All summer red gossips in soft moors sang and danced, and now, when the cold comes, you have to take pencils in your hands. After all, "heating, as there was no, and no". And it is necessary to present at least some arguments of the heating network, calculating the heat received from it, for which it was, after all, "sealed".

When you need to dot the "i"

But a quite reasonable question arises: "How to calculate what is invisible and capable of disappearing in an instant, literally through the window." You should not despair of this struggle with air, it turns out that there are quite intelligible mathematical calculations of the received calories for heating.

Moreover, all these calculations are hidden in the official documents of public utilities. As usual in these institutions, there are several such documents, but the main one is the so-called "Rules for accounting for heat energy and heat carrier". It is he who will help to solve the question - how to calculate Gcal for heating.

Actually, the problem can be solved quite simply and no calculations will be needed if you have a meter of not just water, but hot water... The readings of such a meter are already "crammed" with data on the received heat. By taking readings, you multiply it by the cost rate and get the result.

Basic formula

The situation becomes more complicated if you do not have such a counter. Then you have to be guided by the following formula:

Q = V * (T1 - T2) / 1000

In the formula:

• Q is the amount of heat energy;
• V is the volume of hot water consumption in cubic meters or tons;
• T1 is the hot water temperature in degrees Celsius. More precisely, in the formula, use the temperature, but reduced to the corresponding pressure, the so-called "enthalgy". But in the absence of a better - a corresponding sensor, we simply use the temperature, which is close to enthalgy. Professional heat metering units are able to calculate exactly the enthalgy. Often this temperature is not available for measurement, therefore, they are guided by the constant "from the ZhEKA", which can be different, but usually is 60-65 degrees;
• T2 - temperature cold water in degrees Celsius. This temperature is taken in the cold water pipe of the heating system. Consumers, as a rule, do not have access to this pipeline, therefore it is customary to take constant recommended values ​​depending on the heating season: during the season - 5 degrees; out of season - 15;
• Ratio “1000” allows you to get rid of 10-digit numbers and get the data in gigacalories (not just calories).

As follows from the formula, it is more convenient to use a closed heating system, into which the required volume of water is poured once and in the future it does not flow. But in this case, you are prohibited from using hot water from the system.

Usage closed system forces us to slightly improve the above formula, which already takes the form:

Q = ((V1 * (T1 - T)) - (V2 * (T2 - T))) / 1000

• V1 is the flow rate of the coolant in the supply pipeline, and regardless of whether the coolant is water or steam;
• V2 is the flow rate of the coolant in the return pipeline;
• T1 is the temperature of the coolant at the inlet, in the supply pipeline;
• T2 is the temperature of the coolant at the outlet, in the return pipeline;
• T is the cold water temperature.

Thus, the formula consists of the difference between two factors - the first gives the value of the incoming heat in calories, the second - the value of the output heat.

Helpful advice! As you can see, there is not a lot of mathematics, but calculations still have to be done. Of course, you can immediately rush to your calculator on your mobile phone. But he advises you to create simple formulas in one of the most famous computer office programs - the so-called Microsoft Excel spreadsheet, included in the Microsoft Office package. In Excel, you can not only quickly calculate everything, but also "play" with the initial data, simulate various situations. Moreover, Excel will help you with the construction of graphs of receipt - consumption of heat, and this is a "non-kill" map for a future possible conversation with government bodies.

Alternative options

How exist different ways providing housing with heat by choosing a heat carrier - water or steam, so there are alternative methods for calculating the heat received. Here are two more formulas:

• Q = ((V1 * (T1 - T2)) + (V1 - V2) * (T2 - T)) / 1000

• Q = ((V2 * (T1 - T2)) + (V1 - V2) * (T1 - T)) / 1000

Thus, calculations can be done with your own hands, but it is important to coordinate your actions with the calculations of heat supplying organizations. Their calculation instructions may be completely different from yours.

Helpful advice! Often reference books provide information not in the national system of units of measurement, to which calories belong, but in the international system "C". Therefore, we advise you to remember the conversion factor of kilocalories to kilowatts. It is equal to 850. In other words, 1 kilowatt is equal to 850 kilocalories. From here it is no longer difficult to translate gigacalories, considering that 1 gigacalorie is a million calories.

All counters, and not only the simplest brownies, unfortunately suffer from some measurement error. This is a normal situation, if, of course, the error does not exceed all conceivable limits. To calculate the error (relative, in percent), a special formula is also used:

R = (V1 - V2) / (V1 + V2) * 100,

• V1 and V2 are the previously considered indicators of the coolant flow rate, and
• 100 - conversion factor to percent.

