DETERMINATION OF EMISSION AND BODY BLACK

Thermal radiation is a process of transferring thermal energy through electromagnetic waves. The amount of heat transferred by radiation depends on the properties of the radiating body and its temperature and does not depend on the temperature of the surrounding bodies.

In the general case, the heat flux that enters the body is partially absorbed, partially reflected, and partially passes through the body (Fig. 5.2).

Q=Q A+Q R+Q D ,

Rice. 5.2. Radiant Energy Distribution Diagram

where Q is the heat flux incident on the body;

Q A- the amount of heat absorbed by the body,

Q R- the amount of heat reflected by the body,

Q D is the amount of heat passing through the body.

We divide the right and left parts by the heat flux:

Quantities A, R, D, are called respectively: absorptive, reflective and transmittance of the body.

If a R=D=0, then A=1, i.e. all the heat flux falling on the body is absorbed. Such a body is called completely black.

Bodies that have A=D=0, R=1, i.e. all the heat flux incident on the body is reflected from it, are called white . In this case, if the reflection from the surface obeys the laws of the optics of the body, it is called mirrored - if the reflection is diffuse - absolutely white.

Bodies that have A=R=0 and D=1, i.e. all the flow that falls on the body, passes through it, are called diathermic or completely transparent.

Absolute bodies do not exist in nature, but the concept of such bodies is very useful, especially about a completely black body, since the laws governing its radiation are especially simple, because no radiation is reflected from its surface.

In addition, the concept of a completely black body makes it possible to prove that in nature there are no such bodies that radiate more heat than black ones. For example, in accordance with Kirchhoff's law, the ratio of the emissivity of a body E and its absorbency BUT the same for all bodies and depends only on temperature, for all bodies, including absolutely black, at a given temperature:

.

Since the absorptive power of a perfect black body A o=1, and A 1 and A2 etc. is always less than 1, then it follows from Kirchhoff's law that the limiting emissivity E o has a completely black body. Since there are no absolutely black bodies in nature, the concept of a gray body is introduced, its degree of blackness e, which is the ratio of the emissivity of a gray and black body:

Following Kirchhoff's law and taking into account that A o=1, we can write , whence A=e, i.e. the degree of emissivity characterizes both the relative emissivity and absorptivity of the body. The basic law of radiation, reflecting the dependence of the radiation intensity E o, related to this wavelength range (monochromatic radiation), is Planck's law.

,

where l- wavelength, [m];

From 1\u003d 3.74 × 10 -6 W × m 2, From 2=1.4338×10 -2 m×K;

C1 and From 2 are the first and second Planck constants.

Rice. 5.3. Graphical representation of Planck's law

As can be seen from the graph, a black body radiates at any temperature in a wide range of wavelengths. As the temperature increases, the maximum radiation intensity shifts towards shorter wavelengths. This phenomenon is described by Wien's law:

l max T=2.898×10 -3 m×K,

where lmax is the wavelength corresponding to the maximum radiation intensity.

For values lT>>From 2 instead of Planck's law, you can apply the Rayleigh-Jeans law, which is also called the "law of long-wave radiation":

Radiation intensity referred to the entire wavelength range from l=0 to l=(integral radiation), can be determined from Planck's law by integrating:

where C o\u003d 5.67 W / (m 2 × K 4) - the coefficient of a completely black body. Expression (5.9) is called the Stefan-Boltzmann law, which was established by Boltzmann. For gray bodies, the Stefan-Boltzmann law is written as

. (5.10)

With=C o e is the emissivity of the gray body. Heat exchange by radiation between two surfaces is determined on the basis of the Stefan-Boltzmann law and has the form

, (5.11)

where e PR is the reduced emissivity of two bodies with surfaces H 1 and H 2 ;

. (5.12)

If H 1<<H 2 then the reduced emissivity becomes equal to the emissivity of the surface H 1, i.e. e PR=e 1 . This circumstance is the basis of the method for determining the emissivity and emissivity of gray bodies that are small in size compared to bodies that exchange radiant energy with each other.

. (5.13)

As can be seen from formula (5.13), to determine the degree of emissivity and emissivity With gray body needs to know the surface temperature T W test body, temperature T f environment and radiant heat flux from the surface of the body Q and. Temperatures T W and T f can be measured by known methods, and the radiant heat flux is determined from the following considerations:

The spread of heat from the surface of bodies into the surrounding space occurs through radiation and heat transfer during free convection. Full flow Q from the surface, the body will thus be equal to:

Q = Q L + Q K, whence Q L = Q - Q K ; (5.14)

Q K is the convective component of the heat flux, which can be determined according to Newton's law:

Q K = a K H(tw - tf) (5.15)

On the other hand, the heat transfer coefficient a K can be determined from the expression (see work #3):

a K = Nu f a f /d(5.16)

where Nu f = c(Gr f Pr f)n. (5.17)

The determining temperature in these expressions is the ambient temperature t f .

