hD diagram wet air (Fig. 14.1), proposed in 1918
Fig.14.1. HD-diagram of wet air
L. K. Ramsin, is widely used to solve practical tasks in those areas where wet air serves as a working fluid. At the axis, the ordinates lay the enthalpy H, kJ / kg of wet air, and along the abscissa axis moisture content D, g / kg S.V. For convenience (reduction of the area of \u200b\u200bthe diagram), the abscissa axis is directed at an angle of 135 ° to the axis of the ordinate. On this diagram, instead of the inclined abscissa axis, a horizontal line was carried out on which valid values \u200b\u200bwere applied to the HD diagram of the H \u003d const line - these are cyclone lines, and the line d \u003d const - vertical straight lines.
From the equation
it follows that in the coordinates of HD isotherms are depicted by straight lines. In addition, the curves are applied to the diagram φ \u003d const.
The curve φ \u003d 100% divides the field into two areas and is a kind of border curve: φ< 100% характеризует область ненасы-щенного влажного воздуха (в воздухе содержится перегретый пар); φ > 100% - the area in which the moisture is in the air hour-tile in the drop state;
φ - 100% characterizes saturated wet air.
For the beginning of the reference of the wet air parameters, the point 0 is chosen for which T \u003d 273.15 K, D \u003d 0, H \u003d 0.
Any point on the HD diagram determines the physical condition of the air. For this, two parameters must be specified (for example, φ and t or h u d). The change in the state of the wet wagon is depicted on the process line diagram. Consider a number of examples.
1) The heat heating process occurs at constant moisture content, since the amount of steam in the air in this case does not change. On the HD diagram, this process is depicted Lini 1-2 (Fig.14.2). In this process, the temperature and enthalpy air increases, and reduces it relative humidity.
Fig. 14.2 Image on HD diagram of characteristic processes of air condition change
2) The process of cooling air on the site over the curve φ-100% also flows at constant moisture content (process 1-5). If you continue the cooling process to point 5 "-Nipped on the curve φ-100%, then in this state the wet air will be saturated. The temperature at point 5 is the temperature of the dew point. Further air cooling (below point 5) leads to the condensation of the water part 5) couple.
3) in the process of adiabate air drainage condensation of moisture
It occurs due to the heat of wet air without external heat exchange. This process proceeds with permanent enthalpy (process 1-7), and air moisture content decreases, and its temperature increases.
4) The process of adiabatic humidification of air, accompanying an increase in air moisture content and a decrease in its tempo, depicted on a line diagram 1-4.
The processes of adiabatic humidification and air drainage are widely used to ensure the specified microclima-ta parameters in agricultural industrial premises.
5) The process of dried air at a constant temperature is depicted with a line 1-6, and the process of humidification of air at a constant temperature - line 1-3.
Using a system of equations, comprising 4.9, 4.11, 4.17, as well as a functional connection R N \u003d. f.(t.), L.K. Ramsin built J.-d. Diagram of wet air, which is widely used in the calculations of ventilation and air conditioning systems. This diagram is a graphical relationship between the main air parameters t., , J., d. and R n with a certain barometric air pressure R b.
Building J.-d. Charts are described in detail in the works.
The state of wet air is characterized by a point applied on the field J.-d. Frames limited d. \u003d 0 and curve \u003d 100%.
The position of the point is given by any two parameters of the five, indicated above, as well as dew point temperatures. t. P and wet thermometer t. M. . The exception is combined d. - R P I. d. - t. p, because Each value d. only one table value corresponds R P I. t. p, and combination J. - t. m.
The scheme for determining air parameters for a given point 1 is shown in Fig. one.
Using J.-d. diagram in adj. 4 and scheme in fig. 1, solve specific examples for all 17 possible combinations of the specified initial air parameters, the specific values \u200b\u200bof which are indicated in Table. 7.
Schemes of solutions and the results obtained are shown in Fig. 2.1 ... 2.17. Famous parameters Air highlighted in drawings by thickened lines.
5.2. The angular coefficient of the ray of the process on the J-D diagram
The ability to quickly graphically determine the parameters of wet air is an important, but not the main factor when using J.-d. Charts.
As a result of heating, cooling, drainage or moisture of wet air, its heat-humid state changes. Change processes are depicted on J.-d. A diagram with straight lines that connect points characterizing the initial and final air states.
