Calculation of sections of radiators on the floor space: how

How to calculate the radiator on the floor space - residential or industrial? In this article, we will acquaint the reader with several algorithms of varying complexity and present for reference convenience some reference data. So go.

Stages of calculations

Actually, there are only two of them.

1. Initially, the estimated need for space in thermal power.
2. Then, depending on the specific value of the heat flux (per section, per heater, etc.), the number of the corresponding contour elements is calculated.

Specify: in the network you can find a large number of tables and calculators, directly deriving the number of sections from the area. However, the accuracy of such calculations is usually low, since they completely ignore additional factors that increase or decrease heat losses.

Power calculation

Scheme 1

The simplest scheme is present in Soviet SNiP half a century ago: the power of a heating radiator per room is selected at the rate of 100 watts / 1m2.

The algorithm is clear, extremely simple and inaccurate.

Why?

• Real heat losses are very different for extreme and middle floors, for corner apartments and rooms in the center of the building.
• They depend on the total area of ​​windows and doors, as well as on the structure of the glazing. It is clear that wooden frames with two-glass panes will provide much greater heat loss than a triple glazing.
• In different climatic zones, heat loss will also vary. In -50 C apartment will obviously need more heat than in +5.
• Finally, the selection of a radiator according to the area of ​​the room makes it neglect the height of the ceilings; Meanwhile, the heat consumption in ceilings 2.5 and 4.5 meters high will vary greatly.

Scheme 2

Estimation of thermal power and the calculation of the number of radiator sections by the volume of the room provides significantly greater accuracy.

Here are the instructions for calculating the power:

1. The base amount of heat is estimated at 40 watts / m3.
2. For corner rooms, it increases by 1.2 times, for extreme floors - by 1.3, for private houses - by 1.5.
3. The window adds 100 watts to the room's need for heat, the door to the street - 200.
4. A regional coefficient is introduced. It is taken equal:

Let's take an example with our own hands to find the need for heat in a corner room measuring 4x5x3 meters with one window located in the city of Anapa.

1. The volume of the room is 4 * 5 * 3 = 60 m3.
2. The basic need for heat is estimated at 60 * 40 = 2400 watts.
3. Since the room is angular, we use the coefficient 1.2: 2400 * 1.2 = 2880 watts.
4. The window exacerbates the situation: 2880 + 100 = 2980.
5. The mild climate of Anapa makes its own adjustments: 2980 * 0.8 = 2384 watts.

Scheme 3

Both previous schemes are bad because they ignore the difference between different buildings in terms of wall insulation. Meanwhile, in a modern energy-efficient house with external insulation and in a brick workshop with one-piece glazing, heat loss will be, to put it mildly, different.

Radiators for industrial premises and houses with non-standard insulation can be calculated by the formula Q = V * Dt * k / 860, in which:

• Q - heating circuit power in kilowatts.
• V - heated volume.
• Dt is the calculated temperature delta with the street.

Please note: the room temperature is taken from sanitary standards or technological requirements; street is estimated by the average temperature for the coldest 5 days of winter.

• k - coefficient of warming. Where to get its values?

Let us in this case also accompany the calculation algorithm with an example - we calculate the thermal power that radiators of a production room of 400 sq m with a height of 5 meters, a brick wall thickness of 25 cm and a single glazing should possess. This picture is quite typical for industrial zones.

Let us agree that the temperature of the coldest five-day week is -25 degrees Celsius.

1. For production workshops, the lower limit of the permissible temperature is considered to be +15 C. Thus, Dt = 15 - (-25) = 40.
2. The coefficient of heat insulation is equal to 2.5.
3. The volume of the room is 400 * 5 = 2000 m3.
4. The formula will take the form Q = 2000 * 40 * 2.5 / 860 = 232 kW (with rounding).

Calculation of heating devices

Cast iron, aluminum and bimetallic batteries, steel tubular, panel and plate radiators, as well as convectors are widely used in residential premises for heating.

How to determine the thermal power of each device?

For panels, convectors, non-separable tubular batteries and plates, you can only focus on the characteristics provided by the manufacturer. They are always present in the accompanying documentation or on the manufacturer’s website.

For sectional batteries with a standard (500 mm) vertical size, you can focus on the following values ​​of heat flow:

• Cast iron section - 140-160 watts;
• Aluminum - 180-200;

• Bimetallic - 170-190.

Important point: the rated power is indicated for the 70-degree difference between the radiator and the air in the room. If the difference is half as much, the specific heat transfer will be reduced by the same amount.

So, with the need for a thermal power of 2.3 KW, the aluminum radiator (200 W / section) should have 2300/200 = 12 (with rounding) sections.

A special case

Typical heating radiators for industrial premises are steel all-welded registers. The low price of the material, coupled with high strength makes them much more attractive than other solutions.

Their power can be calculated by the following algorithm:

• For a single horizontal pipe, it is equal to Q = 3.14xD * L * 11.63 * Dt, where D is the diameter of the pipe in meters, L is its length in meters, and Dt is the temperature delta between the room and the coolant.
• In a multi-section horizontal register, the factor 0.9 is used to calculate sections starting from the second.

Thus, a ten-meter single-section register with a diameter of 250 mm, when heated with superheated steam (20 ° C) and at a temperature in the workshop of 15 ° C, will give 3.14 * 0.25 * 10 * 11.63 * (200-15) = 16,889 watts of heat.

Conclusion

As you can see, the applied calculation schemes are relatively simple and quite understandable even for a person who is far from designing heating systems. Additional thematic information can, as usual, be found in the video in this article. Successes!