Nominal losses to the environment. What is thermal pollution of the environment

For reducing heat consumption strict is needed accounting of heat losses in technological equipment and heating networks. Heat losses depend on the type of equipment and pipelines, their correct operation and the type of insulation.

Heat loss (W) is calculated using the formula

Depending on the type of equipment and pipeline, the total thermal resistance is:

for an insulated pipeline with one layer of insulation:

for an insulated pipeline with two layers of insulation:

for technological devices with multilayer flat or cylindrical walls with a diameter of more than 2 m:

for technological devices with multilayer flat or cylindrical walls with a diameter of less than 2 m:

to the inner wall of the pipeline or apparatus and from the outer surface of the wall to environment, W/(m 2 - K); X tr, ?. st, Xj - thermal conductivity, respectively, of the pipeline material, insulation, apparatus walls, i-th layer of the wall, W/(m K); 5 ST. — apparatus wall thickness, m.

The heat transfer coefficient is determined by the formula

or according to the empirical equation

The transfer of heat from the walls of a pipeline or apparatus to the environment is characterized by the coefficient a n [W/(m 2 K)], which is determined by criterion or empirical equations:

according to criterion equations:

Heat transfer coefficients a b i a n are calculated using criterion or empirical equations. If the hot coolant is hot water or condensing steam, then a in > a n, i.e. R B< R H , и величиной R B можно пренебречь. Если горячим теплоносителем является воздух или перегретый пар, то а в [Вт/(м 2 - К)] рассчитывают по критериальным уравнениям:

according to empirical equations:

Thermal insulation of devices and pipelines is made of materials with low thermal conductivity. Well-chosen thermal insulation can reduce heat loss into the surrounding space by 70% or more. In addition, it increases the productivity of thermal installations and improves working conditions.

Thermal insulation of a pipeline consists mainly of one layer, covered on top for strength with a layer of sheet metal (roofing steel, aluminum, etc.), dry plaster made from cement mortars, etc. If a covering layer of metal is used, its thermal resistance can be neglected. If the covering layer is plaster, then its thermal conductivity differs slightly from the thermal conductivity of thermal insulation. In this case, the thickness of the coating layer is, mm: for pipes with a diameter of less than 100 mm - 10; for pipes with a diameter of 100-1000 mm - 15; for pipes with large diameter - 20.

The thickness of the thermal insulation and cover layer should not exceed the maximum thickness, depending on the mass loads on the pipeline and its overall dimensions. In table Table 23 shows the values ​​of the maximum thickness of steam pipeline insulation recommended by the thermal insulation design standards.

Thermal insulation of technological devices can be single-layer or multi-layer. Heat loss through thermal

insulation depends on the type of material. Heat loss in pipelines is calculated per 1 and 100 m of pipeline length, in technological equipment - per 1 m 2 of the surface of the apparatus.

A layer of contaminants on the inner walls of pipelines creates additional thermal resistance to the transfer of heat into the surrounding space. Thermal resistances R (m. K/W) during the movement of some coolants have the following values:

In the pipelines supplying technological solutions to devices and hot coolants to heat exchange units, there are shaped parts in which part of the heat of the flow is lost. Local heat loss (W/m) is determined by the formula

Local resistance coefficients of pipeline fittings have the following values:

When compiling the table. 24 calculation of specific heat losses was carried out for seamless steel pipelines (pressure< 3,93 МПа). При расчете тепловых потерь исходили из следующих данных: тем-

the air temperature in the room was taken to be 20 °C; its speed during free convection is 0.2 m/s; steam pressure - 1x10 5 Pa; water temperature - 50 and 70 °C; thermal insulation is made in one layer of asbestos cord, = 0.15 W/(m K); heat transfer coefficient a„ = 15 W/(m 2 - K).

Example 1. Calculation of specific heat losses in a steam pipeline.

Example 2. Calculation of specific heat losses in a non-insulated pipeline.

Specified conditions

Steel pipeline with a diameter of 108 mm. Nominal diameter d y = 100 mm. Steam temperature 110°C, ambient temperature 18°C. Thermal conductivity of steel X = 45 W/(m K).

The data obtained indicate that the use of thermal insulation reduces heat losses per 1 m of pipeline length by 2.2 times.