Fig. 1. The scheme for determining the parameters of wet air on J.-d. diagram
Table 7.
|
Picture number |
Famous air parameters |
||||||
|
t. 1, ° C |
kJ / kg S.V. |
R p1, kpa |
t. P1, ° C |
t. M1, ° C |
|||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| |
These lines are called rays of processes changes in air condition. The direction of the ray of the process on J.-d. The diagram is determined angular coefficient . If the initial air condition parameters J. 1 I. d. 1, and the final - J. 2 and d. 2, T. corner coefficient expressed by the attitude J./d.Ie:
.
(5.1)
The magnitude of the angular coefficient is measured in the KJ / kg of moisture.
If in equation (29) numerator and denominator multiplied by the mass flow rate of the air G., kg / h, then get:
,
(5.2)
where Q. P is the total amount of heat transmitted when a change in air condition, KJ / h;
W. - The amount of moisture transmitted in the process of changing the condition of air, kg / h.
Depending on the ratio J. and d. The angular coefficient can change its sign and value from 0 to .
In fig. 3 shows the rays of characteristic changes in the state of wet air and the corresponding values \u200b\u200bof the angular coefficient.
1. Wet air with initial parameters J. 1 I. d. 1 heats up with constant moisture content to point 2 parameters, i.e. d. 2 = d. 1 , J. 2 > J. one . The angular coefficient of the ray of the process is:
Fig. 3. Corner coefficient on J.-d. diagram
Such a process is carried out, for example, in surface air heaters, when the temperature and enthalpy of air increase, relative humidity decreases, but the moisture content remains constant.
2. The wet air is simultaneously heated and moistened and acquires parameters of the point 3. The angular coefficient of the beam of the process 3\u003e 0. Such a process proceeds when the dying air assimilates heat and mediating indoors.
3. The wet air is moisturized at a constant temperature to the parameters of the point 4, 4\u003e 0. Almost this process is carried out at moistening of the supply or internal air in a saturated water vapor.
4. The wet air is moistened and heated with an increase in enthalpy to the parameters of the point 5. Since the enthalpy and moisture content of air increase, then 5\u003e 0. Typically, such a process occurs with the direct contact of air with seppe water in irrigation chambers and in cooling towers.
5. Changing the state of wet air occurs at constant enthalpy J. 6 = J. 1 \u003d const. The angular coefficient of such a beam of the process 6 \u003d 0, because J. = 0.
The process of isentalpic humidification of air with circulation water is widely used in air conditioning systems. It is carried out in irrigation chambers or in devices with an irrigated nozzle.
Upon contact with unsaturated wet air with small drops or thin film of water without removal or heat supply from the outside, water as a result of evaporation moisturizes and cools the air, purchasing the temperature of the wet thermometer.
As follows from equation 4.21, in the general case, the angular coefficient of the ray of the process during isentalpine moisture is not equal to zero, because
,
where from w. = 4,186 - specific heat Water, KJ / kg ° C.
A valid isenthalthalpy process, at which \u003d 0 is possible only when t. M. = 0.
6. Wet air is moistened and cooled to point 7. In this case, the angular coefficient 7< 0, т.к. J. 7 – J. 1 0, a d. 7 – d. 1\u003e 0. Such a process proceeds in nozzle irrigation chambers when air contact with cooled water having a temperature above the point of dewing air of the processed air.
7. Wet air is cooled at constant moisture content to point 8 parameters. Since d. = d. 8 – d. 1 \u003d 0, a J. 8 – J. 1 < 0, то 8 \u003d - Air cooling process with d. \u003d Const occurs in surface air coolers at the surface temperature of heat exchange above the temperature of the air dew point when there is no moisture condensation.
8. Wet air is cooled and dried to point 9 parameters. The expression of the angular coefficient in this case has the form:
Cooling with drying occurs in irrigation chambers or in surface air coolers, with a moist air contact with a liquid or solid surface having a temperature below the dew point.
It should be noted that the cooling process with drying during direct contact of air and chilled water is limited by tangent, carried out from point 1 to saturation curve \u003d 100%.
9. Deep drying and air cooling to the parameters of point 10 occurs with direct contact of air with a chilled absorbent, for example, a solution of lithium chloride in irrigation chambers or in devices with an irrigated nozzle. Corner coefficient 10\u003e 0.
