Thermal balance of a steam boiler. Boiler efficiency
Various types boilers have different Efficiency range from 85 to 110%. When choosing boiler equipment, many buyers are interested in how efficiency can exceed 100% and how it is calculated.
In case of electric boilers Efficiency really cannot be higher than 100%. Only boilers running on combustible fuel can have a higher coefficient.
If you remember the school chemistry course, it turns out that with the complete combustion of any fuel, what remains is CO 2 - carbon and H 2 O - water vapor containing energy. During condensation, the energy of the steam increases, that is, additional energy is generated. Based on this, the calorific value of fuel is divided into two concepts: higher and lower specific heat of combustion.
Lowest- represents the heat obtained during the combustion of fuel, when water vapor, along with the energy contained in them, enters the external environment.
Higher calorific value is heat taking into account the energy contained in water vapor.
Officially (in any regulatory documents) Efficiency, both in Russia and in Europe, calculated at the lowest specific heat combustion. But if you still use the heat contained in water vapor, and the calculations are based on the lowest specific heat of combustion, then in this case figures appear that exceed 100%.
Boilers that use the heat of condensation of water vapor are called condensation. And they have an efficiency exceeding 100%.
The difference between the lower and higher heating values of fuel combustion is about 11%. This value is the limit by which the efficiency of boilers can differ.
Main settings
Efficiency can be calculated using two parameters. In Europe, efficiency is usually calculated based on the temperature of the exhaust gases. For example, when burning a kilogram of fuel, a certain amount of kilocalories of heat is obtained, provided that the temperature of the exhaust gases and the ambient temperature are equal.
By measuring the difference between the ambient temperature and the actual temperature of the exhaust gases, it is possible to calculate the boiler efficiency from it.
Roughly speaking, the waste gases escaping into the chimney are subtracted from 100% to arrive at the actual figure.
Calculate correctly
In the USSR, and later in Russia, a fundamentally different calculation method was adopted - the so-called “ reverse balance method" It consists in the fact that heat consumption is determined by the lower calorific value. Then, a heater is placed on the pipe, and the amount of thermal energy that has gone into it is calculated, that is, the amount of energy loss. To calculate efficiency, energy losses are calculated from the total amount of heat.
This approach when determining efficiency gives more accurate indicators. It was adopted as a calculation method because all the bodies of Russian boilers were very poorly thermally insulated, which is why up to 40% of the energy escaped through the walls of the boiler. According to requirements regulatory documents, in Russia it is still customary to consider efficiency using the reverse balance method. Today, this method can be successfully applied to multi-megawatt boilers operating in thermal power plants whose burners never turn off.
Advantages of modern boilers
But this technique is completely inapplicable to modern boilers, since they have a fundamentally different operating scheme. Since the burners of modern boilers operate in automatic mode: they work for 15 minutes, and then stop for 15 minutes until the generated heat is used. The higher the outside temperature, the longer the burner will “stand” and work less. Naturally, in this case we cannot talk about a reverse balance.
Another difference between modern boilers is the presence of thermal insulation. Large manufacturers produce the highest quality units, with better thermal insulation. Heat loss through the walls of such a boiler is no more than 1.5-2%. Buyers often forget about this, believing that the boiler will also heat the room by releasing heat during operation. When purchasing a modern boiler, it is worth remembering that it is not intended for heating a boiler room, and, if necessary, take care of installing heating radiators.
Modern heat preservation technologies
A good steel boiler always has higher efficiency. This is due to the fact that cast iron boilers, unlike steel ones, always have more technological limitations.
Moreover, thanks to the insulation, modern boilers retain heat perfectly. Even two days after it is turned off, the temperature of the boiler body drops by only 20-25 degrees.
The best examples of imported heating equipment are boiler units in which all requirements are correctly taken into account. Therefore, you should not try to “reinvent the wheel” and assemble a boiler from improvised means. After all, you already have a wide selection of the most modern, diverse and carefully thought-out boiler options that will work for a long time and properly, more than meeting all the expectations placed on them and, what is especially pleasant, saving your costs!
