Encyclopedia of Fire Safety

Is a vapor barrier ventilation gap necessary? Do walls made of lightweight blocks need a ventilation gap? Types of longitudinal edges of a plasterboard sheet

Let's say a word about the transformer




For a newbie in power electronics, a transformer is one of the most confusing subjects.
- It is not clear why a Chinese welding machine has a small transformer on an E55 core, produces a current of 160 A and feels great. But in other devices it costs twice as much for the same current and gets incredibly hot.
- It’s not clear: is it necessary to make a gap in the transformer core? Some say it is beneficial, others believe the gap is harmful.
What number of turns is considered optimal? What induction in the core can be considered acceptable? And much more is also not entirely clear.

In this article I will try to clarify frequently arising questions, and the purpose of the article is not to obtain a beautiful and incomprehensible calculation method, but to more fully familiarize the reader with the subject of discussion, so that after reading the article he has a better idea of ​​what can be expected from a transformer, and what to pay attention to when choosing and calculating it. How this will turn out is up to the reader to judge.

Where to begin?



Usually they start with choosing a core to solve a specific problem.
To do this, you need to know something about the material from which the core is made, about the characteristics of the cores made from this material various types, and the more the better. And, of course, you need to imagine the requirements for the transformer: what it will be used for, at what frequency, what power it should deliver to the load, cooling conditions, and, perhaps, something specific.
Just ten years ago, to obtain acceptable results it was necessary to have many formulas and carry out complex calculations. Not everyone wanted to do routine work, and the design of a transformer was most often carried out according to simplified methodology, sometimes at random, and, as a rule, with some reserve, for which they even came up with a name that well reflects the situation - “fright coefficient”. And, of course, this coefficient is included in many recommendations and simplified calculation formulas.
Today the situation is much simpler. All routine calculations are included in programs with a user-friendly interface. Manufacturers of ferrite materials and cores lay out detailed characteristics their products and offer software for selecting and calculating transformers. This allows you to fully use the capabilities of the transformer and use a core of exactly the size that will provide required power, without the coefficient mentioned above.
And you need to start by modeling the circuit in which this transformer is used. From the model you can take almost all the initial data for calculating the transformer. Then you need to decide on the manufacturer of the cores for the transformer and obtain full information about its products.
This article will use modeling in a freely available program and its updating as an example. LTspice IV, and as a core manufacturer - the well-known Russian company EPCOS, which offers the "Ferrite Magnetic Design Tool" program for selecting and calculating its cores

Transformer selection process

We will select and calculate a transformer using the example of using it in a welding power source for a semi-automatic machine, designed for a current of 150 A at a voltage of 40 V, powered by a three-phase network.
The product of an output current of 150 A and an output voltage of 40 V gives the device output power Pout = 6000 W. The efficiency of the output part of the circuit (from transistors to the output) can be taken equal toEfficiency out = 0.98. Then the maximum power supplied to the transformer is
Rtrmax = Pout / Efficiencyout = 6000 W / 0.98 = 6122 W.
We choose the switching frequency of the transistors to be 40 - 50 KHz. In this particular case, it is optimal. To reduce the size of the transformer, the frequency must be increased. But a further increase in frequency leads to an increase in losses in the circuit elements and, when powered from a three-phase network, can lead to electrical breakdown of the insulation in an unpredictable place.
In Russia, the most available type E ferrites are made from N87 material from EPCOS.
Using the Ferrite Magnetic Design Tool program, we will determine the core suitable for our case:

Let us immediately note that the definition will be an estimate, since the program assumes a bridge rectification circuit with one output winding, and in our case, a rectifier with a midpoint and two output windings. As a result, we should expect a slight increase in current density compared to what we included in the program.
The most suitable core is E70/33/32 made of N87 material. But in order for it to transmit a power of 6 kW, it is necessary to increase the current density in the windings to J = 4 A/mm 2, allowing greater copper overheating dTCu[K] and put the transformer in a blower to reduce the thermal resistance Rth[° C/ W] to Rth = 4.5 °C/W.
For correct use core, you need to familiarize yourself with the properties of the N87 material.
From the graph of permeability versus temperature:

it follows that the magnetic permeability first increases to a temperature of 100 ° C, after which it does not increase until a temperature of 160 ° C. In the temperature range from 90° C to 160 ° C changes by no more than 3%. That is, transformer parameters that depend on magnetic permeability in this temperature range are most stable.

