How to make a calculation of a warm floor using the example of a water system
The effectiveness of a warm floor is influenced by many factors. Without taking them into account, even if the system is correctly mounted and the most modern materials are used for its installation, real heat efficiency will not live up to expectations.
For this reason, installation work must be preceded by a competent calculation of the warm floor, and only then can a good result be guaranteed.
Designing a heating system is not cheap, so many home craftsmen make their own calculations. Agree, the idea of reducing the cost of arranging a warm floor seems very tempting.
We will tell you how to create a project, what criteria to consider when choosing the parameters of the heating system and write down a step-by-step calculation procedure. For clarity, we have prepared an example of calculating a warm floor.
The content of the article:
- The initial data for the calculation
- Determination of parameters of a warm floor
- Calculation of the required number of pipes
- We calculate the circulation pump
- Tips for choosing screed thickness
- Conclusions and useful video on the topic
The initial data for the calculation
Initially, a properly planned course of design and installation work will relieve surprises and unpleasant problems in the future.
When calculating a warm floor, it is necessary to proceed from the following data:
- wall material and design features;
- the size of the room in the plan;
- type of finish;
- design of doors, windows and their placement;
- layout of structural elements in plan.
To carry out competent design, it is necessary to take into account the established temperature regime and the possibility of its adjustment.
There are recommendations about the temperature at the floor, providing a comfortable stay in rooms for various purposes:
- 29 ° C - living sector;
- 33 ° C- bath, rooms with a pool and others with a high humidity indicator;
- 35 ° C - cold zones (at entrance doors, external walls, etc.).
Exceeding these values entails overheating of both the system itself and the finish coating with subsequent inevitable damage to the material.
After preliminary calculations, you can choose the heat carrier temperature that is optimal for personal sensations, determine the load on the heating circuit and purchase pumping equipment that perfectly copes with stimulating the coolant movement. It is selected with a margin of coolant flow rate of 20%.
At the design stage, it should be decided whether the underfloor heating will be the main heat supplier or will be used only as an addition to the radiator heating branch. The share of thermal energy losses that he has to compensate depends on this. It can range from 30% to 60% with variations.
The time for heating the water floor depends on the thickness of the elements included in the screed. Water as a coolant is very effective, but the system itself is difficult to install.
Determination of parameters of a warm floor
The purpose of the calculation is to obtain the magnitude of the heat load. The result of this calculation affects the next steps taken. In turn, the average winter temperature in a specific region, the estimated temperature inside the rooms, and the heat transfer coefficient of the ceiling, walls, windows, and doors affect the heat load.
The final result of calculations before floor heating device water type will depend on the availability of additional heating appliances, including heat dissipation of people living in the house and pets. Be sure to take into account the calculation of the presence of infiltration.
One of the important parameters is the configuration of the rooms, so you need a floor plan of the house and the corresponding sections.
Method for calculating heat loss
By defining this parameter, you will find out how much heat the floor must generate for the well-being of people in the room, you can pick up the boiler, pump and floor according to power. In other words: the heat given off by the heating circuits should compensate for the heat loss of the building.
The relationship between these two parameters is expressed by the formula:
Mp = 1.2 x Qwhere
- Mp - required loop power;
- Q - heat loss.
To determine the second indicator, measurements and calculations of the area of windows, doors, floors, external walls are made out. Since the floor will be heated, the area of this enclosing structure is not taken into account. Measurements are made on the outside with the capture of the corners of the building.
The calculation will take into account both the thickness and the coefficient of thermal conductivity of each of the structures. Normative values thermal conductivity coefficient (λ) for the most commonly used materials can be taken from the table.
The calculation of heat loss is performed separately for each element of the building, using the formula:
Q = 1 / R * (tv-tn) * S x (1 + ∑b)where
- R - thermal resistance of the material of which the enclosing structure is made;
- S - area of the structural element;
- tv and tn - the temperature is internal and external, respectively, while the second indicator is taken at the lowest value;
- b - additional heat loss associated with the orientation of the building relative to the cardinal points.
The thermal resistance index (R) is found by dividing the thickness of the structure by the coefficient of thermal conductivity of the material from which it is made.
The value of the coefficient b depends on the orientation of the house:
- 0,1 - north, northwest or northeast;
- 0,05 - west, southeast;
- 0 - south, southwest.
If you consider the issue on any example of calculating a water floor heating, it becomes more understandable.