5.5.4. Scheme of the experimental setup

The experimental setup, the schematic diagram of which is shown in fig. 4 is designed to determine the emissivity of two bodies - copper and aluminum. The investigated bodies are copper (9) and aluminum (10) tubes (elements No. 1 and 2) with a diameter d1=18mm and d2=20mm long L=460mm, arranged horizontally. Inside the tubes there are electric heaters 11 made of nichrome wire, which serve as a heat source. The heat flow is distributed evenly along the length of the pipe. In stationary mode, all the heat generated by the electric heater is transferred through the surface of the pipe to the environment. Total heat dissipation Q from the surface of the pipe is determined by the consumption of electricity. The power consumption of electricity is regulated by an autotransformer and measured by an ammeter and a voltmeter or wattmeter.

Rice. 5.4. Scheme of the experimental setup

To reduce heat loss from the ends of the tubes, heat-insulating plugs (12) are placed. To measure the surface temperature in the walls of each of the tubes, 5 copper-constant thermocouples are laid (No. 1-5 the first pipe and No. 7-11 the second pipe). The thermocouples are connected in turn to the measuring device (13) using the switch (14).

5.5.5. The procedure for conducting experiments and processing the results

Before proceeding with the laboratory work, it is necessary to get acquainted with the theoretical material and the installation device. The work is carried out in two modes.

Table 5.2

Calculation table for work No. 2

No. p / p	Value name	Determination of quantities and design ratios	First mode
Element 1	Element 2
1.	Grasgoff's criterion
a.	Volume expansion coefficient
in.	temperature difference	Dt = tw - tf
with.	Air kinematic viscosity coefficient	nf, m 2 / sec
2.	Nusselt criterion	Nu f = c (Cr f Pr f)n
a.	Prandtl criterion	Pr f
in.	The coefficients are selected from the table. 6.2. (See work No. 3)	c
n
3.	Pipe surface
4.	Heat transfer coefficient
a.	Air thermal conductivity coefficient.	lf
5.	Convective component of heat flow.
6.	The amount of radiant heat flux
7.	Degree of blackness
8.	Emissivity
9.	Average emissivity

After taking measurements in the 1st mode, it is necessary to show the teacher the observation log, and then set the 2nd thermal mode. The steady state thermal regime occurs in about 3-5 minutes. while working on a PC.

On each of the modes it is necessary to produce with an interval of 2-3 minutes. at least 2 temperature measurements on each of the thermocouples and power according to the readings of the voltmeter and ammeter. Record the measurement data in the observation log - table. 5.1. Measurements should be made only in steady state. The results of the calculations are summarized in Table. 5.3. Build graphs based on the data e = f(t) for 2 tested materials. Compare the obtained data with the reference ones (Table 1 – appendices).

The physical parameters of air are taken from Table. 3 applications at defining temperature tf .

The calculation of the work is carried out according to the table. 5.2.

Table 5.3

Journal of observations for papers No. 2, 3, 4

for laminar flow

T
table 6

T (46) thermophysical parameters of dry air

at a pressure of 101.3 10³ Pa

for turbulent conditions

where λ m- thermal conductivity of the gas, for air, the value can be selected from the table. 6; N i- coefficient taking into account the orientation of the surface of the body:

8. Determine the thermal conductivity σ to between the surface of the body and

about environment:

where S n, S in, S b - areas of the lower, upper and side surfaces of the block body, respectively;

S n = S in = L one · L 2 ;S b = 2 L 3 (L 1 +L 2).

For more efficient heat removal, IVEP blocks with finned surfaces are often used. If the designer is faced with the task of performing a thermal calculation for this type of secondary power supply unit, then he needs to additionally determine the effective heat transfer coefficient α eff i of the finned i-th surface, which depends on the design of the ribs and the overheating of the body relative to the environment. α eff i is determined in the same way as in the calculation of radiators (see calculation of radiators, p. 5.5).

After determining the effective heat transfer coefficient α eff i , proceed to the calculation of the thermal conductivity of the entire body σ k, which consists of the sum of the conductivities of the non-finned σ to 0 and finned σ to p surfaces:

G
de σ to 0 is calculated by formula (47), but without taking into account the ribbed surface;

G
de S pi is the area of ​​the base of the ribbed surface; N i is a coefficient that takes into account the orientation of this surface.

9. We calculate the overheating of the housing of the IVEP block in the second approximation θ k0:

G
de To KP - coefficient depending on the perforation of the block body To P; To H1 - coefficient taking into account the atmospheric pressure of the environment.

Chart to determine the ratio To H1, shown in fig. 9, and the coefficient To KP in fig. fourteen.

The perforation coefficient is determined by (11) - (13), and according to the graph shown in fig. eight.

10. Determine the calculation error:

E
If δ ≤ 0.1, then the calculation can be considered complete. Otherwise, you should repeat the calculation of the temperature of the housing of the secondary power supply for a different value θ k, adjusted to the side θ to 0 .

11. Calculate the temperature of the block body:

H
and with this, the first stage of calculating the thermal regime of the IVEP unit is completed.

Stage 2. Determination of the average surface temperature of the heated zone.