10. Wet air is dried, i.e. Gives moisture, with permanent enthalpy to point 11 parameters. The expression of the angular coefficient has the form
.
Such a process can be carried out using solutions of absorbent or solid adsorbents. Note that the real process will have an angular coefficient 11 \u003d 4,186 t. 11, where t. 11 - Final air temperature over a dry thermometer.
From fig. 3. It can be seen that all possible changes in the state of wet air are located on the field J.-d. charts in four sectors whose boundaries are lines d. \u003d Const I. J. \u003d const. In sector I, processes occur with an increase in enthalpy and moisture content, so the values \u200b\u200b\u003e 0. In the II sector, air is drained with an increase in enthalpy and value < 0. В секторе III процессы идут с уменьшением энтальпии и влагосодержания и > 0. In the IV sector, air humidification processes occur with a decrease in enthalpy, so < 0.
The I-D wet air diagram was composed of Professor Leonid Konstantinovich Ramsin in 1918. It graphically connects 5 wet air parameters:
· Specific heat generation (enthalpy) I B.,
· Temperature t.,
· Relative humidity φ ,
· Partial pressure of water vapor p P..
Knowing any two of these parameters, you can define all the others.
The diagram is compiled for a certain barometric pressure.
At the axis of the ordinate (vertical), the values \u200b\u200bof heat-containing (enthalpy) are postponed I S. dry air, on the abscissa axis (horizontal) - moisture content d.. Lines of permanent heat generation (enthalpy) I \u003d const (adiaba) are held at an angle of 135º to the ordinate axis. Lines of permanent moisture content d.\u003d const pass parallel to the axes of the ordinate.
Constant relative humidity curves are also applied φ \u003d Const and at an angle to the axis of the ordinate line isotherm T \u003d const.
Lines φ \u003d 0 I. d.\u003d 0 coincide, because the complete absence of moisture in the air is equally characterized.
Through the intersection point of lines with parameters d.\u003d 0 I. t.\u003d 0 passes line i \u003d 0. The values \u200b\u200bof the heat generation (enthalpy) above this line are positive, below are negative.
The line φ \u003d 100% divides the diagram into two parts. Above the line is the area of \u200b\u200bwet unsaturated air. Line itself φ \u003d 100% corresponds to saturated air - " saturation curve " Below the line is a surrounding air region, " zone Tuman "Where water is in the air of a suspended state in a liquid or solid phase.
I-D charts and schemes for determining wet air parameters for point A.
Basic air treatment processes
And their image on I-D diagram
When considering the process of changing the state of wet air, the following is accepted assumption : air properties change throughout its volume at the same time.
In fact, this is not the case, since the layers closest to hot surfaces will have a temperature higher than deleted. Based on this, it follows that the average values \u200b\u200bof air parameters for the entire volume are accepted as active.
Processing of wet air - i.e., changing its parameters is made by special devices. The following is a description of only the appointment and principle of operation of such devices, without consideration of their design, varieties and installation.
To elementary devices that are tools for exposure to air parameters include:
· Calorifer
· Irrigation (nozzle) chamber (water humidifier)
· Steam humidifier (steam generator)
HEATER
Heater- This isople-banner, changing the temperature of the air without affecting moisture content.
Dry heating
The process is observed only in the heat exchanger (caloriefer).
Air heating occurs at constant moisture content (D \u003d const), since the moisture does not go anywhere, and it is not added to anywhere, since the processed air contacts only with the dry surface of the heat exchanger (Calrifer). Only the number of explicit heat change changes.
At the same time, the process does not change moisture content, the temperature and enthalpy increase, and relative humidity falls ( t 2.>t 1.,I 2.>I 1.,φ 2.<φ 1., d 2.=d 1.\u003d const).
Heat and heat for air heating in the caloriefer:
Q K. = ΔI ∙ G., kj / h \u003d, wt, where
ΔI. - the difference in the heat generations of KJ / kg of air after and to the carrier, respectively;
G. - air flow passing through the calorifer, kg / h
Dry cooling
Air cooling occurs with constant moisture content (D \u003d const), since the moisture does not go anywhere, and it is not added to anywhere, since the air contacts only with the dry surface of the heat exchanger (aircraft). Only the number of explicit heat change changes.