Our specialists will help you choose boiler and related equipment and advise on technical issues!
Contact sales department by phone:
General equation of heat balance of a boiler unit
The ratio connecting the heat input and consumption in a heat generator constitutes its heat balance. The goals of compiling a heat balance of a boiler unit are to determine all incoming and outgoing balance sheet items; efficiency calculation boiler unit, analysis of balance sheet expenditure items in order to establish the reasons for the deterioration of the boiler unit.
In a boiler unit, when fuel is burned, the chemical energy of the fuel is converted into thermal energy combustion products. The released heat of the fuel is used to generate useful heat contained in steam or hot water and to cover heat losses.
In accordance with the law of conservation of energy, there must be equality between the incoming and outgoing heat in the boiler unit, i.e.
For boiler installations, the heat balance is per 1 kg of solid or liquid fuel or 1 m 3 of gas under normal conditions ( 
). The items of income and consumption in the heat balance equation have dimensions MJ/m 3 for gaseous and MJ/kg for solid and liquid fuels.
The heat from fuel combustion entering the boiler unit is also called available heat, it is denoted by .In the general case entrance part The heat balance is written as:
where is the lowest calorific value of solid or liquid fuel per working mass, MJ/kg;
Lower calorific value of gaseous fuel per dry weight, MJ/m 3 ;
Physical heat of fuel;
Physical heat of air;
Heat introduced into the furnace of a boiler with steam.
Let us consider the components of the incoming part of the heat balance. In the calculations, the lowest working heat of combustion is accepted if the temperature of the combustion products leaving the boiler is higher than the condensation temperature of water vapor (usually tg = 110...120 0 C). When cooling combustion products to a temperature at which condensation of water vapor is possible on the heating surface, calculations should be performed taking into account the higher calorific value of fuel combustion
The physical heat of the fuel is equal to:
Where With t – specific heat capacity of the fuel, 
for fuel oil and 
for gas;
t t – fuel temperature, 0 C.
When entering the boiler, solid fuel usually has a low temperature, approaching zero, therefore Q f.t. is small in importance and can be neglected.
To reduce viscosity and improve atomization, fuel oil (liquid fuel) enters the furnace heated to a temperature of 80...120 0 C, so its physical heat is taken into account when performing calculations. In this case, the heat capacity of fuel oil can be determined by the formula:
Accounting Q f.t. is carried out only when burning gaseous fuel with a low calorific value (for example, blast furnace gas) provided it is heated (up to 200...300 0 C). When burning gaseous fuels with a high calorific value (for example, natural gas) there is an increased ratio of the mass of air and gas (approximately 10 1). In this case, the fuel - gas is usually not heated.
Physical heat of air Q f.v. is taken into account only when it is heated outside the boiler due to an external source (for example, in a steam heater or in an autonomous heater when additional fuel is burned in it). In this case, the heat introduced by the air is equal to:
where is the ratio of the amount of air at the entrance to the boiler (air heater) to the theoretically necessary one;
The enthalpy of the theoretically required air heated before the air heater, 
:
,
here the temperature of the heated air in front of the air heater of the boiler unit is 0 C;
Enthalpy of theoretically required cold air, :
The heat introduced into the boiler furnace with steam during steam atomization of fuel oil is taken into account in the form of the formula:
Where G p – steam consumption, kg per 1 kg of fuel (for steam spraying of fuel oil G n = 0.3…0.35 kg/kg);
h n – steam enthalpy, MJ/kg;
2.51 is the approximate value of the enthalpy of water vapor in the combustion products leaving the boiler unit, MJ/kg.
In the absence of heating of fuel and air from external sources, the available heat will be equal to:
The consumption part of the heat balance includes usefully used heat Q floor in the boiler unit, i.e. heat expended to generate steam (or hot water), and different heat losses, i.e.