1. Calculate the conditional specific surface power q from the heated zone of the block according to the formula (19).

2. From the graph in fig. 7 we find in the first approximation the overheating of the heated zone θ h relative to the temperature surrounding the block of the environment.

3. We determine the coefficients of heat transfer by radiation between the lower α zln, upper α zlv and side α zlb surfaces of the heated zone and the body:

where ε P i - reduced degree of emissivity i-th surface of the heated zone and body:

ε s i and S h
i - emissivity and area i th surface of the heated zone.

R is. fifteen

4. For determining temperature t m = ( t k + t 0 +θ h) / 2 and the determining size h i we find the Grashof number Gr hi and Prandtl Pr (formula (43) and Table 6).

5. We calculate the coefficients of convective heat transfer between the heated zone and the body for each surface;

for bottom surface

for top surface

d for side surface

6. We determine the thermal conductivity σ zk between the heated zone and the body:

G
de Toσ is the coefficient taking into account conductive heat transfer:

σ – specific thermal reducibility from the modules to the block body, depends on the pressing force to the body (Fig. 15); in the absence of clamping σ = 240 W / (m 2 K); Sλ is the contact area of ​​the module frame with the block body.

Table 7

Thermophysical properties of materials

7. We calculate the heating of the heated zone θ z0 in the second approximation:

G
de K w - determined according to the graph shown in fig. eleven; K h2 - determined according to the schedule (Fig. 10).

8. Determine the calculation error

E
if δ< 0,1, то расчет окончен. При δ ≥ 0,1 следует повторить расчет для скорректированного значенияθ h.

9. Calculate the temperature of the heated zone

E
step 3. Calculation of the surface temperature of the component included in the IVEP scheme

Here is the sequence of calculation required to determine the temperature of the case of the component installed in the module of the first level of disaggregation.

1. Determine the equivalent thermal conductivity of the module in which the component is located, for example, a microcircuit, for the following options:

in the absence of heat-conducting tires λ equiv = λ P, where λ P is the thermal conductivity of the base material of the board;

in the presence of heat-conducting tires

G de λ w - thermal conductivity of the material of the heat-conducting bus; V P - the volume of the printed circuit board, taking into account the volume of heat-conducting tires; V w is the volume of heat-conducting tires on the printed circuit board; A– surface fill factor of the module board with heat-conducting busbars:

G
de S w is the total area occupied by heat-conducting tires on the printed circuit board.

In table. 7 shows the thermophysical parameters of some materials.

2. Determine the equivalent radius of the microcircuit body:

G
de S o IC - the area of ​​​​the base of the microcircuit.

3. Calculate the heat flux propagation coefficient:

G
de α 1 and α 2 - heat transfer coefficients from the first and second sides of the printed circuit board; for natural heat exchange

δ P
– thickness of the printed circuit board of the module.

4. We determine the desired overheating of the surface of the microcircuit case:

where AT and M- conditional values ​​​​introduced to simplify the recording form: with a one-sided arrangement of microcircuit cases on a printed circuit board AT= 8.5π R 2 VT/K, M= 2; with two-sided arrangement of housings AT= 0,M= 1;To- empirical coefficient: for microcircuit cases, the center of which is less than 3 R,To= 1.14; for microcircuit packages, the center of which is separated from the ends of the printed circuit board at a distance of more than 3 R,To= 1;Toα - heat transfer coefficient from microcircuit cases is determined according to the graph shown in fig. sixteen; To 1 and To 0 – modified Bessel functions; N - number i- x microchip cases located at a distance of no more than 10/ m, i.e r i ≤ 10 m; Δ t c - average volume overheating of air in the block:

Q
ims i - power dissipated i-th microcircuit; S ims i - total surface area i-th microcircuit; δ s i - gap between the microcircuit and the board; λ s i - coefficient of thermal conductivity of the material filling this gap.

5. Determine the surface temperature of the microcircuit case:

P
The above algorithm for calculating the temperature of the microcircuit can be applied to any other discrete component that is part of the secondary power supply unit. In this case, the discrete component can be considered like a microcircuit with a local heat source on the plate, and the corresponding values ​​of the geometric parameters can be entered into equations (60) - (63).

The radiation of solid bodies is surface, and the radiation of gases is volumetric.

Heat transfer by radiation between two flat parallel gray surfaces of solids with temperatures T 0 1 abs and T 0 2 abs (T 1 > T 2) is calculated by the formula

C CR - reduced emissivity;

C 1 - emissivity of the surface of the first body;

C 2 - emissivity of the surface of the second body;

With s = 4.9 kcal / m 2 hour deg 1 - black body emissivity.

In practical calculations, it is more convenient to use the so-called emissivity

=.

Reduced emissivity

In the case when the first body with surface F 1 from all

sides surrounded by the surface F 2 of the second body, the amount of heat transferred is determined by the formula

The reduced emissivity and the reduced emissivity are determined by the formulas

In the case when F 2 >F 1, i.e.

C pr \u003d C 1 and pr = 1 .