The permissible percentage of error in calculating heat is considered to be no more than 2 percent, given that the error of measuring instruments is no more than 1 percent. You can, of course, do with the old proven method, here you don't really need to do any calculations.

Presentation of the received data

The price of all calculations is your confidence in the adequacy of your financial costs to the heat received from the state. Although, in the end, you still will not understand what gcal is in heating. In all honesty, let's say that in many ways this is the magnitude of our sense of self and attitude to life. Certainly, you need to have some base "in numbers" in your head. And it is expressed in what is considered a good norm, when your formula gives 3 Gcal per month for an apartment of 200 square meters. Thus, if the heating season lasts 7 months - 21 Gcal.

But all these values ​​are rather difficult to imagine “in the shower” when warmth is really needed. All these formulas and even the results they give you correctly will not warm you up. They will not explain to you why, even with 4 gcal per month, you still feel warm. And the neighbor has only 2 gcal, but he does not boast and constantly keeps the window open.

There can be only one answer - his atmosphere is also warmed by the warmth of those around him, and you have no one to cuddle with, although "the room is full of people." He gets up in the morning at 6 and runs in any weather to exercise, and you lie to the last under the covers. Warm yourself from the inside, hang a photo of the family on the wall - everyone in summer swimsuits on the beach in Foros, watch more often the video of the last climb to Ai-Petri - everyone is naked, hot, then you won't even feel a couple of hundred calories outside.

Heat energy has several measurement options.

Energy power, which is measured in watts (W, mW and kW), is most often indicated by heating boilers, heaters, etc.

Another unit of energy measurement, gigocalorie (Gcal), can be encountered when installing heat meters.

Also, the supplied heat is sometimes indicated in Gcal, in payment receipts.

And if the calculation is accepted by the management company in one unit, and the meter shows another, it may be necessary to convert Gcal to kW and vice versa on a monthly basis. Having figured out everything once, you can learn how to do it quickly and easily.

When constructing buildings, all measurements and heat engineering calculations are made in giga calories. Utilities also prefer this unit of measure for its closeness to real life and the ability to compute on an industrial scale.

From the school course, we remember that a calorie is the work that is needed to heat 1 gram of water per unit ° C (at a certain atmospheric pressure).

In life, one has to deal with Kcal and Gcal, gigacalorie.

• 1 Kcal = 1,000 Cal.
• 1 Gcal = 1 million Kcal, or 1 Billion. cal.

In receipts for heating, the measurement can be used:

• Gcal;
• Gcal / hour.

In the first case, we mean the supplied heat for a certain period (this can be a month, a year or a day). Gcal / hour is a characteristic of the power of a device or a process (such a unit of measurement can report the performance of a heating device or the rate of heat loss of a building in winter). In receipts, heat is meant, which was released in 1 hour. Then, to recalculate for a day, you need to multiply the number by 24, and for a month by another 30/31.

1 Gcal / hour = 40 m 3 of water, which was heated to 25 ° C in 1 hour.

Also, a gigacalorie can be tied to the volume of fuel (solid or liquid) Gcal / m3. And it shows how much heat can be obtained from a cubic meter of this fuel.

How to translate energy units?

On the Internet, you can really find a huge number of online calculators that convert the required values ​​automatically.

When it comes to figuring out everything, long formulas and proportions are often offered that can repel the average consumer who graduated from school many years ago.

But it is possible to understand everything! You will need to memorize 1 or 2 numbers, an action, and you can easily do offline translation yourself.

How to convert kW to Gcal / h

Key metric for converting data from kilowatts to calories:

1 kW = 0.00086 Gcal / hour

To find out how many Gcal is obtained, you need to multiply the available number of kW by a constant value, 0.00086.

Let's look at an example. Suppose you need to convert 250 kW to calories.

250 kW x 0.00086 = 0.215 Gcal / hour.

(More accurate online calculators will show 0.214961).

Is it the heating season and the batteries are still cold? Don't look for ways to warm up on your own, demand that your rights be respected. Follow the link for information on where to call and what to do if there is no heating.

Conversion of Gcal to kW / h

The opposite situation, when you need to convert Gcal to kW. You need to know how many kW contains 1 Gcal

1 Gcal = 1163 kW.

This means that one gigacalorie of heat will need to be spent to obtain 1163 kilowatts of energy.

