Physical properties of air: density, viscosity, specific heat capacity. Changes in efficiency and tpp with changes in air heating temperature and the proportion of recirculated gases and air preheating

— 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 finned; the latter, in turn, are divided into lamellar and spiral-wound.

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

The heaters come in two models: medium (C) and large (B).

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

Where Q"— heat consumption for heating air, kJ/h (kcal/h); Q- the same, W; 0.278 — conversion factor 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 - air density (at temperature t K), kg/m 3 ]; With— specific heat air, equal to 1 kJ/(kg-K); tk is the air temperature after the air heater, °C; t n— air temperature before the heater, °C.

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

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 maximum permissible concentration of which is more than 100 mg/m3. When designing general ventilation intended to combat gases whose maximum permissible concentration is less than 100 mg/m3, as well as when designing supply ventilation to compensate for air removed through local suction, 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 parameters of category B).

In a room without excess heat, supply air should be supplied with a temperature equal to the internal air temperature tB for of this premises. If there is excess heat, supply air is supplied at a reduced temperature (by 5-8° C). It is not recommended to supply supply air with a temperature below 10° C to the room even in the presence of significant heat generation due to the possibility of colds occurring. The exception is the use of special anemostats.

The required heating surface area of ​​the air heaters Fк m2 is determined by the formula:

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

If the coolant is steam, then the average coolant temperature tav.T. 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:

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

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

By mass velocity we mean the mass of air, kg, passing in 1 s through 1 m2 of the open cross-section of the heater. Mass velocity vp, kg/(cm2), is determined by the formula

The model, brand and number of air heaters are selected based on the open cross-sectional area fL and heating surface FK. After selecting heaters, the mass velocity of air movement is specified based on the actual open cross-sectional area of ​​the heater fD of a given 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" is the heat consumption for heating the air, kJ/h (kcal/h); pv is the density of water equal to 1000 kg/m3, sv is the specific heat capacity of water equal to 4.19 kJ/(kg-K); fTP — open cross-sectional area for coolant passage, m2, tg - temperature hot water in the supply line, °C; t 0 — return water temperature, 0C.

The heat transfer of heaters is affected by the piping scheme. With a parallel pipeline connection scheme, only part of the coolant passes through a separate heater, and with sequential circuit The entire coolant flow passes through each heater.

The resistance of heaters to air passage 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 successive heaters is:

where m is the number of heaters located in series. The calculation ends with checking the thermal performance (heat transfer) of air heaters using the formula

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

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

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

The degree of heating of the surface, and therefore the air, depends, first of all, 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: altitude above sea level;

B: underlying surface;

B: distance from the coasts of oceans and seas.

A – Since air heating occurs from the earth’s surface, the less absolute altitudes terrain, 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 of different color and structure absorb and reflect the sun’s rays differently. The maximum reflectivity is characteristic of 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 water mass The world's oceans, which occupy 2/3 of the Earth's surface, heat up very slowly and cool very slowly due to their high heat capacity. Land heats up quickly and cools quickly, i.e., in order to heat 1 m2 of land and 1 m2 of water surface to the same t, it is necessary to spend different quantities 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 surface of the Earth. In the presence of large quantity water vapor in the air, water droplets reflect, scatter, absorb solar rays and not all of them reach the surface of the planet, its heating decreases.

The highest air temperatures recorded in the regions tropical deserts. IN central regions In the Sahara, for almost 4 months the temperature in the air in the shade is more than 40 o C. At the same time, at the equator, where the angle of incidence of the sun's rays is greatest, the temperature does not exceed +26 o 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 covered with a "blanket" of clouds, then not all infrared rays leave the planet, since the clouds delay them, reflecting them back to the earth's surface.

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

Literature

1. Zubaschenko E.M. Regional physical geography. Earth's climates: teaching aid. Part 1. / E.M. Zubaschenko, V.I. Shmykov, A.Ya. Nemykin, N.V. Polyakova. – Voronezh: VSPU, 2007. – 183 p.

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

Vacuum energy and dark energy, which still exists only in theory. In this article, the first in the “Heating Engineering” section, I will try to use a simple and accessible language using practical example, talk about the most important type 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 thermal engineering as a branch of the science of obtaining, transferring and using thermal energy. Modern thermal 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 “temperature change” of a system.

An external influence on a system, which changes its internal energy, can be the result of heat exchange. Thermal energy, which is acquired or lost by the system as a result of such interaction with the environment, is called amount of heat and is measured in SI units in Joules.

