Thermal calculation of a heating system: how to correctly calculate the load on a system

Alexey Dedyulin
Checked by a specialist: Alexey Dedyulin
Author: Kirill Egorov
Last update: August 2019

Design and thermal calculation of the heating system is an obligatory stage in arranging home heating. The main task of the computing activities is to determine the optimal parameters of the boiler and the radiator system.

Agree, at first glance it might seem that only an engineer can carry out a heat engineering calculation. However, not everything is so complicated. Knowing the algorithm of actions, it will turn out to independently perform the necessary calculations.

The article sets out in detail the calculation procedure and provides all the necessary formulas. For a better understanding, we have prepared an example of thermal calculation for a private house.

Thermal calculation of heating: general order

The classical thermal calculation of the heating system is a consolidated technical document, which includes the mandatory phased standard calculation methods.

But before studying these calculations of the main parameters, you need to decide on the concept of the heating system itself.

The heating system is characterized by forced supply and involuntary heat removal in the room.

The main tasks of calculation and design of the heating system:

  • most reliably determine heat loss;
  • determine the amount and conditions of use of the coolant;
  • select the elements of generation, displacement and heat transfer as precisely as possible.

During construction heating systems It is necessary to initially collect a variety of data on the room / building where the heating system will be used. After performing the calculation of the thermal parameters of the system, analyze the results of arithmetic operations.

Based on the data obtained, the components of the heating system are selected with the subsequent purchase, installation and commissioning.

Classic type of heating
Heating is a multicomponent system for ensuring the approved temperature regime in a room / building. It is a separate part of the communications complex of a modern housing

It is noteworthy that the specified method of thermal calculation allows you to accurately calculate a large number of quantities that specifically describe the future heating system.

As a result of thermal calculation, the following information will be available:

  • number of heat losses, boiler power;
  • the number and type of heat radiators for each room separately;
  • hydraulic characteristics of the pipeline;
  • volume, coolant speed, heat pump power.

Thermal calculation is not a theoretical outline, but quite accurate and reasonable results, which are recommended to be used in practice when selecting components of a heating system.

Room temperature standards

Before making any calculations of the system parameters, it is necessary, at a minimum, to know the order of the expected results, as well as to have standardized characteristics of some tabular quantities that need to be substituted into the formulas or oriented on them.

By performing parameter calculations with such constants, you can be sure of the reliability of the desired dynamic or constant system parameter.

Room temperature
For premises of various purposes, there are reference standards for temperature conditions of residential and non-residential premises. These standards are enshrined in the so-called GOST

For a heating system, one of these global parameters is the room temperature, which should be constant regardless of the season or environmental conditions.

According to the regulations of sanitary standards and regulations, there are differences in temperature relative to the summer and winter periods of the year. The air conditioning system is responsible for the temperature regime of the room in the summer season, the principle of its calculation is described in detail in this article.

But the room temperature in the winter is provided by the heating system. Therefore, we are interested in temperature ranges and their tolerances for deviations for the winter season.

Most regulatory documents specify the following temperature ranges, which allow a person to be comfortable in a room.

For non-residential office premises up to 100 m2:

  • 22-24 ° C - optimal air temperature;
  • 1 ° C - allowable fluctuation.

For office type premises with an area of ​​more than 100 m2 the temperature is 21-23 ° C. For non-residential premises of industrial type, the temperature ranges vary greatly depending on the purpose of the room and the established labor protection standards.

Comfortable temperature
Comfortable room temperature for each person is “own."Someone likes to be very warm in the room, someone comfortable when the room is cool - it's all quite individual

As for residential premises: apartments, private houses, estates, etc., there are certain temperature ranges that can be adjusted depending on the wishes of the residents.

And yet for specific rooms of an apartment and a house we have:

  • 20-22 ° C - residential, including children’s room, tolerance ± 2 ° С -
  • 19-21 ° C - kitchen, toilet, tolerance ± 2 ° C;
  • 24-26 ° C - bath, shower, pool, tolerance ± 1 ° C;
  • 16-18 ° C - corridors, hallways, stairwells, pantries, tolerance + 3 ° C

It is important to note that there are several more basic parameters that affect the temperature in the room and which you need to focus on when calculating the heating system: humidity (40-60%), the concentration of oxygen and carbon dioxide in the air (250: 1), air velocity masses (0.13-0.25 m / s), etc.