In order to reduce heat loss due to radiation, so-called screens are used. The screen is a thin-walled sheet that covers the radiating surface and is located at a short distance from the latter. As a first approximation, convective heat transfer through the air gap between the screen and the radiating surface is not taken into account. Also, the thermal resistance of the wall of the screen itself is always neglected, i.e., the temperatures on its surfaces are considered the same.

For flat parallel screens, the formula for heat transfer by radiation is used with the replacement pr so-called equivalent emissivity

where 12 ,23, etc. - determined by the formula for pr, the reduced degree of emissivity during heat exchange by radiation between the 1st and 2nd surface, between the 2nd and 3rd surface, etc.

When shielding cylindrical bodies (pipes), the equivalent emissivity

The amount of transferred heat Q is calculated by the formula

Gas emission

Radiating gases are triatomic and polyatomic gases. Of greatest practical interest is radiation

CO 2 and H 2 O.

The emission of gases is selective and depends on the size and shape of the gas volume.

The amount of heat transferred by radiation from the gas volume, the components of which are CO 2 and H 2 O, to the surrounding shell, which has the properties of a gray body, is determined by the formula

where T gas is the absolute temperature of the radiating gas volume;

T article - the absolute temperature of the surrounding shell;

= 0,5 (+ 1) - effective degree of emissivity of the shell (at from 0.8 to 1.0);

=
+
- the degree of blackness of the gas, determined by the graphs of fig. 85 and 86 for the average gas temperature;

- the degree of blackness of the gas, determined by the same graphs, but by the temperature st of the shell;

β-correction for the partial pressure of water vapor, determined from the graph in fig. 87.

Degree of blackness of carbon dioxide
and water vapor
depends on the temperature of the gas volume and the effective thickness of the radiating layer ps, where p ata is the partial pressure of the radiating component and sm is the reduced beam length.

The reduced beam length can be approximately determined by the formula

where Vm 3 - volume filled with radiating gas (radiating volume);

Fm 2 - shell surface.

For individual special cases, the reduced beam length is determined by the following formulas:

for the volume of gas in the annular space (s 1 - longitudinal pitch, i.e., the distance between the axes of the pipes in a row; s 2 - transverse pitch, i.e., the pitch between rows; d - pipe diameter)

for a plane-parallel gas layer of infinite length with a thickness

s= 1.8 ;

for cylinder diameter d

Sometimes the concept of the heat transfer coefficient by radiation is introduced α l kcal / m 2 hour deg. This coefficient is determined by the formula

Example. Determine the amount of heat transferred by radiation from an incandescent steel plate, the surface temperature of which is t 1 = 1027 ° C, to another similar plate, the surface temperature of which is t 2 = 27 ° C, located parallel to the first.

Solution. From Appendix 20 we find the degree of blackness of the steel plate (oxidized):
. We define the reduced

emissivity according to the formula

Amount of transferred heat

Example. A steel steam pipeline with a diameter of 300 mm, the temperature of the outer wall of which t 1 \u003d 300 ° C, is laid indoors. In order to reduce heat losses, the steam pipeline is closed with a double cylindrical casing (screen). The first casing with a diameter of 320 mm is made of thin steel sheets ( = 0.82), the second casing with a diameter of 340 mm is made of thin aluminum sheets ( = 0.055). Determine the heat loss per 1 running. m of bare and shielded steam pipeline, as well as the temperature of the aluminum casing. Ignore convective heat transfer. The room temperature is 25°C.

Decision. Let us determine the heat loss by a bare steam pipeline, assuming that the surface of the steam pipeline F 1 is many times smaller than the surface of the walls of the room F 4 . At F 1<

pr = 1 = 0.80

(for oxidized steel).

According to the formula

Now let's determine the heat loss in the presence of screens. We determine the reduced emissivity coefficients:

Equivalent emissivity

The amount of heat transferred by radiation

Thus, as a result of the installation of screens, the heat loss decreased in

To determine the temperature of an aluminum sheet, we make an equation

Solving this equation, we find

Example. A thermocouple was used to measure the temperature of the hot air stream flowing through the channel. Radiant heat exchange occurs between the thermocouple junction and the channel walls (Fig. 88), which distorts the thermocouple readings. To reduce the error in measuring the temperature, the thermocouple is closed with a screen tube 1. Find the actual temperature of the air flow if the thermocouple shows a temperature of t = 200 ° C. The temperature of the inner channel wall t st = 100 ° C. The emissivity of the screen and thermocouple junction are the same and equal to 0.8. The heat transfer coefficient from air to the thermocouple junction α= 40 kcal/m 2 hour deg, and to the screen surface α= 10 kcal/m 2 hour deg.

Solution. Denote the real

(desired) air temperature t in.

Temperature determined by

thermocouple, is the temperature

her junction t.

Let us compose the heat balance equation for the thermocouple junction. The amount of heat received by the junction due to convection is

and the amount of heat given off by radiation from the surface F junction to the surface F e surrounding the thermocouple junction of the screen tube,

where T e is the absolute temperature of the inner surface of the screen tube.

Given that F e >> F, we get
.