If you are not a heating engineer and do not deal with thermal engineering issues on a daily basis, then when you encounter them, sometimes without experience it can be very difficult to quickly understand them. Without experience, it is difficult to even imagine the dimensions of the required values ​​of the amount of heat and thermal power. How many Joules of energy are needed to heat 1000 cubic meters of air from a temperature of -37˚С to +18˚С?.. What power of the heat source is needed to do this in 1 hour?.. Today we can answer these not the most difficult questions “immediately” “Not everyone is an engineer. Sometimes specialists even remember the formulas, but only a few can apply them in practice!

After reading this article to the end, you will be able to easily solve real industrial and everyday problems related to heating and cooling of 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 during various physical processes.

Most known substances can different temperatures and pressure to be in solid, liquid, gaseous or plasma states. Transition from one state of aggregation to another occurs when constant temperature (provided that 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 presented in the figure. It shows the temperature dependence of a substance T on the amount of heat Q, brought to a certain closed system containing a certain mass of a specific substance.

1. A solid that has a temperature T1, heat to temperature Tmelt, 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 the solid is heated to the boiling point (gas formation) Tkp, spending on this amount of heat equal Q3-Q2 .

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

5. On last stage the gas heats up depending on the temperature Tkp up to a certain temperature T2. In this case, the amount of heat consumed will be Q5-Q4. (If we heat the gas to the ionization temperature, the gas will turn into plasma.)

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

Moving in the opposite direction, we will remove the same amount of heat from the substance Q5, having gone 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 loss to the external environment.

Note that a 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 realized that the processes of transitions between aggregate states of matter are characterized by energy consumption at a constant temperature. When heating a substance that is in one unchanged state of aggregation, the temperature rises and thermal energy is also consumed.

Main heat transfer formulas.

The formulas are very simple.

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

1. From the heat consumption side, that is, 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 a substance in J/kg

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

2.1. When fuel burns:

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 – effective current value in A

U – effective voltage value 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, during heating, additionally directly proportional to the temperature difference. Proportionality coefficients ( c , λ , r , q ) for each substance they have their own meanings 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 the source of thermal energy should be - everything is logical.

Calculation of an applied problem in Excel.

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

Conditions of the problem:

We bring 3 tons of rolled metal from a warehouse on the street to the premises of the rolled metal preparation workshop with dimensions of 24m x 15m x 7m. There is ice on the metal rolling stock total mass 20kg. It's -37˚С outside. How much heat is needed to heat the metal to +18˚С; heat the ice, melt it and heat the water to +18˚С; heat the entire volume of air in the room, assuming that the heating was completely turned off before? What power should the heating system have if all of the above must be completed in 1 hour? (Very harsh and almost unrealistic conditions - especially regarding air!)

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

Check out the color formatting of cells and fonts on the “” page.

Initial data:

1. We write the names of the 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:

to 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 melting ice λ enter in J/kg

to cell F6: 330000

5. A lot of substances m We enter in kg respectively for steel and ice

to cell D7: 3000

to cell E7: 20

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

in cells F7 and G7: =E7 =20

We find the mass of air by multiplying the volume of the room by the specific gravity

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

6. Process time t per minute we write only once for steel

to cell D8: 60

The time values ​​for heating the ice, melting it and heating the resulting water are calculated from the condition that all these three processes must be completed in the same amount of time as allotted for heating the metal. 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 during the same allotted time, we read

in cell H8: =D8 =60,0

7. The initial temperature of all substances T1 We put it in ˚C

to cell D9: -37

to cell E9: -37

to cell F9: 0

to cell G9: 0

to cell H9: -37

8. The final temperature of all substances T2 We put it in ˚C

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 regarding clauses 7 and 8.

Calculation results:

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

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

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

We read out the total amount of thermal energy required for all processes

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

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

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

in the 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/hour.

This completes the calculation in Excel.

Conclusions:

Please note that heating air requires more than twice as much energy as heating the same mass of steel.

Heating water costs twice as much energy as heating ice. The melting process consumes many times more energy than the heating process (at a small temperature difference).

Heating water requires ten times more thermal energy than heating steel and four times more than heating air.

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We recalled the concepts of “quantity of heat” and “thermal power”, examined the fundamental formulas of heat transfer, and analyzed a practical example. I hope that my language was simple, clear and interesting.

