Calculation of thermal load on heating, calculation of heat loss of the house

It’s not just about comfort when it comes to keeping your house warm and comfortable during the cold months—it’salsoabout effectiveness and economy. You can maximize the efficiency of your heating system and reduce energy costs by being aware of how much heat your home loses and how much it needs. This is the point at which figuring out your home’s heat loss and thermal load becomes crucial.

Calculating your thermal load entails figuring out how much heat your house needs to stay at a comfortable temperature. It considers a number of variables, including the number of occupants, insulation levels, climate, and house size. You may prevent your heating system from being undersized or oversized and make sure it runs effectively without wasting energy by precisely estimating the thermal load.

However, figuring out how much heat escapes through your home’s doors, windows, walls, and other structural components is part of the process. The quality of the insulation, air leakage, and outside temperature all affect this heat loss. In order to reduce energy waste and maintain a comfortable indoor environment, you can identify areas for improvement, such as adding insulation or sealing drafts, by understanding where and how heat escapes.

Making these calculations can give you important information about the thermal performance of your house, whether you’re building a new house or trying to upgrade your current heating system. Equipped with this knowledge, you can choose the right HVAC system, upgrade your insulation, and take other energy-saving steps to create a more comfortable and energy-efficient living environment.

Averaged calculation and accurate

The following scheme is used to calculate the average given the factors that have been described. A room in 20 kV should receive 2,000 watts of heat flow if one square meter requires 100 watts of heat flow. Out of eight sections, the radiator (commonly bimetallic or aluminum) emits approximately 150 watts. 2,000 divided by 150 yields 13 sections. However, this is a fairly extensive thermal load calculation.

The precise appears a little unsettling. Nothing difficult, really. This is the equation:

• Q1 – type of glazing (ordinary = 1.27, double = 1.0, triple = 0.85);
• Q2 – wall insulation (weak, or absent = 1.27, the wall laid out in 2 bricks = 1.0, modern, high = 0.85);
• Q3 – ratio of the total area of window openings to the floor area (40% = 1.2, 30% = 1.1, 20% – 0.9, 10% = 0.8);
• Q4 -street temperature (the minimum value is taken: -35 o C = 1.5, -25 o C = 1.3, -20 o c = 1.1, -15 o C = 0.9, -10 o C = 0.7);
• Q5 – the number of external walls in the room (all four = 1.4, three = 1.3, corner room = 1.2, one = 1.2);
• Q6 – type of estimated room above the estimated room (cold attic = 1.0, warm attic = 0.9, housing heated room = 0.8);
• Q7 – ceiling height (4.5 m = 1.2, 4.0 m = 1.15, 3.5 m = 1.13.0 m = 1.05, 2.5 m = 1.3).

Any of the methods that have been described can be used to determine an apartment building’s thermal load.

In understanding how to heat and insulate your house effectively, it"s crucial to grasp the concept of thermal load and heat loss calculation. Essentially, calculating the thermal load means figuring out how much heat your home needs to stay comfortably warm during the coldest times. This involves considering factors like the size of your house, the local climate, and the insulation quality. On the other hand, calculating heat loss involves determining how much heat your house loses to the outside environment. This takes into account factors such as insulation effectiveness, air leaks, and the materials your house is made of. Both calculations are essential for designing an efficient heating system and ensuring your home stays cozy while minimizing energy waste and costs.

Features of the selection of radiators

The radiators, panels, "warm" floor, convectors, etc. are the typical parts of the room’s heat supply. D. Radiators are the most prevalent components of the heating system.

A unique hollow design of a modular alloy with excellent heat transfer is called a thermal radiator. Steel, aluminum, ceramics, cast iron, and other alloys are used to make it. The heating radiator’s basic working principle can be boiled down to the coolant’s energy being radiated into the room through its "petals."

The massive cast-iron batteries were replaced with aluminum and bimetallic heating radiators. This product is a widely used instrument for heat radiation in rooms due to its ease of production, high heat transfer, and effective design.

The room’s heating radiators can be calculated using a variety of techniques. The techniques in the list below have been arranged to improve computation accuracy.

