How to calculate the natural ventilation of a residential building

Maintaining adequate ventilation in your house is essential to comfortable living conditions and high indoor air quality. Maintaining fresh and healthy air quality in your home can be achieved with natural ventilation, which depends on the flow of air naturally. It is also a cost-effective solution. We’ll look at how to determine your residential building’s natural ventilation requirements in this article.

Without using mechanical systems like fans or air conditioners, natural ventilation lets fresh air into your home and stale air out. Utilizing natural phenomena like wind and temperature variations, you can establish a consistent airflow throughout your home, minimizing the accumulation of contaminants and moisture.

Your home’s natural ventilation needs are determined by taking into account a number of variables, including the building’s size and layout, the quantity and size of windows and doors, and the climate in your area. You can choose the best natural ventilation techniques for your home by being aware of these factors.

In addition to enhancing indoor air quality, adequate ventilation also lowers the need for artificial heating and cooling by assisting in the regulation of indoor temperature. You and your family can live in a healthier, cozier, and more energy-efficient home by making the most of natural ventilation.

Step 1: Calculate the total floor area of the house.	Example: 150 square meters
Step 2: Determine the ventilation rate required per hour.	Example: 0.5 air changes per hour (ACH)
Step 3: Multiply the total floor area by the ventilation rate.	Example: 150 sqm × 0.5 ACH = 75 cubic meters per hour (m³/h)

Sanitary requirements of regulatory documents

Two primary documents govern the minimum quantity of air that must be supplied to and removed from the cottage’s rooms that have ventilation systems:

1. "Buildings of apartment buildings"-SNiP 31-01-2003, paragraph 9.
2. "Heating, ventilation and air conditioning" – SP 60.13330.2012, mandatory application "K".

The first document lays out the hygienic and sanitary conditions for air exchange in apartment building residential spaces. The ventilation computation ought to be predicated on these data. The air flow rate in volume per unit of time (m3/h) and the hourly multiplicity are the two types of dimensions that are used.

Citation. The number that represents the number of times the air environment in the room will be fully updated in an hour serves as an indicator of the frequency of air exchange.

The following consumption, or the number of air mixture updates (multiplicity), should be provided by supply and exhaust ventilation, depending on the purpose of the room:

• living room, nursery, bedroom – 1 time per hour;
• kitchen with an electric stove – 60 m³/h;
• bathroom, bathroom, 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 during the operation of the equipment is required;
• a boiler room with a heat generator burning natural gas – a three -fold update plus the volume of air required for combustion;
• The pantry, dressing room and other utility rooms – multiplicity 0.2;
• drying or sneaked – 90 m³/h;
• Library, study – 0.5 times within an hour.

Note: When there are no people present or when there is non-functioning equipment, SNiP reduces the demand on general ventilation. In residential settings, the frequency drops to 0.2, and in technical settings, it can reach 0.5. The yearly single-up air update requirement for rooms housing gas-heating plants is still in place.

The document’s page 9 suggests that the hood’s volume and the size of the influx are equal. A little more straightforward and dependent on the number of people in the room for at least two hours are requirements SP 60.13330.2012:

1. If there are 20 m² or more the area of the apartment per 1 living, a fresh influx is provided in the room in a volume of 30 m³/h per 1 person.
2. The volume of supply air is considered by area when 1 tenant accounts for less than 20 squares. The ratio is this: 3 m³ of tributary is supplied to 1 m² of home.
3. If the apartment does not provide for ventilation (there are no windows and opening windows), for each resident it is necessary to apply 60 m³/h of a clean mixture regardless of quadrature.

There is absolutely no contradiction between the listed regulatory requirements found in the two separate documents. First, the general ventilation system’s performance is determined using SNiP 31-01-2003 "Residential buildings."

The outcomes are checked against the Code of Ventilation and Air Conditioning requirements and, if needed, modified. We will examine the computed algorithm using the drawing of a one-story house as an example below.

Determination of air flow rate by multiplicity

In an apartment or a country cottage, the supply and exhaust ventilation are typically calculated individually for each room. The obtained results are summarized to determine the building’s overall air mass consumption. There is a very basic formula used:

• L is the sake volume of supply and exhaust air, m³/h;
• S is the quadrature of the room where ventilation is calculated, m²;
• H – ceilings height, m;
• N is the number of air -air updates for 1 hour (SNiP is regulated).