It does not change moisture content, the temperature and heat-containing (enthalpy) decreases, and relative humidity increases ( t 2.<t 1.,I 2.<I 1.,φ 2.>φ 1., d 2.=d 1.\u003d const).
Cost costs in the caloriefer are determined in order similar to the calculations of heat. At the same time, the negative value of the heat of the ground will mean no heat costs, but the cold.
Dew point
If during dry cooling the process d.\u003d const reaches the lines of relative humidity φ \u003d 100%, then with a further decrease in temperature from the air, moisture begins to stand out, since the water steam condensation occurs.
Dew point - saturated air condition ( φ \u003d 100%) with this moisture content d.. It is at the point of intersection of lines d.\u003d Const I. φ \u003d 100%. Isothermary passing through this point corresponds dew point temperature T Tr..
The essence of the process is that when cooled air containing water vapors in a constant quantity, this temperature occurs, in which steam cannot be held with air and goes into a liquid state.
Cooling with drying
If the temperature of the heat exchanger surface (calorfor) t pov below the temperature point of the dew, then with a further decrease in the air temperature, the process after reaching the dew point further passes along the line φ \u003d 100%. At the same time, steam is condensed and, accordingly, the air moisture content decreases. Also, the enthalpy decreases during the process, and relative humidity reaches a maximum possible value of 100% ( t 2.<t 1.,I 2.<I 1.,φ 1.<φ 2.≈100%, d 2.<d 1.).
Amount of moisture remote from everyone The air kilogram is defined as the difference of moisture content values \u200b\u200bat the dew point and at the end point of the process Δd.=d 2.– d Tr, D Tr \u003d D 1. Water consumption condensed in the caloriefer is determined by the formula: W \u003d G. .
It should be noted that in practice, the process can not go strictly along the line φ \u003d 100%, and along it, with values φ about 95%. At the same time, the final air temperature will be slightly higher than the temperature of the heat exchanger surface (calorfor).
After reading this article, I recommend reading an article about entalpy, hidden cooling capacity and determination of the amount of condensate generated in air conditioning and drying systems:
Good day Dear novice colleagues!
At the very beginning of his professional path, I came across this diagram. At first glance, she may seem terrible, but if you understand the main principles for which it works, you can love it and love: d. In everyday life, it is called an I-D diagram.
In this article, I will try simply (on the fingers) to explain the highlights so that you later pushing out the foundation obtained on your own deepened in this cobweb of air characteristics.
Approximately it looks like in textbooks. Somehow urgently becomes.
I will remove everything too much that I will not be necessary for my explanation and imagine the same diagram in this form:
(To increase the drawing, you must click and then click on it)
All the same, it is still not entirely clear what it is. We will analyze it on 4 elements:
The first element is moisture content (D or D). But before I start a conversation about the humidity of the air as a whole, I would like to agree with something with you.
Let's agree on the shore at once about one concept. Get rid of one firmly fallen in us (at least in me) stereotype about what steam is. Since childhood, I was shown on a boiling pan or a kettle and said, a finger-spinning "smoke" with a finger: "Look! These are couples. " But as many people who are friends with physics, we must understand that "water vapor - gaseous state water . Has no colors, taste and smell. " This is just, H2O molecules in a gaseous condition that are not visible. And the fact that we see that flowing from the kettle is a mixture of water in a gaseous state (pairs) and "droplets of water in the boundary state between liquid and gas", or rather we see the latter (as well as reservations, you can call what we see - fog). As a result, we get that at the moment, there are dry air around each of us (a mixture of oxygen, nitrogen ...) and steam (H2O).
So, moisture content tells us how much this couple is present in the air. On most of the I-D diagrams, this value is measured in [g / kg], i.e. How many grams of steam (H2O in a gaseous condition) is located in one kilogram of air (1 cubic meter of air in your apartment weighs about 1.2 kilograms). In your apartment for comfortable conditions in 1 kilogram of air there should be 7-8 grams of steam.
On the I-D diagram, the moisture content is depicted by vertical lines, and the gradation information is located at the bottom of the diagram:
(To increase the drawing, you must click and then click on it)
The second is important to understand the element - air temperature (T or T). I think there is no need to explain anything. On the majority of the diagrams, this value is measured in degrees Celsius [° C]. On the I-D diagram, the temperature is depicted by inclined lines, and the gradation information is located on the left side of the chart:
(To increase the drawing, you must click and then click on it)The third element of the ID diagram is relative humidity (φ). Relative humidity, this is just the humidity about which we hear from TVs and radio when we listen to the weather forecast. It is measured in percent [%].
