About thermal energy in simple terms! Calculation in Excel applied task.

Pass through a transparent atmosphere without heating it, they reach earth's surface, heat it, and from it the air is subsequently heated.

The degree of surface heating, and hence the air, depends primarily on the latitude of the area.

But at each specific point, it (t o) will also be determined by a number of factors, among which the main ones are:

A: height above sea level;

B: underlying surface;

B: distance from the coasts of oceans and seas.

A - Since the air is heated from the earth's surface, the lower the absolute heights of the area, the higher the air temperature (at the same latitude). In conditions of air unsaturated with water vapor, a pattern is observed: for every 100 meters of altitude, the temperature (t o) decreases by 0.6 o C.

B - Qualitative characteristics of the surface.

B 1 - surfaces different in color and structure absorb and reflect the sun's rays in different ways. The maximum reflectivity is typical for snow and ice, the minimum for dark-colored soils and rocks.

Illumination of the Earth by the sun's rays on the days of the solstices and equinoxes.

B 2 - different surfaces have different heat capacity and heat transfer. So the water mass of the World Ocean, which occupies 2/3 of the Earth's surface, due to the high heat capacity, heats up very slowly and cools very slowly. The land quickly heats up and quickly cools, i.e., in order to heat up to the same t about 1 m 2 of land and 1 m 2 of water surface, it is necessary to spend a different amount of energy.

B - from the coasts to the interior of the continents, the amount of water vapor in the air decreases. The more transparent the atmosphere, the less sunlight is scattered in it, and all the sun's rays reach the Earth's surface. In the presence of a large number water vapor in the air, water droplets reflect, scatter, absorb the sun's rays and not all of them reach the surface of the planet, while heating it decreases.

The highest air temperatures are recorded in areas of tropical deserts. In the central regions of the Sahara, for almost 4 months, t about air in the shade is more than 40 ° C. At the same time, at the equator, where the angle of incidence of the sun's rays is the largest, the temperature does not exceed +26 ° C.

On the other hand, the Earth, as a heated body, radiates energy into space mainly in the long-wave infrared spectrum. If the earth's surface is wrapped in a "blanket" of clouds, then not all infrared rays leave the planet, since the clouds delay them, reflecting back to the earth's surface.

With a clear sky, when there is little water vapor in the atmosphere, the infrared rays emitted by the planet freely go into space, while the earth's surface cools down, which cools down and thereby reduces the air temperature.

Literature

1. Zubashchenko E.M. Regional physical geography. Climates of the Earth: teaching aid. Part 1. / E.M. Zubashchenko, V.I. Shmykov, A.Ya. Nemykin, N.V. Polyakov. - Voronezh: VGPU, 2007. - 183 p.

Mankind knows few types of energy - mechanical energy (kinetic and potential), internal energy (thermal), field energy (gravitational, electromagnetic and nuclear), chemical. Separately, it is worth highlighting the energy of the explosion, ...

Vacuum energy and still existing only in theory - dark energy. In this article, the first in the "Heat engineering" section, I will try in a simple and accessible language, using a practical example, to talk about the most important form of energy in people's lives - about thermal energy and about giving birth to her in time thermal power.

A few words to understand the place of heat engineering as a branch of the science of obtaining, transferring and using thermal energy. Modern heat engineering has emerged from general thermodynamics, which in turn is one of the branches of physics. Thermodynamics is literally “warm” plus “power”. Thus, thermodynamics is the science of the "change in temperature" of a system.

The impact on the system from the outside, in which its internal energy changes, can be the result of heat transfer. Thermal energy, which is gained or lost by the system as a result of such interaction with the environment, is called amount of heat and is measured in the SI system in Joules.

If you are not a heat engineer and do not deal with heat engineering issues on a daily basis, then when you encounter them, sometimes without experience it can be very difficult to quickly figure them out. It is difficult to imagine even the dimensions of the desired values ​​of the amount of heat and heat power without experience. How many Joules of energy is needed to heat 1000 cubic meters of air from a temperature of -37˚С to +18˚С?.. What power of a heat source is needed to do this in 1 hour? Sometimes experts even remember the formulas, but only a few can put them into practice!