An illustration of a computation. 15.75 m2 is the area of a living room in a one-story building with a 3 m ceiling. The multiplicity of N for residential premises is equal to one, per SNiP 31-01-2003 instructions. The air mixture consumption per hour will therefore be L = 15.75 x 3 x 1 = 47.25 m³/h.

A crucial aspect. The installed ventilation equipment determines how much air mixture is removed from the kitchen when using a gas stove. A typical plan looks like this: a natural ventilation system provides one exchange as required by standards, and a domestic kitchen hood emits an extra 100 m3/h.

The dimensions of the ventilation ducts are established, the natural or forced air exchange scheme is developed, and similar computations are performed for every other room (see the example below). This process acceleration through automation will benefit the calculation program.

Online Calculator to help

The program takes into account the multiplicity controlled by SNiP to determine the necessary amount of air. Simply select a room from the selection and enter its dimensions. [WPCC]

Note: The calculator considers only a three-time exchange for boiler houses equipped with a gas heat generator. The quantity of supply air used to burn the fuel needs to be added to the outcome as well.

We find out the air exchange by the number of residents

The simplest formula for calculating a room’s ventilation is prescribed in Appendix "K" of SP 60.13330.2012:

We interpret the names of the given formula:

• L is the desired value of the influx (hood), m³/h;
• M – the volume of the air clean mixture per 1 person., the application "K", m³/h indicated in the table;
• N – the number of people constantly located in the room in question 2 hours a day or more.

One more instance. It makes sense to presume that two family members are going to be residing in the same living room of a one-story building for an extended period of time. Given that each tenant has an area larger than twenty square meters and that ventilation is organized, parameter M is set at 30 m³/h. We calculate the influx amount as follows: l = 30 x 2 = 60 m³/h.

Vital. Keep in mind that the outcome exceeds the multiplicity’s calculated value of 47.25 m³/h. Subsequent computations ought to incorporate the 60 m³/h value.

The formula above cannot be applied if there are so many people residing in the apartment that, on average, each person is given less than 20 m². According to the guidelines, the living room’s and the other rooms’ areas should be multiplied by three m³/h in this instance. Given that the dwelling’s total quadrature is 91.5 m², the ventilation air volume that will be needed is 91.5 x 3 = 274.5 m³/h.

The atmosphere in large halls with three-meter ceilings is evaluated in two ways:

1. If a large number of people are often in the room, calculate the cubature of the supplied air in a specific indicator of 30 m³/h per 1 person.
2. Когда количество посетителей постоянно меняется, вводится понятие обслуживаемой зоны высотой 2 метра от пола. Determine the volume of this space (multiply the area by 2) and provide the required rates required by the norms, as described in the previous section.

To ensure a comfortable and healthy indoor environment, it"s crucial to calculate the natural ventilation of your home effectively. Natural ventilation relies on the natural flow of air through openings in a building. By understanding the principles of natural ventilation and how to calculate it, you can optimize airflow, improve indoor air quality, and reduce the need for mechanical ventilation systems. Key factors to consider include the size and location of openings, wind speed and direction, temperature differentials, and the layout of your home. By following a few simple steps, you can determine the natural ventilation rate of your residential building, ensuring a more energy-efficient and comfortable living space for you and your family.

An example of the calculation and arrangement of ventilation

Consider the drawing above, which shows the layout of a private home with an internal area of 91.5 m² and a height of 3 m. How to use the SNiP method to determine the total amount of hood or inflow into the building:

1. 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.
2. In the children"s room: 21 x 3 x 1 = 63 m³/h.
3. Kitchen: 21 x 3 x 1 + 100 = 163 m³/h.
4. Bathroom – 25 m³/h.
5. Total 47.25 + 47.25 + 63 + 163 + 25 = 345.5 m³/h.

Note: There is no normalized air exchange in the hallway or corridor.

Let’s now verify that the outcomes adhere to the second regulatory document. There are two people in the living room, bedroom, and nursery because there is a family of four residing in the home (two adults and two children). We calculate the air exchange rate in these rooms based on the occupancy rate, which is 2 x 30 = 60 m³/h (per room).