From the hysteresis plots at temperatures of 25 ° C and 100 ° C:


it can be seen that the range of induction at a temperature of 100 ° C is less than at a temperature of 25 ° C. It should be taken into account as the most unfavorable case.

From the graph of losses versus temperature:

It follows that at a temperature of 100 ° C, losses in the core are minimal. The core is adapted to operate at a temperature of 100 ° C. This confirms the need to use the properties of the core at a temperature of 100 ° C when modeling.

The properties of the E70/33/32 core and N87 material at a temperature of 100 ° C are given in the tab:

We use this data to create a model of the power part of the welding current source.

Model file: HB150A40Bl1.asc

Drawing;

The figure shows a model of the power part of the half-bridge power supply circuit semi-automatic welding machine, designed for a current of 150 A at a voltage of 40 V powered from a three-phase network.
The lower part of the figure represents the " " model. ( description of the operation of the protection scheme in .doc format). Resistors R53 - R45 are a model of variable resistor RP2 for setting the cycle-by-cycle protection current, and resistor R56 corresponds to resistor RP1 for setting the magnetizing current limit.
The U5 element called G_Loop is a useful addition to LTspice IV from Valentin Volodin, which allows you to view the transformer hysteresis loop directly in the model.
We will obtain the initial data for calculating the transformer in the most difficult mode for it - at the minimum permissible supply voltage and maximum PWM filling.
The figure below shows the oscillograms: Red - output voltage, blue - output current, green - current in the primary winding of the transformer.

It is also necessary to know the root mean square (RMS) currents in the primary and secondary windings. To do this, we will again use the model. Let us select the current graphs in the primary and secondary windings in steady state:


We move the cursor over the inscriptions one by oneat the top of I(L5) and I(L7) and with the "Ctrl" key pressed, click the left mouse button. In the window that appears we read: the RMS current in the primary winding is equal (rounded)
Irms1 = 34 A,
and in the secondary -
Irms2 = 102 A.
Let us now look at the hysteresis loop in steady state. To do this, click the left mouse button in the label area on the horizontal axis. The insert appears:

Instead of the word "time" in top window we write V(h):

and click "OK".
Now on the model diagram, click on pin “B” of element U5 and observe the hysteresis loop:

On the vertical axis, one volt corresponds to an induction of 1T; on the horizontal axis, one volt corresponds to the field strength in 1 A/m.
From this graph we need to take the induction range, which, as we see, is equal to
dB = 4 00 mT = 0.4 T (from - 200 mT to +200 mT).
Let's return to the Ferrite Magnetic Design Tool program, and on the "Pv vs. f,B,T" tab we will look at the dependence of losses in the core on the induction range B:


Note that at 100 Mt the losses are 14 kW/m3, at 150 mT - 60 kW/m3, at 200 mT - 143 kW/m3, at 300 mT - 443 kW/m3. That is, we have an almost cubic dependence of losses in the core on the induction range. For a value of 400 mT, losses are not even given, but knowing the dependence, one can estimate that they will amount to more than 1000 kW/.m 3. It is clear that such a transformer will not work for a long time. To reduce the induction swing it is necessary either to increase the number of turns in the transformer windings or to increase the conversion frequency. A significant increase in the conversion frequency in our case is undesirable. An increase in the number of turns will lead to an increase in current density and corresponding losses - according to a linear dependence on the number of turns, the induction range also decreases according to a linear dependence, but a decrease in losses due to a decrease in the induction range - according to a cubic dependence. That is, in the case where the losses in the core are significant more losses in wires, increasing the number of turns has a great effect in reducing overall losses.
Let's change the number of turns in the transformer windings in the model:

Model file: HB150A40Bl2.asc

Drawing;

The hysteresis loop in this case looks more encouraging:


The induction range is 280 mT. You can go even further. Let's increase the conversion frequency from 40 kHz to 50 kHz:

Model file: HB150A40Bl3.asc

Drawing;

And the hysteresis loop:


The induction range is
dB = 22 0 mT = 0.22 T (from - 80 mT to +140 mT).
Using the graph on the "Pv vs. f,B,T" tab, we determine the magnetic loss coefficient, which is equal to:
Pv = 180 kW/m 3 .= 180 * 10 3 W/m 3 .
And, taking the core volume value from the core properties tab
Ve = 102000 mm 3 = 0.102 * 10 -3 m 3, we determine the value of magnetic losses in the core:
Pm = Pv * Ve = 180 * 10 3 W/m 3 * 0.102 * 10 -3 m 3 .= 18.4 W.

