# Calculation of underfloor heating: Calculation example for water system

The efficiency of underfloor heating is influenced by many factors. Without taking them into account even if the system is correctly installed, and its unit used the most modern materials, the actual thermal efficiency does not live up to expectations.

For this reason, installation work must be preceded by a competent calculation of a warm floor, and only then can we ensure a good result.

The development of the heating system project is not cheap, so many domestic craftsmen to carry out calculations on their own. Agree, the idea of ​​reducing the resettlement floor heating costs seems very enticing.

We'll show you how to create a project, what criteria to consider when choosing a heating system parameters and write out a step by step method of calculation. For clarity, we have prepared a sample calculation of floor heating.

The content of the article:

• Initial data for calculating
• Defining the parameters of floor heating
• Method of calculation of heat loss
• A concrete example of the calculation
• The necessary heat for heating the air
• Calculating the number of pipes
• We expect the circulating pump
• Guidelines for selecting the thickness of the screed
• Conclusions and useful videos on the topic

## Initial data for calculating

Initially well-planned course of design and installation work will eliminate the unpleasant surprises and problems in the future.

In the calculation of floor heating should be based on the following data:

• Material of walls and their construction;
• size of the room in the plan;
• form a final coating;
• construction of doors, windows and their placement;
• location in terms of structural elements.

To perform a competent design compulsory registration required set temperature and the possibility of its adjustment. To carry out a rough calculation it is assumed that 1 m2 heating system should compensate the heat losses in the 1 kW. If the heating water circuit is used as a supplement to the primary system, then he has to cover only a part of heat

There are recommendations about the temperature in the floor, ensuring a comfortable stay in the premises for various purposes:

• 29 ° C - living sector;
• 33 ° C- bath, room with pool and others with high humidity;
• 35 ° C - the cold zone (at the entry door, exterior walls, etc.).

Exceeding these values ​​entails overheating of both the system and followed by a final coating inevitable damage of the material.

After spending preliminary calculations, it is possible to select the optimum of personal feelings coolant temperature to determine the load on the the heating circuit and purchase pumps, impeccably coping with motion stimulation coolant. His pick with a reserve of heating water flow rate of 20%. Much time is spent on heating screeds capacity of more than 7 cm. Therefore, when constructing water systems tend not to exceed the specified limit. The most appropriate coating for water sexes considered outdoor ceramics, a parquet because of its ultra-low thermal conductivity of warm floors is not laid

At the design stage should decide whether the primary supplier floor heating or heat will be used only as a supplement to the heating radiator branch. On this depends the proportion of heat loss that he will have to reimburse. It could range from 30% to 60% with variations.

water floor heating time is depending on the thickness of elements within the coupler. Water as a coolant is very effective, but the system is complex to install.

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Photo of To perform calculations water floor heating systems primarily make calculations of heat loss, which should compensate for the contour. If this optional system, you take into account the heat losses Calculations are made only for that part of the floor, which will be located a heating coil. Those areas where the pipes are laid, for example, under the furniture, not used in the calculation For the production of calculations needed average temperature of coolant at the outlet of the collector device and the inlet into it of the return For accurate results, you need to know the thermal conductivity of the plan to install the pipes and the approximate length of the heating circuit Radiant floor in a wooden house Variant water circuit layouts The collector and the heating conduit systems the heating surface of copper tubes

## Defining the parameters of floor heating

The purpose of the calculation is to obtain the value of the thermal load. The result of this calculation affects the subsequent steps taken. In turn, the thermal load on the average value affects the winter temperature in a particular region, the estimated temperature inside the rooms, the heat transfer coefficient of the ceiling, walls, windows and doors. The cause of heat loss are poorly insulated walls, windows and doors of the house. The largest percentage of heat lost through ventilation and roof system (+)

The final result of calculations before floor heating device water type will depend on the availability of additional heat sources, including heat residing in the house of people and pets. Be sure to take into account in the calculation of the presence of infiltration.

One important parameter is the shape of the room, so you need a floor plan of the house and the relevant sections.

### Method of calculation of heat loss

Given this option, you will learn how much heat is to produce flooring for a comfortable well-being of people in the room, can pick up the boiler, pump and floor power. In other words, the heat given to heating circuits, to compensate the heat loss from buildings.

The relationship between these two parameters expressed by the formula

Mn = 1.2 x Qwhere

• Mp - required power circuits;
• Q - heat loss.

To determine the second indicator make out measurements and calculating the area of ​​windows, doors, ceilings, external walls. Since the floor will be heated, the area of ​​the enclosing structure is not taken into account. Measurements made on the outside of the building with the capture angle.

The calculation will take into account the thickness and thermal conductivity of each of the structures. Normative values thermal conductivity (Λ) for the most commonly used materials can be taken from the table. From the table, you can take the value of the coefficient for the calculation. It is important to check with the supplier's value of thermal resistance of the material when installed windows of metal (+)

Calculation of heat loss is performed separately for each building element, using the formula:

Q = 1 / R * (ti-ti) * S x (1 + Σb)where

• R - thermal resistance of the material of construction of the cladding;
• S - the area of ​​the component;
• ti and ti - the temperature inside and outside, respectively, said second component taken by the lowest values;
• b - additional heat loss associated with the orientation of the building relative to the cardinal.