After reading this article to the end, you will be able to easily solve real production and household tasks related to heating and cooling various materials. Understanding the physical essence of heat transfer processes and knowledge of simple basic formulas are the main blocks in the foundation of knowledge in heat engineering!

The amount of heat in various physical processes.

Most known substances can be in solid, liquid, gaseous or plasma states at different temperatures and pressures. Transition from one aggregate state to another takes place at constant temperature(provided that the pressure and other parameters do not change environment) and is accompanied by the absorption or release of thermal energy. Despite the fact that 99% of matter in the Universe is in the plasma state, we will not consider this state of aggregation in this article.

Consider the graph shown in the figure. It shows the dependence of the temperature of a substance T on the amount of heat Q, summed up to a certain closed system containing a certain mass of a particular substance.

1. A solid that has a temperature T1, heated to a temperature Tm, spending on this process an amount of heat equal to Q1 .

2. Next, the melting process begins, which occurs at a constant temperature Tpl(melting point). To melt the entire mass of a solid, it is necessary to expend thermal energy in the amount Q2 — Q1 .

3. Next, the liquid resulting from the melting of a solid is heated to the boiling point (gas formation) Tkp, spending on this amount of heat equal to Q3-Q2 .

4. Now at a constant boiling point Tkp liquid boils and evaporates, turning into a gas. To convert the entire mass of liquid into gas, it is necessary to spend thermal energy in quantity Q4-Q3.

5. At the last stage, the gas is heated from the temperature Tkp up to some temperature T2. In this case, the cost of the amount of heat will be Q5-Q4. (If we heat the gas to the ionization temperature, the gas will turn into plasma.)

Thus, heating the original solid from the temperature T1 up to temperature T2 we spent thermal energy in the amount Q5, translating the substance through three states of aggregation.

Moving in the opposite direction, we will remove the same amount of heat from the substance Q5, passing through the stages of condensation, crystallization and cooling from temperature T2 up to temperature T1. Of course, we are considering a closed system without energy losses to the external environment.

Note that the transition from the solid state to the gaseous state is possible, bypassing the liquid phase. This process is called sublimation, and the reverse process is called desublimation.

So, we have understood that the processes of transitions between the aggregate states of a substance are characterized by energy consumption at a constant temperature. When a substance is heated, which is in one unchanged state of aggregation, the temperature rises and thermal energy is also consumed.

The main formulas for heat transfer.

The formulas are very simple.

Quantity of heat Q in J is calculated by the formulas:

1. From the heat consumption side, i.e. from the load side:

1.1. When heating (cooling):

Q = m * c *(T2 -T1)

m – mass of substance in kg

With - specific heat capacity of a substance in J / (kg * K)

1.2. When melting (freezing):

Q = m * λ

λ – specific heat of melting and crystallization of a substance in J/kg

1.3. During boiling, evaporation (condensation):

Q = m * r

r – specific heat of gas formation and condensation of matter in J/kg

2. From the side of heat production, that is, from the side of the source:

2.1. When burning fuel:

Q = m * q

q – specific heat of combustion of fuel in J/kg

2.2. When converting electricity into thermal energy (Joule-Lenz law):

Q =t *I *U =t *R *I ^2=(t /r)*U ^2

t – time in s

I – current value in A

U – r.m.s. voltage in V

R – load resistance in ohms

We conclude that the amount of heat is directly proportional to the mass of the substance during all phase transformations and, when heated, is additionally directly proportional to the temperature difference. Proportionality coefficients ( c , λ , r , q ) for each substance have their own values ​​and are determined empirically (taken from reference books).

Thermal power N in W is the amount of heat transferred to the system in a certain time:

N=Q/t

The faster we want to heat the body to a certain temperature, the greater the power should be the source of thermal energy - everything is logical.

Calculation in Excel applied task.