Specific heat losses, W/m2, in technological equipment for tanning and fulling-felt production are:

Example 3. Calculation of specific heat losses in technological devices.

1. The “Giant” drum is made of larch.

2. Dryer from Hirako Kinzoku.

3. Longboat for dyeing berets. Made of stainless steel [k = 17.5 W/(m-K)]; there is no thermal insulation. Overall dimensions of the longboat 1.5 x 1.4 x 1.4 m. Wall thickness 8 ST = 4 mm. Process temperature t = = 90 °C; air in the workshop / av = 20 °C. Air speed in the workshop v = 0.2 m/s.

The heat transfer coefficient a can be calculated as follows: a = 9.74 + 0.07 At. At /av = 20 °C a is 10-17 W/(m 2 K).

If the surface of the coolant of the apparatus is open, the specific heat losses from this surface (W/m2) are calculated using the formula

Industrial service "Capricorn" (Great Britain) proposes to use the "Alplas" system to reduce heat losses from open surfaces of coolants. The system is based on the use of hollow polypropylene floating balls that almost completely cover the surface of the liquid. Experiments have shown that at a water temperature in an open tank of 90 °C, heat losses when using a layer of balls are reduced by 69.5%, two layers - by 75.5%.

Example 4. Calculation of specific heat losses through the walls of a drying unit.

The walls of the drying unit can be made of various materials. Consider the following wall designs:

1. Two layers of steel 5 ST = 3 mm thick with insulation between them in the form of an asbestos board 5 I = 3 cm thick and thermal conductivity X U = 0.08 W/(m K).

The heat balance of a boiler unit establishes equality between the amount of heat entering the unit and its consumption. Based on the heat balance of the boiler unit, fuel consumption is determined and the coefficient is calculated useful action, which is the most important characteristic energy efficiency of the boiler.

In a boiler unit, the chemically bound energy of the fuel is converted into physical heat of combustible combustion products during the combustion process. This heat is spent on generating and superheating steam or heating water. Due to inevitable losses during heat transfer and energy conversion, the generated product (steam, water, etc.) absorbs only part of the heat. The other part consists of losses, which depend on the efficiency of the organization of energy conversion processes (fuel combustion) and heat transfer to the generated product.

The heat balance of a boiler unit consists of establishing equality between the amount of heat supplied to the unit and the sum of the heat used and heat losses. The heat balance of the boiler unit is compiled for 1 kg of solid or liquid fuel or for 1 m 3 of gas. An equation in which the heat balance of a boiler unit for the steady thermal state of the unit is written in the following form:

Q r / r = Q 1 + ∑Q n

Q p / p = Q 1 + Q 2 + Q 3 + Q 4 + Q 5 + Q 6 (19.3)

Where Q r / r is the heat available; Q 1 - used heat; ∑Q n - total losses; Q 2 - heat loss with exhaust gases; Q 3 - heat loss from chemical underburning; Q 4 - heat loss from mechanical incomplete combustion; Q 5 - heat loss to the environment; Q 6 - heat loss with physical heat of slag.

If each term on the right side of equation (19.3) is divided by Q p/ p and multiplied by 100%, we obtain the second type of equation in which the heat balance of the boiler unit is:

q 1 + q 2 + q 3 + q 4 + q 5 + q 6 = 100% (19.4)

In equation (19.4), the value q 1 represents the gross efficiency of the installation. It does not take into account the energy costs for servicing the boiler installation: driving smoke exhausters, fans, feed pumps and other costs. The “net” efficiency is less than the “gross” efficiency, since it takes into account the energy costs for the installation’s own needs.

The left incoming side of the heat balance equation (19.3) is the sum of the following quantities:

Q p / p = Q p / n + Q v.vn + Q steam + Q physical t (19.5)

where Q B.BH is the heat introduced into the boiler unit with air per 1 kg of fuel. This heat is taken into account when the air is heated outside the boiler unit (for example, in steam or electric heaters installed before the air heater); if the air is heated only in the air heater, then this heat is not taken into account, since it is returned to the furnace of the unit; Q steam - heat introduced into the furnace with blown (nozzle) steam per 1 kg of fuel; Q physical heat - physical heat of 1 kg or 1 m 3 of fuel.

