The average gas consumption for heating a house is 150 m²: an example of calculations and an overview of heat engineering formulas

Financing the heating season is a significant part of the budget spent on housing maintenance. Knowing the price and average gas consumption for heating a house 150 m², you can accurately determine the cost of heating the premises. These calculations are easy to do on their own without paying for the services of heat engineers.

You will learn all about gas consumption standards and methods for calculating gas consumption from our article. We will talk about how much energy is needed to compensate for the heat loss of a house during the heating season. Let’s show you what formulas should be used in calculations.

Heating of country cottages

When calculating the gas flow rate that is needed to heat a house, the most difficult task will be heat loss calculation, which must be fully compensated by the heating system during operation.

The complex of heat losses depends on the climate, the design features of the building, the materials used and the parameters of the ventilation system.

Calculation of compensated heat

The heating system of any building should compensate for its heat loss Q (W) in the cold period. They occur for two reasons:

1. heat transfer through the perimeter of the house;
2. heat loss due to cold air entering through the ventilation system.

Formally, heat loss through the wall and roof Q_tp can be calculated using the following formula:

Q_tp = S * dT / R,

Where:

• S - surface area (m²);
• dT - temperature difference between indoor and outdoor air (° C);
• R - an indicator of resistance to heat transfer of materials (m² * ° C / W).

The last indicator (also called the “coefficient of thermal resistance”) can be taken from the tables attached to building materials or products.

The type of double-glazed window significantly depends on the indicator of heat loss at home. The high price of insulated windows will be justified due to fuel economy

Example. Let the outer wall of the room have an area of ​​12 m²of which 2 m² takes a window.

The indicators of resistance to heat transfer are as follows:

• Aerated concrete blocks D400: R = 3.5.
• Double-glazed window double-chamber with argon “4M1 - 16Ar - 4M1 - 16Ar - 4I": R = 0.75.

In this case, at room temperature “+ 22 ° C”, and outdoor - “–30 ° C” the heat loss of the outer wall of the room will be:

• Q_tp (wall) = 10 * (22 - (- 30)) / 3.5 = 149 W:
• Q_tp (window) = 2 * (22 - (- 30)) / 0.75 = 139 W:
• Q_tp = Q_tp (wall) + Q_tp (window) = 288 watts.

This calculation gives the correct result if there is no uncontrolled air exchange (infiltration).

It can occur in the following cases:

• The presence of structural defects, such as loose fit of window frames to the walls or peeling of insulation material. They must be eliminated.
• Aging of the building, as a result of which chips, cracks or voids form in the masonry. In this case, it is necessary to introduce correction factors in the indicator of heat transfer resistance of materials.

In the same way, it is necessary to determine the heat loss through the roof, if the object is located on the top floor. Through the floor, any significant loss of energy occurs only if there is an unheated, blown-out basement, such as a garage. Heat hardly leaves the earth.

To calculate the heat transfer resistance of multilayer materials, it is necessary to summarize the performance of individual layers. Usually, only the most non-conductive materials are taken for calculations.

Consider the second reason for heat loss - the ventilation of the building. Energy consumption for heating the supply air (Q_at) can be calculated by the formula:

Q_at = L * q * c * dTwhere:

• L - air consumption (m³ / h);
• q - air density (kg / m³);
• c - specific heat of the incoming air (kJ / kg * ° C);
• dT - temperature difference between indoor and outdoor air (° C).

The specific heat of air in the temperature range of interest to us [–50 .. +30 ° С] is equal to 1.01 kJ / kg * ° С or in terms of the required dimension: 0.28 W * h / kg * ° С. Air density depends on temperature and pressure, but for calculations you can take a value of 1.3 kg / m³.

Example. For room 12 m² with the same temperature difference as in the previous example, heat losses due to ventilation work will be:

Q_at = (12 * 3) * 1.3 * 0.28 * (22 - (- 30)) = 681 W.

Designers take air flow according to SNiP 41-01-2003 (in our example 3 m³ / h at 1 m² living room area), but this value can be significantly reduced by the owner of the building.

Total total heat loss of the model room is:

Q = Q_tp + Q_at = 969 watts

To calculate the heat loss per day, week or month, you need to know the average temperature for these periods.

From the above formulas it is clear that the calculation of the volume of gas consumed both for a short period of time and for the entire cold season must be carried out taking into account the climate of the area where the heated object is located. Therefore, it is possible to use well-proven standard solutions only for similar environmental conditions.

To determine similar climatic parameters, you can use the maps of average monthly temperatures in winter. They can be easily found on the Internet.

