Already gave the method of estimating the capacity of the heater, but polls are approximate, suitable only for solving minor problems. Today we decided to reveal the difficulty of the assumption that the losses linearly depend on the temperature difference on both sides of the surface. These are walls, windows, doors, smooth surface of the convector heater. Accordingly, changing the power of heaters, designed to compensate for leaks. This assumption is consistent with the SNiPs, which already contains scientific formulas. How to calculate the power of the heater, if there is no information about radiators, material, wall structure and other parameters useful to professionals( and omitted in the literature).
Heat loss and heater power
Assuming that the temperature in the apartment house is the same, heat does not flow through the floor and interior walls. In the presence of a basement or attic, the reader will have to supplement our findings with their own. Losses through the wall facing the street depend on the difference in temperature in the room and outside. We present a graph in the form of a line, and the slope is determined by the power of the battery, which is unknown in advance.
We see the dependence of heat loss Q on the temperature on the street t, where for convenience the difference between the external and internal temperatures is given. It can be seen that the dependence is linear, and at 20 ºC outside the window the heat losses are zero, and at –40 ºC they are 2x, in real terms, the indicators may vary. This is a typical situation when a philistine is faced with the difficulties of calculating the power of a heater for a room. We conduct consideration under the assumption that the room temperature is 20 ºC( typical value in the documentation, including building codes and instrument manuals).
Assume that at an outdoor temperature of -10 ºC, the battery heats up so that the room is exactly 20 ºC, necessary by theory. Next step:
- The temperature outside the window drops to -14 ºС.
- Put inside an additional oil heater 1.5 kW.
- Temperature returns to normal( 20 ° C).
- In this case, any 4 ºС differences are equal to heat losses of 1.5 kW.
From this case, we calculate the rated power of the battery at the measured temperature under conditions when the indoor climate is set( 20 ºС).
Central heating radiators counterbalance the difference losses of 30 ° C, which means that the energy delivered by the devices is( 30/4) x 1.5 kW = 11.25 kW.
Now know what to do if the degrees behind the window fall to -40.Additional heaters will be required with a total power equal to 11.25 kW of radiators. Notice, we do not take into account the heat released by people: during the experience the room is empty. Or, conversely, sit there family. Then the found 11.25 kW will be equal to the total power of people and batteries at a temperature of 20 ºС.
Summarizing the calculation of the power of heaters for an arbitrary case
But the difficulty is different: there is a certain temperature of the battery, room, street, and you need to calculate the power of the heater. Now we will try to solve this problem, without waiting for the establishment outside the window -14 ºC.For example, in a room of 20 ºC, but it is necessary to find the power of the batteries in order to approximate the result for an arbitrary case of weather and boiler conditions. Here you need to know that the power of the battery depends on the temperature difference between the room and the surface of the radiator. So, we bring a 1.5 W oil heater into the house and see that the room temperature has risen to 23 ºC.This is a bit much, but does not play a special role. It will also be necessary to measure the power of the batteries( outside the window, by agreement, -10 ºC).Suppose a radiator on a surface is 60 ºC.This is a typical value for Europe, in Russia and hotter central heating is in theory, and practitioners give up to 36 degrees Celsius in the cold.
This is interesting! In Russia, the practice of laying pipes with hot water, sometimes heating on the surface of the earth. Not too expensive. Then in good faith they wear thermal insulation, but children, passers-by who are forced to step over, tear off their fur coats.
Let the rated power of the battery at 20 ºС be equal to N, at the same time the difference between the surface and room temperatures is equal to 40. In the new conditions the power will decrease to 37N / 40.We obtain the equality:
( 37N / 40 + 1.5) - N = 3 ºС.