Calculation of heat loss in the house

According to the second law of thermodynamics (school physics), there is no spontaneous transfer of energy from less heated to more heated mini or macro objects. A special case of this law is the “desire” to create temperature equilibrium between two thermodynamic systems.

For example, the first system is an environment with a temperature of -20 ° C, the second system is a building with an internal temperature of + 20 ° C. According to the above law, these two systems will seek to balance through the exchange of energy. This will occur through heat loss from the second system and cooling in the first.

Temperature map
We can definitely say that the ambient temperature depends on the latitude at which the private house is located. And the temperature difference affects the amount of heat leakage from the building (+)

By heat loss is meant the involuntary release of heat (energy) from a certain object (house, apartment). For an ordinary apartment, this process is not so “noticeable” in comparison with a private house, since the apartment is located inside the building and is “adjacent” to other apartments.

In a private house through the external walls, floor, roof, windows and doors, to one degree or another, the heat “leaves”.

Knowing the amount of heat loss for the most adverse weather conditions and the characteristics of these conditions, it is possible to calculate the power of the heating system with high accuracy.

So, the volume of heat leakage from the building is calculated by the following formula:

Q = Qfloor+ Qwall+ Qwindow+ Qroof+ QDoor+ ... + Qiwhere

Qi - the amount of heat loss from the uniform appearance of the shell of the building.

Each component of the formula is calculated by the formula:

Q = S * ΔT / Rwhere

  • Q - heat leakage, V;
  • S - area of ​​a particular type of structure, sq. m;
  • ∆T - the difference in ambient and indoor air temperatures, ° C;
  • R - thermal resistance of a certain type of structure, m2* ° C / W.

The value of thermal resistance for real materials is recommended to be taken from auxiliary tables.

In addition, thermal resistance can be obtained using the following ratio:

R = d / kwhere

  • R - thermal resistance, (m2* K) / W;
  • k - thermal conductivity of the material, W / (m2*TO);
  • d - the thickness of this material, m

In old houses with a damp roofing structure, heat leakage occurs through the upper part of the building, namely through the roof and attic. Carrying out events for ceiling insulation or attic roof insulation solve this problem.

House through a thermal imager
If you insulate the attic space and the roof, then the total heat loss from the house can be significantly reduced

In the house there are several more types of heat loss through cracks in the structures, ventilation system, cooker hood, opening windows and doors. But to take into account their volume does not make sense, since they make up no more than 5% of the total number of major heat leaks.

Determination of boiler power

To maintain the temperature difference between the environment and the temperature inside the house, an independent heating system is needed that maintains the desired temperature in each room of a private house.

The basis of the heating system are different types of boilers: liquid or solid fuel, electric or gas.

The boiler is the central unit of the heating system that generates heat. The main characteristic of the boiler is its power, namely, the conversion rate is the amount of heat per unit time.

After calculating the heat load for heating, we obtain the required rated boiler power.

For an ordinary multi-room apartment, the boiler power is calculated through the area and specific power:

Rboiler= (Spremises*Rspecific)/10where

  • Spremises - total area of ​​the heated room;
  • Rprecocious - specific power relative to climatic conditions.

But this formula does not take into account the heat loss, which is enough in a private house.

There is another relationship that takes this parameter into account:

Rboiler= (Qlosses* S) / 100where

  • Rboiler - boiler power;
  • Qlosses - heat loss;
  • S - heated area.

The rated capacity of the boiler must be increased. A reserve is necessary if it is planned to use a boiler for heating water for the bathroom and kitchen.

Tank boiler
In most heating systems of private houses, it is recommended that you use an expansion tank in which the supply of coolant will be stored. Every private home needs hot water

In order to provide for the boiler power reserve in the last formula, it is necessary to add the safety factor K:

Rboiler= (Qlosses* S * K) / 100where

TO - it will be equal to 1.25, that is, the design capacity of the boiler will be increased by 25%.