In stationary mode, the heat balance for the thermocouple junction will be expressed by the equation

Let us now calculate the heat balance for the screen tube, neglecting the thermal resistance of the tube itself. Heat gain by convection

The heat input due to the radiation of the thermocouple junction is obviously equal to the heat

which in turn is equal to

Heat consumption due to radiation of the outer surface of the screen tube on the surrounding walls of the channel

and since in this case F st >> F e, then
. Thus, the heat balance of the screen tube is expressed by the equation

Usually in this equation the first term in the left hand side is neglected.

parts (by virtue of F e >> F). Then

The joint solution of the equations allows us to determine the desired

Temperature t in

We solve the resulting equations graphically, calculating from them

Temperature t in depending on t e. The point of intersection of the corresponding curves (Fig. 89) determines the temperature t in:

Error in determining the temperature using a thermocouple

Example. Determine the amount of heat transferred by radiation to steel pipes located in the flue of a water-tube steam boiler. The partial pressures of carbon dioxide in water vapor in flue gases are, respectively, p C O 2 = 0.15 atm and p H 2 O = 0.075 atm. Outer diameter of pipes d= 51 mm; their longitudinal pitches 1 = 90 mm and transverse pitches 2 = 70 mm. Gas temperature

n
at the inlet to the gas duct t / \u003d 1000 0 C, and at the outlet of the gas duct t // \u003d 800 0 C. The outdoor temperature

pipe surface is constant

and equal to t st \u003d 230 0 C.

Decision. Preliminary

determine the average temperature

gas flow, which we accept

equal to the design temperature t gas.

Corresponding effective layer thicknesses

According to the charts in Fig. 85 and 86 we find

Correction β for the partial pressure of water vapor (according to Fig. 87) β \u003d 1.06.

According to the formula

Radiant heat transfer coefficient

Example. A mixture of gases moves in a cylindrical steel pipe with an inner diameter d = 0.25 m. Average gas temperature t gas = 1100 0 C. Partial pressure of carbon dioxide

= 0.45 ata. Wall temperature t st \u003d 300 0 C. Determine the amount of heat transferred by radiation per 1 linear meter. m of pipe.

SOLUTION: Reduced beam length

S=0.9d=0.9 0.25=0.225 m.

Effective thickness of the radiating layer

s
\u003d 0.225 0.45 \u003d 0.101 m ata.

According to fig. 85 determined at t= 1100°C
\u003d 0.10: at t \u003d 300 0 С
= 0.095. Since there is no water vapor in the mixture, gas = 0.10 and
= 0,095.

According to the formula

For 1 line m

Tasks

453. Determine the amount of heat radiated by a steel plate at a temperature of t 1 \u003d 600 0 C on a brass sheet of the same size at a temperature of t 2 \u003d 27 0 C, located parallel to the plate. Determine also the coefficient of heat transfer by radiation.

Answer: q 12 \u003d 5840 kcal / m 2 hour; α l \u003d 10.2 kcal / m 2 hour deg.

454. Radiant heat exchange takes place between two parallel planes. Surface having temperature t 1 =

600°C and blackness \u003d 0.64, radiates heat in the amount

q 12 \u003d 1000 kcal / m 2 hour. Determine the temperature of the heat-receiving aluminum rough surface ( = 0,055).

Answer: t 2 \u003d 390 0 C.

455. Determine the amount of heat q 12 kcal / m 2 hour, radiated by the surface of a flat wall to another parallel flat wall. The wall temperatures are respectively equal to t 1 \u003d 227 ° C and t 2 \u003d 27 0 C. The determination is made for four options:

a) C 1 \u003d C 2 \u003d C s \u003d 4.9 kcal / m 2 hour deg 4 (absolutely black surfaces);

b) C 1 \u003d C 2 \u003d 4.3 kcal / m 2 hour deg 4 (matte steel surfaces);

c) C 1 \u003d 4.3 kcal / m 2 hour deg 4 (matte steel surface),

C 2 \u003d 0.3 kcal / m 2 hour deg 4 (tinplate);

d) C 1 \u003d C 2 \u003d 0.3 kcal / m 2 hour deg 4 (tinplate).

Answer: a) q 12 \u003d 2660 kcal / m 2 hour; 6) q 12 \u003d 2080 kcal / m 2 hour;

c) q 12 \u003d 160 kcal / m 2 hour; d) q 12 \u003d 84 kcal / m 2 hour.

456. A steel pipe with a diameter of d = 200 mm and a length of 1 = 5 m is located in a brick room, the width of which is a = 8 m and the height h = 5 m. Determine the heat loss for the pipe by radiation if the surface temperature of the pipe t 1 = 327 ° C, and temperature of the surface of the walls of the room t 2 = 27 ° C.

Answer: Q 12 \u003d 14950 kcal / hour.

457. Solve the previous problem, provided that a) the steel pipe is in a brick corridor with a section of 2 x 1 m and b) the steel pipe is in a brick channel with a section of 350 x 350 mm. The wall temperature in both cases is t 2 = 27 ° C. Compare the results with the answer to the previous problem.

