Ever wonder how your house’s heating system keeps you toasty warm throughout the bitterly cold winter months? It is the outcome of meticulous planning and calculation rather than magic. Comprehending the thermal load on your home’s heating system is essential to guaranteeing maximum effectiveness and comfort. We’ll explore the fundamentals of this thermal load calculation in this post, illuminating the science of maintaining a comfortable temperature in your house.

To begin with, let’s define "thermal load" precisely. In a nutshell, it’s the quantity of heat energy required to keep your house at a specific temperature while accounting for elements like the outside temperature, the amount of insulation, and the type of building materials. A number of intricate but necessary steps must be taken in order to calculate the thermal load and make sure your heating system is capable of handling the load.

An important component affecting a building’s thermal load is its insulation. Insulation functions as a barrier, keeping out unwanted heat in the summer and preventing heat from escaping in the winter. The thermal load on your heating system will decrease with improved home insulation. The kind, thickness, and positioning of the insulation, among other factors, affect how well your house performs thermally overall.

The climate where your house is located is another important factor to take into account when calculating the thermal load. The amount of heat energy needed to maintain indoor comfort levels varies depending on the region and the degree of temperature fluctuations that occur there throughout the year. Heating engineers can precisely determine the thermal load on a building and create a suitable heating system to meet its requirements by considering local climate data.

When determining the thermal load, it’s also critical to take your home’s size and layout into account. Greater heating capacity is usually needed in larger homes in order to maintain consistent temperatures in various areas. Other variables that may affect the total thermal load include the quantity and size of windows, drafts, and air leaks. Heating professionals can identify the most effective heating solution for your particular home by carefully examining these factors.

Factor Considered | Calculation Method |

Area of the building | Measure length × width of each room, sum up |

Insulation quality | Assess thickness and type of insulation material |

Climate | Consider average temperature, wind speed, and humidity |

- Ways to determine the load
- For example, a project of one -story house 100 m²
- We consider the flow rate of heat in quadrature
- Calculation of thermal load by the volume of rooms
- Calculated algorithm according to SNiP
- Determine the heat loss of the walls and roof
- Floor division into zones
- Heating of ventilation air
- The final calculation
- Video on the topic
- Calculation of heating load based on the analysis of the architectural model of the building.
- Nevzorova a. B. Lecture 8 hydraulic calculation of the heating system

## Ways to determine the load

Give a brief definition of the term first. The total amount of heat used by the heating system to bring the space up to temperature during the coldest time of year is known as the thermal load. The value is found in the formulas for the Latin letter Q and is calculated using energy units such as kilowatts, kilocalories, and—less frequently—kilojoules.

Selecting a boiler, water system heaters, and batteries based on power is simple when one is aware of the demand on a private home’s heating system overall and the specific requirements of each room. How is this parameter calculated?

- If the height of the ceilings does not reach 3 m, an enlarged calculation is made by the area of heated rooms.
- With an overlap height of 3 m or more, heat consumption is considered the volume of rooms.
- Determination of heat loss through external fences and costs for heating ventilation air according to SNiP.

Note: Online calculators have become very popular recently and can be found on the pages of many different websites. They make it simple and quick to calculate the amount of thermal energy; further instructions are not needed. Minus: Since the programs are written by non-heating equipment experts, the accuracy of the results needs to be verified.

The first two computed approaches rely on the application of a particular thermal property in connection to the building’s volume or heated area. The algorithm is straightforward, widely used, but it produces results that are extremely similar and ignores the cottage’s level of insulation.

It is far more difficult to consider thermal energy consumption as process engineers do, according to SNiP. The final figures will, with a 95% accuracy rate, represent the true situation, even though we will need to gather a lot of reference data and perform calculations. We will endeavor to streamline the approach and render the load on heating computation as comprehensible as feasible.

In figuring out how much strain a heating system can handle in a building, we delve into a process called thermal load calculation. It"s like sizing up the appetite of a hungry beast before feeding it. We consider factors like the building"s size, insulation quality, outside temperature, and even the number of windows and doors. This helps us understand how much heat the building will lose and how much the heating system needs to compensate for that loss. By crunching these numbers, we ensure the heating system is just right—not too weak to leave us shivering in the cold nor too powerful to waste energy and money. So, it"s all about finding that sweet spot where comfort meets efficiency.

## 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.

## 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.

## Calculated algorithm according to SNiP

Of all the methods now in use, this one is the most accurate. You can choose heating equipment with confidence if you follow our instructions and complete the computation accurately. You can be 100% certain of the outcome. This is how the process appears:

- Measure the quadrature of the outer walls, floors and ceilings separately in each room. Determine the area of windows and entrance doors.
- Calculate thermal losses through all external fences.
- Find out the consumption of thermal energy going to heating ventilation (infiltration) air.
- Summarize the results and get a real heat load indicator.

A crucial aspect. Internal floors are not considered in a two-story cottage because they do not border the outside space.