The heat introduced with air is calculated by the equality

Q B.BH = β V 0 C p (T g.in - T x.in)

where β is the ratio of the amount of air at the inlet to the air heater to the theoretically required one; с р - average volumetric isobaric heat capacity of air; at air temperatures up to 600 K can be considered with p = 1.33 kJ/(m 3 K); T g.v - temperature of heated air, K; T cold air is the temperature of cold air, usually taken to be 300 K.

The heat introduced with steam for atomizing fuel oil (nozzle steam) is found using the formula:

Q pairs = W f (i f - r)

where W f - nozzle steam consumption equal to 0.3 - 0.4 kg/kg; i f - enthalpy of nozzle steam, kJ/kg; r is the heat of vaporization, kJ/kg.

Physical heat of 1 kg of fuel:

Q physical t - s t (T t - 273),

where c t is the heat capacity of the fuel, kJ/(kgK); Tt - fuel temperature, K.

The value of Q physical. t is usually insignificant and is rarely taken into account in calculations. The exceptions are fuel oil and low-calorie combustible gas, for which the value of Q physical t is significant and must be taken into account.

If there is no preheating of air and fuel and steam is not used to atomize fuel, then Q p / p = Q p / n. The heat loss terms in the heat balance equation of the boiler unit are calculated based on the equalities given below.

1. Heat loss with exhaust gases Q 2 (q 2) is determined as the difference between the enthalpy of gases at the outlet of the boiler unit and the air entering the boiler unit (double-air heater), i.e.

where V r is the volume of combustion products of 1 kg of fuel, determined by formula (18.46), m 3 /kg; c р.r, с р.в - average volumetric isobaric heat capacities products of combustion of fuel and air, defined as heat capacities gas mixture(§ 1.3) using tables (see appendix 1); Tx, Tx.in - temperatures of exhaust gases and cold air; a is a coefficient that takes into account losses from mechanical underburning of fuel.

Boiler units and industrial furnaces usually operate under a certain vacuum, which is created by smoke exhausters and a chimney. As a result, through leaks in fences, as well as through inspection hatches, etc. a certain amount of air is sucked in from the atmosphere, the volume of which must be taken into account when calculating Ix.

The enthalpy of all air entering the unit (including suction cups) is determined by the coefficient of excess air at the outlet of the installation α ух = α t + ∆α.

The total air leakage in boiler installations should not exceed ∆α = 0.2 ÷ 0.3.

Of all heat losses, the value of Q 2 is the most significant. The value of Q2 increases with an increase in the excess air coefficient, temperature of exhaust gases, humidity of solid fuel and ballasting of gaseous fuel with non-combustible gases. Reduced air intake and improved combustion quality lead to a slight decrease in heat loss Q 2 . The main determining factor influencing heat loss from flue gases is their temperature. To reduce Tx, the area of ​​heat-using heating surfaces - air heaters and economizers - is increased.

The value of Tx affects not only the efficiency of the unit, but also the capital costs required for installing air heaters or economizers. As Tx decreases, efficiency increases and fuel consumption and costs decrease. However, this increases the area of ​​heat-using surfaces (at low temperature pressures, the heat exchange surface area must be increased; see § 16.1), resulting in increased installation costs and operating costs. Therefore, for newly designed boiler units or other heat-consuming installations, the value of Tx is determined from a technical and economic calculation, which takes into account the influence of Tx not only on efficiency, but also on the amount of capital costs and operating costs.

Another important factor, influencing the choice of Tx, is the sulfur content in the fuel. At low temperature (less than dew point temperature flue gases) condensation of water vapor on the heating surface pipes is possible. When interacting with sulfur dioxide and sulfuric anhydrides, which are present in combustion products, form sulfur dioxide and sulfuric acid. As a result, the heating surfaces are subject to intense corrosion.

Modern boiler units and kilns building materials have Tx = 390 - 470 K. When burning gas and solid fuels with low humidity, Tx - 390 - 400 K, wet coals

Tx = 410 - 420 K, fuel oil Tx = 440 - 460 K.

The humidity of the fuel and non-combustible gaseous impurities are gas-forming ballast, which increases the amount of combustion products obtained during combustion of the fuel. In this case, losses Q 2 increase.