Answer: a) Q 12 \u003d 14900 kcal / hour; b) Q 12 \u003d 14500 kcal / hour.

458. Determine the loss of heat due to radiation in one line. m of steel steam pipeline. The outer diameter of the steam pipeline is d \u003d 0.2 m, its surface temperature t 1 \u003d 310 0 C, and the temperature

ambient air t 2 \u003d 50 0 C. Compare the results of the solution with the answer to problem 442.

Answer: q \u003d 2575 kcal / running. m hour; heat loss due to radiation is 2.36 times greater than heat loss through convective heat transfer.

459. A cast-iron furnace door measuring 500 x 400 mm in a steam boiler has a temperature of t 1 = 540 ° C ( = 0.64). Determine the amount of radiated heat if the temperature in the boiler room t 2 \u003d 35 ° C. Also determine the heat transfer coefficient by radiation.

Answer: Q \u003d 2680 kcal / hour; α l \u003d 2b.5 kcal / m 2 hour deg.

460. Determine the heat transfer by radiation between matte steel parallel surfaces (see problem 455 6), if a screen is placed between them in the form of a thin steel sheet with the same emissivity.

Answer: q 12 \u003d 1040 kcal / m 2 hour.

461. Solve problem 460 provided that a screen is placed between the steel surfaces, consisting of four thin steel sheets with the same emissivity.

Answer: q 12 \u003d 416 kcal / m 2 hour.

462. Solve problem 455 6 provided that a screen of tinplate is placed between the steel surfaces. Compare the result of the solution with the answer to problem 455 6.

Answer: q 12 \u003d 81 kcal / m 2 hour, i.e. the amount of heat transferred decreases by about 25 times.

463. Solve problem 455 6 provided that a screen consisting of two sheets of tinplate is placed between the steel surfaces.

Answer: q 12 \u003d 41.5 kcal / m 2 hour.

464. The furnace of a steam boiler is filled with a fiery torch having a conditional temperature t 1 = 1000 0 C and a conditional degree of blackness = 0.3. Determine the amount of heat radiated through the firing hole of the furnace, closed by a cast-iron door ( \u003d 0.78) as well as the temperature of the door itself, if the temperature in the boiler room t 2 \u003d 30 0 С (the cast-iron door can be considered as a flat screen between the torch and the environment). The degree of emissivity of the environment is taken equal to 1.0.

Answer: q \u003d 25530 kcal / m 2 hour; t dv \u003d b5b ° C.

465. Solve the previous problem, provided that the cast-iron door is equipped with a cast-iron reflector located on the side of the furnace (such a reflector can be considered as a screen).

Answer: q \u003d 19890 kcal / m 2 hour; t dv \u003d 580 ° C.

466. Solve the example on page 225 provided that the thermocouple junction is not protected by a screen tube.

Answer: t in \u003d 230 0 C; the error in determining the temperature is 13%.

467. Solve problem 458 provided that the steam pipeline is surrounded by a sheet steel screen ( = 0.82). Screen diameter d e = 0.3 m. There is air between the steam line and the steel screen. When determining the heat loss due to radiation, the convective heat exchange between the screen and the air is not taken into account. Determine also the temperature of the screen. Compare the results with the answer to problem 458. Answer: q \u003d 1458 kcal / rm. m hour; t e \u003d 199 ° C.

468. Solve the previous problem, taking into account convective heat transfer between the screen and air, assuming the heat transfer coefficient equal to α e = 20 kcal / m 2 hour deg. Compare the result with the answer of problems 458 and 467.

Answer: q \u003d 1890 kcal / running. m hour; t e \u003d 126 ° C.

Indication. When solving problem 468, it is necessary to compose

heat balance equation.

469. A steam pipeline with a diameter of d \u003d 0.2 m (indicated in task 458) is covered with thermal insulation consisting of 5 aluminum foil screens ( = 0.055). The distance between foil layers is = 5 mm. Determine how many times the heat loss by radiation from an insulated steam pipeline is less than the heat loss from an uninsulated steam pipeline. Answer: 127 times less.

470. Determine the coefficient of heat transfer by radiation from flue gases to the walls of hot water pipes of a steam boiler. Outer diameter of pipes d= 44.5 mm, longitudinal pitch of pipes in a row

s 1 = 135 mm, and transverse pitch s 2 = 90 mm. The temperature of the gases at the inlet to the gas duct t / \u003d 900 0 C, and at the outlet t // \u003d 700 ° C. The surface temperature of the pipe walls t st = 300 ° C. The partial pressures of triatomic gases are:
= 0.18 ata and
= 0.08 ata.

Answer: α l 12.8 kcal/m 2 hour deg.

471. Solve the previous problem, provided that the steps of the pipes are reduced to s 1 = 81 mm and s 2 = 65 mm, and the rest of the initial data are left unchanged. Answer: α l \u003d 8 kcal / m 2 hour deg.

