Power units
Power is measured in joules per second, or watts. Along with watts, horsepower is also used. Before the invention of the steam engine, the power of engines was not measured, and, accordingly, there were no generally accepted units of power. When the steam engine began to be used in mines, engineer and inventor James Watt began to improve it. In order to prove that his improvements made the steam engine more productive, he compared its power to the working capacity of horses, since horses have been used by people for many years, and many could easily imagine how much work a horse can do in a certain amount of time. In addition, not all mines used steam engines. On those where they were used, Watt compared the power of the old and new models of the steam engine with the power of one horse, that is, with one horsepower. Watt determined this value experimentally, observing the work of draft horses at the mill. According to his measurements, one horsepower is 746 watts. Now it is believed that this figure is exaggerated, and the horse cannot work in this mode for a long time, but they did not change the unit. Power can be used as a measure of productivity, as increasing power increases the amount of work done per unit of time. Many people realized that it was convenient to have a standardized unit of power, so horsepower became very popular. It began to be used in measuring the power of other devices, especially vehicles. Even though watts have been around for almost as long as horsepower, horsepower is more commonly used in the automotive industry, and it's clearer to many buyers when a car's engine power is listed in those units.
60 watt incandescent lamp
Calculation of heating radiators by area
The easiest way. Calculate the amount of heat required for heating, based on the area of \u200b\u200bthe room in which radiators will be installed. You know the area of \u200b\u200beach room, and the need for heat can be determined according to the building codes of SNiP:
- for an average climatic zone, 60-100W is required for heating 1m 2 of a dwelling;
- for areas above 60 o, 150-200W is required.
Based on these norms, you can calculate how much heat your room will require. If the apartment / house is located in the middle climatic zone, 1600W of heat (16 * 100 = 1600) will be required to heat an area of 16m 2. Since the norms are average, and the weather does not indulge in constancy, we believe that 100W is required. Although, if you live in the south of the middle climatic zone and your winters are mild, consider 60W.
Calculation of heating radiators can be done according to the norms of SNiP
A power reserve in heating is needed, but not very large: with an increase in the amount of power required, the number of radiators increases. And the more radiators, the more coolant in the system. If for those who are connected to central heating this is not critical, then for those who have or plan individual heating, a large volume of the system means large (extra) costs for heating the coolant and a large inertia of the system (the set temperature is maintained less accurately). And the logical question arises: “Why pay more?”
Having calculated the need for heat in the room, we can find out how many sections are required. Each of the heaters can emit a certain amount of heat, which is indicated in the passport. The found heat demand is taken and divided by the radiator power. The result is the required number of sections to make up for losses.
Let's count the number of radiators for the same room. We have determined that we need to allocate 1600W. Let the power of one section be 170W. It turns out 1600/170 \u003d 9.411 pieces.You can round up or down as you wish. You can round it into a smaller one, for example, in the kitchen - there are enough additional heat sources, and into a larger one - it is better in a room with a balcony, a large window or in a corner room.
The system is simple, but the disadvantages are obvious: the height of the ceilings can be different, the material of the walls, windows, insulation and a number of other factors are not taken into account. So the calculation of the number of sections of heating radiators according to SNiP is indicative. You need to make adjustments for accurate results.
Adjustment of results
In order to get a more accurate calculation, you need to take into account as many factors as possible that reduce or increase heat loss. This is what the walls are made of and how well they are insulated, how big the windows are, and what kind of glazing they have, how many walls in the room face the street, etc. To do this, there are coefficients by which you need to multiply the found values \u200b\u200bof the heat loss of the room.
The number of radiators depends on the amount of heat loss
Windows account for 15% to 35% of heat loss. The specific figure depends on the size of the window and how well it is insulated. Therefore, there are two corresponding coefficients:
- ratio of window area to floor area:
- 10% — 0,8
- 20% — 0,9
- 30% — 1,0
- 40% — 1,1
- 50% — 1,2
- glazing:
- three-chamber double-glazed window or argon in a two-chamber double-glazed window - 0.85
- ordinary two-chamber double-glazed window - 1.0
- conventional double frames - 1.27.
Walls and roof
To account for losses, the material of the walls, the degree of thermal insulation, the number of walls facing the street are important. Here are the coefficients for these factors.