1. In the area. N = (s*100)/c, where n is the number of sections, s is the area of the room (M2), C is the heat transfer of one radiator section (Tue, taken from those passport or certificate for the product), 100 W – the amount of heat flow, which is necessary for heating 1 m2 (empirical value). The question arises: how to take into account the height of the ceiling of the room?
2. By volume. N = (s*h*41)/c, where n, s, c is similar. H – the height of the room, 41 W – the amount of heat flow, which is necessary for heating 1 m3 (empirical value).
3. According to the coefficients. N = (100*S*K1*K2*K3*K4*K5*K6*K7)/C, where N, S, C and 100 are similarly. K1 – taking into account the number of chambers in the double -glazed window of the room, K2 – thermal insulation of the walls, K3 – the ratio of the area of the windows to the area of the room, K4 – the average minus temperature in the coldest week of winter, K5 – the number of external walls of the room (which “go out” to the street), K6 – type of room from above, K7 – ceiling height.

When determining the number of sections, this method yields the most accurate results. Naturally, fractional computation results are always rounded to the nearest whole number.

For example, a project of one -story house 100 m²

To provide a comprehensible explanation of all the methods for calculating thermal energy, we suggest using the one-story house depicted in the drawing as an example. Based on the external scenario, the house’s total area is 100 square feet. We enumerate the building’s technical attributes:

• The construction region – a strip of moderate climate (Minsk, Moscow);
• The thickness of external fences is 38 cm, the material is silicate brick;
• External wall insulation – foam 100 mm thick, density – 25 kg/m³;
• floors – concrete on the ground, the basement is absent;
• Overlap – railway slabs insulated from the side of the cold attic with a foam 10 cm;
• windows – standard metal -plastic for 2 glasses, size – 1500 x 1570 mm (h);
• The front door is a metal 100 x 200 cm, insulated from the inside with extruded polystyrene foam 20 mm.

The boiler room is housed in a separate structure, and the cottage has interior partitions that are spaced 12 centimeters apart. The drawing shows the locations of the rooms, and the ceiling heights will be determined using the 2.8 or 3 m computed methodology as explained.

Sequence of calculation of heating systems

The following is the stage of calculations for heating and heating equipment systems, independent of the number of circuits and heated area:

1. It is made taking into account their thermal insulation properties to determine the amount of heat loss in the winter period.
2. The volume and amount of heat necessary for heating in rooms is determined.
3. The result of heat engineering calculation is the thermal balance of the house or the required power of the heating system.
4. Based on the results of the preparation of the thermal balance for each room, the selection of sizes of heating devices is performed, the type of radiators is determined acceptable for this case. Then the devices in the premises in the areas of maximum heat loss should be placed.
5. According to the selected scheme of the heating system (gravitational or pumping, one -pipe or two -pipe, riser or collector) on the construction plans, the pipelines are placed and the determination of the location of the connection to the radiators.
6. In the process of hydraulic calculation, the diameters of each section of the heating system, the flow rate of the coolant and the loss of pressure along the length of the pipelines for friction are determined.
7. According to the results of hydraulic calculation, the selection of circulation pumps is carried out.
8. Depending on the fundamental diagram of the heating system and the main technical data of the system and the maximum working pressure, the volume of the expansion tank is a choice of the volume of the temperature expansion of the coolant to compensate for the temperature expansion.
9. A selection of the type, quantity and power of the heating boilers is performed, as well as a principled circuit of the boiler room with the placement of basic equipment and concomitant systems.

We consider the flow rate of heat in quadrature

The simplest heat calculation is typically used to get an approximate estimate of the heating load. It involves taking the building’s area from the external measurement and multiplying it by 100 watts. As a result, a 100 m² summer house will require 10,000 watts, or 10 kW, of heat. The outcome enables you to select a boiler with a stock coefficient of 1.2–1.3; in this instance, the unit’s power is assumed to be 12.5 kW.

We suggest carrying out more precise computations that account for the development region, the number of windows, and the rooms’ locations. For ceiling heights up to three meters, the following formula should be applied:

Every room is subjected to a separate calculation, after which the outcomes are compiled and multiplied by the regional coefficient. Interpreting formula names:

• Q – the desired load, Tue;
• SPP – quadrature of the room, m²;
• Q is an indicator of the specific thermal characteristic, assigned to the area of the room, W/m²;
• k – coefficient taking into account the climate in the area of residence.