472. In a narrow channel with a cross section of 820 x 20 mm, a mixture of gases of the following composition (by volume) moves: N 2 = 73%; O 2 = 2%; CO 2 \u003d 15%; H 2 O \u003d 10%. The average temperature of the mixture of gases t gas = 900 ° C, the pressure of the mixture p = 1 atm. The channel walls are made of sheet steel. The temperature on the surface of the channel walls t st \u003d 100 ° C. Determine the amount of heat transferred from the gases to the channel walls by radiation. Answer: q \u003d 4000 kcal / m 2 hour.

Objective

Familiarization with the methodology for conducting experiments to determine the degree of blackness of the surface of the body.

Development of skills for conducting experiments.

Exercise

Determine the degree of emissivity ε and the emissivity from the surfaces of 2 different materials (painted copper and polished steel).

Determine the dependence of the change in the degree of emissivity on the surface temperature.

Compare the value of the emissivity of painted copper and polished steel with each other.

Theoretical Introduction

Thermal radiation is a process of transferring thermal energy through electromagnetic waves. The amount of heat transferred by radiation depends on the properties of the radiating body and its temperature and does not depend on the temperature of the surrounding bodies.

In the general case, the heat flux that enters the body is partially absorbed, partially reflected, and partially passes through the body (Fig. 1.1).

Rice. 1.1. Radiant Energy Distribution Diagram

(2)

where - heat flux incident on the body,

- the amount of heat absorbed by the body,

- the amount of heat reflected by the body,

- the amount of heat passing through the body.

We divide the right and left parts by the heat flux:

Quantities
are called respectively: absorptive, reflective and transmittance of the body.

If a
, then
, i.e. all the heat flux falling on the body is absorbed. Such a body is called absolutely black .

Bodies that have
,
those. all the heat flux incident on the body is reflected from it, are called white . In this case, if the reflection from the surface obeys the laws of the optics of the body, it is called mirrored – if the reflection is diffuse – absolutely white .

Bodies that have
,
those. all the heat flux incident on the body passes through it, are called diathermic or completely transparent .

Absolute bodies do not exist in nature, but the concept of such bodies is very useful, especially about a completely black body, since the laws governing its radiation are especially simple, because no radiation is reflected from its surface.

In addition, the concept of a completely black body makes it possible to prove that in nature there are no such bodies that radiate more heat than black ones.

For example, in accordance with Kirchhoff's law, the ratio of the emissivity of a body and its absorbency the same for all bodies and depends only on temperature, for all bodies, including absolutely black, at a given temperature:

(3)

Since the absorptive power of a perfect black body
a and etc. is always less than 1, then it follows from Kirchhoff's law that the limiting emissivity has a completely black body. Since there are no absolutely black bodies in nature, the concept of a gray body is introduced, its degree of blackness ε, which is the ratio of the emissivity of a gray and absolutely black body:

Following Kirchhoff's law and taking into account that
can be written
where
those . the degree of blackness characterizes both the relative emissivity and absorptivity of the body . The basic law of radiation, reflecting the dependence of the radiation intensity
related to this wavelength range (monochromatic radiation) is Planck's law.

(4)

where - wavelength, [m];

;

and are the first and second Planck constants.

On fig. 1.2 this equation is presented graphically.

Rice. 1.2. Graphical representation of Planck's law

As can be seen from the graph, a black body radiates at any temperature in a wide range of wavelengths. As the temperature increases, the maximum radiation intensity shifts towards shorter wavelengths. This phenomenon is described by Wien's law:

Where
is the wavelength corresponding to the maximum radiation intensity.

For values
instead of Planck's law, you can apply the Rayleigh-Jeans law, which is also called the "law of long-wave radiation":

(6)

Radiation intensity, referred to the entire wavelength range from
before
(integral radiation), can be determined from Planck's law by integrating:

where is the black body emissivity. The expression is called the Stefan-Boltzmann law, which was established by Boltzmann. For gray bodies, the Stefan-Boltzmann law is written as:

(8)

is the emissivity of the gray body. Heat exchange by radiation between two surfaces is determined on the basis of the Stefan-Boltzmann law and has the form:

(9)

If a
, then the reduced emissivity becomes equal to the emissivity of the surface , i.e.
. This circumstance is the basis of the method for determining the emissivity and emissivity of gray bodies that are small in size compared to bodies that exchange radiant energy with each other.

(10)

(11)

As can be seen from the formula, the definition of emissivity and emissivity With gray body needs to know the surface temperature test body, temperature environment and radiant heat flux from the surface of the body
. Temperatures and can be measured by known methods. And the radiant heat flux is determined from the following considerations.

The spread of heat from the surface of bodies into the surrounding space occurs through radiation and heat transfer during free convection. Full flow from the surface of the body, thus, will be equal to:

, where
;

- convective component of the heat flow, which can be determined by the Newton-Richmann law:

(12)

On the other hand, the heat transfer coefficient can be determined from the expression:

(13)

the determining temperature in these expressions is the temperature of the boundary layer:

Rice. 2 Scheme of the experimental setup

Legend:

B - switch;

P1, P2 - voltage regulators;

PW1, PW2 - power meters (wattmeters);

NE1, NE2 - heating elements;

IT1, IT2 - temperature meters;

T1, T2, etc. - thermocouples.

