Radiation- R.K.Rajput

THERMAL RADIATION : PROPERTIES AND PROCESSES

763

ENGINEERING HEAT AND MASS TRANSFER

762

where hm, represents wavelength corresponding to maximum spectral blackbody emissive oower at a given temperature T. The eqn. (12.20) is valid for entire spectrum of wavelength for blackbody. Substituting hmm-as 2897.61Tin eqn. (12.19) to obtain maximum spectral emissive power at temperature T, C,T'

Using C1 and C, ;

-

-

12.6.3. Stefan Boltzmann Law The total emissive power of a blackbody E, may be obtained by integrating Planck's distribution eqn. (12.19) over entire wavelength as E, = I:~,,dh

C ,Td h AT)^ [exp (C2/hT)- 11 Tdh = dx, then

= T4

AT = X,

Put

J"o

Example 12.2. One side of metallic plate is insulated, while the other side absorbs a radiant heat flux of 900 W l m 2 . The convectiue heat transfer coefficient between the plate and ambient air is 10 WIm2.K. The surface emissivity of the plate is 0.8. The surrounding and ambient air are at 27°C. Determine the temperature of the plate under steady state conditions. Solution Given :

qr = 900 W/m2, h, = 10 W/m2.K E = 0.8, T_ = 27°C = 300 K. To find : The temperature of the plate. Assumptions : 1. Steady state conditions. 2. One side of the plate is adiabatic. 3. Constant properties. Analysis :The radiant heat flux absorbed by the plate, will be dissipated by convection and radiation. Thus 9,. = hc (T, - TJ + E a (T: - Tm4) Using the numerical values, 900 = 10(Ts- 300) + 0.8 x 5.67 x lo-' (Ts4- 3004) It gives, Ts = 354.8 K. Ans.

Example 12.3. A central heating radiator has a surface temperature of 70°C and heats a room maintained at 20°C. Calculate the contribution of convection and radiation to heat transfer from the radiator. Use following correlation for determination of natural convection coefficient. Nu, = 0.1 18(GrL Pr)II3 The properties of fluid i n the room are p=1.2kglm3, y = 1 . 8 ~ 1 0 ~ k g l m . s , kf=0.026WIm.K. Pr = 0.71 (J.N.T.U., May 2003) Solution Given :Radiation and convection heat transfer from a radiator into a room. Ts = 70°C = 343 K T_ = 20°C = 293 K Properties of fluid and correlation. To find :Natural convection and radiation heat transfer rate. Assumptions : (i) Steady state conditions. ( i i )Blackbody behaviour of radiator. (iii)Constant properties and a = 5.67 x Wlm2.K4. Analysis :The radiation heat transfer rate. \

Its integration yields to

We have

1 n4 T = g0 n n=l "

which gives

The convective heat transfer rate is obtained from Nu, = 0.1l8(Gr, Pr)lJ3

where The constant a is called the Stefan Boltzmann constan,t.


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where

Gr, =

p 2 g P(Ts - Tm)L3 =

(1.2)2 x 9.81 x

1

x

(70 - 2 0 ) ~ ~

p2 293 (1.8 x = 7.44 x 109 ~3 ~~ L NuL = 0.118(7.44 x lo9 L3 x 0 . 7 1 ) =~ 205.35 Then, The average heat transfer coefficient

l2

= h,(T, - Tm)= 5.34(70 - 20) = 267 W/m2. A The total heat transfer rate per unit area is qtOtal= 367 + 267 = 634 W/m2. Ans,

