The task of organized air exchange in the rooms of a house or apartment is to remove excess moisture and exhaust gases by replacing it with fresh air. Accordingly, for the exhaust and inflow device, it is necessary to determine the amount of removed air masses - to calculate the ventilation separately for each room. Calculation methods and norms of air consumption are accepted exclusively according to SNiP.
Sanitary requirements of regulatory documents
The minimum amount of air supplied and removed from the cottage rooms by the ventilation system is regulated by two main documents:
- “Residential multi-apartment buildings” - SNiP 31-01-2003, paragraph 9.
- “Heating, ventilation and air conditioning” - SP 60.13330.2012, mandatory Appendix “K”.
The first document sets out the sanitary and hygienic requirements for air exchange in residential premises of apartment buildings. The ventilation calculation should be based on these data. Two types of dimensions are used - air mass flow rate per unit time (m³ / h) and hourly multiplicity.
Reference. The multiplicity of air exchange is expressed by a figure indicating how many times within 1 hour the air environment of the room is completely updated.
Depending on the purpose of the room, the supply and exhaust ventilation must provide the following flow rate or the number of updates of the air mixture (multiplicity):
- living room, children's room, bedroom - 1 time per hour;
- kitchen with electric stove - 60 m³ / h;
- bathroom, bath, toilet - 25 m³ / h;
- for a furnace with a solid fuel boiler and a kitchen with a gas stove, a multiplicity of 1 plus 100 m³ / h is required during the operation of the equipment;
- a boiler room with a heat generator burning natural gas - a three-fold renewal plus the amount of air required for combustion;
- pantry, dressing room and other utility rooms - multiplicity 0.2;
- drying or laundry - 90 m³ / h;
- library, study - 0.5 times per hour.
Note. SNiP provides for a decrease in the load on general ventilation with idle equipment or no people. In residential premises, the ratio decreases to 0.2, technical - to 0.5. The requirement for rooms where gas-powered installations are located is unchanged - an hourly one-time air renewal.
Paragraph 9 of the document implies that the volume of the hood is equal to the amount of inflow. The requirements of SP 60.13330.2012 are somewhat simpler and depend on the number of people staying in the room for 2 hours or more:
- If there are 20 m² or more of apartment area per 1 resident, a fresh influx of 30 m³ / h per person is provided into the rooms.
- The volume of supply air is considered by area when less than 20 squares fall on 1 tenant. The ratio is: 3 m³ of inflow is supplied per 1 m² of housing.
- If ventilation is not provided in the apartment (there are no window panes and windows that cannot be opened), it is necessary to supply 60 m³ / h of clean mixture to each resident, regardless of quadrature.
The listed regulatory requirements of two different documents do not contradict each other at all. Initially, the performance of the ventilation general exchange system is calculated according to SNiP 31-01-2003 "Residential buildings".
The results are checked against the requirements of the Code of Practice “Ventilation and air conditioning” and are adjusted if necessary. Below we will analyze the calculation algorithm using the example of a one-story house shown in the drawing.
Determination of air flow rate
This typical calculation of supply and exhaust ventilation is performed separately for each room of an apartment or a country cottage. To find out the mass flow rate of the building as a whole, the results are summarized. A fairly simple formula is used:
Explanation of symbols:
- L is the desired volume of supply and exhaust air, m³ / h;
- S - the quadrature of the room where the ventilation is calculated, m²;
- h - ceiling height, m;
- n - the number of updates to the air environment of the room within 1 hour (regulated by SNiP).
Calculation example. The living area of a one-story building with a ceiling height of 3 m is 15.75 m². According to the requirements of SNiP 31-01-2003, the multiplicity n for residential premises is equal to one. Then the hourly flow rate of the air mixture will be L = 15.75 x 3 x 1 = 47.25 m³ / h.
An important point. The determination of the volume of the air mixture removed from the kitchen with a gas stove depends on the installed ventilation equipment. A common scheme looks like this: a one-time exchange according to the standards is provided by a natural ventilation system, and an additional 100 m³ / h is thrown out by a domestic kitchen hood.
Similar calculations are done for all other rooms, a scheme for organizing air exchange (natural or forced) is developed and the dimensions of the ventilation ducts are determined (see the example below). Automate and speed up the process will help the calculation program.
Online calculator to help
The program considers the required amount of air according to the multiplicity regulated by SNiP. Just select the type of room and enter its dimensions.
Note. For boiler houses with a gas heat generator, the calculator takes into account only threefold exchange. The amount of supply air used for fuel combustion must be added to the result additionally.
We find out the air exchange by the number of residents
Appendix "K" SP 60.13330.2012 prescribes to calculate the ventilation of the room according to the simplest formula:
Decipher the notation of the presented formula:
- L is the desired value of the inflow (exhaust), m³ / h;
- m is the volume of the air clean mixture per 1 person indicated in the table of Appendix “K”, m³ / h;
- N - the number of people who are constantly in the room in question 2 hours a day or more.
