# Calculation of water heating: formulas, rules, examples of implementation

For the introduction of a heating installation, where water acts as a circulating substance, it is necessary to first produce accurate hydraulic calculations.

It is impossible to independently calculate water heating (hereinafter - ITS) without using profile programs, because calculations use complex expressions, which cannot be determined using a standard calculator.

## Thermal balance of housing construction

When developing and implementing any type of heating system, it is necessary to know the heat balance (hereinafter - TB). Knowing the thermal power to maintain the temperature in the room, you can choose the right equipment and correctly distribute its load.

In winter, the room bears a certain heat loss (hereinafter - TP). The bulk of the energy goes through the enclosing elements and ventilation openings.Insignificant expenses fall on infiltration, heating of objects, etc.

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TP depend on the layers that make up the enclosing structures (hereinafter - OK). Modern building materials, in particular, insulation, have a low coefficient of thermal conductivity (hereinafter - CT), so that less heat is lost through them. For houses of the same area, but with a different structure OK, heat costs will be different.

In addition to determining the TA, it is important to calculate the TB of the dwelling. The indicator takes into account not only the amount of energy leaving the room, but also the amount of necessary power to maintain certain degree measures in the house.

The most accurate results are provided by specialized programs designed for builders. Thanks to them, it is possible to take into account more factors affecting the TP.

The greatest amount of heat leaves the room through the walls, floor, roof, the least - through the doors, window openings.

With high accuracy it is possible to calculate the TP of a dwelling using formulas.

Total house heating costs are calculated using the equation:

$Q=Q_{ok}+Q_{v}$

In terms of$Q_{ok}$- the amount of heat leaving the room through OK,$Q_{v}$- heat ventilation costs.

Losses through ventilation are taken into account if the air entering the room has a lower temperature.

The calculations are usually taken into account OK, entering one side of the street. These are exterior walls, floor, roof, doors and windows. General TP$Q_{ok}$equal to the sum of TP each OK, that is:

$Q_{ok}=\sum&space;Q_{st}+\sum&space;Q_{okn}+\sum&space;Q_{dv}+\sum&space;Q_{ptl}+\sum&space;Q_{pl}$

In the equation:

• $Q_{st}$- the value of TP walls;
• $Q_{okn}$- TP windows;
• $Q_{dv}$- TP doors;
• $Q_{ptl}$- TP ceiling;
• $Q_{pl}$- TP floor.

If the floor or ceiling has a different structure over the entire area, then TP is calculated for each section separately.

### Calculation of heat loss through OK

For calculations, the following information is required:

• wall structure, materials used, their thickness, CT ;;
• outdoor temperature in the extremely cold five-day winter in the city;
• area OK;
• orientation OK;
• recommended temperature in the dwelling in winter.

To calculate the TP you need to find the total thermal resistance ROK. For this you need to know the thermal resistance R1, R2, R3, ..., Rn of each layer is OK.

The coefficient Rn is calculated by the formula:

$R_{n}=B/k$

In the formula B, the thickness of the layer is OK in mm, k is the CT of each layer.

Total R can be determined by the expression:

$R=\sum&space;R_{n}$

Manufacturers of doors and windows usually indicate the coefficient R in the passport to the product, so it is not necessary to count it separately.

The thermal resistance of windows can not be calculated, since the technical data sheet already contains the necessary information, which simplifies the calculation of TP

The general formula for calculating the TP through OK is as follows.

$Q_{ok}=\sum&space;S\cdot&space;(t_{vnt}-t_{nar})\cdot&space;R\cdot&space;l$

In terms of:

• S - area OK, m2;
• tvnt- the desired room temperature;
• tnar- outdoor air temperature;
• R - coefficient of resistance, calculated separately or taken from the passport of the product;
• l - specifying coefficient taking into account the orientation of the walls relative to the cardinal points.

Calculation of TB allows you to choose the equipment of the required capacity, which eliminates the possibility of the formation of a shortage of heat or its excess. The shortage of heat energy is compensated for by increasing the air flow through the ventilation, and the surplus by installing additional heating equipment.

### Heat consumption of ventilation

The general formula for calculating the TP of ventilation is as follows:

$Q_{v}=0.28\cdot&space;L_{n}\cdot&space;\rho&space;_{vnt}\cdot&space;c\cdot&space;(t_{vnt}-t_{nar})$

In the expression, variables have the following meaning:

• $L_{n}$- the cost of incoming air;
• $\rho&space;_{_{vnt}}$- air density at a certain temperature in the room;
• c is the heat capacity of the air;
• $t_{vnt}$- the temperature in the house;
• $t_{nar}$- outdoor air temperature.