Where Q u.g. – heat loss with exhaust gases;
Q h.n. , Q m.s. – heat loss from chemical and mechanical incomplete combustion of fuel;
Q But. – heat loss from external cooling of the external enclosures of the boiler;
Q f.sh. – loss of slag with physical heat;
Q acc. – consumption (sign “+”) and supply (sign “-”) of heat associated with the unsteady thermal operating conditions of the boiler. At steady thermal state Q acc. = 0.
So the general equation for the heat balance of a boiler unit at steady state thermal mode can be written as:
If both sides of the presented equation are divided by and multiplied by 100%, we get:
Where 
components of the expenditure part of the heat balance, %.
3.1 Heat loss from flue gases
Heat loss with flue gases occurs due to the fact that the physical heat (enthalpy) of gases leaving the boiler at a temperature t u.g. , exceeds the physical heat of the air entering the boiler α u.g. and fuel With T t t. The difference between the enthalpy of exhaust gases and the heat entering the boiler with air from the environment α u.g. , represents the heat loss with exhaust gases, MJ/kg or (MJ/m 3):
.
Heat loss with flue gases usually occupies the main place among the heat losses of the boiler, amounting to 5...12% of the available heat of the fuel. These heat losses depend on the temperature, volume and composition of combustion products, which, in turn, depend on the ballast components of the fuel:
The ratio characterizing the quality of the fuel shows the relative yield of gaseous combustion products (at α = 1) per unit heat of combustion of the fuel and depends on the content of ballast components (moisture) in it W r and ash A r for solid and liquid fuels, nitrogen N 2, carbon dioxide CO 2 and oxygen ABOUT 2 for gaseous fuel). With an increase in the content of ballast components in the fuel, and, consequently, the loss of heat with exhaust gases increases accordingly.
One of the possible ways to reduce heat loss with flue gases is to reduce the coefficient of excess air in the flue gases α ug, which depends on the coefficient of air flow in the furnace and the ballast air sucked into the boiler flues, which are usually under vacuum:
Possibility of reduction α , depends on the type of fuel, the method of its combustion, the type of burners and the crushing device. At favorable conditions By mixing fuel and air, the excess air required for combustion can be reduced. When burning gaseous fuel, the excess air coefficient is taken to be 1.1, when burning fuel oil = 1.1...1.15.
Air suction through the gas path of the boiler can, in the limit, be reduced to zero. However, complete sealing of the places where pipes pass through the lining, sealing of hatches and peepholes is difficult and practically = 0.15..0.3.
Ballast air in combustion products in addition to increasing heat loss Q u.g. also leads to additional energy costs for the smoke exhauster.
Another important factor influencing the value Q t.g., is the temperature of the flue gases t u.g. . Its reduction is achieved by installing heat-using elements (economizer, air heater) in the tail part of the boiler. The lower the temperature of the exhaust gases and, accordingly, the smaller the temperature difference between the gases and the heated working fluid (for example, air), the larger the heating surface area is required to cool the combustion products.
An increase in the temperature of the flue gases leads to an increase in losses from Q u.g. and, consequently, to additional fuel costs to produce the same amount of steam or hot water. Due to this optimal temperature t u.g. is determined on the basis of technical and economic calculations when comparing the finished capital costs for the construction of a heating surface and fuel costs (Fig. 3.).
In addition, when the boiler is operating, the heating surfaces may become contaminated with soot and fuel ash. This leads to a deterioration in the heat exchange of combustion products with the heating surface. At the same time, in order to maintain a given steam output, it is necessary to increase fuel consumption. The drift of heating surfaces also leads to an increase in the resistance of the gas path of the boiler. In this regard, to ensure normal operation of the unit, systematic cleaning of its heating surfaces is required.
3.2 Heat loss from chemical incomplete combustion
Heat loss from chemical incomplete combustion (chemical underburning) occurs when fuel is incompletely burned within the combustion chamber and flammable gaseous components appear in the combustion products - CO, H2, CH4, CmHn, etc. the afterburning of these combustible gases outside fireboxes are almost impossible due to their relatively low temperature.