- brick walls with a thickness of two bricks are considered the norm - 1.0
- insufficient (absent) - 1.27
- good - 0.8
The presence of external walls:
- indoors - no loss, factor 1.0
- one - 1.1
- two - 1.2
- three - 1.3
The amount of heat loss is influenced by whether the room is heated or not located on top. If there is a habitable heated room above (the second floor of the house, another apartment, etc.), the reducing factor is 0.7, if the heated attic is 0.9. It is generally accepted that an unheated attic does not affect the temperature in and (factor 1.0).
It is necessary to take into account the features of the premises and climate in order to correctly calculate the number of radiator sections
If the calculation was carried out by area, and the height of the ceilings is non-standard (a height of 2.7 m is taken as the standard), then a proportional increase / decrease using a coefficient is used. It is considered easy. To do this, divide the actual height of the ceilings in the room by the standard 2.7 m. Get the required coefficient.
Let's calculate for example: let the height of the ceilings be 3.0 m. We get: 3.0m / 2.7m = 1.1. This means that the number of radiator sections, which was calculated by the area for a given room, must be multiplied by 1.1.
All these norms and coefficients were determined for apartments. To take into account the heat loss of the house through the roof and basement / foundation, you need to increase the result by 50%, that is, the coefficient for a private house is 1.5.
climatic factors
You can make adjustments depending on the average temperatures in winter:
Having made all the required adjustments, you will get a more accurate number of radiators required for heating the room, taking into account the parameters of the premises. But these are not all the criteria that affect the power of thermal radiation. There are other technical details, which we will discuss below.
Reasons for translating
Power and current strength are the key characteristics necessary for the competent selection of protective devices for equipment powered by electricity. Protection is needed to prevent melting of the wiring insulation and breakage of the units.
It is clear that the lighting circuit, electric stove and coffee machine need devices with varying degrees of protection against short circuit and overheating. They require a different load to power them. For cables supplying current to devices, the cross section will also be different, i.e. capable of providing a specific type of equipment with the current of the power they require.
Each protective device must operate at the moment of a power surge that is dangerous for the protected type of equipment or a group of technical devices. This means that RCDs and automata should be selected so that during a threat to a low-power device, the network is not completely turned off, but only the branch for which this jump is critical.
On the cases of circuit breakers offered by the distribution network, a number is affixed indicating the value of the maximum allowable current. Naturally, it is indicated in Amps.
But on the electrical appliances that are required to protect these machines, the power they consume is indicated. This is where the need for translation comes in. Despite the fact that the units we are analyzing belong to different current characteristics, the connection between them is direct and rather close.
Voltage is called the potential difference, in other words, the work invested in moving a charge from one point to another. It is expressed in Volts. Potential - this is the energy at each of the points in which the charge is / was.
By current strength is meant the number of amperes passing through the conductor in a specific unit of time. The essence of power is to reflect the speed at which the charge was moving.
Power is expressed in Watts and Kilowatts. It is clear that the second option is used when a too impressive four- or five-digit figure needs to be reduced for ease of perception. To do this, its value is simply divided by a thousand, and the remainder is rounded up as usual.
To power powerful equipment, a higher energy flow rate is needed. The maximum allowable voltage for it is greater than for low-power equipment. The automata selected for it should have a higher trigger limit. Therefore, an accurate selection by load with a well-executed conversion of units is simply necessary.
Calculation of the number of radiators in a private house
If for apartments you can take the average parameters of the heat consumed, since they are designed for the standard dimensions of the room, then in private construction this is wrong. After all, many owners build their houses with ceiling heights exceeding 2.8 meters, in addition, almost all private premises are corner-shaped, so more power will be required to heat them.
In this case, calculations based on the area of \u200b\u200bthe room are not suitable: you need to apply the formula taking into account the volume of the room and make adjustments by applying the coefficients for reducing or increasing heat transfer.
The values of the coefficients are as follows:
- 0,2 - the resulting final power number is multiplied by this indicator if multi-chamber plastic double-glazed windows are installed in the house.
- 1,15 - if the boiler installed in the house is operating at the limit of its capacity. In this case, every 10 degrees of the heated coolant reduces the power of the radiators by 15%.
- 1,8 - the magnification factor to be applied if the room is corner, and there is more than one window in it.
To calculate the power of radiators in a private house, the following formula is used:
- V - the volume of the room;
- 41 - the average power required to heat 1 m2 of a private house.