We now set a sufficiently long simulation time in the model to bring its state closer to the steady state, and again determine the root-mean-square values ​​of the currents in the primary and secondary windings of the transformer:
Irms1 = 34 A,
and in the secondary -
Irms2 = 100 A.
We take from the model the number of turns in the primary and secondary windings of the transformer:
N1 = 12 turns,
N2 = 3 turns,
and determine the total number of ampere turns in the transformer windings:
NI = N1 * Irms1 + 2 * N2 * Irms2 = 12 vit * 34 A + 2 * 3 vit * 100 A = 1008 A*vit.
In the topmost figure, on the Ptrans tab, in the lower left corner in the rectangle, the recommended value for the fill factor of the core window with copper is shown for this core:
fCu = 0.4.
This means that with such a fill factor, the winding must be placed in the core window, taking into account the frame. Let's take this value as a guide to action.
Taking the window cross-section from the core properties tab An = 445 mm 2, we determine the total permissible cross-section of all conductors in the frame window:
SCu = fCu*An
and determine what current density in the conductors must be allowed for this:
J = NI / SCu = NI / fCu * An = 1008 A*vit / 0.4 * 445 mm 2 = 5.7 A*vit/mm 2 .
Dimension means that regardless of the number of turns in the winding, for each square millimeter copper should account for 5.7 A of current.

Now you can move on to the design of the transformer.
Let's return to the very first figure - the Ptrans tab, according to which we estimated the power of the future transformer. It has a parameter Rdc/Rac, which is set to 1. This parameter takes into account the way the windings are wound. If the windings are wound incorrectly, its value increases and the power of the transformer decreases. Research on how to properly wind a transformer has been carried out by many authors; I will only give conclusions from these works.
First - instead of one thick wire for winding high-frequency transformer, it is necessary to use a bundle of thin wires. Because the working temperature assumed to be around 100 ° C, the wire for the harness must be heat-resistant, for example, PET-155. The tourniquet should be slightly twisted, and ideally it should be a LITZ strand twist. In practice, a twist of 10 turns per meter of length is sufficient.
Secondly, next to each layer of the primary winding there should be a layer of the secondary. With this arrangement of windings, currents in adjacent layers flow in opposite directions and magnetic fields, created by them, are subtracted. Accordingly, the total field and the effects caused by it are weakened. harmful effects.
Experience shows that if these conditions are met,at frequencies up to 50 kHz the parameter Rdc/Rac can be considered equal to 1.