When using formula (19.6), it should be borne in mind that the volumes of combustion products are calculated without taking into account the mechanical underburning of fuel. The actual amount of combustion products, taking into account the mechanical incompleteness of combustion, will be less. This circumstance is taken into account by introducing the correction factor a = 1 - p 4 /100 into formula (19.6).

2. Heat loss from chemical underburning Q 3 (q 3). Gases at the exit from the furnace may contain products of incomplete combustion of fuel CO, H 2, CH 4, the heat of combustion of which is not used in the combustion volume and further along the boiler unit path. The total heat of combustion of these gases causes chemical underburning. The causes of chemical underburning may be:

• lack of oxidizing agent (α<; 1);
• poor mixing of fuel with oxidizer (α ≥ 1);
• large excess of air;
• low or excessively high specific energy release in the combustion chamber q v, kW/m 3.

The lack of air leads to the fact that some of the combustible elements of the gaseous products of incomplete combustion of fuel may not burn at all due to the lack of an oxidizer.

Poor mixing of fuel with air causes either a local lack of oxygen in the combustion zone, or, conversely, a large excess of it. A large excess of air causes a decrease in combustion temperature, which reduces the rate of combustion reactions and makes the combustion process unstable.

Low specific heat release in the furnace (q v = BQ p / n /V t, where B is fuel consumption; V T is the volume of the furnace) causes strong heat dissipation in the combustion volume and leads to a decrease in temperature. Excessive qv values ​​also cause the appearance of chemical underburning. This is explained by the fact that the completion of the combustion reaction requires a certain time, and with a significantly increased value of qv, the time the fuel-air mixture remains in the combustion volume (i.e., in the zone of the highest temperatures) turns out to be insufficient and leads to the appearance of combustible components in the gaseous combustion products. In the furnaces of modern boiler units, the permissible qv value reaches 170 - 350 kW/m 3 (see § 19.2).

For newly designed boiler units, qv values ​​are selected according to standard data depending on the type of fuel burned, combustion method and design of the combustion device. During balance tests of operating boiler units, the value of Q 3 is calculated based on gas analysis data.

When burning solid or liquid fuel, the value of Q 3, kJ/kg, can be determined by formula (19.7)

3. Heat loss from mechanical incomplete combustion of fuel Q 4 (g 4). When burning solid fuel, residues (ash, slag) may contain a certain amount of unburned combustible substances (mainly carbon). As a result, the chemically bound energy of the fuel is partially lost.

Heat loss from mechanical incomplete combustion includes heat losses due to:

• failure of small fuel particles through the gaps in the grate Q pr (q pr);
• removal of some part of unburnt fuel with slag and ash Q shl (q shl);
• entrainment of small fuel particles by flue gases Q un (q un)

Q 4 - Q pp + Q un + Q shl

Heat loss q yn takes on large values ​​during flaring of pulverized fuel, as well as when burning non-caking coals in a layer on fixed or movable grates. The value of q un for layered furnaces depends on the apparent specific energy release (heat voltage) of the combustion mirror q R, kW/m 2, i.e. on the amount of released thermal energy per 1 m 2 of burning fuel layer.

The permissible value of q R BQ p / n / R (B - fuel consumption; R - combustion surface area) depends on the type of solid fuel burned, the design of the furnace, the excess air coefficient, etc. In layered furnaces of modern boiler units, the value of q R has values ​​in the range of 800 - 1100 kW/m 2. When calculating boiler units, the values ​​q R, q 4 = q np + q shl + q un are taken according to standard materials. During balance tests, heat loss from mechanical underburning is calculated based on the results of laboratory technical analysis of dry solid residues for their carbon content. Typically for fireboxes with manual fuel loading q 4 = 5 ÷ 10%, and for mechanical and semi-mechanical fireboxes q 4 = 1 ÷ 10%. When burning pulverized fuel in a flare in boiler units of medium and high power, q 4 = 0.5 ÷ 5%.

4. Heat loss to the environment Q 5 (q 5) depends on a large number of factors and mainly on the size and design of the boiler and furnace, the thermal conductivity of the material and the thickness of the lining walls, the thermal performance of the boiler unit, the temperature of the outer layer of the lining and the surrounding air, etc. d.

