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, taking into account the need for rational use of non-renewable energy sources. To this end, international standards are constantly being tightened, limiting the performance of engine start-up and operation at low and even high temperatures. environment. 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 of electronic systems for automatic control of the 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 the 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 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 mode, the engine begins to operate on a homogeneous mixture, which leads to a reduction in emissions of harmful substances in the 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 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 high air temperature 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.
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 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
|
V>55kg/hour |
V>100kg/hour |
|||
|
Q, kJ/sec |
Q, kJ/sec |
|||
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 to the 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
|
Q1, kJ/sec |
Q2, kJ/sec |
|||
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
Karnaukhov V.N. OPTIMIZATION OF HEATING ELEMENT POWER TO MAINTAIN 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 (access date: 02/01/2020). We bring to your attention magazines published by the publishing house "Academy of Natural Sciences"
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 in simple and accessible language, using a practical example, to talk about the most important form 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?.. These are not the most difficult questions Today, not all engineers are able to answer “immediately”. 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 at constant temperature(provided that pressure and other environmental parameters do not change) 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 to Q3-Q2 .
4. Now at a constant boiling point Tkp the 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 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 make an important management decision that 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, otherwise and months, 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 with a total mass of 20 kg on the rolled metal. 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 items 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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The main physical properties air: air density, its dynamic and kinematic viscosity, specific heat capacity, thermal conductivity, thermal diffusivity, Prandtl number and entropy. The properties of air are given in tables depending on the temperature at normal atmospheric pressure.
Air density depending on temperature
A detailed table of dry air density values at various temperatures and normal atmospheric pressure is presented. What is the density of air? The density of air can be determined analytically by dividing its mass by the volume it occupies. under given conditions (pressure, temperature and humidity). You can also calculate its density using the formula of the ideal gas equation of state. To do this, you need to know the absolute pressure and temperature of the air, as well as its gas constant and molar volume. This equation allows you to calculate the dry density of air.
On practice, to find out what the density of air is at different temperatures, it is convenient to use ready-made tables. For example, the table below shows the density of atmospheric air depending on its temperature. Air density in the table is expressed in kilograms per cubic meter and is given in the temperature range from minus 50 to 1200 degrees Celsius at normal atmospheric pressure (101325 Pa).
| t, °С | ρ, kg/m 3 | t, °С | ρ, kg/m 3 | t, °С | ρ, kg/m 3 | t, °С | ρ, kg/m 3 |
|---|---|---|---|---|---|---|---|
| -50 | 1,584 | 20 | 1,205 | 150 | 0,835 | 600 | 0,404 |
| -45 | 1,549 | 30 | 1,165 | 160 | 0,815 | 650 | 0,383 |
| -40 | 1,515 | 40 | 1,128 | 170 | 0,797 | 700 | 0,362 |
| -35 | 1,484 | 50 | 1,093 | 180 | 0,779 | 750 | 0,346 |
| -30 | 1,453 | 60 | 1,06 | 190 | 0,763 | 800 | 0,329 |
| -25 | 1,424 | 70 | 1,029 | 200 | 0,746 | 850 | 0,315 |
| -20 | 1,395 | 80 | 1 | 250 | 0,674 | 900 | 0,301 |
| -15 | 1,369 | 90 | 0,972 | 300 | 0,615 | 950 | 0,289 |
| -10 | 1,342 | 100 | 0,946 | 350 | 0,566 | 1000 | 0,277 |
| -5 | 1,318 | 110 | 0,922 | 400 | 0,524 | 1050 | 0,267 |
| 0 | 1,293 | 120 | 0,898 | 450 | 0,49 | 1100 | 0,257 |
| 10 | 1,247 | 130 | 0,876 | 500 | 0,456 | 1150 | 0,248 |
| 15 | 1,226 | 140 | 0,854 | 550 | 0,43 | 1200 | 0,239 |
At 25°C, air has a density of 1.185 kg/m3. When heated, the air density decreases - the air expands (its specific volume increases). With increasing temperature, for example, to 1200°C, a very low air density is achieved, equal to 0.239 kg/m 3, which is 5 times less than its value at room temperature. IN general case, reduction when heated allows a process such as natural convection to take place and is used, for example, in aeronautics.
