pressure of propane. The absolute vapor pressure of propane at a temperature of -20° C is 2.44526 bar. The pressure in the tank will be equal to:
pressure in the tank (gage or gauge)
1.46526 bar
atmospheric pressure= 0.980 bar =
Absolute_pressure
2.44526 bar
In the column corresponding to the density of the liquid, we find that the density of liquid propane at -20° C will be 554.48 kg/m 3 . Next, we find in the corresponding column the density of saturated vapors, which is equal to 5.60 kg/m3. The liquid enthalpy will be 476.2 kJ/kg, and the vapor enthalpy will be 876.8 kJ/kg. Accordingly, the enthalpy difference will be (876.8 - 476.2) = 400.6 kJ/kg.
A little later we will consider the use of the Mollier diagram in practical calculations to determine the operation of re-liquefaction plants.
Many mushroom pickers are familiar with the expressions “dew point” and “catch condensation on primordia.”
Let's look at the nature of this phenomenon and how to avoid it.
From a school physics course and their own experience, everyone knows that when it gets quite cold outside, fog and dew may form. And when it comes to condensation, most people imagine this phenomenon like this: once the dew point has been reached, water from the condensate will flow from the primordium in streams, or drops will be visible on the growing mushrooms (the word “dew” is associated with drops). However, in most cases, condensation forms in the form of a thin, almost invisible film of water, which evaporates very quickly and is not even noticeable to the touch. Therefore, many are perplexed: what is the danger of this phenomenon if it is not even visible?
There are two such dangers:
- since it occurs almost imperceptibly to the eye, it is impossible to estimate how many times per day the growing primordia were covered with such a film, and what damage it caused them.
It is precisely because of this “invisibility” that many mushroom pickers do not attach importance to the phenomenon of condensation and do not understand the importance of its consequences for the formation of the quality of mushrooms and their yield.
- The water film, which completely covers the surface of the primordia and young mushrooms, does not allow the moisture that accumulates in the cells of the surface layer of the mushroom cap to evaporate. Condensation occurs due to temperature fluctuations in the growing chamber (details below). When the temperature equalizes, a thin layer of condensation from the surface of the cap evaporates and only then does the moisture from the body of the oyster mushroom itself begin to evaporate. If the water in the cells of the mushroom cap stagnates long enough, the cells begin to die. Long-term (or short-term, but periodic) exposure to a water film so inhibits the evaporation of the mushroom bodies’ own moisture that primordia and young mushrooms up to 1 cm in diameter die.
When the primordia become yellow, soft like cotton wool, and leak when pressed, mushroom pickers usually attribute everything to “bacteriosis” or “bad mycelium.” But, as a rule, such death is associated with the development of secondary infections (bacterial or fungal), which develop on primordia and fungi that died from the effects of condensation.
Where does condensation come from, and what temperature fluctuations must there be for the dew point to occur?
To answer, let's look at the Mollier diagram. It was invented to solve problems graphically, instead of cumbersome formulas.
We will consider the simplest situation.
Let's imagine that the humidity in the chamber remains unchanged, but for some reason the temperature begins to drop (for example, water with a temperature lower than normal enters the heat exchanger).
Let’s say the air temperature in the chamber is 15 degrees and the humidity is 89%. On the Mollier diagram, this is blue point A, to which the orange straight line led from number 15. If we continue this straight line upward, we will see that the moisture content in this case will be 9.5 grams of water vapor per 1 m³ of air.
Because we assumed that humidity does not change, i.e. the amount of water in the air has not changed, then when the temperature drops by just 1 degree, the humidity will already be 95%, at 13.5 - 98%.
If we lower a straight line (red) from point A, then at the intersection with the humidity curve of 100% (this is the dew point) we get point B. Drawing a horizontal straight line to the temperature axis, we will see that condensation will begin to fall at a temperature of 13.2.
What does this example tell us?
We see that a decrease in temperature in the zone of formation of young drusen by only 1.8 degrees can cause the phenomenon of moisture condensation. Dew will fall precisely on the primordium, since they always have a temperature 1 degree lower than in the chamber - due to the constant evaporation of their own moisture from the surface of the cap.
Of course, in a real situation, if air comes out of the duct two degrees lower, then it mixes with more warm air in the chamber and the humidity does not rise to 100%, but in the range from 95 to 98%.
But, it should be noted that in addition to temperature fluctuations in the real growing chamber, we also have humidification nozzles that supply moisture in excess, and therefore the moisture content also changes.
As a result, cold air may be supersaturated with water vapor, and when mixed at the outlet of the air duct it will end up in a fog-forming area. Since there is no ideal distribution of air flows, any displacement of the flow can lead to the fact that it is near the growing primordium that the very dew zone is formed that will destroy it. In this case, the primordium growing nearby may not be affected by this zone, and condensation will not fall on it.
The saddest thing about this situation is that, as a rule, the sensors hang only in the chamber itself, and not in the air ducts. Therefore, most mushroom growers do not even suspect that such fluctuations in microclimatic parameters exist in their chamber. Cold air, leaving the air duct, mixes with a large volume of air in the room, and air with “averaged values” in the chamber comes to the sensor, and for mushrooms a comfortable microclimate is important precisely in their growth zone!