Heat losses to the environment at nominal performance are determined according to standard data depending on the power of the boiler unit and the presence of additional heating surfaces (economizer). For steam boilers with steam output up to 2.78 kg/s q 5 - 2 - 4%, up to 16.7 kg/s - q 5 - 1 - 2%, more than 16.7 kg/s - q 5 = 1 - 0 ,5%.

Heat losses to the environment are distributed among the various gas ducts of the boiler unit (furnace, superheater, economizer, etc.) in proportion to the heat given off by the gases in these gas ducts. These losses are taken into account by introducing the heat retention coefficient φ = 1 q 5 /(q 5 + ȵ k.a) where ȵ k.a is the efficiency of the boiler unit.

5. The loss of heat with the physical heat of ash and slag removed from the furnaces Q 6 (q 6) is insignificant, and it should be taken into account only for layer and chamber combustion of multi-ash fuels (such as brown coals, shale), for which it is 1 - 1, 5%.

Heat loss from hot ash and slag q 6,%, is calculated using the formula

where a shl is the proportion of fuel ash in the slag; C shl - heat capacity of slag; T slag - slag temperature.

When flaring pulverized fuel, a sh = 1 - a un (a un is the proportion of fuel ash carried away from the furnace with gases).

For layered fireboxes a sl shl = a shl + a pr (a pr is the share of fuel ash in the “dip”). For dry slag removal, the slag temperature is assumed to be Tsh = 870 K.

With liquid slag removal, which is sometimes observed during flaring of pulverized fuel, T ash = T ash + 100 K (T ash is the temperature of the ash in the liquid-melting state). When burning oil shale in layers, a correction is introduced to the ash content Ar for the carbon dioxide content of carbonates, equal to 0.3 (CO 2), i.e. The ash content is assumed to be equal to AR + 0.3 (CO 2) r / k. If the removed slag is in a liquid state, then the value of q 6 reaches 3%.

In furnaces and dryers used in the building materials industry, in addition to the heat losses considered, it is also necessary to take into account the heating losses of transport devices (for example, trolleys) on which the material is subjected to heat treatment. These losses can reach 4% or more.

Thus, the "gross" efficiency can be defined as

ȵ k.a = g 1 - 100 - ∑q losses (19.9)

We denote the heat absorbed by the generated product (steam, water) as Qк.a, kW, then we have:

for steam boilers

Q 1 = Q k.a = D (i n.n - i p.n) + pD/100 (i - i p.v) (19.10)

for hot water boilers

Q 1 = Q c.a = M in c r.v (T out - T in) (19.11)

Where D is the boiler productivity, kg/s; i p.p - enthalpy of superheated steam (if the boiler produces saturated steam, then instead of i p.v should be put (i p.v) kJ/kg; i p.v - enthalpy of feed water, kJ/kg; p - amount of water removed from the boiler unit in order to maintain the permissible salt content in the boiler water (the so-called continuous blowdown of the boiler), %; i - enthalpy of boiler water, kJ/kg; M in - water flow through the boiler unit, kg/s; c r.v - heat capacity of water , kJ/(kgK); Tout - temperature of hot water at the exit from the boiler; Tin - temperature of water at the entrance to the boiler.

Fuel consumption B, kg/s or m 3 /s, is determined by the formula

B = Q k.a /(Q r / n ȵ k.a) (19.12)

The volume of combustion products (see § 18.5) is determined without taking into account losses from mechanical underburning. Therefore, further calculation of the boiler unit (heat exchange in the furnace, determination of the area of ​​heating surfaces in the flues, air heater and economizer) is carried out according to the estimated amount of fuel B p:

(19.13)

When burning gas and fuel oil, B p = B.

The heat flow Q p through the surface S st of the dryer walls is calculated using the heat transfer equation:

Q p = k*Δt avg *S st,

The heat transfer coefficient k is calculated using the formula for a multilayer wall:

where δ and λ are the thickness and thermal conductivity coefficient of various layers of lining and thermal insulation, respectively.

Let's find the value of the criterion Re:

Re=v*l/υ=2.5 m/s*1.65 m/29*10 -6 m 2 /s=142241

Nu=0.66*Re 0.5 *Pr 0.33 =0.66*142241 0.5 *1.17 0.33 =262.2.