If we compare the density of air relative to , then air is three orders of magnitude lighter - at a temperature of 4°C, the density of water is 1000 kg/m3, and the density of air is 1.27 kg/m3. It is also necessary to note the air density at normal conditions. Normal conditions for gases are those at which their temperature is 0°C and the pressure is equal to normal atmospheric pressure. Thus, according to the table, air density under normal conditions (at NL) is 1.293 kg/m 3.
Dynamic and kinematic viscosity of air at different temperatures
When performing thermal calculations, it is necessary to know the value of air viscosity (viscosity coefficient) at different temperatures. This value is required to calculate the Reynolds, Grashof, and Rayleigh numbers, the values of which determine the flow regime of this gas. The table shows the values of the dynamic coefficients μ and kinematic ν air viscosity in the temperature range from -50 to 1200°C at atmospheric pressure.
The viscosity coefficient of air increases significantly with increasing temperature. For example, the kinematic viscosity of air is equal to 15.06 10 -6 m 2 /s at a temperature of 20°C, and with an increase in temperature to 1200°C, the viscosity of air becomes equal to 233.7 10 -6 m 2 /s, that is, it increases 15.5 times! The dynamic viscosity of air at a temperature of 20°C is 18.1·10 -6 Pa·s.
When air is heated, the values of both kinematic and dynamic viscosity increase. These two quantities are related to each other through the air density, the value of which decreases when this gas is heated. An increase in the kinematic and dynamic viscosity of air (as well as other gases) when heated is associated with a more intense vibration of air molecules around their equilibrium state (according to MKT).
| t, °С | μ·10 6 , Pa·s | ν·10 6, m 2 /s | t, °С | μ·10 6 , Pa·s | ν·10 6, m 2 /s | t, °С | μ·10 6 , Pa·s | ν·10 6, m 2 /s |
|---|---|---|---|---|---|---|---|---|
| -50 | 14,6 | 9,23 | 70 | 20,6 | 20,02 | 350 | 31,4 | 55,46 |
| -45 | 14,9 | 9,64 | 80 | 21,1 | 21,09 | 400 | 33 | 63,09 |
| -40 | 15,2 | 10,04 | 90 | 21,5 | 22,1 | 450 | 34,6 | 69,28 |
| -35 | 15,5 | 10,42 | 100 | 21,9 | 23,13 | 500 | 36,2 | 79,38 |
| -30 | 15,7 | 10,8 | 110 | 22,4 | 24,3 | 550 | 37,7 | 88,14 |
| -25 | 16 | 11,21 | 120 | 22,8 | 25,45 | 600 | 39,1 | 96,89 |
| -20 | 16,2 | 11,61 | 130 | 23,3 | 26,63 | 650 | 40,5 | 106,15 |
| -15 | 16,5 | 12,02 | 140 | 23,7 | 27,8 | 700 | 41,8 | 115,4 |
| -10 | 16,7 | 12,43 | 150 | 24,1 | 28,95 | 750 | 43,1 | 125,1 |
| -5 | 17 | 12,86 | 160 | 24,5 | 30,09 | 800 | 44,3 | 134,8 |
| 0 | 17,2 | 13,28 | 170 | 24,9 | 31,29 | 850 | 45,5 | 145 |
| 10 | 17,6 | 14,16 | 180 | 25,3 | 32,49 | 900 | 46,7 | 155,1 |
| 15 | 17,9 | 14,61 | 190 | 25,7 | 33,67 | 950 | 47,9 | 166,1 |
| 20 | 18,1 | 15,06 | 200 | 26 | 34,85 | 1000 | 49 | 177,1 |
| 30 | 18,6 | 16 | 225 | 26,7 | 37,73 | 1050 | 50,1 | 188,2 |
| 40 | 19,1 | 16,96 | 250 | 27,4 | 40,61 | 1100 | 51,2 | 199,3 |
| 50 | 19,6 | 17,95 | 300 | 29,7 | 48,33 | 1150 | 52,4 | 216,5 |
| 60 | 20,1 | 18,97 | 325 | 30,6 | 51,9 | 1200 | 53,5 | 233,7 |
Note: Be careful! Air viscosity is given to the power of 10 6 .