With the complex geometry of the house and the variety of materials used in its construction and insulation, you can use the services of specialists to calculate the required amount of heat.

Ways to minimize heat loss

The cost of heating a house is a significant part of the cost of its maintenance. Therefore, it is reasonable to perform some types of work aimed at reducing heat loss by ceiling insulationthe walls of the house floor insulation and related designs.

Application insulation schemes outside and inside the house can significantly reduce this figure. This is especially true of old buildings with strong wear of walls and floors. The same polystyrene foam plates allow not only to reduce or completely eliminate freezing, but also minimize air infiltration through the protected coating.

Significant savings can also be achieved if the summer areas of the house, such as the veranda or the attic, are not connected to the heating. In this case, there will be a significant reduction in the perimeter of the heated part of the house.

Using the attic floor only in the summer significantly saves the cost of heating the house in winter. However, in this case, the ceiling of the upper floor should be well insulated

If you strictly follow the ventilation standards of the premises, which are prescribed in SNiP 41-01-2003, then the heat loss from air exchange will be higher than from freezing of the walls and roof of the building. These rules are mandatory for designers and any legal entities if the premises are used for production or the provision of services. However, the residents of the house may, at their discretion, reduce the values ​​indicated in the document.

In addition, heat exchangers can be used to heat cold air coming in from the street, rather than appliances that consume electricity or gas. So an ordinary plate heat exchanger can save more than half the energy, and a more complex device with a coolant can save about 75%.

Calculation of the required volume of gas

Combustible gas must compensate for heat loss. For this, in addition to the heat loss at home, it is necessary to know the amount of energy released during combustion, which depends on the efficiency of the boiler and the calorific value of the mixture.

Boiler selection rule

The choice of heater must be carried out taking into account the heat loss at home. It should be enough for the period when the annual minimum temperature is reached. In the passport or wall mounted gas boiler the “nominal thermal power” parameter, which is measured in kW for household appliances, is responsible for this.

Since any structure has thermal inertia, for calculating the necessary boiler capacity for the minimum temperature, the indicator of the coldest five-day period is usually taken. For a particular area, it can be found in organizations involved in the collection and processing of weather information, or from table 1. SNiP 23-01-99 (column No. 4).

A fragment of table 1 from SNiP 23-01-99. Using it, you can get the necessary data on the climate of the area where the heated object is located

If the boiler power exceeds the indicator sufficient for heating the room, then this does not lead to an increase in gas consumption. In this case, the equipment downtime will be longer.

Sometimes there is a reason to choose a boiler of slightly lower power. Such devices can be much cheaper when buying and operating. However, in this case, it is necessary to have a spare heat source (for example, a heater with a gas generator), which can be used in severe frosts.

The main indicator of the efficiency and efficiency of the boiler is the efficiency. For modern household equipment, it ranges from 88 to 95%. The efficiency is registered in the passport of the device and it is used in calculating gas consumption.

Heat release formula

To correctly calculate the consumption of natural or liquefied gas for heating a house with an area of ​​about 150 m² you need to know another indicator - the calorific value (specific heat of combustion) of the supplied fuel. According to the SI system, it is measured in J / kg for liquefied gas or in J / m³ for natural.

Gas holders (tanks for storing liquefied gas) are characterized in liters. To find out how much fuel will enter it in kilograms, you can apply a ratio of 0.54 kg / 1 l

There are two values ​​of this indicator - lower calorific value (H_l) and higher (H_h) It depends on the humidity and temperature of the fuel. When calculating take an indicator H_l - You need to find it out from the gas supplier.

If there is no such information, then in the calculations you can take the following values:

• for natural gas H_l = 33.5 mJ / m³;
• for liquefied gas H_l = 45.2 mJ / kg.

Given that 1 mJ = 278 W * h, we obtain the following calorific values:

• for natural gas H_l = 9.3 kW * h / m³;
• for liquefied gas H_l = 12.6 kW * h / kg.

Volume of gas consumed over a certain period of time V (m³ or kg) can be calculated using the following formula:

V = Q * E / (H_l * K)where:

• Q - heat loss of the building (kW);
• E - duration of the heating period (h);
• H_l - minimum calorific value of gas (kW * h / m³);
• K - boiler efficiency.

For liquefied gas, dimension H_l equal to kW * h / kg.

Example of calculating gas consumption

For example, take a typical prefabricated frame wooden two-story cottage. Region - Altai Territory, Barnaul.