If ventilation is installed in the building, the parameter$L_{n}$taken from the technical characteristics of the device. If there is no ventilation, then the standard indicator of specific air exchange is taken, equal to 3 m.3at one o'clock. Based on this,$L_{n}$calculated by the formula:

$L_{n}=3\cdot&space;S_{pl}$

In terms of$S_{_{pl}}$- floor area.

2% of all heat losses accounted for infiltration, 18% - for ventilation. If the room is equipped with a ventilation system, then the calculations take into account the TP through ventilation, and do not take infiltration into account

Next, you should calculate the density of air$\rho&space;_{_{vnt}}$at a given room temperature$t_{vnt}$. This can be done by the formula:

$\rho&space;_{vnt}=\frac{353}{273+t_{vnt}}$

Specific heat$c=1.005$.

If the ventilation or infiltration is unorganized, there are gaps or holes in the walls, then the calculation of the TP through the holes should be entrusted to special programs.

## Example of heat balance calculation

Consider a house with a height of 2.5 m, a width of 6 m and a length of 8 m, located in the city of Okha in the Sakhalin region, where in an extremely cold 5-day thermometer thermometer drops to -29 degrees.

As a result of the measurement, the temperature of the soil was set to +5. The recommended temperature inside the structure is +21 degrees.

It is most convenient to draw a diagram of the house on paper, indicating not only the length, width and height of the building, but also orientation relative to the cardinal points, as well as the location, dimensions of windows and doors (+)

The walls of the house in question consist of:

• masonry thickness B = 0.51 m, CT k = 0.64;
• mineral wool B = 0.05 m, k = 0.05;
• facing B = 0.09 m, k = 0.26.

When determining k, it is better to use the tables presented on the manufacturer’s website, or to find information in the product data sheet.

Knowing the thermal conductivity, it is possible to choose the most efficient materials from the point of view of thermal insulation. Based on the above table, it is most advisable to use in the construction of mineral wool and polystyrene foam

The flooring consists of the following layers:

• OSB plates B = 0.1 m, k = 0.13;
• minwats B = 0.05 m, k = 0.047;
• cement screed B = 0.05 m, k = 0.58;
• expanded polystyrene B = 0.06 m, k = 0.043.

In the house there is no basement, and the floor has the same structure over the entire area.

The ceiling consists of layers:

• sheets of drywall B = 0.025 m, k = 0.21;
• insulation B = 0.05 m, k = 0.14;
• roofing overlap B = 0.05 m, k = 0.043.

There are no exits to the attic.

There are only 6 two-chamber windows with I-glass and argon in the house. From the technical passport to the products it is known that R = 0.7. The windows have dimensions of 1.1x1.4 m.

Doors have dimensions of 1x2.2 m, R = 0.36.

### Calculation of heat loss walls

The walls throughout the area consist of three layers. First, we calculate their total thermal resistance using the formula:

$R=\sum&space;R_{n}$

and expression

$R_{n}=B/k$

Given the initial information, we get:

$R_{st}=\frac{0.51}{0.64}+\frac{0.05}{0.05}+\frac{0.09}{0.26}=0.79+1+0.35=2.14$
Having learned R, one can proceed to the calculations of the TS of the north, south, east and west walls.

Additional factors take into account the peculiarities of the location of the walls relative to the cardinal points. Usually in the northern part during the cold weather a “rose of winds” is formed, as a result of which the TA from this side will be higher than from other

Calculate the area of ​​the north wall

$S_{sev.sten}=8\cdot&space;2.5=20$

Then, substituting into the formula

$Q_{ok}=\sum&space;S\cdot&space;(t_{vnt}-t_{nar})\cdot&space;R\cdot&space;l$

and taking into account that l = 1.1, we get:

$Q_{sev.st}=20\cdot&space;(21+29)\cdot&space;1.1\cdot&space;2.14=2354$

South wall area$S_{yuzh.st}=S_{sev,st}=20$. There are no built-in windows or doors in the wall, therefore, taking into account the coefficient l = 1, we get the following TP:

$Q_{yuzh.st}=20\cdot&space;(21+29)\cdot&space;1\cdot&space;2.14=2140$

For the western and eastern walls, the coefficient is l = 1.05.Therefore, you can find the total area of ​​these walls, that is:

$S_{zap.st}+S_{vost.st}=2\cdot&space;2.5\cdot&space;6=30$

6 windows and one door are built into the walls. Calculate the total area of ​​windows and S doors:

$S_{okn}=1.1\cdot&space;1.4\cdot&space;6=9.24$

$S_{dv}=1\cdot&space;2.2=2.2$

Define S walls without taking into account S windows and doors:

$S_{vost+zap}=30-9.24-2.2=18.56$

We calculate the general TP of the eastern and western walls:

$Q_{vost+zap}=18.56\cdot&space;(21+29)\cdot&space;2.14\cdot&space;1.05=2085$

Having obtained the results, we calculate the amount of heat escaping through the walls:

$Q_{st}=Q_{sev.st}+Q_{yuzh.st}+Q_{vost+zap}=2140+2085+2354=6579$

Total total TP walls are 6 kW.

### Calculation of TP windows and doors

The windows are located on the eastern and western walls, therefore, when calculating, the coefficient is l = 1.05. It is known that the structure of all structures is the same and R = 0.7. Using the area values ​​given above, we get:

$Q_{okn}=9.24\cdot&space;(21+29)\cdot&space;1.05\cdot&space;0.7=340$

Knowing that for doors R = 0.36, and S = 2.2, we define them TP:

$Q_{dv}=2.2\cdot&space;(21+29)\cdot&space;1.05\cdot&space;0.36=42$

As a result, 340 W of heat comes out through the windows, and 42 W through the doors.

### Determination of TP floor and ceiling

Obviously, the area of ​​the ceiling and floor will be the same, and is calculated as follows:

$S_{pol}=S_{ptl}=6\cdot&space;8=48$

Calculate the total thermal resistance of the floor with regard to its structure.

$R_{pol}=\frac{0.1}{0.13}+\frac{0.05}{0.047}+\frac{0.05}{0.58}+\frac{0.06}{0.043}=0.77+1.06+0.17+1.40=3.4$

Knowing what the temperature of the soil is tnar= + 5 and taking into account the coefficient l = 1, we calculate the Q of the floor:

$Q_{pol}=48\cdot&space;(21-5)\cdot&space;1\cdot&space;3.4=2611$

Rounding off, we get that floor heat loss is about 3 kW.

In the calculations of TP, it is necessary to take into account layers that affect thermal insulation, for example, concrete, boards, brickwork, insulation, etc. (+)

Define Rptl.

$R_{ptl}=\frac{0.025}{0.21}+\frac{0.05}{0.14}+\frac{0.05}{0.043}=0.12+0.71+0.35=1.18$

Having determined the thermal resistance of the ceiling, we find it Q:

$Q_{ptl}=48\cdot&space;(21+29)\cdot&space;1\cdot&space;1.18=2832$

It follows that through the ceiling and the floor goes almost 6 kW.

### Calculation of TP ventilation

Indoor ventilation is organized, calculated by the formula:

$Q_{v}=0.28\cdot&space;L_{n}\cdot&space;\rho&space;_{vnt}\cdot&space;c\cdot&space;(t_{vnt}-t_{nar})$

Based on the technical characteristics, the specific heat exchange is 3 cubic meters per hour, that is:

$L_{n}=3\cdot&space;48=144$.

To calculate the density, we use the formula:

$\rho&space;_{vnt}=\frac{353}{273+t_{vnt}}$

Design room temperature is +21 degrees.

TP ventilation is not calculated if the system is equipped with a device for heating air

Substituting the known values, we get:

$\rho&space;_{vnt}=\frac{353}{273+21}=1.2$

Substitute the resulting figures in the above formula:

$Q_{v}=0.28\cdot&space;144\cdot&space;\1.2\cdot&space;1.005\cdot&space;(21+29)=2431$

Given the TP for ventilation, the total Q of the building will be:

$Q=7000+6000+3000=16000$.

Translated in kW, we get a total heat loss of 16 kW.

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## Features of the calculation of CBO

After finding the TP indicator, they proceed to the hydraulic calculation (hereinafter referred to as GR), on the basis of which information about:

• optimal diameter of pipes, which, with pressure drops, will be able to pass a given amount of coolant;
• coolant flow in a particular area;
• water speeds;
• resistivity value.

Before starting the calculations, to simplify the calculations, they depict the spatial scheme of the system, in which all its elements are arranged parallel to each other.

The diagram shows the heating system with upper wiring, the movement of the coolant - dead-end (+)

Consider the main stages of the calculation of water heating.

### GR main circulation ring

The method of calculating the GH is based on the assumption that in all risers and branches the temperature drops are the same.