Another example. It is reasonable to assume that in the same living room of a one-story house, two family members stay for a long time. Given that ventilation is organized and each tenant has over 20 squares of area, the parameter m is taken to be 30 m³ / h. We consider the amount of inflow: L = 30 x 2 = 60 m³ / h.
Important. Note that the result obtained is greater than the value determined by the multiplicity (47.25 m³ / h). For further calculations, the figure 60 m³ / h should be included.
If the number of people living in the apartment is so large that each person is allocated less than 20 m² (on average), then the above formula cannot be used. The rules indicate: in this case, the area of the living room and other rooms should be multiplied by 3 m³ / h. Since the total quadrature of the home is 91.5 m², the calculated volume of ventilation air will be 91.5 x 3 = 274.5 m³ / h.
In spacious rooms with high ceilings (from 3 m), updating the atmosphere is considered in two ways:
- If a large number of people are often in the room, calculate the cubic meter of air supplied by the specific indicator of 30 m³ / h per 1 person.
- When the number of visitors is constantly changing, the concept of a service area 2 meters above the floor is introduced. Determine the volume of this space (multiply the area by 2) and provide the required multiplicity, as described in the previous section.
Example of calculation and arrangement of ventilation
As a basis, we take the layout of a private house with an internal area of 91.5 m² and ceilings 3 m high, shown above in the drawing. How to calculate the amount of exhaust / inflow to the whole building according to the SNiP methodology:
- The volume of remote air from the living room and bedroom, which has an equal quadrature, will be 15.75 x 3 x 1 = 47.25 m³ / h.
- In the children's room: 21 x 3 x 1 = 63 m³ / h.
- Kitchen: 21 x 3 x 1 + 100 = 163 m³ / h.
- Bathroom - 25 m³ / h.
- Total 47.25 + 47.25 + 63 + 163 + 25 = 345.5 m³ / h.
Note. Air exchange in the hallway and corridor is not standardized.
Now we check the results for compliance with the second regulatory document. Since the family has a family of 4 people (2 adults + 2 children), 2 people are in the living room, bedroom and children's room for a long time.We recalculate the air exchange in the indicated rooms according to the number of people: 2 x 30 = 60 m³ / h (in each room).
The volume of extracts from the nursery meets the requirements (63 cubic meters per hour), but the values for the bedroom and living room will have to be adjusted. 47.25 m³ / h is not enough for two people, we take 60 cubic meters and again we recalculate the total amount of air exchange: 60 + 60 + 63 + 163 + 25 = 371 m³ / h.
It is equally important to correctly distribute the air flow in the building. In private cottages, it is customary to arrange natural ventilation systems - it is much cheaper and easier to install electric superchargers with air ducts. Add only one element of the forced removal of harmful gases - a cooker hood.
How to organize the natural movement of flows:
- We will provide inflow to all living quarters through automatic valves integrated into the window profile or directly into the outer wall. After all, standard metal-plastic windows are tight.
- In the partition between the kitchen and the bathroom, we will arrange a block of three vertical shafts facing the roof.
- Under the interior doors we provide gaps up to 1 cm wide for air passage.
- We install a cooker hood and connect it to a separate vertical channel. She will take part of the load - she will remove 100 cubic meters of exhaust gas in 1 hour during the cooking process. Remains 371 - 100 = 271 m³ / h.
- We will bring out two mines with bars to the bathroom and kitchen. Pipe sizes and heights are calculated in the last section of this manual.
- Due to the natural draft arising in two channels, air rushes from the nursery, bedroom and hall into the corridor, and then to the exhaust grilles.
Please note: the fresh streams shown on the layout are directed from rooms with clean air to more polluted areas, then are thrown out through the mines.
Calculate the diameters of ventilation ducts
Further calculations are somewhat more complicated, so we will accompany each stage with examples of calculations. The result will be the diameter and height of the ventilation shafts of our one-story building.
We distributed the entire volume of exhaust air into 3 channels: 100 cubic meters. forcibly removes the hood in the kitchen during the period the stove is turned on, the remaining 271 cubic meters leaves naturally in two identical shafts. The flow rate through 1 duct will turn out 271/2 = 135.5 m³ / h. The cross-sectional area of the pipe is determined by the formula:
- F - cross-sectional area of the ventilation duct, m²;
- L - exhaust flow through the shaft, m³ / h;
- ʋ - flow velocity, m / s.
Reference. The air velocity in the channels of natural ventilation is in the range 0.5–1.5 m / s. As the calculated value, we take the average indicator - 1 m / s.
How to calculate the cross section and diameter of one pipe in the example:
- We find the diameter in square meters F = 135.5 / 3600 x 1 = 0.0378 m².