The size of the cottage is 10 x 8.5 m. The slope of the gable roof is 30 °. This project is characterized by a warm attic, a relatively large area of ​​glazing, the absence of a basement and protruding parts of the house

Step 1. We calculate the main parameters of the house for calculating heat loss:

• Floor. In the absence of a purged basement, losses through the floor and foundation can be neglected.
• Window. Double-glazed window double-chamber "4M1 - 16Ar - 4M1 - 16Ar - 4I": R_o = 0.75. Glazing area S_o = 40 m².
• Walls. The area of ​​the longitudinal (side) wall is 10 * 3.5 = 35 m². The area of ​​the transverse (front) wall is 8.5 * 3.5 + 8.5² * tg(30) / 4 = 40 m². Thus, the total perimeter of the building is 150 m², and taking into account the glazing, the desired value S_s = 150 - 40 = 110 m².
• Walls. The main heat-insulating materials are glued beam, 200 mm thick (R_b = 1.27) and basalt insulation, 150 mm thick (R_u = 3.95). The total heat transfer resistance for the wall R_s = R_b + R_u = 5.22.
• Roof. Warming completely repeats the shape of the roof. Roof area without overhangs S_k = 10 * 8.5 / cos (30) = 98 m².
• Roof. The main heat-insulating materials are lining, 12.5 mm thick (R_v = 0.07) and basalt insulation, 200 mm thick (R_u = 5.27). The total heat transfer resistance for the roof R_k = R_v + R_u = 5.34.
• Ventilation. Let the air flow rate be calculated not according to the area of ​​the house, but taking into account the requirements, ensure a value of at least 30 m³ per person per hour. Since the cottage is constantly inhabited by 4 people, then L = 30 * 4 = 120 m³ / h

Step. 2. We calculate the required boiler power. If the equipment has already been purchased, then this step can be skipped.

For our calculations, it is necessary to know only two indicators of a gas boiler: efficiency and rated power. They must be registered in the device passport.

The temperature of the coldest five-day period is “–41 ° C". We will take a comfortable temperature as “+24 ° С". Thus, the average temperature difference over this period will be dT = 65 ° C.

We calculate the heat loss:

• through the windows: Q_o = S_o * dT / R_o = 40 * 65 / 0.75 = 3467 W;
• through the walls: Q_s = S_s * dT / R_s = 110 * 65 / 5.22 = 1370 W;
• through the roof: Q_k = S_k * dT / R_k = 98 * 65 / 5.34 = 1199 W;
• due to ventilation: Q_v = L * q * c * dT = 120 * 1.3 * 0.28 * 65 = 2839 W.

The total heat loss of the whole house during the cold five-day period will be:

Q = Q_o + Q_s + Q_k + Q_v = 3467 + 1370 + 1199 + 2839 = 8875 W.

Thus, for this model house, you can choose a gas boiler with a maximum thermal power parameter of 10-12 kW. If gas is also used to provide hot water, you will have to take a more productive device.

Step 3 We calculate the duration of the heating period and the average heat loss.

Under the cold season, when heating is necessary, understand the season with daily average temperatures below 8-10 ° C. Therefore, for calculations, you can take either columns No. 11-12 or columns No. 13-14 of Table 1 of SNiP 23-01-99.

This choice is left to the owners of the cottage. In this case, there will be no significant difference in annual fuel consumption. In our case, we will dwell on a period with a temperature below “+ 10 ° C”. The duration of this period is 235 days or E = 5640 hours.

With centralized heating, turning on and off the coolant supply is carried out in accordance with the established standards. One of the advantages of a private house is the launch of a heating mode at the request of residents

The heat loss of the house for the average temperature for this period is calculated as in step 2, only the parameter dT = 24 - (- 6.7) = 30.7 ° С. After performing the calculations, we obtain Q = 4192 watts.

Step 4 We calculate the amount of gas consumed.

Let the boiler efficiency K = 0.92. Then the volume of gas consumed (with averaged indicators of the minimum calorific value of the gas mixture) for the cold period of time will be:

• for natural gas: V = Q * E / (H_l * K) = 4192 * 5640 / (9300 * 0.92) = 2763 m³;
• for liquefied gas: V = Q * E / (H_l * K) = 4192 * 5640 / (12600 * 0.92) = 2040 kg.

Knowing the price of gas, you can calculate the financial costs of heating.

Conclusions and useful video on the topic

Reducing gas consumption by eliminating errors associated with home insulation. Real example:

Gas flow at known heat output:

All calculations of heat loss can be carried out independently only when the heat-saving properties of the materials of which the house is built are known. If the building is old, then first of all it is necessary to check it for freezing and eliminate the identified problems.

After that, using the formulas presented in the article, it is possible to calculate the gas flow with high accuracy.

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