To form the bundles, we will choose PET-155 wire with a diameter of 0.56 mm. It is convenient because it has a cross section of 0.25 mm 2. If we reduce it to turns, each turn of the winding from it will add a cross-section Spr = 0.25 mm 2 /vit. Based on the obtained permissible current density J = 5.7 Avit/mm 2, it is possible to calculate how much current should flow per core of this wire:
I 1zh = J * Spr = 5.7 A*vit/mm 2 * 0.25 mm 2 /vit = 1.425 A.
Based on the current values ​​Irms1 = 34 A in the primary winding and Irms2 = 100 A in the secondary windings, we determine the number of cores in the bundles:
n1 = Irms1 / I 1zh = 34 A / 1.425 A = 24 [cores],
n2 = Irms2 / I 1g = 100 A / 1.425 A = 70 [core]. ]
Let's calculate the total number of cores in the cross-section of the core window:
Nzh = 12 turns * 24 cores + 2 * (3 turns * 70 cores) = 288 cores + 420 cores = 708 cores.
Total wire cross-section in the core window:
Sm = 708 cores * 0.25 mm 2 = 177 mm 2
We will find the coefficient of filling the core window with copper by taking the window cross-section from the properties tab An = 445 mm 2 ;
fCu = Sm / An = 177 mm 2 / 445 mm 2 = 0.4 - the value from which we proceeded.
Having accepted average length turn for the E70 frame equal to lв = 0.16 m, we determine the total length of the wire in terms of one core:
lpr =lv * Nzh,
and, knowing the conductivity of copper at a temperature of 100 ° C, p = 0.025 Ohm*mm 2 / m, we determine the total resistance of a single-core wire:
Rpr = r * lpr / Spr = r * lv * Nl/Spr = 0.025 Ohm*mm 2 / m * 0.16 m * 708 cores / 0.25 mm 2 = 11 Ohm.
Based on the fact that the maximum current in one core is equal to I 1zh = 1.425 A, we determine the maximum power loss in the transformer winding:
Prev = I 2 1zh * Rpr = (1.425 A) 2 * 11 Ohm = 22 [W].
Adding to these losses the previously calculated power of magnetic losses Pm = 18.4 W, we obtain the total power of losses in the transformer:
Psum = Pm + Pext = 18.4 W + 22 W = 40.4 W.
The welding machine cannot operate continuously. During the welding process there are pauses during which the machine “rests”. This moment is taken into account by a parameter called PN - load percentage - the ratio of the total welding time over a certain period of time to the duration of this period. Typically, for industrial welding machines, Pn = 0.6 is accepted. Taking into account Mon, the average power losses in the transformer will be equal to:
Rtr = Psum * PN = 40.4 W * 0.6 = 24 W.
If the transformer is not blown, then, taking the thermal resistance Rth = 5.6 ° C/W, as indicated on the Ptrans tab, we obtain the transformer overheating equal to:
Tper = Rtr * Rth = 24 W * 5.6 ° C/W = 134 ° C.
This is a lot, it is necessary to use forced airflow of the transformer. A generalization of data from the Internet on the cooling of ceramic products and conductors shows that when blown, their thermal resistance, depending on the air flow speed, first drops sharply and already at an air flow speed of 2 m/sec is 0.4 - 0.5 of the state rest, then the falling speed decreases, and a flow speed of more than 6 m/sec is impractical. Let's take the reduction factor equal to Kobd = 0.5, which is quite achievable when using a computer fan, and then the expected overheating of the transformer will be:
Tperobd = Rtr * Rth * Kobd = 32 W * 5.6 ° C/W * 0.5 = 67 ° C.
This means that at the maximum permissible temperature environment Tocmax = 40 °C and at full load welding machine The heating temperature of the transformer can reach the value:
Ttrmax = Tormax + Tper = 40 ° C + 67 ° C = 107 ° C.
This combination of conditions is unlikely, but it cannot be excluded. The most reasonable thing would be to install a temperature sensor on the transformer, which will turn off the device when the transformer reaches a temperature of 100 ° C and turn it on again when the transformer cools to a temperature of 90 ° C. Such a sensor will protect the transformer even if the blowing system is disrupted.
Attention should be paid to the fact that the above calculations are made on the assumption that during breaks between welding the transformer does not heat up, but only cools down. But if special measures are not taken to reduce the pulse duration in idle mode, then even in the absence of a welding process, the transformer will be heated by magnetic losses in the core. In the case under consideration, the overheating temperature will be, in the absence of airflow:
Tperxx = Pm * Rth = 18.4 W * 5.6 ° C/W * 0.5 = 103 ° C,
and when blowing:
Tperkhobd = Pm * Rth * Kobd = 18.4 W * 5.6 ° C/W * 0.5 = 57 ° C.
In this case, the calculation should be carried out based on the fact that magnetic losses occur all the time, and losses in the winding wires are added to them during the welding process:
Psum1 = Pm + Pext * PN = 18.4 W + 22 W * 0.6 = 31.6 W.
The overheating temperature of the transformer without blowing will be equal to
Tper1 = Psum1 * Rth = 31.6 W * 5.6 ° C/W = 177 ° C,
and when blowing:
Tper1obd = Psum1 * Rth * Kobd = 31.6 W * 5.6 ° C/W = 88 ° C.

In the last article we talked about polymer film on various surfaces. Today we will take a closer look at how to install a vapor barrier on the ceiling and what materials can be used. Out of habit, polymer films are called vapor barriers, but the essence lies in the functional purpose of the layer not to allow steam to pass through, and a fairly wide range of materials falls under this criterion. Naturally, installation methods also vary.