In life, it is often necessary to make a quick estimated calculation in order to understand whether it makes sense to continue studying a topic, making a project and detailed accurate labor-intensive calculations. Having made a calculation in a few minutes even with an accuracy of ± 30%, you can make an important management decision that will be 100 times cheaper and 1000 times more efficient and, as a result, 100,000 times more efficient than performing an accurate calculation within a week, or even a month, by a group of expensive specialists ...

Conditions of the problem:

In the premises of the shop for the preparation of rolled metal with dimensions of 24m x 15m x 7m, we import rolled metal from a warehouse on the street in the amount of 3 tons. Rolled metal has ice with a total mass of 20 kg. Outside -37˚С. What amount of heat is needed to heat the metal to + 18˚С; heat the ice, melt it and heat the water up to +18˚С; heat the entire volume of air in the room, assuming that the heating was completely turned off before that? What power should the heating system have if all of the above must be completed in 1 hour? (Very rigid and hardly real conditions- especially concerning the air!)

We will perform the calculation in the programMS Excel or in the programOo Calc.

For color formatting of cells and fonts, see the "" page.

Initial data:

1. We write the names of substances:

to cell D3: Steel

to cell E3: Ice

to cell F3: ice/water

to cell G3: Water

to cell G3: Air

2. We enter the names of the processes:

into cells D4, E4, G4, G4: heat

to cell F4: melting

3. Specific heat capacity of substances c in J / (kg * K) we write for steel, ice, water and air, respectively

to cell D5: 460

to cell E5: 2110

to cell G5: 4190

to cell H5: 1005

4. Specific heat of fusion of ice λ in J/kg enter

to cell F6: 330000

5. Mass of substances m in kg we enter, respectively, for steel and ice

to cell D7: 3000

to cell E7: 20

Since the mass does not change when ice turns into water,

in cells F7 and G7: =E7 =20

The mass of air is found by multiplying the volume of the room by the specific gravity

in cell H7: =24*15*7*1.23 =3100

6. Process time t in minutes we write only once for steel

to cell D8: 60

The time values ​​for ice heating, its melting and heating of the resulting water are calculated from the condition that all these three processes must sum up in the same time as the time allotted for heating the metal. We read accordingly

in cell E8: =E12/(($E$12+$F$12+$G$12)/D8) =9,7

in cell F8: =F12/(($E$12+$F$12+$G$12)/D8) =41,0

in cell G8: =G12/(($E$12+$F$12+$G$12)/D8) =9,4

The air should also warm up in the same allotted time, we read

in cell H8: =D8 =60,0

7. The initial temperature of all substances T1 into ˚C we enter

to cell D9: -37

to cell E9: -37

to cell F9: 0

to cell G9: 0

to cell H9: -37

8. Final temperature of all substances T2 into ˚C we enter

to cell D10: 18

to cell E10: 0

to cell F10: 0

to cell G10: 18

to cell H10: 18

I think there shouldn't be any questions on items 7 and 8.

Calculation results:

9. Quantity of heat Q in KJ required for each of the processes we calculate

for steel heating in cell D12: =D7*D5*(D10-D9)/1000 =75900

for heating ice in cell E12: =E7*E5*(E10-E9)/1000 = 1561

for melting ice in cell F12: =F7*F6/1000 = 6600

for water heating in cell G12: =G7*G5*(G10-G9)/1000 = 1508

for air heating in cell H12: =H7*H5*(H10-H9)/1000 = 171330

The total amount of thermal energy required for all processes is read

in merged cell D13E13F13G13H13: =SUM(D12:H12) = 256900

In cells D14, E14, F14, G14, H14, and the combined cell D15E15F15G15H15, the amount of heat is given in an arc unit of measurement - in Gcal (in gigacalories).

10. Thermal power N in kW, required for each of the processes is calculated

for steel heating in cell D16: =D12/(D8*60) =21,083

for heating ice in cell E16: =E12/(E8*60) = 2,686

for melting ice in cell F16: =F12/(F8*60) = 2,686

for water heating in cell G16: =G12/(G8*60) = 2,686

for air heating in cell H16: =H12/(H8*60) = 47,592

The total thermal power required to perform all processes in a time t calculated

in merged cell D17E17F17G17H17: =D13/(D8*60) = 71,361

In cells D18, E18, F18, G18, H18, and the combined cell D19E19F19G19H19, the thermal power is given in an arc unit of measurement - in Gcal / h.