Or vice versa: 1163 kW of energy will be required to obtain one Gcal of heat.

To convert the number of gigocalories you know to kilowatts, you need to multiply the available Gcal by 1163.

0.5 x 1163 = 581.5 kW.

Translation table

A quick translation of round numbers can be done using tables:

Conclusion

So, to make it easier to carry out monthly transfers of heat units, you need to remember a couple of numbers and the action that you need to perform with them.

If there is a reading in kilowatts, it must be multiplied by 0.00086 and you get it in giga calories.

And when the readings are taken in gigacalories, you need to multiply them by 1163 and kilowatts will come out.

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1 kilocalorie (IT) per hour [kcal / h] = 0.001163 kilowatt [kW]

Initial value

Converted value

watt exawatt petawatt terawatt gigawatt megawatt kilowatt hectowatt deciwatt deciwatt sanewatt milliwatt microwatt nanowatt picowatt femtowatt attowatt horsepower horsepower metric horsepower boiler horsepower electrical horsepower pumping horsepower British horsepower British horsepower thermal unit (int.) per hour Brit. thermal unit (IT) per minute Brit. thermal unit (IT) per second Brit. thermal unit (thermochemical) per hour Brit. thermal unit (thermochemical) per minute Brit. thermal unit (thermochemical) per second MBTU (international) per hour Thousand BTU per hour MBTU (international) per hour Million BTU per hour ton of refrigeration kilocalorie (IT) per hour kilocalorie (IT) per minute kilocalorie (IT) second kilocalorie (term) per hour kilocalorie (term) per minute kilocalorie (term) per second calorie (IT) per hour calorie (IT) per minute calorie (IT) per second calorie (term) per hour calorie (therm) per minute calorie (therm) per second foot pound-force per hour foot lbf / minute foot lbf / second pound-foot per hour pound-foot per minute pound-foot per second erg per second kilovolt-ampere volt-ampere newton-meter per second joule per second exjoule per second petajoule per second terajoule per second gigajoule per second megajoule per second kilojoule per second hectojoule per second decjoule per second decijoule per second centijoule per second microjoule per second per second nanojoule per second picojoule per second femtojoule per second attojoule per seconds joule per hour joule per minute kilojoule per hour kilojoule per minute Planck power

Thermal efficiency and fuel efficiency

More about power

General information

In physics, power is the ratio of work to the time it takes to do it. Mechanical work is a quantitative characteristic of the action of force F on the body, as a result of which it moves a distance s... Power can also be defined as the rate at which power is transmitted. In other words, power is a measure of the health of a machine. By measuring the power, you can understand how much and at what speed the work is being done.

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 efficient, he compared its power to the performance of horses, since horses have been used by people for many years, and many could easily imagine how much work a horse could do in a given amount of time. In addition, steam engines were not used in all mines. 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 by observing the work of draft horses at a 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 an indicator of productivity, since as power increases, the amount of work performed per unit of time increases. Many realized that it was convenient to have a standardized power unit, so horsepower became very popular. It began to be used to measure the power of other devices, especially transport. Despite the fact that watts are used almost as long as horsepower, in automotive industry more often horsepower is used, and many buyers understand more clearly when these units indicate the power of an automobile engine.

Household electrical appliances power

Household appliances are usually marked with wattage. Some luminaires 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 luminaire with the socket may be damaged. And the lamp itself at a high temperature in the lamp will not last long. This is mainly a problem with incandescent bulbs. LED, fluorescent and other lamps usually operate at lower wattage at the same brightness and, if used in luminaires designed for incandescent lamps, there is no power problem.

The more the power of the 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 wattage, but also on the type of lamp. The higher the luminous flux of the lamp, the brighter its light looks. For people, it is the high brightness that is important, and not the power consumed by the lamp, therefore, in recent years, alternatives to incandescent lamps are becoming more and more popular. Below are examples of lamp types, their wattage and the luminous flux they generate.

• 450 lumens:
• Incandescent lamp: 40 watts
• Compact fluorescent lamp: 9-13 watts
• LED lamp: 4-9 watts
• 800 lumens:



• Incandescent lamp: 60 watts
• Compact fluorescent lamp: 13-15 watts
• LED lamp: 10-15 watts
• 1600 lumens:
• Incandescent lamp: 100 watts
• Compact fluorescent lamp: 23-30 watts
• LED lamp: 16-20 watts

From these examples, it is obvious that with the same generated luminous flux, LED lamps consume the least energy and are more economical than incandescent lamps. At the time of this writing (2013) the price is LED lamps many times higher than the price of incandescent lamps. Despite this, some countries have banned or are about to ban the sale of incandescent lamps due to their high power.