The basic formula for calculating thermal losses is rather straightforward: given the differences in material composition of windows, walls, and floors, you must determine the amount of energy lost by each type of building structure. The area of glazed openings should be subtracted from the exterior wall quadrature since they allow for a greater heat transfer and are therefore taken into account separately.

As indicated in the diagram, take the width of the rooms, add half the thickness of the inner partition to it, and then measure the outer angle. The objective is to account for the entire surface area that the external fence loses heat over.

### Determine the heat loss of the walls and roof

The following formula can be used to determine the heat flow through a design of the same type, such as a wall:

- The value of heat loss through one fence we designated Qi, Tue;
- A – the square of the wall within the same room, m²;
- TV is a comfortable temperature inside the room, usually +22 ° C is accepted;
- TN – the minimum temperature of street air, which lasts for 5 coldest winter days (take real value for your area);
- R – resistance to the thickness of the external fence transmission of heat, m² ° C/WT.

There is only one ambiguous parameter left in the list above: R. The thickness of the fence and the type of wall structure determine its value. Follow these steps to determine the heat transfer resistance:

- Determine the thickness of the bearing part of the outer wall and separately – a layer of insulation. The letter designation in the formulas is Δ, is considered in meters.
- Find out from the reference tables the coefficients of thermal conductivity of structural materials λ, units of measurement – W/(MºС).
- Alternately substitute the found values into the formula:
- Determine R for each layer of the wall separately, add the results, then use in the first formula.

Proceed to the next room after performing the calculations individually for the windows, walls, and ceilings in the current room. Heat losses through the floors are taken into account separately, as explained below.

Suggestions. The regulatory documentation lists the accurate heat conductivity coefficients for each type of material. This is a set of regulations SP 50.13330.2012 for Russia and DBN in.2.6–31 ~ 2006 for Ukraine. Take note! Use the value of λ listed in column "B" for operating conditions in the computations.

An example of a calculation for our one-story home’s living room, which has ceilings that are three meters high:

- The area of the outer walls along with the windows: (5.04 + 4.04) x 3 = 27.24 m². STATURE OF WINDERS – 1.5 x 1.57 x 2 = 4.71 m². Pure Fence Square: 27.24 – 4.71 = 22.53 m².
- Thermal conductivity λ for laying silicate brick is 0.87 W/(MºС), foam 25 kg/m³ – 0.044 W/(MºС). Thickness – respectively 0.38 and 0.1 m, we count the resistance of heat transfer: r = 0.38 /0.87 + 0.10.044 = 2.71 m² ° C/WT.
- The outer temperature is minus 25 ° C, inside the living room – plus 22 ° C. The difference will be 25 + 22 = 47 ° C.
- Determine heat loss through the walls of the living room: Q = 1/2.71 x 47 x 22.53 = 391 W.

In a similar vein, consideration is given to the overlap and heat flow through the windows. The characteristics of the 22 cm thick reinforced concrete ceiling are found in the normative or reference literature, and the thermal resistance of translucent structures typically indicates the manufacturer:

- R insulated overlap = 0.22 /2.04 + 0.10.044 = 2.38 m² ° C / W, heat loss through the roof – 1/2.38 x 47 x 5.04 x 4.04 = 402 W.
- Losses through window openings: Q = 0.32 x 47 x71 = 70.8 T.

The living room’s overall heat loss (not including the floors) is 391 + 402 + 70.8 = 863.8 T. The results of comparable computations are compiled for the remaining rooms.

Please take note that heat only escapes through the roof and floors and does not come into contact with the exterior of the building along the corridor. Watch the video to see which fences should be considered in the calculated method.

### Floor division into zones

The building in the plan is divided into zones 2 m wide, as indicated by the diagram, to determine the amount of heat lost by floors on the ground. The outside of the building structure is where the first strip begins.

The following is the calculated algorithm:

- Draw a cottage plan, divide 2 m wide into strips. The maximum number of zones is 4.
- Calculate the area of the floor that falls separately into each zone, neglecting interior partitions. Please note: the quadrature in the corners is considered twice (shared in the drawing).
- Using the calculated formula (for convenience we give it again), determine the heat loss in all areas, summarize the numbers obtained.
- The resistance of heat transfer R for zone I is taken equal to 2.1 m² ° C/W, II – 4.3, III – 8.6, the rest of the floor – 14.2 m² ° C/WT.

Note: Starting from the ground level, the first strip is situated on the subterranean portion of the wall if we are discussing a heated basement.

The calculations for floors insulated with mineral wool or polystyrene foam are the same, only to fixed values. R is the insulation layer’s thermal resistance, which is calculated using the δ / λ formula.