Materials with vapor barrier properties

Bitumen mastic can be applied with a brush or roller.

Before you tell us how to lay a vapor barrier on the ceiling, you need to decide on the materials. The ability to retain steam is possessed by:

  • bituminous materials;
  • liquid rubber;
  • polymer films;

The vapor barrier film for the ceiling is attached to a pre-built sheathing, just like foil materials. Liquid rubber, bitumen mastics And roll insulation laid directly on top of the floor, usually made of concrete. Therefore, to decide which vapor barrier is best for the ceiling specifically in your case, you need to start from the presence or absence of sheathing.

Many people believe that a vapor barrier film for the ceiling does not allow moisture to pass through at all, although in fact this is not the case.

Firstly, it is almost impossible to install it so that the layer is completely sealed, and secondly, even the film itself allows a small amount of steam to pass through. Important characteristics:

  • longitudinal and transverse breaking load;
  • resistance to vapor permeation;
  • water resistance;
  • UV resistance.

Laying a vapor barrier on the ceiling only reduces to a minimum the penetration of moisture into the thermal insulation or the ceiling itself. With today's level of technology, there is simply no technical possibility to completely eliminate this process.

Vapor barrier installation methods

The polymer film is secured with a construction stapler.

Installation of ceiling vapor barrier must be considered for each material separately in order to get a complete understanding of installation techniques. Let's start from afar, namely with bitumen materials. Basically, they are positioned as , and also have vapor barrier properties. Such materials are used for insulation basement floor(basement ceiling). Bituminous vapor barrier materials There are two types for the ceiling:

  • mastic;
  • rolls.

Rolls can be ordinary or self-adhesive, which affects the installation method. They are either glued or fused to the work surface. Mastic is used as glue. Even when laying self-adhesive bitumen rolls using the fusing method, it does not hurt to pre-treat the working surface with mastic, although you can do without it. In both cases, the insulation is applied in two layers; if it is rolls, then the joints should be spaced out.

The emergence of more and more new modern materials complicates the question: “Which vapor barrier to choose for the ceiling.”

One of the progressive waterproofing materials that does not allow steam to pass through is liquid rubber.

It consists of two components that, when mixed, form a rubber-like material. It is very elastic and has good adhesion to any surface. Apply using a compressor through a two-jet sprayer. Mixing of components occurs at the intersection of torches in a fraction of a second before the contact of liquid rubber and work surface. Polymerization occurs almost instantly.

We will consider the method of installing a vapor barrier on the ceiling for film and foil materials together, since in both cases the installation is carried out on top of the sheathing. So, the first thing you need is to make the sheathing. Insulation is placed between the guides. A vapor barrier is stretched over the sheathing; it should not sag. The material is attached to wooden blocks construction stapler. Each subsequent tape is laid with an overlap, the joints are taped:

  • for foil materials - aluminum-coated tape;
  • for films - special double-sided tape.

There is a difference between how to lay a film vapor barrier on the ceiling and foil materials, namely which side. The films are placed on either side, since they do not allow steam to pass in both directions. Foil materials are placed with the shiny side inside the room. A finishing finish is installed on top of the vapor barrier.

Is a gap necessary when laying a vapor barrier?

When laying vapor barrier on the sheathing, you need to leave a gap.

One of the most common questions is how to install a vapor barrier on the ceiling: with or without a gap. We are talking about the gap between the film and the insulation, as well as between the film and finishing. Steam moves from a warm environment to a cold one, from a heated room to an unheated one or to the street. Accordingly, the film is placed between the warm environment and the insulation. The steam encounters the insulating layer and, not finding a way out, some of it returns back into the room, and some condenses on the film.

If there is no gap between the vapor barrier and interior decoration walls, then the latter will come into contact with condensed moisture. As a result, mold will appear over time and the finishing material will deteriorate. If there is a gap, moisture will have the opportunity to evaporate, so a buffer air zone is needed in this case.

The gap between the film and the insulation is completely unnecessary, since that tiny part of the moisture that gets into the thermal insulation still moves in the direction away from the vapor barrier. If the thermal insulation cake is made incorrectly and steam is not able to escape from the insulation, then the gap will not affect the situation in any way. The problem can only be solved by eliminating installation errors.

