## Calculation of the number of battery sections

When designing heating systems, a mandatory event is to carry out calculation calculations of heating devices. The result obtained more affects the choice of one or another equipment. heating radiators and heating boilers (if the project is carried out for private houses not connected to the central heating systems).

The most popular are the batteries made in the form of interconnected sections. In this article, it will just be about how to calculate the number of radiator sections.

## We **calculate** the volume of space

For a panel house with a standard ceiling height, as mentioned above, heating is calculated from a 41 W requirement for 1 m3. But if the house is new, there are brick windows, double glazing windows, and the outer wall is isolated, then you need 34 watts per 1 m3.

The formula for calculating the number of radiation sections is as follows: the volume (area multiplied by the ceiling height) is multiplied by 41 or 34 (depending on the type of house), which is divided into the section heater of the manufacturer’s certificate.

For example: **room** area 18 m2, ceilings height 2, 6 m.

The house has a typical panel building. Heat transfer of one radiator section is 170 watts.

18×2,6×41 / 170 = 11.2. So, we need 11 parts of the radiator. This guarantees that the room is not a corner and there is no balcony, otherwise it is better to place 12 pieces.

## We count the batteries in volume

There are norms in SNIP and for **heating** one cubic meter of premises. They are given for different types of buildings:

This calculation of the sections of radiators is similar to the previous one, only now we need not an area, but the volume and norms take others. We multiply the volume by the norm, the resulting number is divided by the power of one section of the radiator (aluminum, bimetallic or cast.iron).

Formula for calculating the number of sections by volume

### Example of calculation by volume

For example, we will **calculate** how many sections are needed in a room with an area of 16 m 2 and the height of the ceiling 3 meters. The building is built of brick. Radiators take the same power: 140 watts:

- We find the volume. 16 m 2 3 m = 48 m 3
- We count the required amount of heat (the norm for brick buildings 34 W). 48 m 3 34 W = 1632 W.
- We determine how many sections are needed. 1632 W / 140 W = 11.66 pcs. Walk round, we get 12 pcs.

Now you know two ways to **calculate** the number of radiators on the room.

## How to **calculate** heat loss?

To completely **calculate** the thermal losses of the room or the whole house, you will need to collect a large amount of information about the structure. The calculations themselves can be performed manually by joint venture 50.13330.2012 or in any online calculator.

- We count the area of the windows, take the area with the frame. If there are two windows in the room, then we fold the total area.
- Measure the total length of the external walls, and then multiply the resulting value by the height of the ceiling.
- Remove the windows area from the walls of the walls.
- We count the floor area to determine thermal losses through infiltration (blowing through technological holes).
- You need to know the type of window: for example, a two.chamber double.glazed window, a regular window with a double frame, etc.D.
- We determine the material of the external walls. For example, brick with insulation of mineral wool.

Thermal losses through internal walls and partitions are usually not taken into account.

- To determine thermal losses through the floor, you need to know the design of the flooring of the first floor: floors on the ground, gender above the technical underground or basement, etc.D.
- To calculate losses through the ceiling, you need to know the structure of the ceiling and its perimeter.

If there is a “warm” attic over the first floor, heated floor, then when calculating for the first floor they do not take into account losses for the ceiling. Energy leaks through the floor are taken into account only on the ground floor. If the heat loss for the attic is calculated, then instead of the ceiling add energy loss through the roof.

In private houses, the greatest heat losses occur on the attic floors, as it comes into contact with the roof. The smallest power is required to warm the rooms on the second floor if the “warm” attic is located above them. On the ground floor usually colder due to the front door and losses through the floors.

## Steel **radiator**

The design of panel devices differs from sectional. Batteries are made of stamped steel sheets 1 1.2 mm, pre.trimmed to the desired size. To choose the **radiator** of the required power, you need to find out the heat transfer of 1 meter of the length of the panel cooked from the sheets.