The power of household electrical appliances may differ depending on the manufacturer, and is not always the same during the operation of the appliance. Below are the approximate capacities of some household appliances.



• Household air conditioners for cooling a residential building, split system: 20-40 kilowatts
• Monoblock window air conditioners: 1-2 kilowatts
• Ovens: 2.1-3.6 kilowatts
• Washers and dryers: 2-3.5 kilowatts
• Dishwashers: 1.8-2.3 kilowatts
• Electric kettles: 1-2 kilowatts
• Microwaves: 0.65-1.2 kilowatts
• Refrigerators: 0.25-1 kilowatts
• Toasters: 0.7-0.9 kilowatts

Power in sports

Performance can be judged by power not only for machines, but also for people and animals. For example, the power at which a basketball player throws the ball is calculated by measuring the force she applies to the ball, the distance the ball flew, and the time that force was applied. There are sites that allow you to calculate the work and power during physical exercise... The user selects the type of exercise, enters height, weight, exercise duration, after which the program calculates the power. For example, according to one of these calculators, the power of a person who is 170 centimeters tall and weighs 70 kilograms, who did 50 push-ups in 10 minutes, is 39.5 watts. Athletes sometimes use devices to measure the power at which muscles are 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 technology to medicine. For example, they can be used to determine the power of a car engine. Several basic types of dynamometers are used to measure the power of vehicles. In order to determine the engine power using dynamometers alone, it is necessary to remove the engine from the car and connect it to the dynamometer. In other dynamometers, the force to be measured is transmitted directly from the wheel of the vehicle. In this case, the car's engine drives the wheels through the transmission, which, in turn, rotate the rollers of the dynamometer, which measures the engine power under various road conditions.

Dynamometers are also used in sports and medicine. The most common type of dynamometer for this purpose is isokinetic. Typically, this is a sensor-based gym equipment connected to a computer. These sensors measure the strength and power of the entire body or specific muscle groups. The dynamometer can be programmed to issue alarms and warnings if the power has exceeded 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 at 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 severe, then the results deteriorate due to the overload of the body. Exercise during some exercise, such as cycling or swimming, is influenced by many factors the environment such as road conditions or wind conditions. Such a load is difficult to measure, however, you can find out with what power the body resists this load, and then change the exercise pattern, depending on the desired load.

Do you find it difficult to translate a unit of measurement from one language to another? Colleagues are ready to help you. Post a question to TCTerms and you will receive an answer within a few minutes.

This article is the seventh publication of the series "Myths of Housing and Public Utilities", dedicated to debunking. Myths and false theories, widespread in the housing and communal services of Russia, contribute to the growth of social tension, the development of "" between consumers and performers utilities leading to extremely negative consequences in the housing industry. The articles of the cycle are recommended, first of all, for consumers of housing and communal services (HUS), however, specialists in housing and communal services may also find something useful in them. In addition, the dissemination of publications from the series "Myths of Housing and Utilities" among consumers of housing and communal services can contribute to a deeper understanding of the housing and communal services sector by residents of apartment buildings, which leads to the development of constructive interaction between consumers and providers of utilities. A complete list of articles in the cycle "Myths of housing and communal services" is available

**************************************************

This article discusses a somewhat unusual question, which, nevertheless, as practice shows, worries quite a significant part of the consumers of utilities, namely: why is the unit of measurement for the standard of consumption of utilities for heating is "Gcal / sq. Meter"? Failure to understand this issue led to the advancement of an unreasonable hypothesis that the alleged unit of measurement for the rate of heat energy consumption for heating was chosen incorrectly. The considered assumption leads to the emergence of some myths and false theories of the housing sector, which are refuted in this publication. Additionally, the article explains what is a communal heating service and how this service is technically provided.

The essence of false theory

It should be noted right away that the incorrect assumptions analyzed in the publication are relevant for cases where there are no heating meters - that is, for those situations when it is used in the calculations.