An illustration of a computation In a country home’s living room:

- The quadrature of the zone I is equal (5.04 + 4.04) x 2 = 18.16 m², plot II – 3.04 x 2 = 6.08 m². The rest of the zones do not get into the living room.
- Energy consumption for the 1st zone will be 1/2.1 x 47 x 18.16 = 406.4 watts, on the second – 1/4.3 x 47 x 6.08 = 66.5 W.
- The value of the heat flow through the floors of the living room – 406.4 + 66.5 = 473 W.

Eliminating typical heat loss in the concerned room is now simple: 863.8 + 473 – 1.34 kW = 1336.8 W, rounded.

### Heating of ventilation air

Natural ventilation is installed in the great majority of private homes and flats. The parishes of windows, doors, and supply holes allow street air to enter. The heating system heats the incoming cold mass, using more energy in the process. How to determine how many of these losses there are:

- Since the calculation of infiltration is too complicated, regulatory documents allow the release of 3 m³ of air per hour for each meter square area of the dwelling. The total consumption of supply air l is simple: the quadrature of the room is multiplied by 3.
- L is the volume, and we need mass M air flow. Learn it by multiplying by gas density taken from the table.
- Air mass M is substituted in the formula of the school course of physics, which allows you to determine the amount of energy spent.

We determine how much heat is needed using a long, suffering living room as an example, measuring 15.75 m². The influx’s mass is 47.25 x 1.422 = 67.2 kg/h, and its volume is 15.75 x 3 = 47.25 m³/h. Taking into account that the air’s heat capacity (represented by the letter C) is equal to 0.28 W / (kg ºΡ), we can calculate the energy consumption as follows: qntent = 0.28 x 67.2 x 47 = 884 watts. The figure is pretty impressive, as you can see, so it is definitely necessary to consider the heating of the air masses.

The total of all previously acquired data determines the final computation of the building’s heat loss plus the heat consumption for ventilation. Specifically, the load on the living room’s heating system will equal 0.88 + 1.34 = 2.22 kW. In a similar manner, the energy costs associated with the cottage are computed and totaled into a single figure.

### The final calculation

It might be interesting to see the outcome throughout the same-story house if the sheer number of formulas hasn’t made your brain explode. In the earlier examples, we completed the majority of the work; the remaining task is to proceed through additional rooms and determine the building’s overall external heat loss. first data discovered:

- Thermal resistance of the walls – 2.71, windows – 0.32, ceilings – 2.38 m² ° C/W;
- ceiling height – 3 m;
- R for the front door insulated with extruded polystyrene foam is 0.65 m² ° C/W;
- Internal temperature – 22, external – minus 25 ° C.

We suggest creating a table in Exel to streamline the computations, and then bringing the intermediate and final results into that table.

Following completion of the computations and table filling, the premises arrived at the following thermal energy cost values:

- The living room is 2.22 kW;
- Kitchen – 2.536 kW;
- The hallway is 745 watts;
- Corridor – 586 W;
- bathroom – 676 watts;
- The bedroom is 2.22 kW;
- Children"s – 2.536 kW.

A private home with a 100 m² area had a final load on its heating system of 11.518 W, or 11.6 kW, rounded. Notably, there is a literal 5% difference in the result compared to close calculation methods.

Regulatory documents, however, state that the final amount needs to be multiplied by the coefficient of 1.1 unaccounted heat loss that results from the building’s orientation with respect to the cardinal points, wind loads, and other factors. As a result, the total output is 12.76 kW. The following information regarding the engineering method is provided in the video:

Comprehending the calculation of thermal load on a building’s heating system is essential to guaranteeing maximum comfort and energy economy. Engineers can calculate how much heat is needed to maintain desired indoor temperatures by evaluating variables such as building size, insulation levels, climate, and heat loss through walls, windows, and roofs.

Although there are many intricate formulas and factors to take into account when calculating thermal load, the fundamental idea is to balance heat gains and losses. Heat losses happen through conduction, convection, and radiation, whereas heat gains are produced by sunlight, appliances, and human occupants. Engineers can create heating systems that satisfy a building’s requirements without wasting energy by taking these factors into consideration.

Knowing the building envelope’s insulating qualities is essential for calculating thermal load. Reduced heat loss from well-insulated windows, roofs, and walls lessens the total thermal load on the heating system. By reducing the need for excessive heating, this not only increases comfort but also reduces energy costs.

The temperature has a big impact on how much heat a building’s heating system can handle. To keep a comfortable temperature inside, milder climates might require less heating, while colder climates might require more. To precisely size heating systems, engineers take into account regional climate data, such as average heating degree days and variations in outdoor temperature.

All things considered, precisely estimating the thermal load on a building’s heating system is necessary for creating effective and efficient heating solutions. Engineers can make sure that heating systems are appropriately sized to meet the demands of the occupants while minimizing energy consumption and costs by taking into account factors like building size, insulation, climate, and heat loss.

## Video on the topic

### Calculation of heating load based on the analysis of the architectural model of the building.

### Nevzorova a. B. Lecture 8 hydraulic calculation of the heating system

**What type of heating you would like to have in your home?**