The calculation algorithm is as follows:

1. In the diagram shown, taking into account heat losses, heat loads are applied to the heaters and risers.
2. On the basis of the scheme, choose the main circulating ring (hereinafter - FCC). The peculiarity of this ring is that in it the circulation pressure per unit length of the ring takes the smallest value.
3. FCC is divided into areas with constant heat consumption. For each site indicate the number, heat load, diameter and length.

In a vertical one-pipe type system, the fcc is taken as the ring through which the most loaded riser passes during dead-end or passing water movement along highways.

In vertical systems of two-pipe type, the HCC passes through the lower heating device, which has a maximum load in case of dead-end or following water movement

In the horizontal one-pipe type system, the fcc should have the lowest circulating pressure and a unit of the length of the ring. For systems with natural circulation, the situation is similar.

With a GH of vertical risers of a single pipe type, flowing, flow-adjustable risers, having in their composition unified units, are considered as a single contour. For risers with closing sections produce separation, taking into account the distribution of water in the pipeline of each instrument node.

Water consumption in a given area is calculated by the formula:

$G_{kont}=\frac{3.6\cdot&space;Q_{kont}\cdot&space;\beta&space;_{1}\cdot&space;\beta&space;_{2}}{(t_{r}-t_{o})\cdot&space;c}$

In the expression of literal characters take the following values:

$Q_{kont}$- heat load circuit;

$\beta&space;_{1},&space;\beta&space;_{2}$- additional table coefficients taking into account the heat transfer in the room;

c is the heat capacity of water, equal to 4.187;

$t_{r}$- water temperature in the supply line;

$t_{o}$- water temperature in the return line.

Having determined the diameter and amount of water, it is necessary to find out the speed of its movement and the value of specific resistance R.All calculations are most conveniently carried out using special programs.

### GH secondary circulating ring

After the main ring GR, the pressure in the small circulation ring is determined through its nearest risers, taking into account that pressure losses can differ by no more than 15% with a dead-end scheme and no more than 5% with passing.

If it is not possible to link the pressure loss, install a throttle washer, the diameter of which is calculated using software methods.

Let's return to the plan of the house located above. By calculation, it was found that to maintain the heat balance will require 16 kW of energy. In this house there are 6 rooms for different purposes - a living room, a bathroom, a kitchen, a bedroom, a corridor, an anteroom.

Based on the dimensions of the structure, it is possible to calculate the volume V:

V = 6 * 8 * 2.5 = 120 m3

Next you need to find the amount of heat output per m3. To do this, Q must be divided by the volume found, that is:

P = 16000/120 = 133 W per m3

Next, you need to determine how much heat capacity will be required for one room. In the diagram, the area of ​​each room is already calculated. Determine the volume:

• the bathroom is 4.19 * 2.5 = 10.47;
• living room - 13.83 * 2.5 = 34.58;
• kitchen - 9.43 * 2.5 = 23.58;
• bedroom - 10.33 * 2.5 = 25.83;
• corridor - 4.10 * 2.5 = 10.25;
• entrance hall - 5.8 * 2.5 = 14.5.

The calculations also need to take into account the premises in which there are no heating batteries, for example, a corridor.

The corridor is heated in a passive way, heat will enter it due to the circulation of thermal air during the movement of people, through doorways, etc.

Determine the required amount of heat for each room by multiplying the volume of the room by the indicator R. We obtain the required power:

• for the bathroom: 10.47 * 133 = 1392 W;
• for the living room: 34.58 * 133 = 4599 W;
• for the kitchen: 23.58 * 133 = 3136 W;
• for the bedroom: 25.83 * 133 = 3435 W;
• for the corridor: 10.25 * 133 = 1363 W;
• for the hallway: 14.5 * 133 = 1889 W.

We proceed to the calculation of radiator batteries. We will use aluminum radiators, whose height is 60 cm, power at a temperature of 70 is equal to 150 watts. Calculate the required number of radiator batteries.

• Bathroom: 1392: 150 = 10
• living room: 4599: 150 = 31
• Kitchen: 3136: 150 = 21
• Bedroom: 3435: 150 = 23
• entrance hall: 1889: 150 = 13

Total will require 98 radiator batteries.

## Conclusions and useful video on the topic

In the video you can see an example of calculation of water heating, which is carried out by means of the Valtec program:

Hydraulic calculations are best carried out with the help of special programs that guarantee high accuracy of calculations, taking into account all the nuances of the design.