As a point of reference. The coefficient K is assumed to be unity if a private residence is situated in a zone of moderate climate. K = 0.7 is utilized in the southern regions, while 1 is used in the northern ones. 5-2.

The indicator Q = 100 W/m² in an approximation of the general quadrature calculation. This method ignores variations in the number of light openings and the locations of the rooms. Compared to the nearby corner bedroom with windows, the cottage’s interior corridor will lose a lot less heat. We suggest accepting the following value for the specific thermal characteristic Q:

• for rooms with one outer wall and window (or door) Q = 100 W/m²;
• Corner rooms with one light opening – 120 W/m²;
• The same, with two windows – 130 W/m².

Clearly illustrated on the building plan is how to select the appropriate value Q. This is how the calculation appears in our example:

Q is equivalent to (15.75 x 130 + 21 x 120 + 5 x 100 + 7 x 100 + 6 x 100 + 15.75 x 130 + 21 x 120) x 1 = 10935 W ≈ 11 kW.

As you can see, the revised computations produced a different outcome, indicating that 1 kW more thermal energy is required to heat a specific 100 m² home. The amount accounts for the heat used to heat the outside air that enters the house through the walls and openings (infiltration).

Calculation of thermal load by the volume of rooms

The preceding option cannot be used because the outcome will be incorrect when the distance between the floors and the ceiling reaches 3 m or more. In these situations, the specific enlarged indicators of heat flow per 1 m³ of the room volume are regarded as the heating load.

The only difference in the area S parameter is the volume, or v: The formula and calculation algorithm stay the same.

Consequently, the cubicature of each room serves as another accepted indicator of the specific consumption Q.

• room inside the building or with one outer wall and window – 35 W/m³;
• A corner room with one window – 40 W/m³;
• The same, with two light openings – 45 W/m³.

Note: The formula uses the same regional coefficients K, but with different values.

For instance, using a 3 m ceiling height, we calculate the load on our cottage’s heating system.

Q is equal to 11182 W ≈ 11.2 kW (47.25 x 45 + 63 x 40 + 15 x 35 + 21 x 35 + 18 x 35 + 47.25 x 45 + 63 x 40) x 1.

It is evident that the heating system’s needed thermal power increased by 200 W in comparison to the earlier computation. Accepting a room’s height of 2.7–2.8 meters and accounting for energy costs through the cubature will result in numbers that are roughly equal. In other words, the technique works well for larger-scale heat loss calculations in rooms of any height.

Gcal calculation formula for heating

Gcal: What is it? Gcal, or gigacaloria, is the unit of measurement used to calculate thermal energy. You can compute the Gcal on your own, but only if you have read up on some thermal energy-related topics beforehand. Take a look at the article General computation information, which includes a formula to determine Gcal.

What is Gcal?

One gram of water needs a specific amount of energy, or calories, to heat it to one degree. In atmospheric pressure conditions, this requirement is met.

A large value, Gcal, is used in thermal energy calculations. One billion calories is equal to one gigacaloria.

According to the Ministry of Fuel and Energy’s document, this has been in use since 1995.

In Russia, 0.9342 Gcal is the average monthly consumption per square meter. Depending on the local weather, this value can change in each region more or less.

If gigacaloria were converted to common values, what would that mean?

1. 1 gigacaloria equals 1162.2 kilowatt hours.
2. In order to heat 1 thousand tons of water to a temperature of +1 degrees, 1 gigacalorium will be required.

Gcal in apartment buildings

Thermal calculations in multi-apartment buildings use gigacalories. You can compute the account for payment of heating if you know the precise amount of heat energy left in the house. For instance, you will have to pay for central heating according to the size of the heated room if a shared or individual heat device is not installed in the home.

The horizontal type, also known as the sequential or collector type, is assumed if the thermal counter is installed. In this version, the system inside the apartment is decided by the residents, and two risers for the supply and reverse pipe are made in the apartment. New homes use these kinds of schemes.