The causes of chemical incomplete combustion may be:
general lack of air;
poor mixture formation, especially on initial stages fuel combustion;
· low temperature in the combustion chamber, especially in the area of fuel combustion;
· insufficient residence time of fuel within the combustion chamber, during which chemical reaction combustion cannot be completed completely.
If there is a sufficient amount of air for complete combustion of the fuel and good mixture formation, the losses depend on the volumetric density of heat release in the furnace, MW/m3:
Where IN– fuel consumption, kg/s;
V t – volume of the firebox, m3.
 Rice. 14.9 Dependence of heat loss on chemical incompleteness of combustion q x.n, %, from the volumetric density of heat release in the furnace qv, MW/m 3 . | The nature of the dependence is presented in Fig. 4. . In the area of low values (left side of the curve), i.e. at low fuel consumption B, losses increase due to a decrease in the temperature level in the combustion chamber. An increase in the volumetric density of heat release (with an increase in fuel consumption) leads to an increase in the temperature level in the furnace and a decrease |
However, upon reaching a certain level with a further increase in fuel consumption (the right side of the curve), losses begin to increase again, which is associated with a decrease in the residence time of gases in the furnace volume and, therefore, the impossibility of completing the combustion reaction.
The optimal value at which losses are minimal depends on the type of fuel, the method of its combustion and the design of the furnace. For modern combustion devices, heat loss from chemical incomplete combustion is 0...2% at 
.when burning solid and liquid fuels:
when burning gaseous fuel:
When developing measures to reduce the value, it should be borne in mind that if conditions exist for the appearance of products incomplete combustion First of all, CO is formed as the most difficult to burn component, and then H 2 and other gases. It follows from this that if there is no CO in the combustion products, then there is no H 2 in them.
Coefficient useful action boiler unit
Efficiency factor boiler unit is the ratio of useful heat consumed to produce steam (or hot water) to the available heat of the boiler unit. However, not all the useful heat generated by the boiler unit is sent to consumers; part of the heat is spent on its own needs. Taking this into account, the efficiency of a boiler unit is distinguished by the heat generated (efficiency - gross) and by the heat released (efficiency - net).
The difference between the generated and released heat is used to determine the consumption for auxiliary needs. Not only heat is consumed for its own needs, but also electrical energy (for example, to drive a smoke exhauster, fan, feed pumps, fuel supply mechanisms), i.e. consumption for own needs includes the consumption of all types of energy spent on the production of steam or hot water.
So, gross efficiency of a boiler unit characterizes the degree of its technical perfection, and net efficiency characterizes commercial profitability.
Efficiency - gross boiler unit can be determined either by the direct balance equation or by the reverse balance equation.
According to the direct balance equation:
For example, in the production of water vapor, the useful heat used is ( see question 2) :
Then 
From the presented expression, you can obtain a formula for determining the required fuel consumption, kg/s (m 3 /s):
According to the reverse balance equation:
The determination of gross efficiency using the direct balance equation is carried out mainly when reporting for a separate period (ten-day, month), and according to the reverse balance equation - when testing boiler units. Calculating efficiency using reverse balance is much more accurate, since the errors in measuring heat losses are smaller than in determining fuel consumption.
Net efficiency is determined by the expression:
where is the energy consumption for own needs, %.
Thus, to improve the efficiency of boiler units, it is not enough to strive to reduce heat losses; it is also necessary to reduce in every possible way the costs of heating and electrical energy for own needs, which amount on average to 3...5% of the heat available to the boiler unit. The efficiency of the boiler unit depends on its load. To build the dependence, you need to subtract sequentially from 100% all losses of the boiler unit, which depend on the load, i.e. 
The efficiency of a boiler unit or the efficiency of a boiler unit is the ratio of the amount of heat used in the boiler unit to the amount of heat expended in the fuel. Part of the steam produced in the boiler unit is directly spent on its own needs, for example, on feed pumps, blower fans, smoke exhausters, and blowing heating surfaces. Taking these costs into account, the concept is introduced Boiler unit net efficiency.