We suggest using the simplest methodology based on the technical data of a serious German manufacturer of panel water radiators Kermi. What is the essence: stamped batteries are unified, types of products differ among themselves by the number of heating panels and heat exchangers. The classification of radiators looks like this:

- Type 10. one.panel device without additional ribs;
- Type 11. 1 panel 1 sheet of corrugated metal;
- Type 12. two panels plus 1 sheet of nuts;
- Type 20. a battery for 2 heating plates, convection nutrition is not provided;
- Type 22. two.panel radiator with 2 sheets that increase the heat exchange area.

### How to turn different radiator valves off

Note. There are also type 33 heaters (3 panels 3 ribs), but such products are less in demand due to increased thickness and price. The most “running” model is type 22.

So, panel stamped devices of any brand differ only in mounting dimensions. The calculation of **heating** radiators is reduced to the choice of a suitable type, then the length of the battery for a specific room is calculated in height and heat transfer. The algorithm is as follows:

- Determine the initial data listed at the beginning of the article.
- Choose the type and height of the
**heating**device. The most common options are products with a height of 30, 40 and 50 cm, type 22. - Use the presented table, which indicates the heat transfer Q (W/1 m. P.) Kermi radiators of different types and sizes depending on operating conditions. Start with the left column. find the corresponding temperature of the room, then the coolant, then the height and the type of battery. In the cell at the intersection of the line and the column you will find a power of 1 meter of the radiator.
- The amount of energy needed for
**heating**is divided into Q. find out the meter of the**radiator**of a given height. - By the catalog, select the water
**heating**device of the corresponding length. If necessary (for example, the battery came out too long) break this size 2-3 devices.

An example of calculation. Determine the dimensions of a steel

radiatorfor the sameroom15.75 m²: heat loss. 2048 W, air temperature. 22 degrees, coolant. 65 ° C. Take standard batteries 500 mm high, type 22. By the table we find Q = 1461 W, find out the total length of the panel 2048 /1461 = 1.4 m. From the catalog of any manufacturer, we select the nearest larger option. 1 heater 1.5 m or 2 devices for 0.7 m.

The end of the first table is the heat transfer 1 m of the length of the radiators “Kermmi”

Advice. Our instruction is 100% true for Kermi products. When buying radiators of another brand (especially Chinese), the length of the panel should be taken with a margin of 10-15%.

### The parameters of the worn out the room

The area of the room | m 2 |

Ceiling height | |

The number of external walls of the room | |

The coefficient of thermal insulation of the walls | |

Accounting for the type of room located above | |

The number of windows | |

Coefficient taking into account the glazing of window openings | |

Average temperature on the street in winter |

Heat transfer 1 section | operating pressure | Sent pressure pressure | Squatience 1 section | Mass 1 section | |

aluminum, with an interaceal distance of 500 mm | 183 T | 20 bar | 30 bar | 0.27 l | 1.45 kg |

aluminum, with an interaceal distance of 350 mm | 139 T | 20 bar | 30 bar | 0.19 l | 1.2 kg |

bimetalic, with an interdose distance of 500 mm | 204 watts | 20 bar | 30 bar | 0.2 l | 1.92 kg |

bimetalic, with an interdose distance of 350 mm | 136 watts | 20 bar | 30 bar | 0.18 l | 1.36 kg |

Cast iron, with an interdose distance of 500 mm | 160 watts | 9 bar | 15 bar | 1.45 l | 7.12 kg |

Cast iron, with an interdose distance of 300 mm | 140 watts | 9 bar | 15 bar | 1.1 l | 5.4 kg |

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## What is an aluminum radiator

Strictly speaking, an aluminum **radiator** is of two types:

Structurally, such a radiator is a pipe collected in the likeness of an accordion along which hot water flows. Flat elements are connected to the pipe, which are heated by the coolant and heat the air in the **room**.

### Cooling Load Calculation. Cold Room hvac

Description of the advantages and disadvantages of each type of **radiator** goes beyond this article, however, you can point out several important factors. Unlike traditional cast.iron ones, aluminum batteries are heated primarily due to convection: heated air rushes up, and its place is a fresh portion of cold. Due to this process, it is possible to heat the room much faster.

