Calculation of water heating: formulas, rules, examples of implementation
Using water as a coolant in a heating system is one of the most popular options for providing your home with heat in the cold season. You only need to properly design and then complete the installation of the system. Otherwise, heating will be ineffective at high fuel costs, which, you see, is extremely uninteresting at today's energy prices.
It is impossible to independently calculate water heating (hereinafter referred to as CBO) without using specialized programs, because the calculations use complex expressions, the values of which cannot be determined using a conventional calculator. In this article, we will analyze in detail the algorithm for performing calculations, give the applicable formulas, considering the course of the calculations using a specific example.
Supplemented material will be supplemented with tables with values and reference indicators that are needed during the calculations, thematic photos and a video in which a clear example of calculation by using the program is shown.
The content of the article:
- Calculation of the heat balance of housing
- Heat balance calculation example
- Features of the calculation of CBO
- Conclusions and useful video on the topic
Calculation of the heat balance of housing
For the introduction of a heating installation, where water acts as a circulating substance, it is necessary to first make accurate hydraulic calculations.
When developing, 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 suffers certain heat losses (hereinafter - TP). The bulk of the energy goes through the enclosing elements and ventilation openings. Insignificant expenses are for infiltration, heating of objects, etc.
TP depend on the layers of which the enclosing structures consist (hereinafter - OK). Modern building materials, in particular insulation, have a low thermal conductivity coefficient (hereinafter referred to as CT), due to which less heat is expelled through them. For houses of the same area, but with a different OK structure, the heat costs will differ.
In addition to determining TP, it is important to calculate the TB of a home. 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 TP.
With high accuracy, you can calculate the TP of the home using formulas.
The total heat consumption of the house is calculated by the equation:
Q = Qok + Qv,
Where Qok - the amount of heat leaving the room through OK; Qv - thermal ventilation costs.
Losses through ventilation are taken into account if the air entering the room has a lower temperature.
The calculations usually take into account OK, entering one side of the street. These are external walls, floor, roof, doors and windows.
General TP Qok equal to the sum of TP of each OK, that is:
Qok = ∑Qst + ∑Qokn + ∑Qdv + ∑Qptl + ∑Qpl,
- Qst - the value of TP walls;
- Qokn - TP windows;
- Qdv - TP doors;
- Qptl - TP ceiling;
- Qpl - TP floor.
If the floor or ceiling has an unequal structure over the entire area, then the TP is calculated for each site separately.
Calculation of heat loss through OK
For calculations, the following information is required:
- wall structure, materials used, their thickness, CT;
- the outside temperature on an extremely cold five-day winter in the city;
- OK area;
- orientation OK;
- Recommended home temperature in winter.
To calculate the TP, you need to find the total thermal resistance ROK. To do this, find out the thermal resistance R1, R2, R3, ..., Rn each layer is OK.
Coefficient Rn calculated by the formula:
Rn = B / k,
In the formula: B - layer thickness OK in mm, k - CT of each layer.
The total R can be determined by the expression:
R = ∑Rn
Manufacturers of doors and windows usually indicate the coefficient R in the passport to the product, so there is no need to calculate it separately.
The general formula for calculating TP through OK is as follows:
Qok = ∑S × (tvnt - tnar) × R × l,
In the expression:
- S - area OK, m2;
- tvnt - desired room temperature;
- tnar - outdoor air temperature;
- R - resistance coefficient, calculated separately or taken from the product passport;
- l - a refinement 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 likelihood of a heat deficit or its excess. The deficit of thermal energy is compensated by increasing the air flow through the ventilation, the excess - by installing additional heating equipment.
Thermal ventilation costs
The general formula for calculating ventilation TP is as follows:
Qv = 0.28 × Ln × pvnt × c × (tvnt - tnar),
Variables have the following meanings in an expression:
- Ln - incoming air costs;
- pvnt - air density at a certain temperature in the room;
- c - heat capacity of air;
- tvnt - temperature in the house;
- tnar - outdoor air temperature.
If ventilation is installed in the building, then parameter Ln taken from the technical characteristics of the device. If there is no ventilation, then a standard indicator of specific air exchange equal to 3 m is taken3 in hour.
Based on this, Ln calculated by the formula:
Ln = 3 × Spl,
In expression Spl - floor area.
Next, calculate the air density pvnt at a given temperature tvnt.
You can do this by the formula:
pvnt = 353 / (273 + tvnt),
Specific heat capacity c = 1.0005.
If ventilation or infiltration is unorganized, there are cracks or holes in the walls, then the calculation of the TP through the holes should be entrusted to special programs.
In our other article, we gave a detailed example of heat engineering calculation buildings with specific examples and formulas.
Heat balance calculation example
Consider a house 2.5 m high, 6 m wide and 8 m long, located in the city of Okha in the Sakhalin Region, where the thermometer thermometer drops to -29 degrees in an extremely cold 5-day period.
As a result of the measurement, the soil temperature was set to +5. The recommended temperature inside the structure is +21 degrees.
The walls of the house in question consist of:
- brickwork with a thickness of B = 0.51 m, CT k = 0.64;
- mineral wool B = 0.05 m, k = 0.05;
- Facings 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 technical passport of the product.