Heat used in the boiler unit to produce steam or hot water,
Where IN - hourly fuel consumption, kg/h (m3/h);
D- hourly productivity of the boiler unit, kg/hour;
q k.a - the amount of heat transferred to water in the boiler unit to convert it into steam or to produce hot water and referred to 1 kg of steam or water, kJ/kg (kcal/kg);
ŋ k.a - efficiency of the boiler unit.
For a boiler unit that produces saturated steam
Where i" - enthalpy of saturated steam;
i p.v - enthalpy of feed water;
q pr- amount of heat removed from the boiler unit with blowdown water, kJ/kg (kcal/kg); usually q pr= (0.01-0.02) · i", Where i" - heat content of water at temperature t n.
For a hot water boiler unit that produces hot water
Where i 1 - enthalpy of water entering the boiler; i 2 is the enthalpy of water leaving the boiler.
If the amount of steam produced and its enthalpy are known, as well as the hourly fuel consumption and the heat of combustion of the fuel, then the efficiency of the boiler unit can be determined, %:
For modern boiler units the value q 1, depending on the steam output of the boiler unit, the temperature of the exhaust gases, the type of fuel burned and the method of its combustion, can vary within a very wide range from 75 to 80% for boiler units of small capacity, in which solid fuel is burned in layered furnaces, and up to 91-95 % for large boiler units with flaring fuel combustion. The highest efficiencies are obtained for boiler units operating on liquid and gaseous fuels.
For boiler units of small capacity, heat loss ranges from 20 to 25%, and for large ones from 5 to 9%. The main heat losses are losses with flue gases q 2
Example.
Determine the efficiency of the boiler unit and estimate the heat losses of the boiler unit with a steam capacity of Q = 10 tons/hour with steam parameters: pressure P= 1.4 MPa (14 kgf/cm2) and temperature t = 197.3°C. Hourly fuel consumption 1500 kg, feed water temperature 100°C, fuel combustion heat Q p n = 20647 kJ/kg (4916 kcal/kg). The heat losses of the boiler unit are assessed using the average values given in the relevant sections. Sizeq PR ( amount of heat removed from the boiler unit with blowdown water) take equal to 0.According to the table and specified steam parameters: pressure R and temperature t we find its enthalpy ~ 2790 kJ/kg (666 kcal/kg). At 100°C the heat content of the feed water will be approximately 419 kJ/kg (100 kcal/kg). Therefore, the heat received by 1 kg of steam according to the formula isq To
. A= 2790 - 419 = 2371 kJ/kg ( q To . a = 666 - 100 = 566 kcal/kg).The efficiency of the boiler unit according to the formula
The amount of heat loss
Σ q i = 100 - ŋ k.a = 100 - 76.8 = 23.2%. Based on averages q 2 ,q 3 , q 4 given in § Heat balance of the boiler unit, we find q 2 = 12,5%, q 3 = 1%, q 4 = 6.25%. Consequently, the amount of losses in environment q 5 = Σ qi- q 2 - q 3 - q 4 = 23,2 - 12,5 - 1 - 6,25 = 3,45%. ,BOILER EFFICIENCY COEFFICIENT
(Boiler efficiency) - the ratio of the amount of heat transferred to the boiler water to convert it into steam during combustion 1 kg fuel, to value calorific value fuel, i.e. the amount of heat that is released when complete combustion 1 kg fuel. The efficiency of boilers reaches values of the order of 0.60-0.85.
Samoilov K. I. Marine Dictionary. - M.-L.: State Naval Publishing House of the NKVMF of the USSR, 1941
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Create a cozy and comfortable atmosphere in country house quite simple - you just need to properly equip the heating system. The main component of an efficient and reliable heating system is the boiler. In the article below, we will talk about how to calculate the efficiency of a boiler, what factors influence it, and how to increase the efficiency of heating equipment in a particular home.