Indicator thermal resistance (R) found by dividing the thickness of the construction material for the thermal conductivity, from which it is made.

The coefficient b is dependent on the orientation of house:

• 0,1 - north, north-west or north-east;
• 0,05 - west, south-east;
• 0 - south, south-west.

If you consider any example of calculation of the water floor heating, it becomes easier to understand.

### A concrete example of the calculation

Assume house wall for intermittent stay 20 cm thick, made of concrete blocks. The total area of ​​the boundary walls minus windows and doorways 60 m². The outside temperature -25 ° C, the internal +20 ° C, the structure is oriented in the south-east.

Given that the thermal conductivity of blocks λ = 0,3 W / (m ° * C) to calculate the heat loss through the walls: R = 0,2 / 0,3 = 0,67 m² ° C / W.

Observed and heat loss through the layer of plaster. If its thickness is 20 mm, the Rsht. = 0.02 / 0.3 = 0.07 m² ° C / W. The sum of these two figures gives the value of heat loss through the walls: 0.67 + 0.07 = 0.74 m² ° C / W.

With all of the source data, substitute them in the formula and is obtained with such heat losses of the room walls: Q = 1 / 0.74 * (20 - (-25)) * 60 * (1 + 0.05) = 3831.08 Watts.

Likewise calculated heat loss through the remaining walling: windows, doors, roof. Heat given off heating loops may be insufficient to heat the air inside the house to the desired values, if their power is low. When excess power will be a coolant overruns

To determine the heat loss through the ceiling take its thermal resistance equal to the value for the proposed or existing insulation type: R = 0,18 / 0,041 = 4,39 m ² ° C / W.

Area ceiling identical to the area of ​​the floor and is 70 m². Substituting these values ​​into the formula to give the heat loss through the upper guard structure: Q sweat. = 1 / 4.39 * (20 - (-25)) * 70 * (1 + 0.05) = 753.42 watts.

To determine the heat loss through the window surface, it is necessary to calculate their area. In the presence of 4-windows width of 1.5 m and a height of 1.4 m and their total area is: 4 * 1.5 * 1.4 = 8.4 meters.

If the manufacturer indicates separately for thermal resistance and glass Profile - 0.5 m² and 0.56 ° C / W, respectively, then Rokon = 0.5 * 90 + 0.56 * 10) / 100 = 0.56 m² ° C / Tues. Here, 90 and 10 - the share attributable to each display element.

Based on the data, still further calculations: Qokon = 1 / 0.56 * (20 - (-25)) * 8.4 * (1 + 0.05) = 708.75 watts.

The front door has an area of ​​0.95 * 2.04 = 1.938 m². Then Rdv. = 0.06 / 0.14 = 0.43 m² ° C / W. Q bits. = 1 / 0.43 * (20 - (-25)) * 1.938 * (1 + 0.05) = 212.95 watts. Since the exterior doors are opened often, through them, a large amount of heat is lost. Therefore, it is important to ensure that they are tight closure

As a result, heat loss amount: Q = 3831,08 +753,42 + 708,75 + 212,95 + 7406,25 = watts.

This result was further added 10% air infiltration, then Q = 7406,25 + 740,6 = 8146,85 watts.

We can now determine the power and heat sex: Mn = 1 * = 8146.85 9776.22 watts or 9.8 kW.

### The necessary heat for heating the air

If the house It is equipped with a ventilation system, Then some of the heat released by the source must be spent for heating the incoming outside air.

For the calculation formula is used:

QB. = C * m * (ti-ti)where

• c = 0.28 kg⁰S and designates the specific heat of air mass;
• m symbol denotes the mass flow of fresh air in kg.

Prepared latest parameter by multiplying the total volume of air equal to the volume of all Improvement provided that the air is updated every hour, at a density that varies depending on the temperature. The graph displayed air density dependence on its temperature. The data necessary for calculating the amount of heat required for heating the air mass flowing into the house as a result of the forced ventilation ()

If the building is supplied 400 m3/ H, then m = 1.422 * 400 = 568.8 kg / h. QB. = 0.28 * 568.8 * 45 = 7166.88 Watts.

In this case, the required thermal power will greatly increase the floor.

## Calculating the number of pipes

For floor heating device with a water choosing different methods of pipelayingCharacterized by their form: three species of snake - the snake itself, corner, a double and a snail. In an assembled circuit washes to meet a combination of different forms. Sometimes the central floor area, select "snail" and for the edges - odnin kind of "snake". "Snail" - a rational choice for the bulk of premises with a simple geometry. In areas severely stretched or having complex shape better to use "snake" (+)

The distance between the pipes is called a step. By choosing this option you need to satisfy two requirements: foot feet should not feel the temperature difference on the individual areas of sex, and use pipes need as efficiently as possible.