Concrete calculation example
Let's say the walls of the house for non-permanent residence, 20 cm thick, are made of aerated concrete blocks. The total area of the enclosing walls minus the window and door openings is 60 m². Outside temperature is -25 ° С, internal + 20 ° С, the construction is oriented to the southeast.
Given that the thermal conductivity of the blocks is λ = 0.3 W / (m ° * C), we can calculate the heat loss through the walls: R = 0.2 / 0.3 = 0.67 m² ° C / W.
Heat loss is also observed through the stucco layer. If its thickness is 20 mm, then Rpcs. = 0.02 / 0.3 = 0.07 m² ° C / W. The sum of these two indicators will give the value of heat loss through the walls: 0.67 + 0.07 = 0.74 m² ° C / W.
Having all the initial data, substitute them into the formula and get the heat loss of the room with such walls: Q = 1 / 0.74 * (20 - (-25)) * 60 * (1 + 0.05) = 3831.08 W.
In the same way, heat losses are calculated through the remaining enclosing structures: windows, doorways, and roofing.
To determine the heat loss through the ceiling, take its thermal resistance equal to the value for the planned or existing type of insulation: R = 0.18 / 0.041 = 4.39 m² ° C / W.
The ceiling area is identical to the floor area and is 70 m². Substituting these values in the formula, heat loss is obtained through the upper enclosing structure: Q sweat. = 1 / 4.39 * (20 - (-25)) * 70 * (1 + 0.05) = 753.42 W.
To determine the heat loss through the surface of the windows, you need to calculate their area. If there are 4 windows with a width of 1.5 m and a height of 1.4 m, their total area will be: 4 * 1.5 * 1.4 = 8.4 m².
If the manufacturer indicates separately the thermal resistance for the double-glazed window and profile - 0.5 and 0.56 m² ° C / W, respectively, then Rokon = 0.5 * 90 + 0.56 * 10) / 100 = 0.56 m² ° C / Tue Here 90 and 10 are the shares attributable to each window element.
Based on the data obtained, further calculations continue: Q window = 1 / 0.56 * (20 - (-25)) * 8.4 * (1 + 0.05) = 708.75 watts.
The outer door has an area of 0.95 * 2.04 = 1.938 m². Then Rdv. = 0.06 / 0.14 = 0.43 m² ° C / W. Q dv. = 1 / 0.43 * (20 - (-25)) * 1.938 * (1 + 0.05) = 212.95 W.
As a result, heat losses will be: Q = 3831.08 +753.42 + 708.75 + 212.95 + 7406.25 = W.
An additional 10% for air infiltration is added to this result, then Q = 7406.25 + 740.6 = 8146.85 watts.
Now you can determine the thermal power of the floor: Mp = 1, * 8146.85 = 9776.22 W or 9.8 kW.
The necessary heat for heating the air
If the house equipped with ventilation system, then some part of the heat generated by the source should be spent on heating the air coming from outside.
For calculation, apply the formula:
Qc. = c * m * (tv – tn)where
- c = 0.28 kg⁰С and denotes the heat capacity of the air mass;
- m The symbol indicates the mass flow rate of outdoor air in kg.
The last parameter is obtained by multiplying the total volume of air equal to the volume of all rooms, provided that the air is updated every hour by a density that varies depending on temperature.
If the building enters 400 m3/ h, then m = 400 * 1.422 = 568.8 kg / h. Qc. = 0.28 * 568.8 * 45 = 7166.88 watts.
In this case, the required thermal power of the floor will increase significantly.
Calculation of the required number of pipes
For the device of the floor with water heating, different pipe laying methodscharacterized by their shape: a snake of three species - the actual snake, angular, double and snail. In one mounted circuit, a combination of different shapes can be found. Sometimes a snail is chosen for the central floor zone and one of the snake species is selected for the edges.
The distance between the pipes is called the step. When choosing this parameter, two requirements must be met: the foot of the foot should not feel the temperature difference in individual zones of the floor, and the pipes should be used as efficiently as possible.
For boundary areas of the floor, a pitch of 100 mm is recommended. In other areas, you can make a choice of pitch in the range from 150 to 300 mm.
To calculate the length of the pipe, there is a simple formula:
L = S / N * 1.1where
- S - area of the contour;
- N - laying step;
- 1,1 - margin for bending 10%.
To the final value add a piece of pipe laid from the collector to the wiring of the warm circuit both on the return and on the flow.
- area - 10 m²;
- collector distance - 6 m;
- laying pitch - 0.15 m.
The solution to the problem is simple: 10 / 0.15 * 1.1 + (6 * 2) = 85.3 m.