How to choose a boiler
Of course, in order to determine how efficient a particular hot water boiler will be, it is necessary to determine its efficiency (efficiency factor). This indicator represents the ratio of the heat used to heat the room to the total amount of thermal energy generated.
The formula for calculating efficiency looks like this:
ɳ=(Q 1 ÷Q ri),
where Q 1 is heat used efficiently;
Q ri – total amount of heat released.
What is the relationship between boiler efficiency and load
At first glance it may seem that the more fuel is burned, the better the boiler works. However, this is not quite true. Dependence of efficiency the boiler from the load manifests itself just the opposite. The more fuel is burned, the more thermal energy is released. At the same time, the level of heat loss also increases, since chimney highly heated flue gases escape. Consequently, fuel is consumed inefficiently.
The situation develops in a similar way in cases where the heating boiler operates at reduced power. If it falls short of the recommended values by more than 15%, the fuel will not burn completely, and the amount flue gases will increase. As a result, the efficiency of the boiler will drop quite significantly. This is why you should adhere to the recommended boiler power levels - they are designed to operate the equipment as efficiently as possible.
Calculation of efficiency taking into account various factors
The above formula is not entirely suitable for assessing the efficiency of equipment, since it is very difficult to accurately calculate the boiler efficiency taking into account only two indicators. In practice, a different, more complete formula is used during the design process, since not all of the heat generated is used to heat the water in the heating circuit. A certain amount of heat is lost during the operation of the boiler.
A more accurate calculation of boiler efficiency is made using the following formula:
ɳ=100-(q 2 +q 3 +q 4 +q 5 +q 6), in which
q 2 – heat loss from escaping flammable gases;
q 3 – heat loss as a result of incomplete combustion of combustion products;
q 4 – heat loss due to underburning of fuel and ash precipitation;
q 5 – losses caused external cooling device;
q 6 – heat loss along with slag removed from the furnace.
Heat loss when removing flammable gases
The most significant heat losses occur as a result of the evacuation of flammable gases into the chimney (q 2). The efficiency of the boiler largely depends on the combustion temperature of the fuel. The optimal temperature pressure at the cold end of the water heater is achieved when heated to 70-110 ℃.
When the temperature of the exhaust combustible gases drops by 12-15 ℃, the efficiency of the water heating boiler increases by 1%. However, in order to reduce the temperature of the exhaust combustion products, it is necessary to increase the size of the heated surfaces, and, therefore, the entire structure as a whole. Moreover, when cooling carbon monoxide the risk of low-temperature corrosion increases.
Among other things, the temperature of carbon monoxide also depends on the quality and type of fuel, as well as the heating of the air entering the firebox. The temperatures of incoming air and exiting combustion products depend on the type of fuel.
To calculate the heat loss rate with flue gases, use the following formula:
Q 2 = (T 1 -T 3) × (A 2 ÷ (21-O 2) + B), where
T 1 – temperature of evacuated flammable gases at the point behind the superheater;
T 3 – temperature of air entering the furnace;
21 – oxygen concentration in the air;
O 2 – amount of oxygen in the exhaust combustion products at the control point;
A 2 and B are coefficients from a special table that depend on the type of fuel.
Chemical underburning as a source of heat loss
Indicator q 3 is used when calculating efficiency gas boiler heating, for example, or in cases where fuel oil is used. For gas boilers, the value of q 3 is 0.1-0.2%. With a slight excess of air during combustion, this figure is 0.15%, and with a significant excess of air it is not taken into account at all. However, when burning a mixture of gases of different temperatures, the value of q 3 = 0.4-0.5%.
If the heating equipment runs on solid fuel, the indicator q 4 is taken into account. In particular, for anthracite coal the value of q 4 = 4-6%, semi-anthracite is characterized by 3-4% heat loss, but when burning hard coal, only 1.5-2% heat loss is formed. For liquid slag removal of burned low-reaction coal, the value of q4 can be considered minimal. But when removing slag in solid form, heat loss will increase to the maximum limit.
Heat loss due to external cooling
Such heat losses q5 usually amount to no more than 0.5%, and as the power of the heating equipment increases, they are reduced even more.