Calculation example
If there is a room of 20 m2 (4 × 5 m - the length of the walls) with a ceiling height of 3 meters, then its volume is easy to calculate:
The resulting value is multiplied by the power accepted according to the norms:
60 × 41 \u003d 2460 W - so much heat is required to heat the area in question.
The calculation of the number of radiators is as follows (given that one section of the radiator emits an average of 160 W, and their exact data depends on the material from which the batteries are made):
Let's assume that you need 16 sections in total, that is, you need to purchase 4 radiators with 4 sections for each wall or 2 with 8 sections. In this case, one should not forget about the adjustment coefficients.
Calculation of the number of batteries per 1 m2
The area of each room where radiators will be installed can be found in the property documents or measured independently.The heat demand for each room can be found in building codes, where it is stated that for heating 1m2 in a certain area of residence, you will need:
- for harsh climatic conditions (temperature reaches below -60 0С) - 150-200 W;
- for the middle band - 60-100 watts.
To calculate, you need to multiply the area (P) by the value of the heat demand. Considering these data, as an example, we will give a calculation for the climate of the middle zone. To sufficiently heat a room of 16 m2, you need to apply the calculation:
The highest value of power consumption was taken, since the weather is changeable, and it is better to provide a small power reserve so that you do not freeze later in winter.
Next, the number of battery sections (N) is calculated - the resulting value is divided by the heat that one section emits. It is assumed that one section emits 170 W, based on this, the calculation is carried out:
It is better to round up - 10 pieces. But for some rooms it is more appropriate to round down, for example, for a kitchen that has additional heat sources. Then there will be 9 sections.
Calculations can be carried out according to another formula, which is similar to the above calculations:
- N is the number of sections;
- S is the area of the room;
- P - heat transfer of one section.
So, N=16/170*100, hence N=9.4
plan heating calculation
Published on 11/13/2014 | Author admin
In order to calculate any heating as accurately as possible, it is necessary to calculate the total heat loss of the house. But, speaking very approximately, the power of any main heating system is based on the calculated value of 100 W / m 2 of the heated area. As a rule, this power is laid with a margin of 15-20%. That is, the total (peak) heating power of a house with an area of 100 m 2 will be equal to: 12 kW (100 W * 1.2 * 100 m 2). Does this mean that the energy consumption of the infrared heating system will be 12 kWh? Not! Since the principle of operation of infrared heating is fundamentally different from traditional heating systems that use a coolant heated by a boiler (water or toxic antifreeze) and batteries to heat the air in the room.
Let us consider in detail the operation of an infrared heating system using the example of PLEN film electric heaters produced by ESB-Technologies. Suppose that in our house of 100 m 2 there are 5 rooms, 3 of which are on the 1st floor and 2 rooms on the second floor. The rooms have an area of 20 m 2 each. Therefore, on the first floor in each room it is necessary to install PLEN heaters with a capacity of: 20 m 2 * 120 W = 2.4 kW. Knowing that the specific power of PLEN is 175 W / m 2. it is easy to calculate that we need PLEN: 2 400 W / 175 W \u003d 13.71 m 2. That is, in each room on the first floor we place approximately 14 m 2 of PLEN, but it is better to take with a margin of 15 m 2. We get the coverage ratio: 15/20 = 75%. Finally, we have: 15 m 2 PLEN in each room and, accordingly, the peak power of the first floor: 15 m 2 * 175 W * 3 \u003d 7 875 W.
Will the consumption be 7.8 kWh? Definitely NO! Firstly, PLEN heaters operate under the control of thermostats that control the air temperature in the room, and to maintain the established comfortable temperature, they will be switched on periodically. From one hour, their work time will be approximately 10 minutes (depending on the heat loss of the house, that is, its insulation). Secondly, thermostats are installed in each separate room and are switched on independently of each other. In this case, we will take the inclusion non-synchronization coefficient as 0.7-0.8. That is, the peak load on the network at the time of switching on will be: 7.8 kW * 0.75 = 5.85 kW. This value is important for calculating the cross section of the supply cable. It follows from the above that with a load at the moment of switching on equal to 5.85 kW and an operating time of 10 min / h, the average hourly electricity consumption of the first floor will be: 5.85 kW / 60 * 10 \u003d 975 W / h. With an area of the first floor equal to 60 m 2, we obtain the specific energy consumption of the PLEN system: 975 W / 60 \u003d 16.25 W / m 2 of the heated area.