It is difficult to clearly formulate false theories following from the hypothesis about the wrong choice of the unit for measuring the heating consumption rate. The consequences of such a hypothesis are, for example, statements:
⁃ « The volume of the coolant is measured in cubic meters, heat energy is in gigacalories, which means that the standard for heating consumption should be in Gcal / cubic meter!»;
⁃ « Utility heating is consumed to heat the space of an apartment, and this space is measured in cubic meters, not square meters! It is illegal to use the area in the calculations, the volume must be used!»;
⁃ « Fuel for preparing hot water used for heating can be measured either in units of volume (cubic meter) or in units of weight (kg), but not in units of area (square meter). The standards are calculated illegally, incorrectly!»;
⁃ « It is absolutely unclear in relation to what area the standard is calculated - to the area of ​​the battery, to the cross-sectional area of ​​the supply pipeline, to the area land plot, on which the house stands, to the area of ​​the walls of this house or, perhaps, to the area of ​​its roof. It is only clear that it is impossible to use the area of ​​premises in the calculations, since in a multi-storey building the premises are located one above the other, and in fact their area is used in calculations many times - approximately as many times as there are floors in the building».

Various conclusions can follow from the above statements, some of which boil down to the phrase “ It's all wrong, I won't pay", And in addition to the same phrase, some also contain some logical arguments, among which the following can be distinguished:
1) since the denominator of the unit of measurement of the standard indicates a lower degree of magnitude (square) than it should be (cube), that is, the applied denominator is less than the one to be applied, then the value of the standard, according to the rules of mathematics, is overestimated (the smaller the denominator of the fraction, the greater the value the fraction itself);
2) an incorrectly selected standard unit of measurement implies additional mathematical actions before being substituted into formulas 2, 2 (1), 2 (2), 2 (3) of Appendix 2 of the Rules for the provision of utilities to owners and users of premises in apartment buildings and residential houses approved by the PP of the Russian Federation of 05/06/2011 N354 (hereinafter - Rules 354) values ​​NT (standard for the consumption of utility services for heating) and TT (tariff for thermal energy).

As such preliminary transformations, actions that do not stand up to criticism are proposed, for example * :
⁃ The NT value is equal to the square of the standard approved by the subject of the Russian Federation, since the denominator of the unit of measurement indicates “ square meter";
⁃ The TT value is equal to the product of the tariff by the standard, that is, TT is not a tariff for heat energy, but a certain specific cost of heat energy consumed for heating one square meter;
⁃ Other transformations, the logic of which could not be comprehended at all, even when trying to apply the most incredible and fantastic schemes, calculations, theories.

Since an apartment building consists of a set of residential and non-residential premises and places common use(common property), while the common property on the basis of the right of common share ownership belongs to the owners of individual premises of the house, the entire volume of heat energy entering the house is consumed by the owners of the premises of such a house. Consequently, the payment for the heat consumed for heating should be made by the owners of the apartment buildings. And here the question arises - how to distribute the cost of the entire volume of heat energy consumed by an apartment building among the owners of the premises of this apartment building?

Guided by quite logical conclusions that the consumption of heat energy in each specific room depends on the size of such a room, the Government of the Russian Federation has established a procedure for distributing the amount of heat energy consumed by the whole house among the premises of such a house in proportion to the area of ​​these premises. This is provided for both by Rules 354 (the distribution of readings of a general house heating meter in proportion to the shares of the area of ​​premises of specific owners in the total area of ​​all premises of the house in the property), and Rules 306 when establishing a standard for heating consumption.

Clause 18 of Appendix 1 to Regulation 306 provides:
« 18. The standard of consumption of utility services for heating in residential and non-residential premises (Gcal per 1 sq. M of the total area of ​​all residential and non-residential premises in apartment building or a residential building per month) is determined by the following formula (formula 18):

where:
- the amount of heat energy consumed during one heating period by apartment buildings that are not equipped with collective (common building) heat energy metering devices, or residential buildings, not equipped with individual metering devices for heat energy (Gcal), determined by formula 19;
- the total area of ​​all residential and non-residential premises in apartment buildings or the total area of ​​residential buildings (sq. M);
- a period equal to the duration of the heating period (the number of calendar months, including incomplete ones, in the heating period)».

Thus, it is precisely this formula that stipulates that the standard for the consumption of utility services for heating is measured precisely in Gcal / sq. Meter, which, among other things, is directly established by subparagraph "e" of paragraph 7 of Rules 306:
« 7. When choosing a unit of measurement of standards for the consumption of utilities, the following indicators are used:
f) with regard to heating:
in residential premises - Gcal per 1 sq. meter the total area of ​​all premises in an apartment building or residential building».

Based on the foregoing, the standard for the consumption of utility services for heating is equal to the amount of heat consumed in an apartment building for 1 square meter the area of ​​premises owned per month during the heating period (when choosing a payment method, it is applied evenly throughout the year).