FEDERAL AGENCY FOR EDUCATION

STATE EDUCATIONAL INSTITUTION OF HIGHER

PROFESSIONAL EDUCATION

"IVANOVSK STATE ENERGY UNIVERSITY

named after V.I. LENIN"

Department of Theoretical Foundations of Heat Engineering

Determination of the integral emissivity of a solid body

Guidelines for performing laboratory work

Ivanovo 2006

Compiled by V.V. Bukhmirov

THOSE. Sozinova

Editor D.V. Rakutina

The guidelines are intended for students studying in the specialties of the heat engineering profile 140101, 140103, 140104, 140106 and 220301 and studying the course “Heat and Mass Transfer” or “Heat Engineering”.

The guidelines contain a description of the experimental setup, the methodology for conducting the experiment, as well as the calculation formulas necessary for processing the results of the experiment.

The methodological guidelines were approved by the cycle methodological commission of the TEF.

Reviewer

Department of Theoretical Foundations of Heat Engineering, Ivanovo State Power Engineering University

1. Task

1. Experimentally determine the integral degree of blackness of a thin tungsten filament.

2. Compare the results of the experiment with reference data.

2. Brief information from the theory of radiative heat transfer

Thermal radiation (radiative heat transfer) is a method of heat transfer in space, carried out as a result of the propagation of electromagnetic waves, the energy of which, when interacting with a substance, is converted into heat. Radiative heat transfer is associated with a double transformation of energy: initially, the internal energy of the body is converted into the energy of electromagnetic radiation, and then, after the transfer of energy in space by electromagnetic waves, the second transition of radiant energy into the internal energy of another body occurs.

The thermal radiation of a substance depends on the temperature of the body (the degree of heating of the substance).

The energy of thermal radiation incident on a body can be absorbed, reflected by the body or pass through it. A body that absorbs all the radiant energy incident on it is called an absolutely black body (blackbody). Note that at a given temperature, the blackbody and radiates the maximum possible amount of energy.

The flux density of the body's own radiation is called it emissivity. This radiation parameter within an elementary section of wavelengths is called the spectral own flux density radiation or spectral emissivity of the body. The emissivity of a black body, depending on temperature, obeys the Stefan-Boltzmann law:

, (1)

where  0 \u003d 5.6710 -8 W / (m 2 K 4) - Stefan-Boltzmann constant; \u003d 5.67 W / (m 2 K 4) - black body emissivity; T is the surface temperature of a completely black body, K.

Absolutely black bodies do not exist in nature. A body whose radiation spectrum is similar to the radiation spectrum of a completely black body and the spectral density of the radiation flux (E ) is the same fraction   of the spectral density of the radiation flux of a completely black body (E 0,λ) is called gray body:

, (2)

where   is the spectral emissivity.

After integrating expression (2) over the entire emission spectrum (
) we get:

, (3)

where E is the emissivity of the gray body; E 0 is the emissivity of the blackbody;  is the integral degree of blackness of the gray body.

From the last formula (3), taking into account the Stefan-Boltzmann law, an expression follows for calculating the flux density of its own radiation (radiance) of a gray body:

where
- gray body emissivity, W / (m 2 K 4); T is body temperature, K.

The value of the integral degree of emissivity depends on the physical properties of the body, its temperature and the roughness of the body surface. The integral degree of emissivity is determined experimentally.

In laboratory work, the integral emissivity of tungsten is found by studying the radiation heat exchange between a heated tungsten filament (body 1) and the walls of a glass container (body 2) filled with water (Fig. 1).

Rice. 1. Scheme of radiative heat transfer in the experiment:

1 - heated thread; 2 - the inner surface of the glass container; 3 - water

The resulting heat flux received by the glass container can be calculated by the formula:

, (6)

where  pr is the reduced emissivity in the system of two bodies;  1 and  2 are the integral degrees of emissivity of the first and second bodies; T 1 and T 2, F 1 and F 2 - absolute temperatures and areas of heat exchange surfaces of the first and second body;  12 and  21 - angular coefficients of radiation, which show what fraction of the hemispherical radiation energy falls from one body to another.

Using the properties of slope coefficients, it is easy to show that
, a
. Substituting the values ​​of the slope coefficients into formula (6), we obtain

. (7)

Since the surface area of ​​the tungsten filament (body 1) is much smaller than the area of ​​the shell surrounding it (body 2), the slope  21 tends to zero:

F 1 F 2
 21 \u003d F 1 / F 2 0 or
. (8)

Taking into account the last conclusion, it follows from formula (7) that the reduced emissivity of the system of two bodies depicted in Fig. 1 is determined only by the radiation properties of the filament surface:

 pr  1 or
. (9)

In this case, the formula for calculating the resulting heat flux perceived by a glass container with water takes the form:

from which follows the expression for determining the integral degree of blackness of a tungsten filament:

, (11)

where
is the surface area of ​​the tungsten filament: d - the diameter and length of the thread.

The emissivity of a tungsten filament is calculated by the obvious formula:

. (12)