The situation with condensation becomes even more unpredictable when the humidification nozzles are not located in the air ducts themselves, but are hung around the chamber. Then the incoming air can dry out the mushrooms, and the nozzles that suddenly turn on can form a continuous film of water on the cap.
Important conclusions follow from all this:
1. Even minor temperature fluctuations of 1.5-2 degrees can cause the formation of condensation and the death of mushrooms.
2. If you do not have the opportunity to avoid microclimate fluctuations, then you will have to lower the humidity to the lowest possible values (at a temperature of +15 degrees, the humidity should be at least 80-83%), then it is less likely that the air will become completely saturated with moisture when lowering temperature.
3. If in the chamber most of the primordia have already passed the phlox stage* and are larger than 1-1.5 cm, then the danger of mushroom death from condensation decreases due to the growth of the cap and, accordingly, the evaporation surface area.
Then the humidity can be raised to optimal (87-89%) so that the mushroom is denser and heavier.
But do this gradually, no more than 2% per day - since as a result of a sharp increase in humidity, you can again get the phenomenon of moisture condensation on the mushrooms.
* The phlox stage (see photo) is the stage of development of the primordia, when the division into individual mushrooms occurs, but the primordia itself still resembles a ball. Outwardly, it looks like a flower with the same name.
4. It is mandatory to have humidity and temperature sensors not only in the oyster mushroom growing chamber, but also in the primordium growth zone and in the air ducts themselves, to record temperature and humidity fluctuations.
5. Any air humidification (as well as its reheating and cooling) in the chamber itself unacceptable!
6. The presence of automation helps to avoid both fluctuations in temperature and humidity, and the death of mushrooms for this reason. A program that controls and coordinates the influence of microclimate parameters must be written specifically for oyster mushroom growth chambers.
I-d diagram for beginners (ID diagram of the state of humid air for dummies) March 15th, 2013
Original taken from Mrcynognathus in I-d diagram for beginners (ID diagram of the state of humid air for dummies)
Good day, dear beginning colleagues!
At the very beginning of its professional path I came across this diagram. At first glance, it may seem scary, but if you understand the main principles by which it works, you can fall in love with it: D. In everyday life it is called an i-d diagram.
In this article, I will try to simply (on fingers) explain the main points, so that you can then, starting from the foundation obtained, independently delve into this web of air characteristics.
This is roughly what it looks like in textbooks. It's getting kind of creepy.
I will remove all the unnecessary things that I will not need for my explanation and present the i-d diagram in this form:
(to enlarge the picture, click and then click on it again)
It’s still not entirely clear what it is. Let's break it down into 4 elements:
The first element is moisture content (D or d). But before I start talking about air humidity in general, I would like to agree on something with you.
Let's agree “on the shore” on one concept right away. Let's get rid of one stereotype that is firmly ingrained in us (at least in me) about what steam is. Since childhood, they pointed me to a boiling pan or kettle and said, pointing a finger at the “smoke” pouring out of the vessel: “Look!” This is steam.” But like many people who are friends with physics, we must understand that “Water vapor is a gaseous state water. Doesn't have colors, taste and smell.” It's just H2O molecules in gaseous state, which are not visible. And what we see coming out of the kettle is a mixture of water in a gaseous state (steam) and “water droplets in a borderline state between liquid and gas,” or rather, we see the latter. As a result, we get that in this moment, around each of us there is dry air (a mixture of oxygen, nitrogen...) and steam (H2O).
So, moisture content tells us how much of this vapor is present in the air. On most i-d diagrams, this value is measured in [g/kg], i.e. how many grams of steam (H2O in gaseous state) are in one kilogram of air (1 cubic meter of air in your apartment weighs about 1.2 kilograms). In your apartment, for comfortable conditions, 1 kilogram of air should contain 7-8 grams of steam.
On the i-d diagram, moisture content is depicted by vertical lines, and gradation information is located at the bottom of the diagram:
(to enlarge the picture, click and then click on it again)
The second important element to understand is air temperature (T or t). I think there is no need to explain anything here. On most ID charts, this value is measured in degrees Celsius [°C]. On the i-d diagram, temperature is depicted by inclined lines, and information about the gradation is located on the left side of the diagram:
(to enlarge the picture, click and then click on it again)
The third element of the ID diagram is relative humidity (φ). Relative humidity is exactly the humidity that we hear about on TV and radio when we listen to the weather forecast. It is measured in percentage [%].
A reasonable question arises: “What is the difference between relative humidity and moisture content?” I will answer this question step by step:
First stage:
Air can hold a certain amount of steam. Air has a certain “steam carrying capacity”. For example, in your room, a kilogram of air can “take on board” no more than 15 grams of steam.
Let's assume that your room is comfortable, and each kilogram of air in your room contains 8 grams of steam, and each kilogram of air can contain 15 grams of steam. As a result, we get that there is 53.3% of the maximum possible vapor in the air, i.e. relative air humidity - 53.3%.