Calculation examples

As indicated, we will give an example of calculation according to the correct method and according to the methods proposed by false theoreticians. To calculate the cost of heating, we will accept the following conditions:

Suppose that the heating consumption standard is approved at 0.022 Gcal / sq. Meter, the heat energy tariff is approved at 2500 rubles / Gcal, the area of ​​the i-th room is assumed to be 50 square meters. To simplify the calculation, we will accept the conditions that payment for heating is carried out, and there is no technical possibility in the house of installing a general-house metering device for heat energy for heating.

In this case, the amount of payment for the utility service for heating in the i-th residential building not equipped with an individual metering device for heat energy and the amount of payment for the utility service for heating in i-th residential or non-residential premises in an apartment building, which is not equipped with a collective (common building) metering device for heat energy, when making payment during the heating period, it is determined by formula 2:

Pi = Si× NT× TT,

where:
Si is the total area of ​​the i-th premise (residential or non-residential) in an apartment building or the total area of ​​a residential building;
NT is the standard for the consumption of communal heating services;
TT is the heat tariff established in accordance with the legislation of the Russian Federation.

The following calculation will be correct (and widely used) for the example under consideration:
Si = 50 square meters
NT = 0.022 Gcal / square meter
TT = 2500 RUB / Gcal

Pi = Si × NT × TT = 50 × 0.022 × 2500 = 2750 rubles

Let's check the calculation by dimensions:
"Square meter"× "Gcal / sq. Meter"× × "Rub. / Gcal" = ("Gcal" in the first factor and "Gcal" in the denominator of the second factor are reduced) = "rub."

The dimensions are the same, the cost of the Pi heating service is measured in rubles. The resulting calculation result: 2750 rubles.

Now let's calculate using the methods proposed by pseudo-theoreticians:

1) The NT value is equal to the square of the standard approved by the subject of the Russian Federation:
Si = 50 square meters
NT = 0.022 Gcal / square meter × 0.022 Gcal / square meter = 0.000484 (Gcal / square meter) ²
TT = 2500 RUB / Gcal

Pi = Si × NT × TT = 50 × 0.000484 × 2500 = 60.5

As can be seen from the presented calculation, the cost of heating was equal to 60 rubles 50 kopecks. The attractiveness of this method lies precisely in the fact that the cost of heating is not 2750 rubles, but only 60 rubles 50 kopecks. How correct is this method and how correct is the calculation result obtained from its application? To answer this question, it is necessary to carry out some transformations admissible by mathematics, namely: we will carry out the calculation not in giga calories, but in mega calories, respectively, transforming all the values ​​used in the calculations:

Si = 50 square meters
NT = 22 Mcal / square meter × 22 Mcal / square meter = 484 (Mcal / square meter) ²
TT = 2.5 rubles / Mcal

Pi = Si × NT × TT = 50 × 484 × 2.500 = 60500

And what do we get as a result? The cost of heating is already 60,500 rubles! We note right away that if the correct method is applied, mathematical transformations should not affect the result in any way:
(Si = 50 square meters
NT = 0.022 Gcal / square meter = 22 Mcal / square meter
TT = 2500 rubles / Gcal = 2.5 rubles / Mcal

Pi = Si× NT× TT = 50× 22 × 2.5 = 2750 rubles)

And if, in the method proposed by pseudo-theoreticians, the calculation is carried out not even in megacalories, but in calories, then:

Si = 50 square meters
NT = 22,000,000 cal / sqm × 22,000,000 cal / sqm = 484,000,000,000,000 (cal / sqm) ²
TT = 0.0000025 rubles / cal

Pi = Si × NT × TT = 50 × 484,000,000,000,000 × 0.0000025 = 60,500,000,000

That is, heating a room with an area of ​​50 square meters costs 60.5 billion rubles a month!

In fact, of course, the considered method is incorrect, the results of its application do not correspond to reality. Additionally, we will check the calculation by dimensions:

"Square meter"× "Gcal / sq. Meter"× "Gcal / sq. Meter"× "RUB / Gcal" = ("square meter" in the first factor and "square meter" in the denominator of the second factor are reduced) = "Gcal"× "Gcal / sq. Meter"× "RUB / Gcal" = ("Gcal" in the first factor and "Gcal" in the denominator of the third factor are reduced) = "Gcal / square meter"× "rub."