The

QcOnv

Example 12.4. A hot water radiator of overall dimensions 2 x 1 x 0.2 m is used to heat the room at 18°C. The surface temperature of radiator is 60°C and its surface is black. The actual surface of the radiator is 2.5 times the area of its envelope for convection for which the convection coefficient is given by h , = 1.3(AT)II3 W/m2.K. Calculate the rate of heat loss from the radiator by convection and radiation. Solution Given :Radiation and convection heat transfer from a radiator. Radiator Dimensions : H = 2 m, L = l m , w=0.2m Tm = 18°C = 291 K T, = 60°C = 333 K, For convection A, = 2.5 kad h , = l . ' ( ~ T ) l ' ~. To find :Rate of heat transfer by convection and radiation. Assumptions : ( i )The radiator as a blackbody. ( i i ) Steady state conditions. (iii)Uniform heat transfer coefficient. Analysis :The area of the radiator ; kad = 2{2 x 1 + 2 x 0.2 + 1x 0.21 = 5.2 m2 Radiation heat transfer rate ; x (3334- 2914) = 1511.2 W Qrad= Arad0 (Ts4- Tm4)= 5.2 x 5.67 x The convection heat transfer coefficient, h , = 1.3(AT)lI3= 1.3(60 - 18)ll3= 4.51 W/m2.K The convection heat transfer area, A, = 2.5 x Arad= 2.5 x 5.2 = 13 m2 The convection heat transfer rate ; Qconv= h , x As(T, - TJ = 4.51 x 13 x (60 - 18) = 2462.24 W The total heat transfer rate from the radiator, Q = Qrad+ Qc0, = 1511.2 + 2462.24 = 3973.67 W. Ans. Example 12.5.-A pipe carrying steam runs i n a large room and exposed to air at 30°C. The pipe surface temperature is 200°C. Diameter of the pipe is 20 cm. If the total heat loss per metre length of the pipe is 1.9193 k W / m , determine the emissivity of the pipe surface.


Given that N u , = 0.53(GrDPr)Ii4 and air properties at 115°C are k f = 0.03306 W1m.K

v = 24.93 x 1 0 d m 2 / s ,

765

Pr = 0.687.

Solution Given :A steam pipe exposed to a large room. Ts = 2OO0C, Tm= 30°C D = 20 cm = 0.2 m, L = 1m, Q = 1.9193 W/m Air properties and relation for NuD. To find :Emissivity of the pipe surface. Assumptions : ( i )Pipe surface is diffuse and gray. (ii) Room and air at uniform temperature of 30°C. Analysis :The steam pipe dissipates heat to room by convection and radiation, i.e., Q = Qconv + Qrad

Q

- =h

L

(nD) (T, - TJ

+ E 0 (nD) ( T , ~- Tm4)

...(1)

Calculation of h in natural convection

Grashof number Gr, =

g P(Ts - 7?,)D3

v

1 (200 - 30) (0.2)~ = 9.81x -x = 55.326 x 388 (24.93 x 10-y2

lo6

The Rayleigh number Ra, = Gr,, Pr = 55.326 x lo6 x 0.687 = 38.009 x 10" The Nusselt number NuD = 0.53(GrDPrI1l4= 0.53(38.009 x lO6)lI4 = 41.61 Convection coefficient,

Using numerical values in eqn. (1) 1919.3 W/m = (6.88 W/m2.K)(n: x 0.2 m) (200 - 30) (K) + E x (5.67 x W/m2.K4)x (n: x 0.2 m) (4734- 3034) (K4) 1919.3 - 734.76 E= = 0.798 0.8. Ans. 1482.94 The emissivity of the pipe surface is 0.8. Example 12.6. Calculate the following quantities for a n industrial furnace (blackbody) emitting radiation at 2650°C. ii) Spectral emissive power at A = 1.2 pm, ( i i ) Wavelength at which the emissive power is maximum, (iii)Maximum spectral emissive power,

-

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767


( i v ) Total emissive power, ( v ) Total emissive power of the furnace, i f it is treated as gray and diffuse body with an emissivity of 0.9.