Because of this, homeowners are able to independently control how much thermal energy they use, choosing between savings and comfort.

The following is the adjustment:

1. Due to the throttle of heating batteries, the heating device is limited, therefore, the temperature in it decreases, and the heat energy consumption is reduced.
2. Installation of a common thermostat on a reverse pipe. In this version, the flow rate of the working fluid is determined by the temperature in the apartment and if it increases, the consumption decreases, and if it decreases, then the consumption increases.

Gcal in private houses

The cost of heat energy with each fuel type is what the residents of a private home are primarily interested in when we discuss Gcal. As a result, we take into account some costs for 1 Gcal for different kinds of fuel:

• Natural gas – 3300 rubles;
• Liquefied gas – 520 rubles;
• Coal – 550 rubles;
• Pellets – 1800 rubles;
• Diesel fuel – 3270 rubles;
• Electricity – 4300 rubles.

The cost may change based on the area, and it’s important to keep in mind that fuel prices rise on a regular basis.

General information about calculations Gcal

Special calculations are required in order to calculate the Gcal, and these calculations are governed by special regulatory acts. The computation is performed by tools that can interpret any confusing points and clarify the steps involved in calculating the Gcal.

Having a personal device will help you stay away from issues and unnecessary expenses. For you, it’s sufficient to take the indicators out of the counter each month and multiply the result by the tariff. The cost of using the heating must be covered by the amount received.

Heat meters

To compute thermal energy, you must be aware of the following details:

1. The temperature of the fluid at the entrance and output of a certain section of the line.
2. Fluid consumption that moves through heating devices.

Heat meters can be used to measure consumption. There are two kinds of heat meters:

1. Wang -shaped counters. Such devices are used to account for thermal energy, as well as hot water consumption. The difference between such meters and devices for cold water accounting is the material from which the impeller is made. In such devices, it is most resistant to high temperatures. The principle of operation is similar in two devices:

• The accounting device is transmitted by the impeller;
• The impeller begins with the movement of the working fluid;
• The transfer is made without direct interaction, and with the help of a permanent magnet.

These devices have a low threshold despite having a simple design. Additionally, they have dependable defense against reading distortion. The external magnetic field prevents the impeller from braking through the use of an antimagnetic screen.

1. Devices with the registrar of differences. Such counters work according to the Bernoulli law, which claims that the speed of fluid or gas flow is inversely proportional to its static movement. If the pressure is recorded by two sensors, then you can easily determine the consumption in real time. The counter involves electronics in the structure. Almost all models provide information about the flow and temperature of the working fluid, and also determine the consumption of thermal energy. You can tune work manually using a PC. You can connect the device to the PC via the port.

Many locals are curious about how to figure out how much Gcal is needed to heat an open heating system where hot water selection is available. Concurrently, the pressure sensors are mounted on the supplies and return pipe. The amount of warm water used for household needs will be indicated by the difference in the working fluid flow.

The following heating formula must be used for heating if you do not have a separate device: Q is equal to V * (T1-T2) / 1000, where

1. Q – the total volume of heat energy.
2. V is the volume of hot water consumption. Measured in tons or cubic meters.
3. T1 is the temperature of hot water, which is measured in degrees Celsius. In this calculation, it is better to take into account the temperature that will be characteristic of a specific working pressure. This indicator has the name – enthalpie. If there is no necessary sensor, then take the temperature that will be similar to enthalpy. Typically, the average indicator of this temperature is in the range of 60-65 degrees Celsius.
4. T2 is the temperature of cold water, which is measured in degrees Celsius. As you know, getting to the pipeline with cold water is not easy, so such values are determined by constant values. They, in turn, depend on climatic conditions outside the house. For example, in the cold season, such a value can be 5 degrees, and in the warm time, when there is no heating, it can reach 15 degrees.
5. 1000 is a coefficient, thanks to which you can get a response in gigacalories. Such a value will be more accurate than in ordinary calories.

A different method of calculating gigacalories is used in closed heating systems. You must apply the following formula to determine the Gcal in the closed heating system: Where Q is equal to ((V1 * (T1 – T)) – (V2 * (T2 – T)))) / 1000.