This indicator is related to the calculation of the steam output of the boiler plant:
- Provided the steam output D is within the range of 42-250 kg/s, the value of heat loss q5=(60÷D)×0.5÷lgD;
- If the value of steam production D exceeds 250 kg/s, the level of heat loss is considered equal to 0.2%.
Amount of heat loss from slag removal
The heat loss value q6 is only significant for liquid slag removal. But in cases where slag is removed from the combustion chamber solid fuel, heat loss q6 is taken into account when calculating the efficiency of heating boilers only in cases where they are more than 2.5Q.
How to calculate the efficiency of a solid fuel boiler
Even with an ideally designed design and high-quality fuel, the efficiency of heating boilers cannot reach 100%. Their work is necessarily associated with certain heat losses caused by both the type of fuel burned and the number of external factors and conditions. To understand how calculating the efficiency of a solid fuel boiler looks like in practice, let’s give an example.
For example, heat loss from removing slag from the fuel chamber will be:
q 6 =(A shl ×Z l ×A r)÷Q ri,
where A slag is the relative value of the slag removed from the furnace to the volume of loaded fuel. If the boiler is used correctly, the share of combustion waste in the form of ash is 5-20%, then given value may be equal to 80-95%.
З l – thermodynamic potential of ash at a temperature of 600 ℃ in normal conditions equal to 133.8 kcal/kg.
A p is the ash content of the fuel, which is calculated based on the total mass of the fuel. IN various types of fuel, the ash content ranges from 5% to 45%.
Q ri is the minimum amount of thermal energy that is generated during fuel combustion. Depending on the type of fuel, the heat capacity ranges from 2500-5400 kcal/kg.
IN in this case taking into account the indicated values of heat loss q 6 will be 0.1-2.3%.
The value of q5 will depend on the power and design performance of the heating boiler. Job modern installations With low power, which are very often used to heat private houses, is usually associated with heat losses of this type in the range of 2.5-3.5%.
Heat loss associated with mechanical underburning of solid fuel q 4 largely depends on its type, as well as on design features boiler They range from 3-11%. This is worth considering if you are looking for a way to make your boiler work more efficiently.
Chemical underburning The amount of fuel usually depends on the concentration of air in the combustible mixture. Such heat losses q 3 are usually equal to 0.5-1%.
Highest percentage heat loss q 2 is associated with the loss of heat along with flammable gases. This indicator is influenced by the quality and type of fuel, the degree of heating of combustible gases, as well as operating conditions and design of the heating boiler. With an optimal thermal design of 150 ℃, evacuable carbon monoxide should be heated to a temperature of 280 ℃. In this case, this value of heat loss will be equal to 9-22%.
If all the listed loss values are summed up, we get the efficiency value ɳ=100-(9+0.5+3+2.5+0.1)=84.9%.
This means that a modern boiler can only operate at 85-90% power. Everything else goes to ensure the combustion process.
Note that achieving such high values is not easy. To do this, you need to competently approach the selection of fuel and provide equipment with optimal conditions. Manufacturers usually indicate what load the boiler should operate with. In this case, it is desirable that most of the time it is set to an economical load level.
To operate the boiler with maximum efficiency, it must be used taking into account the following rules:
- Periodic cleaning of the boiler is required;
- it is important to control the intensity of combustion and the completeness of fuel combustion;
- it is necessary to calculate the draft taking into account the pressure of the supplied air;
- calculation of the ash fraction is necessary.
The quality of solid fuel combustion is positively affected by the calculation of optimal draft taking into account the air pressure supplied to the boiler and the rate of carbon monoxide evacuation. However, as air pressure increases, more heat is removed into the chimney along with combustion products. But too little pressure and limited access of air into the fuel chamber leads to a decrease in combustion intensity and greater ash formation.
If you have a heating boiler installed at home, pay attention to our recommendations for increasing its efficiency. You can not only save on fuel, but also achieve a comfortable microclimate in your home.