This completes the calculation in Excel.

Conclusions:

Note that it takes more than twice as much energy to heat air as it does to heat the same mass of steel.

When heating water, the energy costs are twice as much as when heating ice. The melting process consumes many times more energy than the heating process (with a small temperature difference).

Heating water consumes ten times more heat energy than heating steel and four times more than heating air.

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We remembered the concepts of “amount of heat” and “thermal power”, considered the fundamental formulas for heat transfer, and analyzed a practical example. I hope that my language was simple, understandable and interesting.

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- devices used for heating air in supply ventilation systems, air conditioning systems, air heating, as well as in drying installations.

According to the type of coolant, heaters can be fire, water, steam and electric. .

The most widespread at present are water and steam heaters, which are divided into smooth-tube and ribbed ones; the latter, in turn, are divided into lamellar and spiral-wound.

Distinguish between single-pass and multi-pass heaters. In single-pass, the coolant moves through the tubes in one direction, and in multi-pass, it changes the direction of movement several times due to the presence of partitions in the collector covers (Fig. XII.1).

Heaters perform two models: medium (C) and large (B).

The heat consumption for heating the air is determined by the formulas:

Where Q"— heat consumption for air heating, kJ/h (kcal/h); Q- the same, W; 0.278 is the conversion factor from kJ/h to W; G- mass amount of heated air, kg / h, equal to Lp [here L- volumetric amount of heated air, m 3 / h; p is the air density (at a temperature tK), kg / m 3]; With- specific heat capacity of air, equal to 1 kJ / (kg-K); t k - air temperature after the heater, ° С; t n— air temperature before the air heater, °C.

For heaters of the first stage of heating, the temperature tn is equal to the temperature of the outside air.

The outside air temperature is assumed to be equal to the calculated ventilation temperature (climate parameters of category A) when designing general ventilation designed to combat excess moisture, heat and gases, the MPC of which is more than 100 mg / m3. When designing general ventilation designed to combat gases whose MPC is less than 100 mg / m3, as well as when designing supply ventilation to compensate for air removed through local exhausts, process hoods or pneumatic transport systems, the outside air temperature is assumed to be equal to the calculated outside temperature tn for heating design (climate category B parameters).

In a room without heat surpluses, supply air with a temperature equal to the indoor air temperature tВ for this room should be supplied. In the presence of excess heat, supply air is supplied from low temperature(at 5-8 ° C). Supply air with a temperature below 10°C is not recommended to be supplied to the room even in the presence of significant heat emissions due to the possibility of colds. The exception is the use of special anemostats.

The required surface area for heating heaters Fк m2, is determined by the formula:

Where Q— heat consumption for air heating, W (kcal/h); TO- heat transfer coefficient of the heater, W / (m 2 -K) [kcal / (h-m 2 - ° C)]; t cf.T. — average temperature coolant, 0 С; t r.v. is the average temperature of the heated air passing through the heater, °C, equal to (t n + t c)/2.

If the coolant is steam, then the average temperature of the coolant tav.T. is equal to the saturation temperature at the corresponding vapor pressure.

For water temperature tav.T. is defined as the arithmetic mean of the hot and return water temperatures:

The safety factor 1.1-1.2 takes into account the heat loss for air cooling in the air ducts.

The heat transfer coefficient of heaters K depends on the type of coolant, the mass air velocity vp through the heater, geometric dimensions and design features heaters, the speed of water movement through the tubes of the heater.