Results

From our article today we learned that vapor barrier is functional purpose layer that bitumen mastics and rolled materials, liquid rubber, polymer films and foil materials. We looked at how to attach a vapor barrier to the ceiling:

  • bituminous materials and liquid rubber applied directly to the ceiling (usually concrete);
  • polymer films and foil materials are attached to the sheathing on top of the insulation, and protect the thermal insulation from moisture getting into it.

When installing film and foil materials, you need to leave a gap between the vapor barrier and the interior decoration, but there is no need for a gap between the vapor barrier and the insulation.

Ventilation gap in frame house- this is a moment that often raises many questions among people who are involved in insulating their own home. These questions arise for a reason, since the need for a ventilation gap is a factor that has a huge number of nuances, which we will talk about in today’s article.

The gap itself is the space that is located between the sheathing and the wall of the house. Implemented similar solution by means of bars that are attached on top of the windproof membrane and on the external finishing elements. For example, the same siding is always attached to bars that make the facade ventilated. A special film is often used as insulation, with the help of which the house, in fact, is completely wrapped.

Many will rightly ask, is it really not possible to just take and attach the sheathing directly to the wall? Do they just line up and form an ideal area for installing sheathing? In fact, there are a number of rules that determine the necessity or unnecessaryness of organizing a ventilation facade. Let's figure out whether a ventilation gap is needed in a frame house?

When is a ventilation gap (vent gap) needed in a frame house?

So, if you are thinking about whether a ventilation gap is needed in the facade of your carcass house, pay attention to the following list:

  • When wet If the insulation material loses its properties when wet, then a gap is necessary, otherwise all work, for example, on insulating a home, will be completely in vain
  • Steam Permeation The material from which the walls of your home are made allows steam to pass through into the outer layer. Here, without organizing free space between the surface of the walls and insulation, it is simply necessary.
  • Preventing excess moisture One of the most common questions is the following: is there a need for a ventilation gap between vapor barriers? If the finish is a vapor barrier or moisture-condensing material, it must be constantly ventilated so that excess water is not retained in its structure.

As for the last point, the list of similar models includes the following types of cladding: vinyl and metal siding, profiled sheets. If they are sewn tightly on flat wall, then the remaining accumulated water will have nowhere to go. As a result, materials quickly lose their properties and also begin to deteriorate externally.

Is there a need for a ventilation gap between siding and OSB?

When answering the question of whether a ventilation gap is needed between the siding and OSB (from English - OSB), it is also necessary to mention its need. As already stated, siding is a product that insulates vapor and OSB board completely consists of wood shavings, which easily accumulate residual moisture and can quickly deteriorate under its influence.

Additional reasons to use a ventilation gap

Let's look at a few more mandatory points when clearance is a necessary aspect:

  • Preventing rot and cracks Wall material under decorative layer prone to deformation and deterioration when exposed to moisture. To prevent rot and cracks from forming, just ventilate the surface, and everything will be fine.
  • Preventing condensation The material of the decorative layer may contribute to the formation of condensation. This excess water must be removed immediately.

For example, if the walls of your house are made of wood, then an increased level of moisture will negatively affect the condition of the material. Wood swells, begins to rot, and microorganisms and bacteria can easily settle inside it. Of course, a small amount of moisture will collect inside, but not on the wall, but on a special metal layer, from which the liquid begins to evaporate and be carried away with the wind.

Is there a need for a ventilation gap in the floor? No

Here you need to take into account several factors that determine whether you need to make a gap in the floor:

  • If both floors of your house are heated, then a gap is not necessary If only the 1st floor is heated, then it is enough to lay a vapor barrier on its side to prevent condensation from forming in the ceilings.
  • The ventilation gap must be attached only to the finished floor!

When answering the question of whether a ventilation gap is needed in the ceiling, it should be noted that in other cases this idea is purely optional and also depends on the material chosen for insulating the floor. If it absorbs moisture, then ventilation is simply necessary.