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According to the International Energy Agency, the priority for reducing carbon dioxide emissions from cars is to improve their 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 rational use non-renewable energy sources. For this purpose, international standards are constantly being tightened, limiting the performance of engine starting and operation in conditions of low and even high ambient temperatures. The article discusses the issue of fuel efficiency of internal combustion engines depending on temperature, pressure, and humidity of the surrounding air. The results of a study on maintaining a constant temperature in the intake manifold of an 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

optimal air temperature in the intake manifold

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

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

3. Lenin I.M. Theory of automobile and tractor engines. M.: graduate 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 of 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 into cars. In particular, to the creation electronic systems automatic control engine, transmission chassis And additional equipment. The use of electronic engine control systems (ESC) makes it possible to reduce fuel consumption and exhaust gas toxicity while simultaneously increasing engine power, increasing throttle response and cold start reliability. Modern ECS combine the functions of controlling fuel injection and the operation of the ignition system. To implement program control, the control unit records the dependence of the injection duration (amount of fuel supplied) on the load and engine speed. The dependence is specified 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 control system is used all over the world because selecting data from ready-made tables is a faster process than performing calculations using a computer. The values ​​obtained from the tables are adjusted by the car’s on-board computers depending on the signals from the throttle position sensors, air temperature, air pressure and density. The main difference between 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 allows you to reduce fuel consumption on various vehicles by up to 20%.

Low fuel consumption is achieved through different organization of the two main operating modes of the internal combustion engine: low load mode and high load mode. In this case, the engine in the first mode operates with a non-uniform mixture, a large excess of air and late fuel injection, due to which charge stratification is achieved from a mixture of air, fuel and remaining exhaust gases, as a result of which it operates on a lean mixture. At high load conditions, the engine starts to run on a homogeneous mixture, which leads to reduced emissions harmful substances in exhaust gases. Emission toxicity when using ESCs in diesel engines at start-up can be reduced by various glow plugs. The ECU receives information about intake air temperature, pressure, fuel consumption and crankshaft position. The control unit processes information from the sensors and, using characteristic maps, produces the value of the fuel supply advance angle. In order to take into account changes in the density of 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, an instantaneous change in air density occurs and, as a result, a decrease or increase in the flow of oxygen into the combustion chamber.

Purpose, objectives and research method

At the Tyumen State Oil and Gas University, research was carried out to maintain a constant temperature in the intake manifold of the internal combustion engines of KAMAZ-740, YaMZ-236 and D4FB (1.6 CRDi) of the Kia Sid, MZR2.3-L3T - Mazda CX7. At the same time, temperature fluctuations air mass 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 amount 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 air heating temperature in the intake manifold to 67 ˚C leads to a decrease in the intensity of heat exchange in engines, that is, a decrease in ΔT and an increase in the filling factor. ηv (Fig. 1)

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

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

However, heating the air to more than 67 ˚С does not lead to an increase in ηv due to the fact that the air density decreases. The experimental data obtained showed that the air in naturally aspirated diesel engines during operation has a temperature range of ΔТ=23÷36˚С. Tests have confirmed that for internal combustion engines operating on liquid fuel, the difference in the value of the filling coefficient ηv, calculated from the conditions that the fresh charge is air or an air-fuel mixture, is insignificant and amounts to 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 - effective engine power).

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

Increased air density.

Changes in air viscosity.

The nature of air flow into the combustion chamber.

Numerous studies have proven that heat air in the intake manifold increases fuel consumption slightly. In 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 optimal temperature for 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 was maintained in the intake manifold of the internal combustion engine. At the first stage, the amount of heat required to heat air weighing 1 kg at constant temperature and air pressure is determined, for this we assume: 1. Ambient air temperature t1 = -40˚C. 2. Temperature in the intake manifold t2=+70˚С.

We find the amount of heat required using the equation:

(2)

where CP is the mass heat capacity of air at constant pressure, determined from the table and for air at temperatures from 0 to 200 ˚С.

Amount of heat for greater mass air is determined by the formula:

where n is the volume of air in kg required for heating during engine operation.

When the internal combustion engine operates at speeds above 5000 rpm, air consumption passenger cars reaches 55-60 kg/hour, and cargo - 100 kg/hour. 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 time in seconds.

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

N=1.7 kW - heating element power for passenger cars and with an air flow rate of more than 100 kg/hour 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, T0 - , ρ0 - air density, Rв - 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 calculations of the amount of heat required to heat air weighing 1 kg with an average air flow rate for passenger cars more than V = 55 kg/hour and for trucks - more than V = 100 kg/hour 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

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

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

table 2

Calculation of heating time for different volumes of air

Time is determined by the formula τsec=Q/N at outside air temperature >-40˚С, Q1 at air flow V>55 kg/hour and Q2- V>100 kg/hour

Further, according to Table 2, a graph is drawn for the time of heating the air to +70 ˚C in the internal combustion engine manifold at different heater power. The graph shows that, regardless of the heating time, when the heater power increases, the heating time for different volumes of air equalizes.