The mass velocity is understood as the mass of air, kg, passing through 1 m2 of the living section of the air heater in 1 s. Mass velocity vp, kg/(cm2), is determined by the formula

According to the area of ​​​​the open section fЖ and the heating surface FK, the model, brand and number of heaters are selected. After choosing the heaters, the mass air velocity is specified according to the actual area of ​​​​the open section of the heater fD of this model:

where A, A 1 , n, n 1 and T- coefficients and exponents, depending on the design of the heater

The speed of water movement in the heater tubes ω, m/s, is determined by the formula:

where Q "- heat consumption for heating air, kJ / h (kcal / h); p - water density, equal to 1000 kg / m3, sv - specific heat capacity of water, equal to 4.19 kJ / (kg-K); fTP - open area for the passage of the coolant, m2, tg - temperature hot water in the supply line, ° С; t 0 - return water temperature, 0С.

The heat transfer of heaters is affected by the scheme of tying them with pipelines. With a parallel scheme for connecting pipelines, only part of the coolant passes through a separate heater, and with a sequential scheme, the entire flow of the coolant passes through each heater.

The resistance of heaters to the passage of air p, Pa, is expressed by the following formula:

where B and z are the coefficient and exponent, which depend on the design of the heater.

The resistance of the heaters located in series is equal to:

where m is the number of successively located heaters. The calculation ends with a check of the heat output (heat transfer) of the heaters according to the formula

where QK - heat transfer of heaters, W (kcal / h); QK - the same, kJ/h, 3.6 - conversion factor W to kJ/h FK - heating surface area of ​​heaters, m2, taken as a result of calculation of heaters of this type; K - heat transfer coefficient of heaters, W/(m2-K) [kcal/(h-m2-°C)]; tav.v - the average temperature of the heated air passing through the heater, °C; tav. T is the average temperature of the coolant, °С.

When selecting heaters, the margin for the estimated heating surface area is taken in the range of 15 - 20%, for the resistance to air passage - 10% and for the resistance to water movement - 20%.

1

The International Energy Agency estimates that the priority for reducing car carbon emissions is to improve fuel efficiency. The task of reducing CO2 emissions by increasing the fuel efficiency of vehicles is one of the priorities for the world community, given the need for the rational use of non-renewable energy sources. To this end, international standards are constantly tightened, limiting the performance of engine start-up and operation at low and even high ambient temperatures. The article deals with the issue of fuel efficiency of internal combustion engines depending on the temperature, pressure, humidity of the ambient air. The results of a study on maintaining a constant temperature in the intake manifold of the internal combustion engine in order to save fuel and determine the optimal power of the heating element are presented.

heating element power

ambient temperature

air heating

fuel economy

optimum air temperature in the intake manifold

1. Car engines. V.M. Arkhangelsky [and others]; resp. ed. M.S. Hovah. M.: Mashinostroenie, 1977. 591 p.

2. Karnaukhov V.N., Karnaukhova I.V. Determination of the filling factor in the internal combustion engine // Transport and transport-technological systems, materials of the International Scientific and Technical Conference, Tyumen, April 16, 2014. Tyumen: Tyumen State University Publishing House, 2014.

3. Lenin I.M. Theory of automobile and tractor engines. M.: Higher school, 1976. 364 p.

4. Yutt V.E. Electrical equipment of cars. M: Publishing House Hot Line-Telecom, 2009. 440 p.

5. Yutt V.E., Ruzavin G.E. Electronic control systems for internal combustion engines and methods for their diagnosis. M.: Publishing House Hot Line-Telecom, 2007. 104 p.

Introduction

The development of electronics and microprocessor technology has led to its widespread introduction to cars. In particular, to the creation of electronic systems for automatic control of the engine, transmission, chassis and additional equipment. The use of electronic systems for engine control (ECS) makes it possible to reduce fuel consumption and exhaust gas toxicity with a simultaneous increase in engine power, increase acceleration and cold start reliability. Modern ESUs combine the functions of fuel injection control and the operation of the ignition system. To implement program control in the control unit, the dependence of the injection duration (the amount of fuel supplied) on the load and engine speed is recorded. The dependence is given in the form of a table developed on the basis of comprehensive tests of an engine of a similar model. Similar tables are used to determine the ignition angle. This engine management system is used all over the world, because the selection of data from ready-made tables is a faster process than performing calculations using a computer. The values ​​obtained from the tables are corrected by the on-board computers of the vehicles depending on the signals from the throttle position sensors, air temperature, air pressure and density. The main difference of this system, used in modern cars, is the absence of a rigid mechanical connection between throttle valve and the accelerator pedal that controls it. Compared to traditional systems, ESU can reduce fuel consumption on various vehicles by up to 20%.