When a ventilation gap is not needed

Below are a few cases where this construction aspect does not need to be implemented:

  • If the walls of the house are made of concrete If the walls of your house are made, for example, of concrete, then you don’t need to make a ventilation gap, because this material does not allow steam to pass from the room to the outside. Consequently, there will be nothing to ventilate.
  • If there is a vapor barrier inside the room If with inside If the room has a vapor barrier installed, then the gap also does not need to be organized. Excess moisture simply will not come out through the wall, so there is no need to dry it.
  • If the walls are treated with plaster If your walls are treated e.g. facade plaster, then the gap is not needed. In case outer material the treatment allows steam to pass through well; no additional measures are required to ventilate the casing.

Installation example without ventilation gap

As a small example Let's look at an example of installation without the need for a ventilation gap:

  • At the beginning there is a wall
  • Insulation
  • Special reinforcing mesh
  • Mushroom dowel used for fastening
  • Facade plaster

Thus, any amounts of steam that penetrate the structure of the insulation will be immediately removed through the layer of plaster, as well as through vapor-permeable paint. As you may have noticed, there are no gaps between the insulation and the decoration layer.

We answer the question why a ventilation gap is needed

The gap is necessary for air convection, which can dry out excess moisture and have a positive effect on the safety building materials. The very idea of ​​this procedure is based on the laws of physics. Ever since school we have known that warm air always goes up and cold always goes down. Consequently, it is always in a circulating state, which prevents liquid from settling on surfaces. In the upper part, for example, of the siding, perforations are always made, through which steam escapes out and does not stagnate. Everything is very simple!

A house made of porous blocks cannot be left without a moisture-resistant finish - it must be plastered, lined with bricks (unless provided additional insulation, then without a gap) or mount curtain façade. Photo: Wienerberger

In multi-layer walls with insulation mineral wool a ventilation layer is necessary, since the dew point is usually located at the junction of the insulation with the masonry or in the thickness of the insulation, and its insulating properties sharply deteriorate when moistened. Photo: YUKAR

Today the market offers a huge variety construction technologies, and this often results in confusion. For example, the thesis has become widespread, according to which the vapor permeability of layers in the wall should increase towards the street: only in this way will it be possible to avoid over-wetting the wall with water vapor from the premises. Sometimes it is interpreted as follows: if the outer layer of the wall is made of a denser material, then between it and the masonry of porous blocks there must be a ventilated air gap.

Often a gap is left in any walls with brick cladding. However, for example, masonry made of lightweight polystyrene concrete blocks practically does not allow steam to pass through, which means there is no need for a ventilation layer. Photo: DOK-52

When used for clinker finishing, a ventilation gap is usually necessary, since this material has a low vapor transmission coefficient. Photo: Klienkerhause

Meanwhile, building codes mention a ventilated layer only in connection with, but in the general case, protection from waterlogging of walls “should be ensured by designing enclosing structures with a vapor permeability resistance of internal layers of at least the required value determined by calculation...” (SP 50.13330.2012, P. 8.1). The normal humidity regime of three-layer walls of high-rise buildings is achieved due to the fact that the inner layer of reinforced concrete has a high resistance to vapor transmission.

Common mistake builders: there is a gap, but it is not ventilated. Photo: MSK

The problem is that some multilayer masonry structures used in low-rise housing construction physical properties closer to . A classic example is a wall made of (one block) lined with clinker. Its inner layer has a vapor permeability resistance (R p) equal to approximately 2.7 m 2 h Pa/mg, and the outer layer is about 3.5 m 2 h Pa/mg (R p = δ/μ, where δ - layer thickness, μ - coefficient of vapor permeability of the material). Accordingly, there is a possibility that the increase in humidity in foam concrete will exceed the tolerances (6% by weight during the heating period). This can affect the microclimate in the building and the service life of the walls, so it makes sense to lay a wall of such a design with a ventilated layer.

In such a design (with insulation with sheets of extruded polystyrene foam) there is simply no room for a ventilation gap. However, EPS will interfere gas silicate blocks dry, so many builders recommend vapor barriering such a wall from the side of the room. Photo: SK-159

In the case of a wall made of Porotherm blocks (and analogues) and conventional slotted facing bricks the vapor permeability indicators of the inner and outer layers of masonry will differ insignificantly, therefore ventilation gap It will be more likely to be harmful, since it will reduce the strength of the wall and require an increase in the width of the base part of the foundation.