1. Q is the previous volume of thermal energy;
2. V1 is the heat expenditure parameter in the supply pipe. There can be water vapor or ordinary water as a heat source.
3. V2 – the volume of water flow in the dialing pipe;
4. T1 – temperature in the pipe of the heat carrier;
5. T2 – temperature at the output of the pipe;
6. T – cold water temperature.

This formula uses two parameters to calculate thermal energy for heating: the first represents the heat entering the system, and the second is the heat parameter after the return pipe removes the heat carrier.

Calculation of heat loads

The formula determines the building’s volume based on an external measurement, the temperature of the interior and exterior air, and the calculated (maximum) consumption for heating the building, W.

Where W/(M 3 • 0 C) denotes thermal losses through the external fences of the building unit of volume at the difference in internal and external temperatures, and QOV is the specific thermal characteristic of buildings (heating characteristics). DT is 1̔s.

Table 4 indicates that V is the building’s volume in the external measurement, expressed in m 3;

TVP is a heated room’s average interior air temperature;

The district’s climate in the city of Tomsk is as follows:

For the purpose of designing heating systems, the expected outdoor temperature is 0 with

The TNV ventilation temperature that was computed was 0 with

The lowest month in TNHM’s average outdoor temperature was 0 with

During the heating period, the average outdoor temperature was zero with

The estimated temperature of the interior air at TNO = -40 0 C is for TV = 20 0 C in residential buildings, 22 0 C in kindergartens, and 18 0 C in schools.

B: The adjustment factor based on the outside air temperature excellent between -30 °C (this modification is only applied to residential buildings) and TVR = 18 0C; in other situations, the adjustment is made in accordance with the table.

All public and residential buildings shall be accepted as being of a fixed size.

The volumes and unique heating properties of the building’s end and regular sections are considered when calculating the thermal loads of residential buildings.

Where the end and private portions of the building have different heating characteristics;

VT, VR stand for the private and end section volumes, respectively;

– quantity of regular sections;

A summary table of thermal loads is compiled for four characteristic modes, namely the most winter, the average for the coldest month, medium-dodes, and summer, after the calculated (maximum) heat load has been determined. The formula is used to recalculate the thermal loads in other modes.

Calculation of thermal loads at maximum winter mode

Calculated (maximum) energy use to heat the structure, W,

The amount of work required to heat a five-story, five-section home, W

Where – the particular heating properties of the building’s end section, W/(M 3 0 C);

– the unique heating property of the building’s private area, W/(M 3 0 C);

VT, or the end section’s volume,

VP stands for volume of the regular section.

The amount of energy required to heat a nine-story, six-section home, W

Where – the particular heating properties of the building’s end section, W/(M 3 0 C);

– the unique heating property of the building’s private area, W/(M 3 0 C);

VT, or the end section’s volume,

VP stands for volume of the regular section.

The burden on the school’s heating system, W

Where – the school’s unique heating characteristics, WT/(m 3 0 C);

The kindergarten’s heating load on Tuesday

Where – the kindergarten’s unique heating characteristics, W/(m 3 0 C);

Total kW of heating load for public and residential buildings

12376.835 kW is W.

The formula WT is used to calculate the amount of heat required for ventilation in public buildings.

Where W/(M 3 • 0 C), or the heat consumption of 1 m 3 of the ventilated volume of the building in an external measure with a difference in air temperatures inside the ventilated room and outside air in 1 about C, is the specific heat consumption for ventilation (specific ventilation characteristics of buildings);

V, measured in m^3, is the ventilated building’s outer volume;

The average internal air temperature is known as TVP.

TNV = -25 ˔s is the estimated outdoor temperature for ventilation systems.

We calculate the school’s ventilation flow rate, vt

Where the school’s unique heat consumption for ventilation (the building’s unique ventilation characteristics), WT/(M 3 • 0 C)

V, measured in m^3, is the ventilated building’s outer volume;

We calculate the kindergarten’s ventilation flow rate, vt.

Where the school’s unique heat consumption for ventilation (the building’s unique ventilation characteristics), WT/(M 3 • 0 C)

V, measured in m^3, is the ventilated building’s outer volume;

The ventilation system’s overall heat consumption, WT

Loads for hot water supplies

The formula, vt, is used to calculate the social-legged heat flow, Tue, for the hot water supply in residential and public buildings.