Low fuel consumption is achieved by different organization of the two main modes of operation of the internal combustion engine: low load mode and high load mode. In this case, the engine in the first mode operates with a heterogeneous mixture, a large excess of air and late fuel injection, due to which charge separation from a mixture of air, fuel and remaining exhaust gases is achieved, as a result of which it runs on a lean mixture. In high load mode, the engine starts to work on a homogeneous mixture, which leads to a decrease in emissions of harmful substances in the exhaust gases. The emission toxicity of ESA diesel engines at start-up can be reduced by various glow plugs. The ESU receives information about intake air temperature, pressure, fuel consumption and crankshaft position. The control unit processes information from the sensors and, using characteristic maps, gives the value of the fuel supply advance angle. In order to take into account the change in the density of the incoming air when its temperature changes, the flow sensor is equipped with a thermistor. But as a result of fluctuations in temperature and air pressure in the intake manifold, despite the above sensors, there is an instantaneous change in air density and, as a result, a decrease or increase in oxygen supply to the combustion chamber.

Purpose, objectives and research method

Studies were carried out at the Tyumen State Oil and Gas University in order to maintain a constant temperature in the intake manifold of the internal combustion engine KAMAZ-740, YaMZ-236 and D4FB (1.6 CRDi) of the Kia Sid, MZR2.3-L3T - Mazda CX7. Wherein temperature fluctuations air masses were taken into account by temperature sensors. Ensuring normal (optimal) air temperature in the intake manifold must be carried out under all possible operating conditions: starting a cold engine, operating at low and high loads, when operating at low ambient temperatures.

In modern high-speed engines, the total value of heat transfer turns out to be insignificant and amounts to about 1% of the total amount of heat released during fuel combustion. An increase in the temperature of air heating in the intake manifold to 67 ˚С leads to a decrease in the intensity of heat transfer in engines, that is, a decrease in ΔТ and an increase in the filling factor. ηv (Fig. 1)

where ΔT is the air temperature difference in the intake manifold (˚K), Tp is the air heating temperature in the intake manifold, Tv is the air temperature in the intake manifold.

Rice. 1. Graph of the effect of air heating temperature on the filling factor (on the example of the KAMAZ-740 engine)

However, air heating above 67 ˚С does not lead to an increase in ηv due to the fact that the air density decreases. The obtained experimental data showed that the air in diesel engines without pressurization during its operation has a temperature range ΔТ=23÷36˚С. Tests have confirmed that for internal combustion engines operating on liquid fuel, the difference in the value of the filling factor ηv, calculated from the conditions that the fresh charge is air or an air-fuel mixture, is insignificant and is less than 0.5%, therefore, for all types of engines, ηv is determined by air.

Changes in temperature, pressure and air humidity affect the power of any engine and fluctuate in the range Ne=10÷15% (Ne is the effective engine power).

The increase in aerodynamic air resistance in the intake manifold is explained by the following parameters:

Increased air density.

Change in air viscosity.

The nature of the air entering the combustion chamber.

Numerous studies have shown that high air temperature in the intake manifold increases fuel consumption slightly. At the same time, low temperature increases its consumption by up to 15-20%, so the studies were carried out at an outside air temperature of -40 ˚С and its heating to +70 ˚С in the intake manifold. The optimum fuel consumption is the air temperature in the intake manifold 15÷67 ˚С.

Research results and analysis

During the tests, the power of the heating element was determined to ensure that a certain temperature is maintained in the intake manifold of the internal combustion engine. At the first stage, the amount of heat required to heat 1 kg air at a constant temperature and air pressure is determined, for this we will take: 1. Ambient air temperature t1=-40˚C. 2. Temperature in the intake manifold t2=+70˚С.