As you can see, the dimension "rub." as a result, it does not work, which confirms the incorrectness of the proposed calculation.

2) The value of TT is equal to the product of the tariff approved by the constituent entity of the Russian Federation by the consumption standard:
Si = 50 square meters
NT = 0.022 Gcal / square meter
TT = 2500 rubles / Gcal × 0.022 Gcal / square meter = 550 rubles / square meter

Pi = Si × NT × TT = 50 × 0.022 × 550 = 60.5

The calculation according to the specified method gives exactly the same result as the first incorrect method considered. The second applied method can be refuted in the same way as the first: convert gigacalories to mega- (or kilo-) calories and check the calculation in terms of dimensions.

conclusions

The myth of the wrong choice " Gcal / sq. Meter»As a unit of measurement of the standard of consumption of utility services for heating is refuted. Moreover, the consistency and validity of the use of just such a unit of measurement has been proven. The incorrectness of the methods proposed by pseudo-theoreticians has been proven, their calculations are refuted by the elementary rules of mathematics.

It should be noted that the overwhelming majority of false theories and myths of the housing sector are aimed at proving that the amount of payment presented to owners for payment is overstated - this fact contributes to the "vitality" of such theories, their dissemination and the growth of their supporters. It is quite reasonable that consumers of any service want to minimize their costs, but attempts to use false theories and myths do not lead to any savings, but are aimed only at, at introducing into the minds of consumers the idea that they are being deceived, they are being unjustifiably charged with money funds. Obviously, the courts and supervisory bodies authorized to deal with conflict situations between the contractors and consumers of public services will not be guided by false theories and myths, therefore, there will be no savings and no other positive consequences either for the consumers themselves or for other participants in housing relations. maybe.

Let's start with the concepts of "work" and "power". Work is part of the internal energy expended by a person or a machine over a period of time. In the process of such work, a person or a machine warms up, generating heat. Therefore, both internal energy and the amount of released or absorbed heat, as well as work, are measured in the same units - joules (J), kilojoules (kJ) or megajoules (MJ).

The faster work is done or heat is released, the more intensively the internal energy is consumed. By a measure of this intensity is the power, measured in watts(W), kilowatts (kW), megawatts (MW), and gigawatts (GW). Power is the work done per unit of time (whether it be the work of a motor or the work of an electric current). Thermal power is the amount of heat transferred per unit of time to the heat carrier (water, oil) from the combustion of fuel (gas, fuel oil) in the boiler.

Calorie was introduced back in 1772 Swedish experimental physicist Johann Wilke as a unit of measurement for heat. Currently, a unit multiple of a calorie, a gigacalorie (Gcal), is actively used in such spheres of life as utilities, heating systems and heat power engineering. Its derivative is also used - a gigacalorie per hour (Gcal / h), which characterizes the rate of heat release or heat absorption by one or another equipment. Now let's try to calculate what one calorie is equal to.

Back in school, in physics lessons, we were taught that in order to heat any substance, it must be given a certain amount of heat. There was even such a formula Q = c * m * ∆t, where Q means an unknown amount of heat, m is the mass of the heated substance, c is the specific heat of this substance, and ∆t is the temperature difference by which the substance is heated. So, a calorie is called a non-systemic unit of heat, defined as "the amount of heat spent on heating 1 gram of water per 1 degree Celsius at an atmospheric pressure of 101325 Pa."

Since heat is measured in joules, then using the above formula, we find out what is 1 calorie (cal) in joules... To do this, take from the physics reference book the value of the specific heat capacity of water under normal conditions (atmospheric pressure p = 101325 Pa, temperature t = 20 ° C): c = 4183 J / (kg * ° C). Then one calorie will be equal to:

• 1 cal = 4183 [J / (kg * ° C)] * 0.001 kg * 1 ° C = 4.183 J.

However, the calorie value depends on the heating temperature, so its value is not constant. For practical purposes, the so-called international calorie, or simply calorie, is used, which is 4.1868 J.

Memo 1

• 1 cal = 4.1868 J, 1 kcal = 1000 cal, 1 Gcal = 1 billion cal = 4186800000 J = 4186.8 MJ;
• 1 J = 0.2388 cal, 1 MJ = 1 million J = 238845.8966 cal = 238.8459 kcal;
• 1 Gcal / h = 277777.7778 cal / s = 277.7778 kcal / s = 1163000 J / s = 1.163 MJ / s.