An increase in consumption of 1.5 kW( and a decrease in heating output) gave a temperature increase of 3 ºС.It turns out that 1.5 - 0.075N kW gives an increase of 3 ºС.And from the operating point at normal temperature inside( 20 ºС) to zero( outside temperature 20 ºС) there is a length of 30 ºС.It turns out that:
N = 10( 1.5 - 0.075N), we find the desired value. It turned out about 8.57 kW.This is battery power. Now, knowing the rating, let's build the characteristics for arbitrary temperatures of the battery, the room and the street. For example, in the winter -14 ºC, central heating does not pull, you need to bring the situation back to normal( 20 ºC) indoors. Notice, do not indicate the temperature in the room.8.57 kW are equal to 30 ºC on the temperature scale, so add 8.57 / 30 x 4 kW = 1.15 kW.This means that it is necessary to calculate the power of the oil heater so that it goes no less than this figure, but it is not necessary to exceed the value too much so as not to get out of the desired climate zone. Consequently, we go to the store and take the device with three modes, the first should produce 1.15 kW of heat.
Devices more useful in the cold. For example, at -40 ºC you will need twice as much as radiators give out, which is 17 kW.For electrical distribution panel at the site. Install a gas convector with a coaxial line breaking the wall out. Hybrid options are also possible: part of the blow will be taken by the Underfloor heating, and the rest will fall on gas heating. We believe that now readers understand how to calculate the power of the convector heater.
An example of power calculation for the generalized case of
Suppose there is the same room, but inside 17ºС, and the surface of the radiator, for example, 55 ºС.Outside the frost is -10 ºС, and we achieve the nominal room value( 20 ºС), and the temperature of the central heating radiator is at worst 50 ºC.Let us find the maximum power of the heater, which stretched the worst described case at a temperature outside the window of -30 ºС.First of all, we find the battery power at a surface temperature of 55 ºС and a room temperature of 17 ºС.Already shown how to act in this case, now we will show in practice. We take an oil heater of 1.5 kW, wait until the room enters the mode, and measure the temperature difference. Let for simplicity we get the same 3 ºC.According to the schedule we find the required proportion:
( 1.5 +( 55 - 20) /( 55 - 17) N) - N = 3 ºC.
From the operating point to the intersection of the graph with the horizontal axis, the distance in degrees is 27. The result is:
N = 9( 1.5 - 0.078N), from where we find watts. It turned out 7.9 kW.This is the power of the central heating radiator at a temperature difference of 38 ° C( between the surface of the battery and the room).In the worst case, this differentiation will be less and will be 30. The resulting power will decrease proportionally and will be 6.23 kW.We build the graph for this case in the same way as in the picture. We recall the value of heat loss at 27ºС with a zero point. This is 7.9 kW.We bring the problem to the one solved above, for which we find heat losses at -10 ºC outside and at room temperature 20 ºС.It turns out 30 ºС difference. Therefore, we divide 7.9 by 0.9 and we get 8.77 kW.To keep the room at a given level in these conditions, we add to the batteries the difference( 8.77 - 6.23) = 2.54 kW.
At a temperature outside the window -30 º C conditions will become tougher. We solve the problem as shown above to search for the result. Concerning the already existing heat losses of 8.77 kW, an additional 2/3 of the specified number will be added, amounting to 5.78 kW.The total power of the heaters will exceed the energy of the radiators and will be 5.78 + 2.54 kW = 8.32 kW.It is clear that due to electricity, this result is unlikely, therefore, an infrared fireplace with blue fuel or a similar device is required.
Now, similarly, readers will calculate the power of an infrared heater of any type. The only thing is that the story was conducted in such a way as to warm up the room, but if it is required to give heat exclusively to a specific sector, divide the area by the floor meter and multiply the number in watts by a factor less than one. Get a more modest number. They say that infrared heaters help to save. It is more difficult to calculate the power of a gas heater, since it also heats by convection. In this case, it is necessary to properly position the equipment to obtain the proper effect. For orientation, we use the algorithm given above as a starting point for further research.
Calculations allow an error, but it is realistic to estimate the power required for an apartment. It is important to wait for the temperature to reach the mode, to make measurements as accurately as possible.