There is a reasonable question: "What is the difference between relative humidity from moisture content?" I will answer this question in stages:
First stage:
Air is able to accommodate a certain amount of steam. The air has a certain "steam loading." For example, in your room a kilogram of air can "take onto your board" no more than 15 grams of steam.
Suppose that in your room comfortable, and in every kilogram of air located in your room, there are 8 grams of steam, and accommodate every kilogram of air in itself can be 15 grams of steam. As a result, we get 53.3% steam in the air from the maximum possible, i.e. Relative humidity of air - 53.3%.
Second phase:
Air capacity is different at different temperatures. The higher the air temperature, the greater the steam it can accommodate the lower the temperature, the less capacity.
Suppose that we started the air in your room with a conventional heater with +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 - 29.6% steam from the maximum possible, i.e. Relative air humidity - 29.6%.
The same with cooling. If we cool the air to +11 degrees, then we will get a "loading capacity" equal to 8.2 grams of steam per kilogram of air and relative humidity equal to 97.6%.
Note that moisture in the air was the same amount - 8 grams, and the relative humidity jumped from 29.6% to 97.6%. It happened because of the flow racing.
When you hear about the weather on the radio, where they say that the street is minus 20 degrees and humidity 80%, then it means that there are about 0.3 grams of steam in the air. To get to you in the apartment, this air heats up to +20 and the relative humidity of such air becomes 2%, and this is very dry air (in fact, in the apartment in the winter, moisture keeps at a level of 10-30% thanks to the highlights of moisture from San nodes, from Kitchens and from people, but also below the comfort parameters).
Third stage:
What happens if we omit the temperature to this level when the "loading capacity" of the air will be lower than the amount of steam in the air? For example, up to +5 degrees, where air capacity is 5.5 grams / kilograms. The part of the gaseous H2O, which does not fit in the "body" (we have 2.5 grams), it will begin to turn into a liquid, i.e. in water. In everyday life, this process is particularly visible when the windows are fought due to the fact that the glass temperature is lower than the average temperature in the room, there is little space in the air and steam, turning into a liquid, settles on the glasses.
On the diagram, the relative humidity is depicted with curved lines, and the gradation information is located on the lines themselves:
(To increase the drawing, you must click and then click on it)
The fourth element ID diagram - enthalpy (I or I). In Entalpy, the energy component of the heat-woofer state of the air is laid. With further study (outside this article, for example, in my article about enthalpy ) it is worth paying special attention to it when it comes to the drainage and moisturizing air. But so far we will not sharpen special attention on this element. Enthalpy is measured in [KJ / kg]. The enthalpy diagram is depicted by inclined lines, and the gradation information is located on the chart itself (or on the left and in the upper part of the diagram).
The state of wet air on a psychometric diagram is determined using the two specified parameters. If we choose any temperature over a dry thermometer and any temperature over a wet thermometer, then the intersection point of these lines on the diagram is a point indicating the condition of air under these temperatures. The condition of the air at this point is indicated completely definitely.
When a certain air condition was found on the diagram, all other air parameters can be determined using J-D Chart .
Example 1.
t \u003d 35 ° C , and dew point temperature Tr. equal t T TR \u003d 12 ° С What is the temperature of the wet thermometer?
Decision see Figure 6.
On the temperature scale, we find the numerical value of the dew point temperature t T TR \u003d 12 ° С and spend the line isotherm φ \u003d 100% . Get a point with dew point parameters - TR .
From this point d \u003d Const. t \u003d 35 ° C .
We get a desired point BUT
From the point BUT We carry out the line of permanent heat generation - J \u003d Const. before crossing the line of relative humidity φ \u003d 100% .
Get a wet thermometer point - TM
From the resulting point - TM We carry out the line isotherm - t \u003d Const. before intersection with the temperature scale.
We read the desired numerical value of the temperature of the wet thermometer - TM Points BUT that is equal
t T.M. \u003d 20.08 ° C.
Example 2.
If the humid air temperature over a dry thermometer is equal to t \u003d 35 ° C , and dew point temperature t T TR \u003d 12 ° С What is the relative humidity?