Important:

  1. A gap in the masonry becomes meaningless if entrances and exits from it are not provided. At the bottom of the wall, immediately above the plinth, it is required to be built into the facing masonry ventilation grates, the total area of ​​which must be at least 1/5 of the horizontal sectional area of ​​the gap. Usually, 10x20 cm gratings are installed in increments of 2–3 m (alas, the gratings are not always available and require periodic replacement). In the upper part, the gap is not laid or filled with mortar, but is covered with a polymer masonry mesh, or even better - with perforated panels made of galvanized steel with a polymer coating.
  2. The ventilation gap must be at least 30 mm wide. It should not be confused with the technological one (about 10 mm), which is left to level the brick cladding and is usually filled with mortar during the laying process.
  3. There is no need for a ventilated layer if the walls are covered from the inside with a vapor barrier film followed by finishing

One of the last stages of working with gypsum boards is joining and sealing the seams of the sheets. This is a rather difficult and responsible moment, because incorrect installation jeopardizes the reliability and durability of all your new, just made repairs - cracks may appear in the wall where the seams are. It not only spoils appearance, but also negatively affects the strength of the wall. Therefore, beginners have a lot of doubts about joining sheets of drywall. The most important issue is the gap between the sheets of drywall. But more on that later, but now let’s figure out how to join the sheets together.

Types of longitudinal edges of a plasterboard sheet

Each sheet of drywall has two types of edges: transverse and longitudinal. The first one is not of particular interest to us now - it is always straight, without a layer of cardboard and paper, and for all types of drywall, including waterproof and fire-resistant. It happens longitudinally:

  • Straight (PC markings can be seen on the sheet). This edge does not provide for sealing the joint and is more suitable for “black” finishing. Most often it is present not on drywall, but on sheets of gypsum fiber
  • Semicircular, thinned on the front side (marking – PLUK). It occurs much more often than others. Sealing seams - putty, using serpyanka
  • Beveled (its marking is UK). A rather labor-intensive process of sealing seams in three stages. A prerequisite is treatment with serpyanka. Second most popular drywall edge
  • Rounded (marking of this type is ZK). No joint tape required during installation
  • Semicircular (marked on the sheet - PLC). Work will be required in two stages, but without serpyanka, with the condition that the putty will be of good quality
  • Folded (marking of such sheets is FC). More common on gypsum fiber sheets, like straight edge

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These options can be found in stores. The most common are sheets with PLUK and UK edges. Their main advantage is that there is no need to treat the seams additionally before applying putty.

During the repair, you will need to cut sheets to a given size. In this case, you also need to make an edge - thin the sheet in the right place. This is done with a specially designed tool that removes unnecessary plaster and creates the necessary relief. If this tool is not at hand, use a wallpaper knife; it must be sharp. Remove a couple of millimeters, maintaining an angle of forty-five degrees.

Most main question for beginners - do you need to leave a gap between the sheets of drywall? Yes, because plasterboard sheets, like any other material, tend to expand from heat and swell from moisture. The gap in this situation will help prevent the deformed sheet from leading the rest.

How to properly join drywall

As in any other job, you need to know a certain technology. The first thing you should not forget about is that under no circumstances should you do docking by weight. The place where the edges are joined must be where the frame is located. This applies to all types of docking. Secondly, the arrangement of cut and whole sheets should alternate, as in chess.

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When fastening in two layers, it is necessary to shift the sheets of the second layer by 60 cm in relation to the first. You should start with half, cut along a line running along the sheet.

If the joint is located in a corner, one sheet is attached to the profile, then the second is attached to standing nearby. Only later on external corner put on a perforated corner specially designed for this purpose. The internal one is simply covered with putty. The gap should not exceed 10 mm.

How much gap should be left between sheets of drywall during a normal connection? Experts say that it should be about 7 mm, between the ceiling and the gypsum board - no more than 5, and between the floor and drywall - a gap of 1 cm.

How to seal joints

After joining, there is one more important part left - sealing the seams. Putty will help us with this. Following the instructions, dilute the gypsum base in water. In order for your repair to be durable and reliable, you first need to take care of the quality of the seams, and therefore the putty itself. In addition to this, we need a spatula; a regular 15-centimeter construction spatula will do.

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