Second phase:
Air capacity varies depending on different temperatures. The higher the air temperature, the more steam it can contain; the lower the temperature, the lower the capacity.
Let's assume that we heated the air in your room with a conventional heater from +20 degrees to +30 degrees, but the amount of steam in each kilogram of air remained the same - 8 grams. At +30 degrees, the air can “take on board” up to 27 grams of steam, as a result, in our heated air there is 29.6% of the maximum possible steam, i.e. relative air humidity - 29.6%.
The same goes for cooling. If we cool the air to +11 degrees, we get a “carrying capacity” of 8.2 grams of steam per kilogram of air and a relative humidity of 97.6%.
Note that there was the same amount of moisture in the air - 8 grams, and the relative humidity jumped from 29.6% to 97.6%. This happened due to temperature fluctuations.
When you hear about the weather on the radio in winter, where they say that it is minus 20 degrees outside and the humidity is 80%, this means that there is about 0.3 grams of steam in the air. When this air enters your apartment, it heats up to +20 and the relative humidity of such air becomes equal to 2%, and this is very dry air (in fact, in an apartment in winter the humidity is kept at 20-30% due to moisture released from the bathrooms and from people, but that is also below the comfort parameters).
Third stage:
What happens if we lower the temperature to a level where the “carrying capacity” of the air is lower than the amount of vapor in the air? For example, up to +5 degrees, where the air capacity is 5.5 grams/kilogram. That part of gaseous H2O that does not fit into the “body” (for us it is 2.5 grams) will begin to turn into liquid, i.e. in water. In everyday life, this process is especially clearly visible when windows fog up due to the fact that the glass temperature is lower than average temperature in the room, so much so that there is little room for moisture in the air and steam, turning into liquid, settles on the glass.
On an i-d diagram, relative humidity is depicted by curved lines, and gradation information is located on the lines themselves:(to enlarge the picture, click and then click on it again)
Fourth elementID diagrams - enthalpy (I ori). Enthalpy contains the energy component of the heat and humidity state of the air. With further study (beyond the scope of this article), it is worth paying special attention to it when it comes to dehumidifying and humidifying the air. But for now special attention We will not focus on this element. Enthalpy is measured in [kJ/kg]. On an i-d diagram, enthalpy is depicted as slanted lines, and gradation information is located on the graph itself (or on the left and at the top of the diagram):
(to enlarge the picture, click and then click on it again)
Then everything is simple! The chart is easy to use! Let's take, for example, your comfortable room, in which the temperature is +20°C and the relative humidity is 50%. We find the intersection of these two lines (temperature and humidity) and see how many grams of steam are in our air.
We heat the air to +30°C - the line goes up, because... There is still the same amount of moisture in the air, but only the temperature increases. We put an end to it and see what the relative humidity turns out to be - it turned out to be 27.5%.
We cool the air to 5 degrees - again we draw a vertical line down, and in the region of +9.5 ° C we come across a line of 100% relative humidity. This point is called the “dew point” and at this point (theoretically, since in practice deposition begins a little earlier) condensation begins to form. We cannot move lower along a vertical straight line (as before), because at this point the “carrying capacity” of air at a temperature of +9.5°C is maximum. But we need to cool the air to +5°C, so we continue to move along the relative humidity line (shown in the figure below) until we reach a slanted straight line of +5°C. As a result, our final point was at the intersection of the +5°C temperature line and the 100% relative humidity line. Let's see how much steam is left in our air - 5.4 grams in one kilogram of air. And the remaining 2.6 grams were released. Our air has become dry.(to enlarge the picture, click and then click on it again)
Other processes that can be performed with air using various devices (dehumidification, cooling, humidification, heating...) can be found in textbooks.
Besides the dew point, another important point is the “wet bulb temperature”. This temperature is actively used in the calculation of cooling towers. Roughly speaking, this is the point to which the temperature of an object can drop if we wrap this object in a wet rag and begin to “blow” intensively on it, for example, using a fan. The human thermoregulation system operates on this principle.
How to find this point? For these purposes we will need enthalpy lines. Let's take our comfortable room again, find the point of intersection of the temperature line +20°C and relative humidity 50%. From this point it is necessary to draw a line parallel to the enthalpy lines up to the 100% humidity line (as in the figure below). The point of intersection of the enthalpy line and the relative humidity line will be the point of the wet thermometer. In our case, from this point we can find out what is in our room, so we can cool the object to a temperature of +14°C.(to enlarge the picture, click and then click on it again)
The process ray (angular coefficient, heat-moisture ratio, ε) is constructed in order to determine the change in air from the simultaneous release of heat and moisture by a certain source(s). Usually this source is a person. An obvious thing, but understanding processes i-d diagrams will help to detect a possible arithmetic error, if one occurs. For example, if you plot a beam on a diagram and, under normal conditions and the presence of people, your moisture content or temperature decreases, then it’s worth thinking about and checking the calculations.
In this article, much is simplified for a better understanding of the diagram at the initial stage of studying it. More accurate, more detailed and more scientific information must be sought in educational literature.
P. S. In some sources