Heat transfer coefficient α from the drying agent to the inner surface of the walls:

α 1 =Nu* λ/l=262.2*3.53*10 -2 W/(m*K)/1.65 m=5.61 W/m 2 *K.

The total heat transfer coefficient by convection and radiation from the outer wall to the surrounding air:

α 2 =9.74+0.07*(t st -t c),

where t cf is the temperature of the outer wall, t st =40 0 C,

t in – ambient air temperature, t in = 20 0 C,

α 2 =9.74+0.07*(40 0 C-20 0 C)=11.14 W/m 2 *K.

Based on the temperature of the gases, we select the thickness of the lining (Table 3.1)

linings –

fireclay – 125 mm

steel – 20 mm

chamotte – 1.05 W/m*K

steel - 46.5 W/m*K

We find the heat transfer coefficient:

We determine the surface of the wall S st:

S st =π*d*l=3.14*1.6 m*8 m=40.2 m 2 ,

Q p =2.581 W/(m 2 *K)*89 0 C*40.2 m 2 =9234 W.

The specific heat loss to the environment is determined by the formula:

where W is the mass of moisture removed from the dried material in 1 s.

q p =9234 W/0.061 kg/s=151377.05 W*s/kg.

2.3. Calculation of heater for air drying

The total amount of heat Q 0 is calculated using the formula:

Q 0 =L*(I 1 -I 0)

Q 0 =2.46 kg/s *(159 kJ/kg +3.35 kJ/kg)=399.381 kW

Let's calculate the average temperature difference using the formula of the logarithmic equation:

where Δt m =t 1 -t 2n

Δt b =t 1 -t 2k

t 1 - temperature of heating steam (equal to the saturation temperature of steam at a given pressure).

At a pressure of 5.5 atm. t 1 =154.6 0 C (st. 550)

t 2н, t 2к - air temperature at the entrance to the calorimeter and exit from it, t 2к =150 0 С; t 2n = -7.7 0 C.

Δt b =154.6 0 C+7.7 0 C=162.3 0 C,

Δt m =154.6 0 С-150 0 С=4.6 0 С,

The heat exchange surface S t of the calorimeter is determined by the heat transfer equation:

S t =Q 0 /k Δt avg.,

where k is the heat transfer coefficient, which for finned heaters is applied depending on the air mass velocity ρ*v. Let ρ*v =3 kg/m 2 *s; then k=30 W/m 2 *k.

Find the required number n of heater sections:

n k. =S t / S s,

where S с is the heat transfer surface of the section.

Let's take a finned heater:

Since the actual number of sections is chosen with a 15-20% margin, then n. =6.23+6.23*0.15=7.2≈8 sections.

The mass velocity of air in the heater is calculated:

where L is the flow rate of absolutely dry air,

Table of contents of the topic "Regulation of metabolism and energy. Rational nutrition. Basal metabolism. Body temperature and its regulation.":
1. Energy expenditure of the body under conditions of physical activity. Physical activity rate. Work increase.
2. Regulation of metabolism and energy. Metabolic regulation center. Modulators.
3. Blood glucose concentration. Scheme for regulating glucose concentration. Hypoglycemia. Hypoglycemic coma. Hunger.
4. Nutrition. Nutritional norm. The ratio of proteins, fats and carbohydrates. Energy value. Calorie content.
5. Diet of pregnant and lactating women. Baby food ration. Distribution of daily ration. Alimentary fiber.
6. Rational nutrition as a factor in maintaining and strengthening health. Healthy lifestyle. Meal regimen.
7. Body temperature and its regulation. Homeothermic. Poikilothermic. Isothermy. Heterothermic organisms.
8. Normal body temperature. Homeothermic nucleus. Poikilothermic shell. Comfort temperature. Human body temperature.
9. Heat production. Primary heat. Endogenous thermoregulation. Secondary heat. Contractile thermogenesis. Non-contractile thermogenesis.

There are the following ways for the body to release heat: into the environment: radiation, heat conduction, convection And evaporation.