Solution see Figure 7.
t \u003d 35 ° C and spend the line isotherm - t \u003d Const. .
t T TR \u003d 12 ° С and spend the line isotherm - t \u003d Const. before crossing the line of relative humidity φ \u003d 100% .
Get a dew point - TR .
From this point - TR We carry out the line of constant moisture content - d \u003d Const. t \u003d 35 ° C .
It will be the desired point BUT whose parameters were set.
The desired relative humidity at this point will be equal
φ a \u003d 25%.
Example 3.
If the humid air temperature over a dry thermometer is equal to t \u003d 35 ° C , and dew point temperature t T TR \u003d 12 ° С What is the air enthalpy?
Solution see Figure 8.
On the temperature scale, we find the numerical temperature value of the dry thermometer - t \u003d 35 ° C and spend the line isotherm - t \u003d Const. .
On the temperature scale, we find the numerical value of the temperature point of the dew - t T TR \u003d 12 ° С and spend the line isotherm - t \u003d Const. before crossing the line of relative humidity φ \u003d 100% .
Get a dew point - TR
From this point - TR We carry out the line of constant moisture content - d \u003d Const. before crossing the line of isotherm on the dry thermometer t \u003d 35 ° C .
It will be the desired point BUT whose parameters were set. The desired heat-containing or enthalpy at this point will be equal to
J a \u003d 57.55 kJ / kg.
Example 4.
When air conditioning related to its cooling (warm period of the year), we are mainly interested in determining the amount of heat that should be applied to sufficiently cool the air to maintain the calculated parameters of the microclimate in the room. When air conditioning associated with its heating (cold period of the year), the outer air must be heated to provide the calculated conditions in the workspace area.
Suppose, for example, that the outer air temperature over the wet thermometer is equal to t H T.M \u003d 24 ° C , and in the air-conditioned room it is necessary to maintain t b t.m \u003d 19 ° С Wet thermometer.
The total amount of heat that needs to be removed from 1 kg of dry air is determined by the following procedure.
See Figure 9.
Enhaulpia of outdoor air when t H T.M \u003d 24 ° C on wet thermometer is equal
p \u003d. J H \u003d 71.63 kJ / per 1 kg of dry air.
Internal air enthalpy at t b tm \u003d 19 ° C on wet thermometer is equal
J B \u003d 53.86 kJ / per 1 kg of dry air.
The difference of enthalpy between the outer and the inner air is:
JN - JV \u003d 71.63 - 53.86 \u003d 17.77 kJ / kg.
Based on this, the total amount of heat that should be allocated during air cooling with t H T.M \u003d 24 ° C on a wet thermometer to t b t.m \u003d 19 ° С on a wet thermometer, equal Q \u003d 17.77 kJ per 1 kg of dry air that is equal 4.23 kcal or 4.91 W per 1 kg of dry air.
Example 5.
During the heating season, it is necessary to heat the outer air with t n \u003d - 10 ° С on dry thermometer and with t H T.M \u003d - 12.5 ° C wet thermometer to internal air temperature t B \u003d 20 ° C on dry thermometer and t b t.m \u003d 11 ° С Wet thermometer. Determine the amount of dry heat, which must be added to 1 kg of dry air.
Decision see Figure 10.
On the J-D diagram By two known parameters - on the temperature of the dry thermometer t n \u003d - 10 ° С and on the temperature of the wet thermometer t H T.M \u003d - 12.5 ° C Determine the outdoor air point based on the temperature of the dry thermometer t n \u003d - 10 ° С and from the outdoor temperature - N. .
Accordingly, we determine the point of internal air - IN .
Read the heat generation - external air enthalpy - N. which will be equal
J H \u003d - 9.1 kJ / per 1 kg of dry air.
Accordingly, heat-containing - internal air enthalpy - IN will be equal
J B \u003d 31,66 kJ / per 1 kg of dry air
The difference of enthalpy of internal and outdoor air is equal to:
ΔJ \u003d J B - J H \u003d 31.66 - (-9.1) \u003d 40.76 kJ / kg.
This change in the amount of heat is a change in the amount of heat of only dry air, because There is no change in its moisture content.
Dry or explicit warm - warmwhich is added or removed from the air without changing the aggregate state state (only temperature changes).