The flooring consists of the following layers:
- OSB-plates B = 0.1 m, k = 0.13;
- mineral wool B = 0.05 m, k = 0.047;
- cement screed B = 0.05 m, k = 0.58;
- polystyrene foam B = 0.06 m, k = 0.043.
There is no basement in the house, and the floor has the same structure over the entire area.
The ceiling consists of layers:
- drywall sheets B = 0.025 m, k = 0.21;
- insulation B = 0.05 m, k = 0.14;
- roofing slab B = 0.05 m, k = 0.043.
There are no exits to the attic.
The house has only 6 double-chamber windows with I-glass and argon. From the technical passport for the products it is known that R = 0.7. Windows have dimensions 1.1x1.4 m.
Doors have dimensions of 1x2.2 m, indicator R = 0.36.
Step # 1 - calculation of wall heat loss
Walls over the entire area consist of three layers. First, we calculate their total thermal resistance.
Why use the formula:
R = ∑Rn,
Rn = B / k
Given the initial information, we get:
Rst = 0.51/0.64 + 0.05/0.05 + 0.09/0.26 = 0.79 +1 + 0.35 = 2.14
Having learned R, we can begin to calculate the TP of the northern, southern, eastern and western walls.
We calculate the area of the north wall:
Ssev.sten = 8 × 2.5 = 20
Then, substituting into the formula Qok = ∑S × (tvnt - tnar) × R × l and considering that l = 1.1, we get:
Qsev.sten = 20 × (21 + 29) × 1.1 × 2.14 = 2354
South Wall Area Syuch.st = Ssev.st = 20.
There are no built-in windows or doors in the wall, therefore, given the coefficient l = 1, we obtain the following TP:
Qyuch.st = 20 × (21 +29) × 1 × 2.14 = 2140
For the western and eastern walls, the coefficient l = 1.05. Therefore, you can find the total area of these walls, that is:
Szap.st + Svost.st = 2 × 2.5 × 6 = 30
6 windows and one door are built into the walls. We calculate the total area of windows and S doors:
Sokn = 1.1 × 1.4 × 6 = 9.24
Sdv = 1 × 2.2 = 2.2
Define S walls excluding S windows and doors:
Svost + zap = 30 – 9.24 – 2.2 = 18.56
We calculate the total TP of the eastern and western walls:
Qvost + zap =18.56 × (21 +29) × 2.14 × 1.05 = 2085
After receiving the results, we calculate the amount of heat leaving through the walls:
Qst = Qsev.st + Qyuch.st + Qvost + zap = 2140 + 2085 + 2354 = 6579
Total total TP of the walls is 6 kW.
Step # 2 - calculating TP windows and doors
The windows are located on the eastern and western walls, therefore, when calculating the coefficient l = 1.05. It is known that the structure of all structures is the same and R = 0.7.
Using the values of the area above, we get:
Qokn = 9.24 × (21 +29) × 1.05 × 0.7 = 340
Knowing that for doors R = 0.36, and S = 2.2, we define their TP:
Qdv = 2.2 × (21 +29) × 1.05 × 0.36 = 42
As a result, 340 W of heat goes out through the windows, and 42 W through the doors.
Step # 3 - determining the TP of the floor and ceiling
Obviously, the area of the ceiling and floor will be the same, and is calculated as follows:
Spol = Sptl = 6 × 8 = 48
We calculate the total thermal resistance of the floor, taking into account its structure.
Rpol = 0.1/0.13 + 0.05/0.047 + 0.05/0.58 + 0.06/0.043 = 0.77 + 1.06 + 0.17 + 1.40 = 3.4
Knowing that soil temperature tnar= + 5 and taking into account the coefficient l = 1, we calculate the floor Q:
Qpol = 48 × (21 – 5) × 1 × 3.4 = 2611
Rounding, we get that the heat loss of the floor is about 3 kW.
Determine the thermal resistance of the ceiling Rptl and its Q:
- Rptl = 0.025/0.21 + 0.05/0.14 + 0.05/0.043 = 0.12 + 0.71 + 0.35 = 1.18
- Qptl = 48 × (21 +29) × 1 × 1.18 = 2832
It follows that almost 6 kW leaves through the ceiling and floor.
Step # 4 - calculate ventilation TP
Indoor ventilation is organized, calculated by the formula:
Qv = 0.28 × Ln × pvnt × c × (tvnt - tnar)
Based on the technical characteristics, the specific heat transfer is 3 cubic meters per hour, that is:
Ln = 3 × 48 = 144.
To calculate the density, we use the formula:
pvnt = 353 / (273 + tvnt).
The calculated room temperature is +21 degrees.
Substituting the known values, we obtain:
pvnt = 353/(273+21) = 1.2
We substitute the figures obtained in the above formula:
Qv = 0.28 × 144 × 1.2 × 1.005 × (21 – 29) = 2431
Given TP for ventilation, the total Q of the building will be:
Q = 7000 + 6000 + 3000 = 16000.
Converting to kW, we obtain a total heat loss of 16 kW.