To this it is worth adding a slight weight and ease of installation of aluminum products, as well as their relative cheapness.

## Heat transfer of one section

Today the range of radiators is large. With the external similarity of the majority, thermal indicators can differ significantly. They depend on the material from which they are made, on the size, the thickness of the walls, the internal section and on how well the design is thought out.

Therefore, to say for sure how many kW in 1 section of aluminum (cast iron bimetallic) radiator can only be said in relation to each model. These data are indicated by the manufacturer. After all, there is a significant difference in size: some of them are tall and narrow, others are low and deep. The power of the section of the same height of the same manufacturer, but of different models, may differ by 15-25 watts (see the table below Style 500 and Style Plus 500). Different manufacturers can have even more tangible differences.

Technical characteristics of some bimetallic radiators. Please note that the thermal power of the same sections may have a tangible difference

Nevertheless, for a preliminary assessment of how many sections of the batteries are needed for heating the premises, the middle of the heat capacity for each type of radiator was brought out in the middle. They can be used for approximate calculations (data for batteries with an interax distance of 50 cm are given):

precisely, how many kW in one section of the radiator of bimetallic, aluminum or cast.iron you can when you select the model and decide on the dimensions. The difference in cast.iron batteries can be very large. They are with thin or thick walls, which is why their thermal power changes significantly. Above are the average values for the batteries of the usual shape (accordion) and loved ones. The radiators in the “retro” style of thermal power are many times lower.

These are the technical characteristics of cast.iron radiators of the Turkish company Demir Dokum. The difference is more than solid. She can be even more

Based on these values and average norms in SNiP, the average number of radiator sections per 1 m 2:

How to **calculate** the number of radiator sections according to these data? It’s still easier. If you know the area of the **room**, divide it into the coefficient. For example, **room** 16 m 2. For its heating you will need:

- bimetallic 16 m 2 / 1.8 m 2 = 8.88 pcs, round. 9 pcs.
- aluminum 16 m 2 /2 m 2 = 8 pcs.
- cast iron 16 m 2 / 1.4 m 2 = 11.4 pcs, round. 12 pcs.

These calculations are only approximate. On them you can roughly evaluate the costs of acquiring **heating** devices. You can accurately calculate the number of radiators on the room by choosing a model, and then after counting the amount depending on what temperature of the coolant in your system.

## Calculation of radiator sections depending on real conditions

Once again, we draw your attention to the fact that the thermal power of one section of the battery is indicated for ideal conditions. So much heat will give out the battery if at the entrance its coolant has a temperature of 90 ° C, at the output of 70 ° C, the **room** is maintained 20 ° C. That is, the temperature pressure of the system (also called the “system of the system”) will be 70 ° C. What to do if in your system above 70 ° C at the entrance to it happens? or the temperature in the room is 23 ° C? Recalculate the declared capacity.

To do this, you need to **calculate** the temperature pressure of your heating system. For example, you have 70 ° C at the output of 60 ° C, and in the room you need a temperature of 23 ° C. We find the delta of your system: this is the arithmetic average at the entrance and output, minus the temperature in the room.

The formula for calculating the temperature pressure of the heating system

For our case, it turns out: (70 ° C 60 ° C)/2. 23 ° C = 42 ° C. Delta for such conditions 42 ° C. Next, we find this value in the recalculation table (located below) and the declared power multiply by this coefficient. Teach the power that this section for your conditions can give out.

We find in columns tinted in blue, a line with a 42 ° C delta. It corresponds to a coefficient of 0.51. Now we calculate, thermal power 1 section of the radiator for our case. For example, the declared capacity of 185 watts, using the found coefficient, we get: 185 W 0.51 = 94.35 W. Almost two times less. This power must be substituted when you are calculating the sections of radiators. Only taking into account individual parameters in the room will be heat.