As for the second floor, it will be heated by more than half from the first floor, so the installed power of 70-80 W / m 2 of the heated area is enough for it. We get: 40 m 2 * 75 W = 3 kW. We divide this value by 175 W and get 17 m 2 PLEN. We take 18 m 2 for good measure (after all, we need to heat 2 rooms).In each room, we install 9 m 2 of PLEN, which is equal to 45% of the area of the heated room. Considering the coefficient of non-synchronization of the inclusion of thermostats and the fact that the second floor is heated by about 70-80% from the first, we get that the PLEN of the second floor will turn on only in severe frosts and then for a short time. Its specific energy consumption will be no more than 20-30% of the first floor and, accordingly, equal to 16.25 * 0.25 = 4 W / h per 1 m 2 of the heated area.
Let's calculate the total average hourly consumption of the PLEN heating system for the whole house:
- First floor: 16.25*60=975 W/h. Let's round this figure up to 1 kW / h.
- Second floor: 4*40=160 W/h. Let's round it up to 200 Wh.
- In total, we get 1.2 kW / h.
At a tariff of 2 rubles / kW, the average heating costs will be: 1.2 kW * 2 rubles * 24 hours * 30.5 days = 1,756.8 rubles per month. Of course, this is an average amount, which will vary depending on the outside temperature and the value set on the thermostat.
Posted in Articles
Consumers of electricity in the house
The Decree of the Government of the Russian Federation No. 334 “On improving the procedure for technical connection of consumers to electrical networks” dated April 21, 2009 states that an individual can connect up to 15 kW to his house. Based on this figure, we will make a calculation, but how many kilowatts for the house will be enough for us. To calculate, you need to know how much electricity each electrical appliance in the house consumes.
Table of power of household electrical appliances
The table of power of household electrical appliances shows the approximate figures for electricity consumption. Energy consumption depends on the power of the devices and the frequency of their use.
| Electrical appliance | Power consumption, W |
| Appliances | |
| Electric kettle | 900-2200 |
| coffee machine | 1000-1200 |
| Toaster | 700-1500 |
| Dishwasher | 1800–2750 |
| Electric stove | 1900–4500 |
| Microwave | 800–1200 |
| Electric meat grinder | 700–1500 |
| Refrigerator | 300–800 |
| Radio | 20–50 |
| TV set | 70–350 |
| Music Center | 200–500 |
| Computer | 300–600 |
| Oven | 1100–2500 |
| electric lamp | 10–150 |
| Iron | 700–1700 |
| air purifier | 50–300 |
| Heaters | 1000–2500 |
| A vacuum cleaner | 500–2100 |
| Boiler | 1100–2000 |
| Instantaneous water heater | 4000–6500 |
| hair dryer | 500–2100 |
| washing machine | 1800–2700 |
| Air conditioner | 1400–3100 |
| Fan | 20–200 |
| power tools | |
| Drill | 500–1800 |
| Perforator | 700–2200 |
| Circular saw | 700–1900 |
| Electric planer | 500– 900 |
| Electric jigsaw | 350– 750 |
| Grinding machine | 900–2200 |
| A circular saw | 850–1600 |
Let's do a little calculation based on the data in the table of power consumption of household electrical appliances. For example, in our house there will be a minimum set of electrical appliances: lighting (150 W), refrigerator (500 W), microwave (1000 W), washing machine (2000 W), TV (200 W), computer (500 W), iron (1200 W), vacuum cleaner (1200 W), dishwasher (2000 W). In total, these devices will consume 8750 W, and given that these devices will almost never turn on at once, the received power can be divided in half.
Power in sports
It is possible to evaluate work using power not only for machines, but also for people and animals. For example, the power with which a basketball player throws a ball is calculated by measuring the force she applies to the ball, the distance the ball has traveled, and the time that force has been applied. There are websites that allow you to calculate work and power during exercise. The user selects the type of exercise, enters the height, weight, duration of the exercise, after which the program calculates the power. For example, according to one of these calculators, the power of a person with a height of 170 centimeters and a weight of 70 kilograms, who did 50 push-ups in 10 minutes, is 39.5 watts. Athletes sometimes use devices to measure the amount of power a muscle is working during exercise. This information helps determine how effective their chosen exercise program is.