Thus, the boiler capacity provides the opportunity to maintain the standard air temperature in the rooms of the building, as well as to have an initial and additional volume of hot water in the house.

Features of the selection of radiators

Standard components for providing heat in a room are radiators, panels, underfloor heating systems, convectors, etc. The most common parts of a heating system are radiators.

A heat radiator is a special hollow design of a modular type made of alloy with high heat dissipation. It is made of steel, aluminum, cast iron, ceramics and other alloys. The principle of operation of the heating radiator is reduced to the emission of energy from the coolant into the space of the room through the “petals”.

Multi-section heating radiator
Aluminum and bimetal heating radiator replaced massive cast-iron batteries. Ease of production, high heat dissipation, successful construction and design made this product a popular and widespread tool for radiating heat in a room.

There are several techniques calculation of heating radiators in the room. The following list of methods is sorted in order of increasing accuracy.

Calculation options:

  1. By area. N = (S * 100) / C, where N is the number of sections, S is the area of ​​the room (m2), C - heat transfer from one section of the radiator (W, taken from the passport or product certificate), 100 W - the amount of heat flow that is required 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. N - room height, 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 similar. k1 - taking into account the number of cameras in a double-glazed window of a room’s window, k2 - thermal insulation of walls, k3 - ratio of the area of ​​windows to the area of ​​the room, k4 - average minus temperature in the coldest week of winter, k5 - number of external walls of the room (which “go out” onto the street), k6 - type of room on top, k7 - ceiling height.

This is the most accurate option for calculating the number of sections. Naturally, the fractional results of calculations are always rounded to the next integer.

Hydraulic calculation of water supply

Of course, the “picture” of calculating heat for heating cannot be complete without calculating such characteristics as the volume and speed of the coolant.In most cases, the coolant is ordinary water in a liquid or gaseous state of aggregation.

Piping system
The actual volume of the coolant is recommended to be calculated by summing all the cavities in the heating system. When using a single-circuit boiler, this is the best option. When using double-circuit boilers in the heating system, it is necessary to take into account the costs of hot water for hygienic and other domestic purposes

Calculation of the volume of water heated by a double-circuit boiler to provide residents with hot water and heating the coolant is made by summing the internal volume of the heating circuit and the actual needs of users in heated water.

The volume of hot water in the heating system is calculated by the formula:

W = k * Pwhere

  • W - volume of heat carrier;
  • P - power of the heating boiler;
  • k - power factor (the number of liters per unit of power is 13.5, the range is 10-15 liters).

As a result, the final formula looks like this:

W = 13.5 * P

The coolant speed is the final dynamic assessment of the heating system, which characterizes the rate of fluid circulation in the system.

This value helps to evaluate the type and diameter of the pipeline:

V = (0.86 * P * μ) / ΔTwhere

  • P - boiler power;
  • μ - boiler efficiency;
  • ∆T - temperature difference between the supplied water and the return water.

Using the foregoing methods hydraulic calculation, it will be possible to obtain real parameters, which are the “foundation” of the future heating system.

Thermal Calculation Example

As an example of heat calculation, there is an ordinary 1-storey house with four living rooms, a kitchen, a bathroom, a “winter garden” and utility rooms.

Facade of a private house
The foundation is made of monolithic reinforced concrete slab (20 cm), the external walls are concrete (25 cm) with plaster, the roof is covered with wooden beams, the roof is metal tile and mineral wool (10 cm)

Denote the initial parameters of the house, necessary for the calculations.

Dimensions of the building:

  • floor height - 3 m;
  • a small window of the front and rear of the building 1470 * 1420 mm;
  • large window of the facade 2080 * 1420 mm;
  • entrance doors 2000 * 900 mm;
  • rear doors (exit to the terrace) 2000 * 1400 (700 + 700) mm.

Total width of the building 9.5 m2, length 16 m2. Only living rooms (4 pcs.), A bathroom and a kitchen will be heated.