Latent heat - Heat, going to change the aggregate state of the steam without a change in temperature. The temperature point of the dew refers to the moisture content of the air.
When the dew point change changes, there is a change in moisture content, i.e. In other words, moisture content can only be changed when the dew point temperature changes. It should be noted therefore that if the temperature of the dew point remains constant, then the moisture content also does not change.
Example 6.
Air that has initial parameters t n \u003d 24 ° С on dry thermometer and t h t.m \u003d 14 ° C on the wet thermometer, should be conditioned so that its final parameters of steel are equal t K \u003d 24 ° C on dry thermometer and t T T.M \u003d 21 ° C Wet thermometer. It is necessary to determine the number of added hidden heat, as well as the amount of moisture added.
Solution see Figure 11.
On the temperature scale, we find the numerical temperature value of the dry thermometer - t n \u003d 24 ° С , and spend the line isotherm - t \u003d Const. .
Similarly, on the temperature scale, we find the numerical value of the temperature of the wet thermometer - t H T.M. \u003d 14 ° С , carry out the line isotherm - t \u003d Const. .
Crossing line isotherm - t H T.M. \u003d 14 ° С with a linen of relative humidity - φ \u003d 100% Gives a wet air thermometer point with initial specified parameters - point M.T. (H) .
From this point, we carry a line of permanent heat generation - enthalpy - J \u003d Const. before the intersection with isotherm t n \u003d 24 ° С .
We get a point by J-D diagram with initial wet air parameters - point N. , T read the numerical meaning of enthalpy
J n \u003d 39.31 kJ / per 1 kg of dry air.
Similarly, we do to determine the point of wet air on J-D diagram with finite parameters - point TO .
Numerical value of enthalpy at point TO will be equal
J K \u003d 60.56 kJ / per 1 kg of dry air.
In this case, to the air with the initial parameters at the point N. It is necessary to add hidden heat to the final air parameters on the point TO .
Determine the amount of hidden heat
ΔJ \u003d J K - J H \u003d 60.56 - 39.31 \u003d 21.25 kJ / kg.
We spend from the starting point - point N. and endpoint - point TO Vertical lines of constant moisture content - d \u003d Const. , and read the values \u200b\u200bof absolute air humidity at these points:
J n \u003d 5.95 g / per 1 kg of dry air;
J K \u003d 14.4 g / per 1 kg of dry air.
Taking the difference in absolute air humidity
Δd \u003d D to -D H \u003d 14.4 - 5.95 \u003d 8.45 g / per 1 kg of dry air
we obtain the amount of moisture added by 1 kg of dry air.
Changing the amount of heat is a change in the number only hidden heat, because There is no change in air temperature over a dry thermometer.
Outer air at temperatures t n \u003d 35 ° С on dry thermometer and t H T.M. \u003d 24 ° С on wet thermometer - point N. must be mixed with recycling air having parameters t p \u003d 18 ° C for dry thermometer and φ p \u003d 10% relative humidity - point R.
The mixture should consist of 25% outdoor air and 75% recycling air. Determine the final temperatures of the air mixture over dry and wet thermometers.
Solution see Figure 12.
Apply on J-d chart Points N. and R According to the source data.
Connect points n and p straight line - line of the mixture.
On the line of mixes NR Determine the point of the mix FROM Based on the relation that the mixture should consist of 25% outdoor air and 75% recycling air. To do this from the point R Singing a segment equal to 25% of the total line of the mixture NR . We get the point of the mix FROM .
The remaining length of the cut SN equal to 75% of the mixture line length NR .
From a point with a constant temperature line t \u003d Const. and on the scale of temperatures read the temperature of the mixture point t C \u003d 22.4 ° C on a dry thermometer.
From the point FROM We carry out the line of permanent heat generation J \u003d Const. before crossing the line of relative humidity φ \u003d 100% and get a temperature point to the thermometer wet t c t.m. Mixtures. To obtain a numerical value from this point, we carry out a constant temperature line and on the temperature scale, we determine the numerical value of the temperature of the humid thermometer of the mixture, which is equal t c t.m. \u003d 12 ° С .
If necessary, on J-D diagram You can define all the missing parameters of the mixture:
- heat-containing, equal J C \u003d 33.92 kJ / kg ;
- moisture content, equal d C \u003d 4.51 g / kg ;
- relative humidity φ c \u003d 27% .