- From the school formula for the area of the circle, we determine the diameter of the channel D = 0.22 m. We select the nearest larger duct from the standard series - Ø225 mm.
- If we are talking about a brick shaft laid inside the wall, then the size of the ventilation duct 140 x 270 mm fits the found section (a good match, F = 0.0378 sq. M.).
The diameter of the exhaust pipe for a household hood is considered in a similar way, only the flow rate pumped by the fan is taken more - 3 m / s. F = 100/3600 x 3 = 0.009 m² or Ø110 mm.
We select the height of the pipes
The next step is to determine the traction force that occurs inside the exhaust unit at a given elevation. The parameter is called the available gravitational pressure and is expressed in Pascals (Pa). Settlement formula:
- p is the gravitational pressure in the channel, Pa;
- H - elevation difference between the outlet of the ventilation grill and the ventilation duct cut above the roof, m;
- ρvozd - air density of the room, we accept 1.2 kg / m³ at home temperature +20 ° С.
The calculation method is based on the selection of the required height. First, determine how much you are ready to raise the exhaust pipes above the roof without affecting the appearance of the building, then substitute the height value in the formula.
Example. We take a height difference of 4 m and obtain a thrust pressure p = 9.81 x 4 (1.27 - 1.2) = 2.75 Pa.
Now a difficult stage is coming - the aerodynamic calculation of branch ducts. The task is to find out the resistance of the duct to the gas flow and compare the result with the available pressure (2.75 Pa). If the pressure loss is greater, the pipe will have to increase or increase the bore diameter.
The aerodynamic resistance of the duct is calculated by the formula:
- Δp - total pressure loss in the mine;
- R is the specific friction resistance of the passing stream, Pa / m;
- H - channel height, m;
- ∑ξ is the sum of local resistance coefficients;
- Pv - dynamic pressure, Pa.
We show by example how the resistance value is considered:
- We find the value of the dynamic pressure according to the formula Pv = 1.2 x 1² / 2 = 0.6 Pa.
- We find the friction resistance R according to the table, focusing on the dynamic pressure indicators of 0.6 Pa, a flow velocity of 1 m / s and an air duct diameter of 225 mm. R = 0.078 Pa / m (indicated by a green circle).
- The local resistance of the exhaust shaft is the louvre grille and 90 ° upward bend. The coefficients ξ of these parts are constant values equal to 1.2 and 0.4, respectively. The sum ξ = 1.2 + 0.4 = 1.6.
- Final calculation: Δp = 0.078 Pa / mx 4 m + 1.6 x 0.6 Pa = 1.27 Pa.
Now we compare the calculated pressure formed in the air duct and the resulting resistance. The traction force p = 2.75 Pa is much greater than the pressure loss (resistance) Δp = 1.27 Pa, a shaft 4 meters high is too high, it makes no sense to build such a one.
Since the numbers differ by half (roughly), we shorten the ventilation duct to 2 m and recalculate again:
- Available pressure p = 9.81 x 2 (1.27 - 1.2) = 1.37 Pa.
- The resistivity R and local coefficients ξ remain the same.
- Δp = 0.078 Pa / mx 2 m + 1.6 x 0.6 Pa = 1.15 Pa.
The natural draft pressure of 1.37 Pa exceeds the system resistance Δp = 1.15 Pa, which means that a two-meter-high shaft will work properly for natural exhaust and will provide the necessary flow rate of the removed gases.
Comment. It is not necessary to shorten the duct to 1 m, the ratio will change in the other direction: p = 0.69 Pa, Δp = 1.04 Pa, the traction force is not enough.
The ventilation channel Ø225 mm can be divided into 2 smaller pipes, but not by diameter, but by section. We get 2 round ventilation ducts of 150-160 mm, as done in the photo. The height of both shafts remains unchanged - 2 m.
How to simplify the task - tips
You could make sure that the calculations and the organization of air exchange in the building are rather complicated issues. We tried to explain the technique in the most accessible form, but the calculations still look cumbersome for the average user. We give some recommendations for a simplified solution to the problem:
- The first 3 stages will have to go through in any case - to find out the volume of discharged air, develop a flow pattern and calculate the diameters of the exhaust ducts.
- Take a flow velocity of not more than 1 m / s and determine the cross-section of the channels from it. Aerodynamics is not necessary to overcome - correctly calculate the diameters and simply bring the air ducts to a height of at least 2 meters above the intake grilles.
- Try to use plastic pipes inside the building - thanks to the smooth walls, they practically do not resist the movement of gases.
- Vent ducts laid in a cold attic must be insulated.
- Do not block the exits of the mines with fans, as is customary to do in the toilets of apartments. The impeller will not allow the natural hood to function properly.
For the inflow, install adjustable wall valves in the rooms, get rid of all the cracks where cold air can enter the house uncontrollably.