The amount of heat required is found by the equation:

(2)

where СР is the mass heat capacity of air at constant pressure, determined according to the table and for air at a temperature from 0 to 200 ˚С.

The amount of heat for a larger mass of air is determined by the formula:

where n is the volume of air in kg required for heating when the engine is running.

When the internal combustion engine operates at a speed of more than 5000 rpm, the air consumption of passenger cars reaches 55-60 kg/h, and trucks - 100 kg/h. Then:

The heater power is determined by the formula:

where Q is the amount of heat spent on heating the air in J, N is the power of the heating element in W, τ is the time in sec.

It is necessary to determine the power of the heating element per second, so the formula will take the form:

N=1.7 kW - power of the heating element for cars and at an air flow rate of more than 100 kg/h for trucks - N=3.1 kW.

(5)

where Ttr is the temperature in the inlet pipeline, Ptr is the pressure in Pa in the inlet pipeline, Т0 - , ρ0 is the air density, Rv is the universal gas constant of air.

Substituting formula (5) into formula (2), we obtain:

(6)

(7)

The heater power per second is determined by formula (4) taking into account formula (5):

(8)

The results of calculating the amount of heat required to heat the air with a mass of 1 kg with an average air consumption for cars more than V = 55 kg / h and for trucks - more than V = 100 kg / h are presented in table 1.

Table 1

Table for determining the amount of heat for heating the air in the intake manifold depending on the outside air temperature

	V>55kg/hour		V>100kg/hour
		Q, kJ/s		Q, kJ/s

Based on the data in Table 1, a graph (Fig. 2) was constructed for the amount of heat Q per second spent on heating the air up to optimal temperature. The graph shows that the higher the air temperature, the less heat is needed to maintain the optimum temperature in the intake manifold, regardless of the volume of air.

Rice. 2. The amount of heat Q per second spent on heating the air to the optimum temperature

table 2

Calculation of the heating time for different volumes of air

	Q1, kJ/s		Q2, kJ/s

The time is determined by the formula τsec=Q/N at outdoor temperature >-40˚С, Q1 at air flow rate V>55 kg/h and Q2- V>100 kg/h

Further, according to Table 2, a graph of the time for heating the air to +70 ˚С in the ICE manifold is plotted at different heater power. The graph shows that, regardless of the heating time, when the heater power is increased, the heating time for different air volumes is equalized.

Rice. 3. Time of air heating up to +70 ˚С.

Conclusion

Based on calculations and experiments, it has been established that the most economical is the use of heaters of variable power to maintain a given temperature in the intake manifold in order to obtain fuel savings of up to 25-30%.

Reviewers:

Reznik L.G., Doctor of Technical Sciences, Professor of the Department "Operation of Road Transport" FGBO UVPO "Tyumen State Oil and Gas University", Tyumen.

Merdanov Sh.M., Doctor of Technical Sciences, Professor, Head of the Department "Transport and Technological Systems" FGBO UVPO "Tyumen State Oil and Gas University", Tyumen.

Zakharov N.S., Doctor of Technical Sciences, Professor, current member Russian Academy transport, head of the department "Service of cars and technological machines" FGBO UVPO "Tyumen State Oil and Gas University", Tyumen.

Bibliographic link

Karnaukhov V.N. OPTIMIZATION OF THE POWER OF THE HEATING ELEMENT TO MAINTAIN THE OPTIMUM AIR TEMPERATURE IN THE ICE INTAKE MANIFOLD // Contemporary Issues science and education. - 2014. - No. 3.;
URL: http://science-education.ru/ru/article/view?id=13575 (date of access: 01.02.2020). We bring to your attention the journals published by the publishing house "Academy of Natural History"

Change in flue gas recirculation . Gas recirculation is widely used to expand the range of superheated steam temperature control and allows maintaining the superheated steam temperature even at low loads of the boiler unit. Recently, flue gas recirculation is also gaining popularity as a method to reduce NO x formation. It is also used to recirculate the flue gases into the air stream before the burners, which is more effective in terms of suppressing the formation of NO x .