The nursery’s hood volume (63 cubes per hour) meets the minimum, but the living room and bedroom’s volumes will need to be changed. Recalculate the total air exchange value using 60 cubes because two people cannot achieve the required 47.25 m³/h. 60 + 60 + 63 + 163 + 25 = 371 m³/h/h.

The appropriate distribution of air flows within the building is equally significant. It is common practice in private cottages to set up natural ventilation systems; installing air ducts for electrical superchargers is both less expensive and simpler in this regard. Include the kitchen hood as the only component of the forced removal of dangerous gases.

Arrangement strategies for the organic flow of flows:

1. We will provide an influx in all living rooms through automatic valves built into the window profile or directly into the outer wall. After all, standard metal -plastic windows are sealed.
2. In the partition between the kitchen and the bathroom, we will arrange a block of three vertical mines extending to the roof.
3. Under the interior doors, we will provide gaps up to 1 cm wide for air passage.
4. Install the kitchen hood and connect to a separate vertical channel. She will take on part of the load – will remove 100 cubic meters of exhaust gases in 1 hour in the process of cooking. Will remain 371 – 100 = 271 m³/h.
5. Two mines we will bring out with bars to the bathroom and the kitchen. The dimensions of the pipes and height are calculated in the last section of this leadership.
6. Due to the natural traction that occurs in two channels, the air will rush from the nursery, the bedroom and the hall into the corridor, and then to the exhaust lattices.

Note that the fresh flows shown in the layout are thrown out through the mines after being sent from rooms with clean air environments to more polluted zones.

Watch the following video to learn more about how natural ventilation is organized:

We calculate the diameters of the ventilations

Since subsequent calculations are a little more involved, we provide examples of the calculations for each step. The diameter and height of our one-story building’s ventilation mines will be the end result.

We divided the 100 m cube exhaust air volume among 3 channels. When the kitchen hood is forcibly removed while the plate is being turned on, 271 cubic meters of natural gas escape into two identical mines. Utilization via a single air duct will result in 271 /2 = 135.5 m³/h. The following formula determines the pipe’s cross-sectional area:

• F is the cross -sectional area of the ventilation system, m²;
• L – expenditure of hood through the mine, m³/h;
• ʋ – flow speed, m/s.

Citation. The range of air speed in natural ventilation channels is between 0.5 and 1.5 m/s. We accept the average indicator of 1 m/s as a computed value.

How to determine one pipe’s diameter and section in the example:

1. Find the size of the diameter in square meters f = 135.5 /3600 x 1 = 0.0378 m².
2. From the school formula of the area of the circle, we determine the diameter of the channel d = 0.22 m. Choose the nearest larger air duct from the standard row – Ø225 mm.
3. If we are talking about a brick shaft inherently inside the wall, then the size of the ventrical is suitable for the found section of 140 x 270 mm (successful coincidence, F = 0.0378 m. sq.).

A household hood’s dialing pipe diameter is comparable; the fan’s pumped flow is deemed to be faster than three meters per second. F is equal to 100 /3600 x 3, or Ø110 mm, or 0.009 m².

We select the height of the pipes

Finding the traction force that develops inside the exhaust block at a specific height differential is the next step. The parameter is expressed in Pascals (PA) and is known as the located gravitational pressure. referred to as formula

• P – gravitational pressure in the channel, PA;
• N – the height difference between the output of the ventilation grill and the cut of the ventilation above the roof, m;
• ρ water – the air density of the room, we take 1.2 kg/m³ at home temperature +20 ° C.

The process for calculating is predicated on determining the necessary height. Determine how high you are willing to raise the extract pipes above the roof without compromising the building’s aesthetic first, and then enter the height value into the formula.

For instance. With a height differential of 4 m, the thrust pressure is calculated as follows: p = 9.81 x 4 (1.27 – 1.2) = 2.75 PA.

The hardest part will soon be here: calculating the withdrawal channels’ aerodynamics. The task is to measure the gas flow duct’s resistance and compare the result to the pressure (2.75 PA) that has been located. The pipe will need to be expanded or increased by the passing diameter if the pressure turns out to be higher.

The following formula is used to determine the duct’s aerodynamic resistance:

• ΔP – total pressure losses in the mine;
• R is the specific resistance of the friction of the passing flow, pa/m;
• H – the height of the canal, m;
• ∑ Answers – the sum of local resistance coefficients;
• PV – dynamic pressure, PA.