Gigacalories or kilowatts

Let's finally figure out what is the difference between these units of measurement. Suppose we have a heating device, for example, a kettle. Take 1 liter of cold tap water (temperature t1 = 15 ° C) and boil it (heat it to t2 = 100 ° C). Electric power kettle - P = 1.5 kW. How much heat will the water absorb? To find out, we apply the familiar formula, while taking into account that the mass of 1 liter of water m = 1 kg: Q = 4183 [J / (kg * ° C)] * 1 kg * (100 ° C-15 ° C) = 355555 J = 84922.8528 cal≈85 kcal.

How long does it take for a kettle to boil? Let all the energy of the electric current go to heating the water. Then we will find the unknown time using the energy balance: "The energy consumed by the kettle is equal to the energy absorbed by the water (excluding losses)." The energy consumed by the kettle during the time τ is equal to P * τ. The energy absorbed by water is equal to Q. Then, based on the balance, we obtain P * τ = Q. Hence, the heating time of the kettle will be: τ = Q / P = 355555 J / 1500 W ≈ 237 s ≈ 4 min. The amount of heat transferred by the kettle to the water per unit of time is its thermal power. In our case, it will amount to Q / τ = 84922.8528 cal / 237 s≈358 cal / s = 0.0012888 Gcal / h.

Thus, kW and Gcal / h are units of power, and Gcal and MJ are units of heat and energy. How can such calculations be applied in practice? If we receive a receipt for the payment of heating, then we pay for the heat that the supplying organization supplies us through the pipes. This heat is taken into account in gigacalories, that is, in the amount of heat consumed by us during the billing period. Do I need to convert this unit to joules? Of course not, because we're just paying for a specific number of gigacalories.

However, it is often necessary to choose certain heating devices for a house or apartment, for example, an air conditioner, a radiator, a boiler or a gas boiler. In this connection, it is required to know in advance the thermal power required to heat the room. Knowing this power, you can select the appropriate device. It can be specified both in kW and in Gcal / h, as well as in BTU / h units (British Thermal Unit - British Thermal Unit, h - hour). The following guide will help you convert kW to Gcal / h, kW to BTU / h, Gcal to kWh and BTU to kWh.

Memo 2

• one W = one J / s = 0.2388459 cal / s = 859.8452 cal / h = 0.8598 kcal / h;
• one kW = one kJ / s = 1000 J / s = 238.8459 cal / s = 859845.2279 cal / h = 0.00085984523 Gcal / h;
• one MW = one MJ / s = one million J / s = 1000 kW = 238845.8966 cal / s = 0.85984523 Gcal / h;
• one Gcal / h = one billion cal / h = 1163000 W = 1163 kW = 1.163 MW = 3968156 BTU / h;
• one BTU / h = 0.2931 W = 0.0700017 cal / s = 252.0062 cal / h = 0.2520062 kcal / h;
• one W = 3.412 BTU / h, one kW = 3412 BTU / h, one MW = 3412000 BTU / h.

How is BTU / h defined and what is it used for? 1 BTU is the amount of heat required to heat 1 pound of water 1 ° Fahrenheit (° F). This unit is mainly used to indicate the heat output of installations such as air conditioners.

Calculation examples

So we come to the most important thing. How to convert one value to another using the above ratios? It's not all that difficult. Let's look at some examples.

Example 1

The thermal power of the boiler is 30 kW. What is its equivalent power, expressed in Gcal / h?

Solution. Since 1 kW = 0.00085984523 Gcal / h, then 30 kW = 30 * 0.00085984523 Gcal / h = 0.0257953569 Gcal / h.

Example 2

It is estimated that an air conditioner with a capacity of at least 2.5 kW is required to cool an office. An air conditioner with a capacity of 8000 BTU / h was chosen for the purchase. Is there enough air conditioner capacity to cool the office?

Solution. Since 1 BTU / h = 0.2931 W, then 8000 BTU / h = 2344.8 W = 2.3448 kW. This value is less than the calculated 2.5 kW, therefore the selected air conditioner is not suitable for installation.

Example 3

The heat supply organization supplied 0.9 Gcal of heat per month. How much power does a radiator need to install so that it produces the same amount of heat per month?

Solution. Suppose that heat was supplied to the house evenly within one month (30 days), therefore, the heat output supplied by the boiler house can be found by dividing the total amount of heat by the number of hours in a month: P = 0.9 Gcal / (30 * 24 h) = 0.00125 Gcal / h. This power in terms of kilowatts will be equal to P = 1163 kW * 0.00125 = 1.45375 kW.

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