Specific heat capacity of air at temperatures from -50 to 1200°C
A table of the specific heat capacity of air at various temperatures is presented. The heat capacity in the table is given at constant pressure (isobaric heat capacity of air) in the temperature range from minus 50 to 1200°C for air in a dry state. What is the specific heat capacity of air? The specific heat capacity determines the amount of heat that must be supplied to one kilogram of air at constant pressure to increase its temperature by 1 degree. For example, at 20°C, to heat 1 kg of this gas by 1°C in an isobaric process, 1005 J of heat is required.
The specific heat capacity of air increases with increasing temperature. However, the dependence of the mass heat capacity of air on temperature is not linear. In the range from -50 to 120°C, its value practically does not change - under these conditions, the average heat capacity of air is 1010 J/(kg deg). According to the table, it can be seen that temperature begins to have a significant effect from a value of 130°C. However, air temperature affects its specific heat capacity much less than its viscosity. Thus, when heated from 0 to 1200°C, the heat capacity of air increases only 1.2 times - from 1005 to 1210 J/(kg deg).
It should be noted that the heat capacity humid air higher than dry. If we compare air, it is obvious that water has a higher value and the water content in air leads to an increase in specific heat capacity.
| t, °С | C p , J/(kg deg) | t, °С | C p , J/(kg deg) | t, °С | C p , J/(kg deg) | t, °С | C p , J/(kg deg) |
|---|---|---|---|---|---|---|---|
| -50 | 1013 | 20 | 1005 | 150 | 1015 | 600 | 1114 |
| -45 | 1013 | 30 | 1005 | 160 | 1017 | 650 | 1125 |
| -40 | 1013 | 40 | 1005 | 170 | 1020 | 700 | 1135 |
| -35 | 1013 | 50 | 1005 | 180 | 1022 | 750 | 1146 |
| -30 | 1013 | 60 | 1005 | 190 | 1024 | 800 | 1156 |
| -25 | 1011 | 70 | 1009 | 200 | 1026 | 850 | 1164 |
| -20 | 1009 | 80 | 1009 | 250 | 1037 | 900 | 1172 |
| -15 | 1009 | 90 | 1009 | 300 | 1047 | 950 | 1179 |
| -10 | 1009 | 100 | 1009 | 350 | 1058 | 1000 | 1185 |
| -5 | 1007 | 110 | 1009 | 400 | 1068 | 1050 | 1191 |
| 0 | 1005 | 120 | 1009 | 450 | 1081 | 1100 | 1197 |
| 10 | 1005 | 130 | 1011 | 500 | 1093 | 1150 | 1204 |
| 15 | 1005 | 140 | 1013 | 550 | 1104 | 1200 | 1210 |
Thermal conductivity, thermal diffusivity, Prandtl number of air
The table presents such physical properties of atmospheric air as thermal conductivity, thermal diffusivity and its Prandtl number depending on temperature. Thermophysical properties of air are given in the range from -50 to 1200°C for dry air. According to the table, it can be seen that the indicated properties of air depend significantly on temperature and the temperature dependence of the considered properties of this gas is different.
Passing 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 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 lower the absolute altitude of the area, the higher the air temperature (at one 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. Thus, the water mass of the World Ocean, which occupies 2/3 of the Earth’s surface, heats up very slowly and cools very slowly due to its 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 it scatters sun rays, 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 most high temperatures air recorded in areas 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
- 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.