Rice. 3. Time to heat the air to a temperature of +70 ˚С.

Conclusion

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

Reviewers:

Reznik L.G., Doctor of Technical Sciences, Professor of the Department of “Operation of Motor Transport” of the Federal State Educational Institution of the Educational Institution of Higher Professional Education “Tyumen State Oil and Gas University”, Tyumen.

Merdanov Sh.M., Doctor of Technical Sciences, Professor, Head of the Department of Transport and Technological Systems, Federal State Educational Institution of Higher Educational Institutions 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 automobiles and technological machines” of the Federal State Educational Institution of Higher Educational Institution “Tyumen State Oil and Gas University”, Tyumen.

Bibliographic link

When is the sun hotter - when is it higher above your head or when is it lower?

The sun is hotter when it is higher. In this case, the sun's rays fall at a right angle, or close to a right angle.

What types of rotation of the Earth do you know?

The Earth rotates around its axis and around the Sun.

Why does the cycle of day and night occur on Earth?

The change of day and night is the result of the axial rotation of the Earth.

Determine how the angle of incidence of the sun's rays differs on June 22 and December 22 at parallels 23.5° N. w. and Yu. sh.; on parallels 66.5° N. w. and Yu. w.

On June 22, the angle of incidence of the sun's rays at parallel 23.50 N. latitude. 900, S. – 430. At parallel 66.50 N. – 470, 66.50 S. – sliding angle.

On December 22, the angle of incidence of the sun's rays at the parallel is 23.50 N. 430, S. – 900. At parallel 66.50 N. – sliding angle, 66.50 S. – 470.

Think about why the warmest and coldest months are not June and December, when the sun's rays have the greatest and smallest angles of incidence on earth's surface.

Atmospheric air is heated by the earth's surface. Therefore, in June the earth's surface warms up, and the temperature reaches its maximum in July. The same thing happens in winter. In December the earth's surface cools down. The air cools down in January.

Define:

average daily temperature based on four measurements per day: -8°C, -4°C, +3°C, +1°C.

The average daily temperature is -20C.

average annual temperature Moscow, using table data.

The average annual temperature is 50C.

Determine the daily temperature range for the thermometer readings in Figure 110, c.

The temperature amplitude in the figure is 180C.

Determine how many degrees the annual amplitude in Krasnoyarsk is greater than in St. Petersburg, if the average temperature in July in Krasnoyarsk is +19°C, and in January - -17°C; in St. Petersburg +18°C and -8°C, respectively.

The temperature range in Krasnoyarsk is 360C.

The temperature range in St. Petersburg is 260C.

The temperature range in Krasnoyarsk is 100C greater.

Questions and tasks

1. How does atmospheric air heat up?

By transmitting the sun's rays, the atmosphere hardly heats up from them. The earth's surface heats up and itself becomes a source of heat. It is from this that the atmospheric air is heated.

2. How many degrees does the temperature in the troposphere decrease with every 100 m rise?

As you rise upward, every kilometer the air temperature drops by 6 0C. So, by 0.60 for every 100 m.

3. Calculate the air temperature outside the aircraft if the flight altitude is 7 km and the temperature at the Earth’s surface is +200C.

The temperature during an ascent of 7 km will drop by 420. This means that the temperature outside the plane will be -220.

4. Is it possible to find a glacier in the mountains at an altitude of 2500 m in the summer if the temperature at the foot of the mountains is +250C?

The temperature at an altitude of 2500 m will be +100C. A glacier will not be found at an altitude of 2500 m.

5. How and why does the air temperature change during the day?

During the day, the sun's rays illuminate the earth's surface and warm it, which also heats the air. At night, the supply of solar energy stops, and the surface along with the air gradually cools down. The sun is highest above the horizon at noon. This is when the most solar energy comes in. However, the highest temperature is observed 2-3 hours after noon, since it takes time to transfer heat from the Earth's surface to the troposphere. The lowest temperature occurs before sunrise.

6. What determines the difference in heating of the Earth’s surface throughout the year?

Over the course of a year, in the same area, the sun's rays fall on the surface in different ways. When the angle of incidence of the rays is more vertical, the surface receives more solar energy, the air temperature rises and summer begins. When the sun's rays are more inclined, the surface heats up weakly. The air temperature drops at this time, and winter comes. The warmest month in the Northern Hemisphere is July, and the coldest month is January. In the Southern Hemisphere it is the other way around: the coldest month of the year is July, and the warmest month is January.