Features of the calculation of CBO
After finding the TP indicator, they proceed to hydraulic calculation (hereinafter - GR).
Based on it, information is obtained on the following indicators:
- the optimum diameter of the pipes, which, when the pressure drops, will be able to pass a given amount of coolant;
- coolant flow in a certain area;
- water speed;
- resistivity value.
Before starting the calculations, to simplify the calculations, they depict a spatial diagram of the system on which all its elements are arranged parallel to each other.
Consider the main stages of water heating calculations.
GR of the main circulation ring
The GR calculation methodology is based on the assumption that in all risers and branches the temperature differences are the same.
The calculation algorithm is as follows:
- In the diagram shown, taking into account heat loss, heat loads are applied to heating appliances, risers.
- Based on the scheme, choose the main circulation ring (hereinafter - HCC). The peculiarity of this ring is that in it the circulation pressure per unit length of the ring takes the least value.
- HCC is divided into sections with constant heat consumption. For each section indicate the number, thermal load, diameter and length.
In the vertical single-tube system, the ring through which the most loaded riser passes when the water flows in a dead end or along the mains passes through is taken as the fcc. We talked in more detail about linking circulation rings in a single-tube system and choosing the main one in the next article. We separately paid attention to the order of calculations, using a specific example for clarity.
In a horizontal system of a single-tube type, the fcc must have the lowest circulation pressure and a unit of ring length. For systems with natural circulation The situation is similar.
With GR risers of a vertical system of a single-tube type, flow-through, flow-adjustable risers, having unified nodes in their composition, are considered as a single circuit. For risers with closing sections, separation is made, taking into account the distribution of water in the pipeline of each instrument node.
Water consumption at a given site is calculated by the formula:
Gkont = (3.6 × Qkont × β1 × β2) / ((tr - t0) × c)
In the expression, alphabetic characters take the following meanings:
- Qkont - thermal load of the circuit;
- β1, β2 - additional tabular coefficients taking into account the heat transfer in the room;
- c - the heat capacity of water is 4.187;
- tr - water temperature in the supply line;
- t0 - water temperature in the return line.
Having determined the diameter and amount of water, it is necessary to know the speed of its movement and the value of the resistivity R. All calculations are most conveniently carried out using special programs.
GH of the secondary circulation ring
After GR of the main ring, the pressure in the small circulation ring formed through its closest risers is determined, taking into account that pressure losses can differ by no more than 15% with a deadlock and no more than 5% with a passing one.
If it is not possible to relate the pressure loss, install a throttle washer, the diameter of which is calculated using software methods.
Calculation of radiator batteries
Let's go back to the plan of the house located above. Through calculations, it was found that 16 kW of energy would be required to maintain the heat balance. In this house there are 6 premises for various purposes - a living room, a bathroom, a kitchen, a bedroom, a corridor, an entrance hall.
Based on the dimensions of the structure, you can calculate the volume V:
V = 6 × 8 × 2.5 = 120 m3
Next, you need to find the amount of thermal power per m3. To do this, Q must be divided by the found volume, that is:
P = 16000/120 = 133 W per m3
Next, you need to determine how much heat power is required for one room. In the diagram, the area of each room has already been calculated.
Define the volume:
- a bathroom – 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;
- the corridor – 4.10×2.5=10.25;
- hallway – 5.8×2.5=14.5.
In the calculations, you also need to consider rooms in which there are no heating batteries, for example, a corridor.
Determine the required amount of heat for each room, multiplying the volume of the room by the indicator R.
We get the required power:
- for the bathroom - 10.47 × 133 = 1392 W;
- for the living room - 34.58 × 133 = 4599 W;
- for 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 150 watts.
We calculate the required number of radiator batteries:
- a bathroom – 1392/150=10;
- living room – 4599/150=31;
- kitchen – 3136/150=21;
- bedroom – 3435/150=23;
- hallway – 1889/150=13.
Total required: 10 + 31 + 21 + 23 + 13 = 98 radiator batteries.
Our site also has other articles in which we examined in detail the procedure for performing thermal calculation of the heating system, step-by-step calculation of the power of radiators and heating pipes. And if your system assumes the presence of warm floors, then you will need to perform additional calculations.
All these issues are covered in more detail in our following articles:
- Thermal calculation of a heating system: how to correctly calculate the load on a system
- Calculation of heating radiators: how to calculate the required number and power of batteries
- Pipe volume calculation: calculation principles and calculation rules in liters and cubic meters
- How to make a calculation of a warm floor using the example of a water system
- Calculation of pipes for underfloor heating: types of pipes, methods and step of laying + calculation of flow
Conclusions and useful video on the topic
In the video you can see an example of calculating water heating, which is carried out by means of the Valtec program:
Hydraulic calculations are best carried out using special programs that guarantee high accuracy of calculations, take into account all the nuances of the design.
Do you specialize in calculating heating systems using water as a coolant and want to supplement our article with useful formulas, share professional secrets?
Or maybe you want to focus on additional calculations or point out inaccuracies in our calculations? Please write your comments and recommendations in the block under the article.