Where C = 4.187 is the water’s heat capacity;

T is the quantity (people) in units of measurement;

A: the daily rate of hot water consumption, expressed as kg (l) per unit measurement at a temperature of tg = 55 ° C;

In residential structures, a = 105 l/day per individual

A = 8 l/day per person for school,

Plantata for children = 30 l/day per individual,

Tx is the temperature of cold tap water; it is acceptable at 5 O S during the heating season and 15 O C during the summer;

Coefficient in subscriber systems that accounts for hot water cooling is 1.2.

The formula determines the social-legged heat stream, Tue, for the hot water supply of residential buildings.

Tuesday, conditional heat stream for the school’s and kindergarten’s hot water supply

The overall demand for hot water, WT

Recalculation of thermal loads to other modes

Recalculating thermal loads in other modes: the formula is used to determine the average for the coldest month, the medium-heat month, and the summer.

Using this dependence, we calculate the average heating load for residential buildings during a cold month, vt.

Where, according to Appendix 1 [2]; tx.m is the average temperature of the coldest month;

We calculate WT, the typical heating load for public buildings during a cold month.

The entire heating load for public and residential buildings in a cold month, vt

For residential buildings, we calculate the thermal load for the medium-feeding period.

What location is it?P is the heating period’s average temperature (Appendix 1 [2]);

For public buildings, we calculate the thermal load during the medium-feeding phase.

The entire heating load for public and residential buildings during the medium-feeding phase, TT

We calculate the average ventilation load (vt) for public buildings during a cold month.

The entire average ventilation load for public buildings during a cold month, WT

We calculate the average ventilation load for the school and kindergarten during the middle feeding period, vt.

The entire average ventilation load for public buildings during the medium-feeding period, WT

Loads for hot water supplies

The heat flux required to prepare hot water will drop in the summer and can be found using the formula.

Where KS is a coefficient that accounts for the drop in summer water flow relative to winter flow. When data are lacking, KS = 0.8 is accepted;

We calculate the typical load on the hot water system for residential buildings during the summer.

We calculate the typical load on the hot water system for public buildings in the summer, WT

We calculate the kW load on the DHW during the summer.

Table 3 will be updated with the computation results.

Table 3: Thermal load summary table

Maintaining a cozy and energy-efficient home requires an understanding of your house’s thermal load and heat loss. Homeowners can make educated decisions about their insulation and heating systems and ultimately save money and lessen their impact on the environment by accurately calculating these factors.

Your home’s thermal load is influenced by a number of factors, including the building materials, insulation, occupancy patterns, and climate. You can determine how much heat energy is needed to keep a room at a comfortable temperature even in the winter by evaluating these factors.

Finding the places in your home where energy is escaping is made easier by calculating heat loss. Doors, windows, roofs, and poorly insulated walls are common places for heat loss. Homeowners can ensure optimal heating system performance and avoid energy waste by addressing these weak points.

The accuracy of these computations is now easier than ever thanks to modern technology. Homeowners can evaluate their thermal load and heat loss efficiently and rapidly with the help of online calculators and software. These tools provide insightful information about enhancing energy efficiency while accounting for a variety of factors.

There are several long-term advantages to putting strategies in place to lessen heat loss and thermal load. Reduce utility costs, improve indoor comfort, and support a more sustainable environment by making investments in high-quality insulation, caulking air leaks, and switching to energy-efficient heating systems.

In conclusion, the first steps to creating a more comfortable and energy-efficient living space are figuring out how much heat your home loses and how much heat it receives. Homeowners can take charge of their heating and insulation systems, saving money and lowering their carbon footprint in the process, by using the tools that are readily available and putting good strategies into practice.

Video on the topic

Calculation of heat loss at home. Video course

Fast calculation of heat loss at home

Calculation of heat loss at home

How to choose the power of the boiler and radiators through the calculation of heat loss of the house through fences

Lecture 5. Calculation of heat loss of the premises

Calculation of heating of a private house Part 3 Calculation of heat loss and capacity of the boiler in the Valtec program