We’ll provide an illustration of how the resistance value is taken into account here:

1. Find the value of dynamic pressure according to the formula pv = 1.2 x 1² / 2 = 0.6 PA.
2. The resistance from friction R is found according to the table, focusing on the indicators of dynamic pressure 0.6 Pa, flow speed 1 m/s and air duct diameter 225 mm. R = 0.078 PA/m (indicated by a green circle).
3. Local resistances of an exhaust shaft are a blind grate, a branch of 90 ° and an umbrella at the end of the pipe. The coefficients of these parts are constant values equal to 1.20.4 and 1.3, respectively. Sum ξ = 1.2 + 0.4 + 1.3 = 2.9.
4. Final calculation: δp = 0.078 PA/m x 4 m + 2.9 x 0.6 PA = 2.05 PA.

Compare the resistance that results from the air duct’s calculated pressure formation. It is useless to build a mine 4 meters high because the traction power (p = 2.75 PA) is greater than the pressure loss (resistance) (ΔP = 2.05 PA).

We will now reduce the ventilation to 3 m and recalculate once more:

1. Passed pressure p = 9.81 x 3 (1.27 – 1.2) = 2.06 PA.
2. Specific resistance R and local coefficients cil remain the same.
3. Δp = 0.078 PA/m x 3 m + 2.9 x 0.6 PA = 1.97 PA.

The natural thrust of 2.06 Pa Naportas surpasses the system resistance of Δp = 1.97 PA, indicating that the three-meter-high mine will function consistently for a natural hood and will supply the required amount of remote gas consumption.

Significant observation. The difference was only 2.06 – 1.97 = 0.09 PA between the thrust strength and the duct resistance. The height of the pipe in our example is better to accept with a margin of 3.5 m so that the hood operates steadily in any weather.

The Ø225 mm ventilation channel can be split into two smaller pipes based on section rather than diameter. Two spherical venti—150–160 mm—as seen in the picture are obtained. Both of the mines are still 3.5 meters high.

How to simplify the task – tips

It is possible to ensure that the calculations and the arrangement of the building’s air exchange system are fairly challenging. Although we made every effort to make the technique as clear as possible, the calculations still appear complicated to the average user. We offer some suggestions for a condensed resolution to the issue:

1. The first 3 stages will have to go through in any case – find out the volume of the thrown air, develop a stream movement scheme and calculate the diameters of the exhaust ducts.
2. Take the flow rate of no more than 1 m/s and determine the cross section of the channels on it. It is not necessary to overcome aerodynamics – calculate the diameters correctly and simply take the air pipelines to a height of at least 3 meters above the foster grilles.
3. Try to use plastic pipes inside the building – thanks to the smooth walls, they practically do not resist gases.
4. Ventilations laid in a cold attic, be sure to insulate.
5. Do not overlap the mine outputs with fans, as is customary to do in the toilets of apartments. The impeller will not allow the natural extract to function normally.

Install adjustable wall valves in each room to control the influx of cold air, and seal up any gaps that allow it to enter the house uncontrollably.

Determining your home’s natural ventilation is crucial to keeping your interior atmosphere cozy and healthful. Without depending entirely on mechanical systems, you can guarantee sufficient airflow throughout your home by knowing the fundamentals of natural ventilation and how to calculate it.

Knowing the idea of air change rate (ACH) is essential to calculating natural ventilation. The amount of fresh air replaced in a space per hour is measured by ACH. You can make sure that indoor air quality stays high and that residents are comfortable by finding your home’s air conditioning consumption (ACH).

The size and placement of windows, home design, and regional climate all have an impact on a residential building’s ability to naturally ventilate. You can estimate the natural ventilation rate of your home and make any necessary adjustments to improve airflow by considering these factors and utilizing basic formulas.

In addition to maintaining indoor air quality, proper natural ventilation lowers energy costs and carbon emissions by minimizing the need for mechanical heating and cooling systems. You and your family can live in a more sustainable, comfortable, and healthy environment by putting effective natural ventilation strategies into practice.

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Video on the topic

How to do ventilation in a private house ? Selection and calculation. Hood in the house. Air duct for ventilation

Video 25 Calculation of the natural exhaust ventilation system