The introduction of relatively cold recirculated gases into the lower part of the furnace leads to a decrease in the heat absorption of the radiant heating surfaces and to an increase in the temperature of the gases at the furnace outlet and in the convective gas ducts, including the temperature of the flue gases. An increase in the total flow of flue gases in the section of the gas path before the selection of gases for recirculation contributes to an increase in the heat transfer coefficients and heat absorption of convective heating surfaces.

Rice. 2.29. Changes in steam temperature (curve 1), hot air temperature (curve 2) and flue gas losses (curve 3) depending on the share of flue gas recirculation r.

On fig. 2.29 shows the characteristics of the TP-230-2 boiler unit with a change in the proportion of gas recirculation to the lower part of the furnace. Here the share of recycling

where V rc is the volume of gases taken off for recirculation; V r - the volume of gases at the point of selection for recirculation without taking into account V rc. As can be seen, an increase in the share of recirculation by every 10% leads to an increase in the flue gas temperature by 3–4°C, Vr - by 0.2%, steam temperature - by 15 ° C, and the nature of the dependence is almost linear. These ratios are not unambiguous for all boiler units. Their value depends on the temperature of the recirculated gases (the place of gas intake) and the method of introducing them. The discharge of recirculated gases into the upper part of the furnace does not affect the operation of the furnace, but leads to a significant decrease in the temperature of the gases in the area of ​​the superheater and, as a result, to a decrease in the temperature of the superheated steam, although the volume of combustion products increases. Discharge of gases into the upper part of the furnace can be used to protect the superheater from impact. high temperature gases and reduce superheater slagging.

Of course, the use of gas recirculation leads to a decrease not only in efficiency. gross, but also efficiency net of the boiler unit, as it causes an increase in electricity consumption for own needs.

Rice. 2.30. Dependence of heat losses with mechanical underburning on the temperature of hot air.

Hot air temperature change. The change in hot air temperature is the result of a change in the operating mode of the air heater due to the influence of factors such as changes in temperature difference, heat transfer coefficient, gas or air flow. Increasing the temperature of the hot air increases, albeit slightly, the level of heat release in the furnace. The hot air temperature has a significant effect on the characteristics of boiler units operating on fuel with a low volatile output. A decrease in ^ r.v in this case worsens the conditions for fuel ignition, the mode of drying and grinding of the fuel, leads to a decrease in the temperature of the air mixture at the inlet to the burners, which can cause an increase in losses with mechanical underburning (see Fig. 2.30).

. Changing the air preheating temperature. Air preheating in front of the air heater is used to increase the temperature of the wall of its heating surfaces in order to reduce the corrosive effect of flue gases on them, especially when burning high-sulfur fuels. According to PTE, when burning sulphurous fuel oil, the air temperature in front of tubular air heaters must not be lower than 110 ° C, and in front of regenerative ones - not lower than 70 ° C.

Pre-heating of air can be carried out by recirculating hot air to the inlet of blast fans, however, in this case, the efficiency of the boiler unit decreases due to an increase in electricity consumption for the blast and an increase in the temperature of the flue gases. Therefore, it is advisable to heat the air above 50°C in heaters operating on selective steam or hot water.

Air preheating entails a decrease in the heat absorption of the air heater due to a decrease in temperature difference, the temperature of the flue gases and the heat loss increase. Air preheating also requires additional energy costs for air supply to the air heater. Depending on the level and method of air preheating, for every 10° C of air preheating, the efficiency gross changes by about 0.15-0.25%, and the temperature of the flue gases - by 3-4.5 ° C.

Since the share of heat taken for air preheating in relation to the heat output of boiler units is quite large (2-3.5%), the choice of the optimal air heating scheme has great importance.

Cold air

Rice. 2.31. Scheme of two-stage air heating in heaters with network water and selective steam:

1 - network heaters; 2 - the first stage of air heating with network water of the heating system; 3 - the second stage of air heating pzrom; 4 - pump for supplying return network water to heaters; 5 - network water for air heating (scheme for summer period); 6 - network water for air heating (scheme for the winter period).