House plan
To accurately calculate the heat loss on the walls from the area of ​​the external walls, you need to subtract the area of ​​all windows and doors - this is a completely different type of material with its thermal resistance

We start by calculating the areas of homogeneous materials:

  • floor area - 152 m2;
  • roof area - 180 m2 , considering the height of the attic 1.3 m and the width of the run - 4 m;
  • window area - 3 * 1.47 * 1.42 + 2.08 * 1.42 = 9.22 m2;
  • door area - 2 * 0.9 + 2 * 2 * 1.4 = 7.4 m2.

The area of ​​the outer walls will be 51 * 3-9.22-7.4 = 136.38 m2.

We proceed to the calculation of heat loss on each material:

  • Qfloor= S * ΔT * k / d = 152 * 20 * 0.2 / 1.7 = 357.65 W;
  • Qroof= 180 * 40 * 0.1 / 0.05 = 14400 W;
  • Qwindow= 9.22 * 40 * 0.36 / 0.5 = 265.54 W;
  • Qthe door= 7.4 * 40 * 0.15 / 0.75 = 59.2 W;

As well as Qwall equivalent to 136.38 * 40 * 0.25 / 0.3 = 4546. The sum of all heat losses will be 19628.4 watts.

As a result, we calculate the power of the boiler: Pboiler= Qlosses* Sheating_room* K / 100 = 19628.4 * (10.4 + 10.4 + 13.5 + 27.9 + 14.1 + 7.4) * 1.25 / 100 = 19628.4 * 83.7 * 1.25 / 100 = 20536.2 = 21 kW.

We will calculate the number of sections of radiators for one of the rooms. For all others, the calculations are similar. For example, the corner room (left, lower corner of the diagram) is 10.4 m2.

Therefore, N = (100 * k1 * k2 * k3 * k4 * k5 * k6 * k7) / C = (100 * 10.4 * 1.0 * 1.0 * 0.9 * 1.3 * 1.2 * 1.0 * 1.05) /180=8.5176=9.

For this room, 9 sections of a heating radiator with a heat transfer of 180 watts are needed.

We turn to the calculation of the amount of coolant in the system - W = 13.5 * P = 13.5 * 21 = 283.5 liters. So, the coolant speed will be: V = (0.86 * P * μ) / ∆T = (0.86 * 21000 * 0.9) /20=812.7 l.

As a result, a full revolution of the total volume of coolant in the system will be equivalent to 2.87 times in one hour.

A selection of articles on thermal calculation will help determine the exact parameters of the elements of the heating system:

  1. Calculation of the heating system of a private house: rules and examples of calculation
  2. Thermotechnical calculation of a building: specifics and formulas for computing + practical examples

Conclusions and useful video on the topic

A simple calculation of the heating system for a private house is presented in the following review:

All the subtleties and generally accepted methods for calculating the heat loss of a building are shown below:

Another option for calculating heat leaks in a typical private house:

This video talks about the features of the circulation of an energy carrier for heating a home:

The thermal calculation of the heating system is individual in nature, it must be done correctly and accurately. The more accurate the calculations will be made, the less will have to overpay the owners of a country house during operation.

Do you have experience in performing thermal calculation of the heating system? Or have questions about the topic? Please share your opinion and leave comments. The feedback block is located below.

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Visitors Comments
  1. In my opinion, not every average person can do such calculations. Many people prefer to simply pay money to professionals working in this field and get a finished result. But as for the reduction of heat loss, here everyone needs to think personally and insulate their home. Now there is a fairly wide selection of materials for a variety of wallets.

    • Anatoly78

      I would argue with you. You know, at first it also seemed to me that there was no way to figure it out, a huge number of formulas and concepts that I did not know before. But I still decided to try. And you know, if you sit and go a little deeper into the analysis, there is nothing, in general, complicated. Eyes are afraid, as they say!
      I do not have any special education, but I believe that a man in the house should do everything with his own hands (if possible, of course)

  2. Alexei

    Thank you very much for the article, I’ll say: everything is very clear on the basis of school physics. I am an electronics engineer, I am engaged in automation of boiler houses and other systems, later I began to install heating and water supply systems, I want to study the whole principle of work and calculation myself, a very useful article. Thank.