For border areas recommend the use of sex in step 100 mm. On other sections can be made within the pitch range from 150 to 300 mm. Important insulation floor. On the first floor, its thickness should reach at least 100 mm. For this purpose, mineral wool or polystyrene foam extrusion

a simple formula for calculating the length of the tube:

L = S / N * 1.1where

• S - contour area;
• N - a step of laying;
• 1,1 - reserve for bends 10%.

To the total value added length of pipe laid by the collector wirings to warm circuit as return pipe, and the feed.

Example of calculation.

Initial values:

• area - 10 m²;
• distance to the collector - 6 m;
• stacking step - 0.15 m.

Solution of the problem is simple: 10 / 0.15 * 1.1 + (6 * 2) = 85.3 m.

Using metal pipes of up to 100 m, often choose the diameter 16 or 20 mm. When the length of the pipe 120-125 m its cross section should be 20 mm².

Single circuit design is only suitable for areas with limited space. The floor in large rooms divided into several circuits in a ratio of 1: 2 - design length should exceed 2 times the width.

The calculated value of previously - is the length pipes for floor generally. However, to complete the picture you want to select an individual loop length.

This parameter affects the hydraulic circuit resistance determined selected diameter pipe and the water volume supplied per unit time. If these factors is neglected, the pressure loss will be so large that no pump does not cause coolant to circulate. Determining the flow pipe, depending on the selected stacking step

The contours of the same length - is a perfect case, but in practice are rare, because the area of ​​the premises very different purpose and different length of lead circuits to the same value just inexpedient. Professionals allow the difference in length of the pipes 30 to 40%.

The diameter and the collector node mixing throughput capacity determined allowable number of loops connected thereto. In the passport in the mixing unit always possible to find the value of the thermal load to which it is designed.

Suppose the capacity factor (Kvs) Equal to 2.23 m3/ч. With such a specific ratio pump model withstand the load of 10 to 15 watts.

To determine the number of circuits, it is necessary to calculate the heat load of each. If the area occupied by the warm floor is 10 m² and heat 1 m², the indicator Kvs is 80 W, the 10 * 80 = 800 watts. Hence, the mixing assembly 15 can provide 000/800 = 18.8 Improvement or loops 10 m².

These figures are maximum, and use them only in theory, but in reality the figure should be reduced by at least 2, then 18 - 2 = 16 loops.

It is necessary in the selection of mixing junction (collector) look, if he has such a number of conclusions.

Verifying the selection tube diameter

To check whether it was chosen cross-section of the pipe, you can use the formula:

υ = 4 * Q * 10ᶾ / n * d²

When the speed value found corresponds to the cross section of pipes is selected correctly. Regulatory documents permit a maximum speed of 3 m / sec. with a diameter up to 0.25 m, but the optimum value is 0.8 m / sec., since the growth of its magnitude increases noise effect in the pipeline.

## We expect the circulating pump

To the system obtained economically, it is necessary pick up the circulation pumpProviding the necessary pressure and the optimal water flow in the circuits. Passports pumps usually indicate the pressure in the loop length and the greatest total flow of coolant in all loops.

At the head is influenced by hydraulic losses:

Δ h = L * Q² / k1where

• L - length of the contour;
• Q - water consumption l / sec;
• k1 - coefficient characterizing the losses in the system, the indicator can be obtained from the reference table on hydraulics or passport equipment.

Knowing the value of the pressure, flow rate calculated in the system:

Q = k * √Hwhere

k - a flow coefficient. Professionals take consumption for every 10 m² house 0.3-0.4 l / s. Among the warm water floor is a specific role for the circulation pump. Only unit with a capacity 20% higher than the actual flow will be able to overcome the resistance in the pipes

The figures relating to the pressure and flow rate indicated in the passport, should not be taken literally - is the maximum, but in fact they are influenced by the length, the network geometry. At too large a head of decreased contour length or diameter of pipe increases.

## Guidelines for selecting the thickness of the screed

The directories you can find information about what the minimum screed thickness of 30 mm. When the room is high enough, a tie enclose insulation that increases thermal efficiency, to give a heating circuit.

The most popular material for the substrate is extruded polystyrene. It has heat resistance is significantly lower than that of concrete.

When the device ties to counterbalance the linear expansion of the concrete, the perimeter of the room decorate damper tape. It is important to choose the correct thickness. Experts advise at room space not exceeding 100 m², 5 mm to arrange the compensating layer.

If the area value greater due to the length exceeding 10 m, the thickness is calculated by the formula:

b = 0,55 * Lwhere

L - a room length in meters.

## Conclusions and useful videos on the topic

On the calculation and installation of warm hydraulic floor this video:

The video provided practical recommendations on the floor laying. The information will help to avoid mistakes that usually allow fans:

The calculation makes it possible to design a "warm floor" system with optimum performance. It is permissible to mount the heating, using published data and recommendations.

It will work, but professionals are advised to still take the time to calculate that in the end the system to use less energy.

You have experience in the calculation of floor heating and preparation of the project of the heating circuit? Or have questions on the subject? Please share your opinion and leave comments. #### Water heated floors under tiles: installation and connection systemsUnderfloor Heating System

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