Using metal-plastic pipes up to 100 m in length, most often choose a diameter of 16 or 20 mm. With a pipe length of 120-125 m, its cross-section should be 20 mm².
The single-circuit design is only suitable for rooms with a small area. The floor in large rooms is divided into several circuits in a ratio of 1: 2 - the length of the structure should exceed the width by 2 times.
The previously calculated value is the length pipes for the floor generally. However, to complete the picture, you need to highlight the length of a separate contour.
This parameter is affected by the hydraulic resistance of the circuit, determined by the diameter of the selected pipes and the volume of water supplied per unit time. If these factors are neglected, the pressure loss will be so large that no pump will cause the coolant to circulate.
Contours of the same length - this is an ideal case, but rarely encountered in practice, because the area of premises for different purposes is very different and it is simply not practical to bring the length of the contours to one value. Professionals allow a difference in pipe length from 30 to 40%.
The value of the diameter of the collector and the throughput of the mixing unit determines the permissible number of loops connected to it. In the passport to the mixing unit, you can always find the value of the heat load for which it is designed.
Assume bandwidth ratio (Kvs) is 2.23 m3/ h With this coefficient, certain pump models can withstand a load of 10 to 15 watts.
To determine the number of circuits, you need to calculate the thermal load of each. If the area occupied by the heated floor is 10 m², and the heat transfer is 1 m², then the indicator Kvs is 80 watts, then 10 * 80 = 800 watts. This means that the mixing unit will be able to provide 15,000/800 = 18.8 rooms or circuits with an area of 10 m².
These indicators are maximum, and they can be applied only theoretically, but in reality the figure needs to be reduced by at least 2, then 18 - 2 = 16 contours.
Need for selection mixing unit (collector) see if he has so many conclusions.
Checking the correctness of the selection of the diameter of the pipes
To check whether the pipe section was correctly selected, you can use the formula:
υ = 4 * Q * 10ᶾ / n * d²
When the speed corresponds to the found value, the pipe section is selected correctly. Regulatory documents allow a maximum speed of 3 m / s. with a diameter of up to 0.25 m, but the optimum value is 0.8 m / s., since with an increase in its value, the noise effect in the pipeline increases.
Additional information on the calculation of underfloor heating pipes is given in this article.
We calculate the circulation pump
For the system to be economical, you need pick up the circulation pumpproviding the necessary pressure and optimal flow rate in the circuits. In the passports of the pumps usually indicate the pressure in the circuit of the longest length and the total flow of coolant in all loops.
The pressure is influenced by hydraulic losses:
∆ h = L * Q² / k1where
- L - the length of the contour;
- Q - water consumption l / s;
- k1 - coefficient characterizing losses in the system, the indicator can be taken from the hydraulic reference tables or from the equipment passport.
Knowing the pressure, calculate the flow in the system:
Q = k * √Hwhere
k Is the flow coefficient. Professionals accept the consumption for every 10 m² of the house in the range of 0.3-0.4 l / s.
The figures relating to the pressure and flow rates indicated in the passport cannot be taken literally - this is the maximum, but in fact they are influenced by the length and geometry of the network. If the pressure is too high, reduce the length of the circuit or increase the diameter of the pipes.
Tips for choosing screed thickness
In the directories you can find information that the minimum thickness of the screed is 30 mm. When the room is quite high, a heater is placed under the screed, which increases the efficiency of using the heat given off by the heating circuit.
The most popular substrate material is extruded polystyrene foam. Its heat transfer resistance is significantly lower than that of concrete.
When installing screeds, in order to balance the linear expansion of concrete, the perimeter of the room is formed with a damper tape. It is important to choose its thickness correctly. Experts advise that with a room area not exceeding 100 m², arrange a 5 mm compensating layer.
If the area is larger due to the length exceeding 10 m, the thickness is calculated by the formula:
b = 0.55 * Lwhere
L - this is the length of the room in m.
Conclusions and useful video on the topic
About the calculation and installation of a warm hydraulic floor, this video footage:
The video provides practical recommendations for laying the floor. Information will help to avoid mistakes that lovers usually make:
Calculation makes it possible to design a "warm floor" system with optimal performance. It is permissible to install heating using passport data and recommendations.
It will work, but professionals advise all the same to spend time on the calculation, so that in the end the system consumes less energy.
Do you have experience in calculating a warm floor and preparing a heating circuit project? Or have questions about the topic? Please share your opinion and leave comments.