Average gas consumption for heating a house of 150 m²: example calculations and review of thermal formulas

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

You will learn everything about gas consumption standards and methods for calculating blue fuel consumption from the article we presented. We will tell you how much energy is required to compensate for heat losses at home during the heating season. We will tell you what formulas should be used in the calculations.

Heating of country cottages

When calculating the gas consumption needed to heat a house, the most difficult task will be heat loss calculation, which the heating system must fully compensate during operation.

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

Calculation of the compensated amount of heat

The heating system of any building must compensate for its heat loss Q (W) during the cold period of time. They happen for two reasons:

1. heat exchange through the perimeter of the house;
2. heat loss as a result of 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 room and street air (°C);
• R – indicator of resistance to heat transfer of materials (m² * °C / W).

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

The heat loss rate of the house significantly depends on the type of double-glazed window. The high price of insulated windows will be justified due to fuel savings

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

The heat transfer resistance indicators are as follows:

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

In this case, at room temperature “+22°С”, and street temperature – “–30°С”, the heat loss from 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 W.

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

It may occur in the following cases:

• The presence of structural defects, such as loose fitting of window frames to the walls or peeling of insulating material. They need to be eliminated.
• Aging of a building, resulting in chips, cracks or voids in the masonry. In this case, it is necessary to introduce correction factors into the heat transfer resistance of materials.

In the same way, it is necessary to determine heat loss through the roof if the object is located on the top floor. Through the floor, any significant energy loss occurs only if there is an unheated, ventilated basement space, such as a garage. Almost no heat goes into the ground.

To calculate the heat transfer resistance index of multilayer materials, it is necessary to sum up the indicators of individual layers. Usually, only the most non-thermal conductive materials are taken for calculations

Let's consider the second reason for heat loss - building ventilation. Energy consumption for heating the supply air (Q_V) can be calculated using the formula:

Q_V = L * q * c * dT, Where:

• L – air flow (m³ / h);
• q – air density (kg/m³);
• c – specific heat capacity of incoming air (kJ/kg *°C);
• dT – temperature difference between room and street air (°C).

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

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

Q_V = (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.

In total, the total heat loss of the model room is:

Q = Q_tp + Q_V = 969 W.

To calculate 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 facility is located.Therefore, well-proven standard solutions can only be used for similar natural conditions.

To determine similar climatic parameters, you can use 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 home makes up a significant portion of the cost of maintaining it. Therefore, it is reasonable to carry out some types of work aimed at reducing heat loss by ceiling insulation, house walls, floor insulation and related structures.

Application external insulation schemes and from inside the house can significantly reduce this figure. This is especially true for old buildings with severe wear and tear on walls and ceilings. The same polystyrene foam boards can not only 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 verandas or the attic floor, are not connected to 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 must be well insulated

If you strictly follow the standards for ventilation of premises, which are prescribed in SNiP 41-01-2003, then 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 provision of services. However, residents of the house can, at their discretion, reduce the values ​​​​specified in the document.

In addition, to heat the cold air coming from the street, you can use heat exchangers, rather than devices that consume electricity or gas. Thus, an ordinary plate heat exchanger can save more than half of the energy, and a more complex device with coolant can save about 75%.

Calculation of the required volume of gas

The burned gas must compensate for the heat loss. To do this, in addition to the heat loss of the house, 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 made taking into account the heat loss of the house. It should be enough for the period when the annual minimum temperatures are reached. In the passport of the floor or wall-mounted gas boiler The parameter “nominal thermal power” is responsible for this, which is measured in kW for household appliances.

Since any structure has thermal inertia, to calculate the required boiler power, the minimum temperature is usually taken from the coldest five-day period. For a specific area, it can be found in organizations involved in the collection and processing of meteorological information, or from Table 1. SNiP 23-01-99 (column No. 4).

Fragment of table 1 from SNiP 01/23/99. Using it, you can obtain the necessary data on the climate of the area where the heated facility is located

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

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

The main indicator of the efficiency and economy of a boiler is the efficiency factor. For modern household equipment it ranges from 88 to 95%. The efficiency is stated in the device passport and is used when 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 It is necessary to find out one more 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 tanks (containers for storing liquefied gas) are characterized in liters. To find out how much fuel will be included in it in kilograms, you can use the ratio 0.54 kg / 1 l

There are two values ​​for this indicator - lower calorific value (H_l) and highest (H_h). It depends on the humidity and temperature of the fuel. When calculating, take the indicator H_l – you need to find this out from your gas supplier.

If there is no such information, then the following values ​​can be taken in calculations:

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

Taking into account the fact 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.

The 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, let's 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 angle of the gable roof is 30°. This project is characterized by a warm attic, a relatively large glazing area, the absence of a basement and protruding parts of the house

Step 1. Let's calculate the main parameters of the house to calculate heat loss:

• Floor. In the absence of a ventilated basement, losses through the floor and foundation can be neglected.
• Window. Double-glazed unit “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 (facade) wall is 8.5 * 3.5 + 8.5² * tg(30) / 4 = 40 m². Thus, the total perimeter area of ​​the building is 150 m², and taking into account glazing the desired value S_s = 150 – 40 = 110 m².
• Walls. The main thermal insulation materials are laminated timber, 200 mm thick (R_b = 1.27) and basalt insulation, 150 mm thick (R_u = 3.95). Total heat transfer resistance for a wall R_s = R_b + R_u = 5.22.
• Roof. The insulation completely follows the shape of the roof. Roof area without overhangs S_k = 10 * 8.5 / cos (30) = 98 m².
• Roof. The main thermal insulation materials are lining, 12.5 mm thick (R_v = 0.07) and basalt insulation, 200 mm thick (R_u = 5.27). Total heat transfer resistance for a roof R_k = R_v + R_u = 5.34.
• Ventilation. Let the air flow be calculated not according to the area of ​​the house, but taking into account the requirements to ensure a value of at least 30 m³ per person per hour. Since 4 people permanently live in the cottage, then L = 30 * 4 = 120 m³ / h.

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

For our calculations, we need 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”. Let’s take a comfortable temperature as “+24 °C”. Thus, the average temperature difference over this period will be dT = 65 °C.

Let's calculate heat loss:

• through windows: Q_o = S_o * dT / R_o = 40 * 65 / 0.75 = 3467 W;
• through 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 entire 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 supply, then you will have to take a more productive device.

Step 3. Let's calculate the duration of the heating period and average heat loss.

The cold season, when heating is necessary, is understood as a season with average daily 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 remains with the owners of the cottage. In this case, there will be no significant difference in annual fuel consumption. In our case, we will focus on the period with temperatures 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 occurs according to established standards. One of the advantages of a private house is the start of the heating mode at the request of the residents

The heat loss of the house for the average temperature over this period is calculated in the same way as in step 2, only the parameter dT = 24 – (– 6.7) = 30.7 °С. After carrying out the calculations we get Q = 4192 W.

Step 4. Let's calculate the volume of gas consumed.

Let the boiler efficiency K = 0.92. Then the volume of gas consumed (with average indicators of the minimum calorific value of the gas mixture) during 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 gas prices, 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 consumption at known thermal power:

All calculations of heat loss can be carried out independently only when the heat-saving properties of the materials from 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 this, using the formulas presented in the article, you can calculate gas consumption with high accuracy.

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