Radiation- this is a method of transferring heat to the environment by the surface of the human body in the form of electromagnetic waves in the infrared range (a = 5-20 microns). The amount of heat dissipated by the body into the environment by radiation is proportional to the surface area of ​​the radiation and the difference between the average temperatures of the skin and the environment. Radiation surface area is the total surface area of ​​those parts of the body that come into contact with air. At an ambient temperature of 20 °C and a relative air humidity of 40-60%, the adult human body dissipates by radiation about 40-50% of the total heat given off. Heat transfer by radiation increases as the ambient temperature decreases and decreases as it increases. Under conditions of constant ambient temperature, radiation from the body surface increases as the skin temperature increases and decreases as it decreases. If the average temperatures of the surface of the skin and the environment are equalized (the temperature difference becomes zero), heat transfer by radiation becomes impossible. It is possible to reduce the heat transfer of the body by radiation by reducing the surface area of ​​the radiation (“curling the body into a ball”). If the ambient temperature exceeds the average skin temperature, the human body, absorbing infrared rays emitted by surrounding objects, warms up.

Rice. 13.4. Types of heat transfer. The ways in which the body transfers heat to the external environment can be conditionally divided into “wet” heat transfer, associated with the evaporation of sweat and moisture from the skin and mucous membranes, and “dry” heat transfer, which is not associated with the loss of fluid.

Thermal conduction- a method of heat transfer that occurs during contact or contact of the human body with other physical bodies. The amount of heat given off by the body to the environment in this way is proportional to the difference in the average temperatures of the contacting bodies, the area of ​​the contacting surfaces, the time of thermal contact and the thermal conductivity of the contacting body. Dry air and adipose tissue are characterized by low thermal conductivity and are heat insulators. The use of clothing made from fabrics containing a large number of small, stationary air “bubbles” between the fibers (for example, woolen fabrics) allows the human body to reduce heat dissipation through conduction. Humid air and water saturated with water vapor are characterized by high thermal conductivity. Therefore, a person’s stay in an environment with high humidity and low temperature is accompanied by increased heat loss from the body. Wet clothing also loses its insulating properties.

Convection- a method of heat transfer from the body, carried out by transferring heat by moving air (water) particles. To dissipate heat by convection, a flow of air with a lower temperature than the temperature of the skin is required over the surface of the body. In this case, the layer of air in contact with the skin heats up, reduces its density, rises and is replaced by colder and more dense air. Under conditions when the air temperature is 20 °C and the relative humidity is 40-60%, the body of an adult dissipates about 25-30% of heat into the environment by conduction and convection (basic convection). As the speed of air flow (wind, ventilation) increases, the intensity of heat transfer (forced convection) also increases significantly.

Heat release from the body by heat conduction, convection And out of the way meanings, called together "dry" heat transfer, becomes ineffective when the average temperatures of the body surface and the environment are equalized.

Heat transfer by evaporation- this is the body’s way of dissipating heat into the environment due to its expenditure on the evaporation of sweat or moisture from the surface of the skin and moisture from the mucous membranes of the respiratory tract (“wet” heat transfer). In humans, sweat is constantly secreted by the sweat glands of the skin (“palpable,” or glandular, loss of water), and the mucous membranes of the respiratory tract are moisturized (“imperceptible” loss of water) (Fig. 13.4). In this case, the “perceptible” loss of water by the body has a more significant impact on the total amount of heat given off by evaporation than the “imperceptible” one.

At an external temperature of about 20 "C, the evaporation of moisture is about 36 g/h. Since 0.58 kcal of thermal energy is spent on the evaporation of 1 g of water in a person, it is easy to calculate that through evaporation the body of an adult person releases about 20% of all dissipated heat. Increasing external temperature, performing physical work, staying in heat-insulating clothing for a long time increase sweating and it can increase to 500-2000 g/h. If the external temperature exceeds the average skin temperature, then the body cannot release it into the external environment heat by radiation, convection and heat conduction. The body under these conditions begins to absorb heat from the outside, and the only way to dissipate heat is to increase the evaporation of moisture from the surface of the body. Such evaporation is possible as long as the ambient air humidity remains less than 100%. With intense sweating, high humidity and low air speed, when drops of sweat, without having time to evaporate, merge and flow from the surface of the body, heat transfer by evaporation becomes less effective.