Dynamometers
To measure power, special devices are used - dynamometers. They can also measure torque and force.Dynamometers are used in various industries, from engineering to medicine. For example, they can be used to determine the power of a car engine. To measure the power of cars, several main types of dynamometers are used. In order to determine the power of the engine using dynamometers alone, it is necessary to remove the engine from the car and attach it to the dynamometer. In other dynamometers, the force for measurement is transmitted directly from the wheel of the car. In this case, the car's engine through the transmission drives the wheels, which, in turn, rotate the rollers of the dynamometer, which measures the power of the engine under various road conditions.
This dynamometer measures the torque as well as the power of the car's powertrain.
Dynamometers are also used in sports and medicine. The most common type of dynamometer for this purpose is isokinetic. Usually this is a sports simulator with sensors connected to a computer. These sensors measure the strength and power of the entire body or individual muscle groups. The dynamometer can be programmed to give signals and warnings if the power exceeds a certain value
This is especially important for people with injuries during the rehabilitation period, when it is necessary not to overload the body.
According to some provisions of the theory of sports, the greatest sports development occurs under a certain load, individual for each athlete. If the load is not heavy enough, the athlete gets used to it and does not develop his abilities. If, on the contrary, it is too heavy, then the results deteriorate due to overload of the body. Physical activity during some activities, such as cycling or swimming, depends on many environmental factors, such as road conditions or wind. Such a load is difficult to measure, but you can find out with what power the body counteracts this load, and then change the exercise scheme, depending on the desired load.
Article author: Kateryna Yuri
Power of household electrical appliances
On household electrical appliances, the power is usually indicated. Some lamps limit the power of the bulbs that can be used in them, for example, no more than 60 watts. This is because higher wattage bulbs generate a lot of heat and the bulb holder can be damaged. And the lamp itself at a high temperature in the lamp will not last long. This is mainly a problem with incandescent lamps. LED, fluorescent and other lamps generally operate at lower wattage at the same brightness and if used in luminaires designed for incandescent lamps there are no wattage problems.
The greater the power of the electrical appliance, the higher the energy consumption and the cost of using the appliance. Therefore, manufacturers are constantly improving electrical appliances and lamps. The luminous flux of lamps, measured in lumens, depends on the power, but also on the type of lamps. The greater the luminous flux of the lamp, the brighter its light looks. For people, it is high brightness that is important, and not the power consumed by the llama, so recently alternatives to incandescent lamps have become increasingly popular. Below are examples of types of lamps, their power and the luminous flux they create.
How many kilowatts is needed to heat a house
The main consumers of electricity in homes are lighting, cooking, heating and hot water.
During the cold period, it is important to pay attention to the heating of the house. Electric heating in the house can be of several types:
- water (batteries and boiler);
- purely electric (convector, warm floor);
- combined (warm floor, batteries and boiler).
Let's look at electric heating options and electricity consumption.
- Heating with a boiler. If you plan to install an electric boiler, then the choice should fall on a three-phase boiler.The boiler system equally divides the electrical load into phases. Manufacturers produce boilers with different capacities. To choose it correctly, you can make a simplified calculation, divide the area of \u200b\u200bthe house by 10. For example, if the house has an area of \u200b\u200b120 m2, then a 12 kW boiler will be needed for heating. To save on electricity, you need to establish a two-tariff mode of using electricity. Then at night the boiler will work at an economical rate. Also, in addition to the electric boiler, you need to install a buffer tank, which will accumulate warm water at night and distribute it to heating appliances during the day.
- Convector heating. As a rule, convectors are installed under windows and connected directly to a power outlet. Their number should correspond to the presence of windows in the room. Experts recommend calculating the total amount for the power consumption of all heating devices and equally distributing it over all three phases. For example, heating of one floor can be connected to the first one. To another phase, the entire second floor. To the third phase, attach the kitchen and bathroom.Today, the convectors have advanced features. So you can set the desired temperature and choose the time for heating. To save money, you can set the time and date of the convector. The device is equipped with the possibility of a “multi-tariff”, which includes a heater, at the required power or at a reduced rate (after 23:00 and before 08:00). The energy calculation for convectors is similar to the boiler in the previous paragraph.
- Heating with underfloor heating. A very convenient option for heating, as you can set the desired temperature for each room. It is not recommended to install a warm floor at the place where furniture, a refrigerator, as well as a bathroom are installed. As calculations show, a house of 90 m2 with an installed convector and underfloor heating, on one floor, consumes from 5.5 to 9 kW of electricity.