Solution Given :Average solar constant for determination of temperature of sun (Q/A), = 1353 W/m2, D = 1.392 x lo6 km, s = 1.496 x lo8 km

Solution Given :An industrial furnace as blackbody radiating at T = 2650°C = 2923 K. To find : ( i ) E,, at h = 1.2 pm

( i v ) Eb (Total emissive power) ( v )E Eb. Analysis :(i)The spectral emissive power at h = 1.2 pm : The Planck's distribution law, eqn. (12.19)

Fig. 12.9

To find :The temperature of sun. Assumptions : ( i )The negligible emissive power of earth in comparison of sun. ( i i )Due to large distance, all rays of sun falling on the earth's surface. (iii) Sun has spherical surface. Analysis :The energy radiated by sun (blackbody)

where Substituting the values,

I%

=

3.742 x l o 8 ( 1 . 2 1 [ex.[ ~

1.438 x l o 4 1.2 x 2923

1

= 2.53 x

lo6 W/m2.pm.

Ans.

Qsun

(ii) The wavelength at which the emissive power is maximum : Using Wien's displacement law, eqn. (12.20) hm, T = 2897.6 pm.K 2897.6 'max = 2923 - 0.9913 p . Ans. (iii)Maximum spectral emissive power, by eqn. (12.21) Ebhmm= 12.87 x 10-lo T5 W/m2.pm = 12.87 x 10-lo x (2923)5= 2.746 x lo8 W/m2.p. Ans. ( i v ) Total emissive power

= 4.139 x lo6 W/m2. Ans. ( v ) Total emissive power with E = 0.9 E = & E b = ~ a T 0.9 4 =~ 4 . 1 3 lo6 9 ~ = 3.725 x lo6 W/m2. Ans.

Example 12.7. The average solar radiation flux on the earth's atmosphere is 1353 W / m 2 and it is known as solar constant. Calculate the temperature of sun (a blackbody), 1.392 x 106 k m in diameter, when it has mean distance of 1.496 x l o 8 k m from the earth's atmosphere.

= Asun0 Tkn = n: (DS,J2

0 T,4Un

...( 1 ) = n: x (1.392 x lo9 m)2 x 5.67 x 104 ,T : = 3.45 x 1011,T : The sun is considered as source at a distance s = 1.496 x 1011 m from earth's surface.

- 11

Mean area, which receives solar radiation A = 47ts2 = 4.n x (1.456 x The solar flux incidence on the earth is q = -Q sun A

or

T,;

or 1353 =

= 1.102 x 1015 or T,,,

m2

= 2.812 x

3.45 x lo1' 2.812 x

T'&

= 5762.2 K. Ans.

Example 12.8. Calculate the equilibrium temperature for a plat% exposed to a solar flux of 700 W / m 2and convection environment at 25OC, with convection coefficient of 10 W/m2.K. If the plate is coated with asun= 0.12 ;uplate= 0.9. ( a ) White paint : (N.M.U., Nov. 2000) = 0.95. ( b )Flat black paint : a,,, = 0.96 ;aPlate Solution Given :A plate exposed to solar flux and convection environment ; Solar flux = 700 W/m2, T_ = 25°C = 298 K, h = 10 W/m2.K, ( a )a,,, = 0.12 ; a,,, = 0.9 ( b )a,,, = 0.96 ; a,,,,, = 0.95. T o find :The equilibrium temperature in above two cases.

768


Assumptions : 1. Steady state conditions. 2. Plate surface is gray, opaque and diffused. Analysis :Making the energy balance for the plate : Solar flux on the plate = Convection flux + Radiation flux (Emissive power) Considering T is the temperature of plate, then Qsun as,, = h(T - T_) + E (T4- Tm4) A

or

or

( a )When the plate is coated with white paint : asun= 0.12 ; uplate= 0.9 or = 0.9 (By Kirchhoff 's law) J x (T - 2984) 0.12 x 700 = 10 x (T - 298) + 0.9 x 5.67 x 84 = 10 T - 2980 + 5.103 x 104 p- 402.43 T4 + 195.963 x lo6 T - 67.929 x lo9 = 0 It is a non linear equation and its numerical (Newton Raphson method) solution gives ; T = 303.40 K or 30.4"C. Ans. ( b )When plate is coated with black paint : a,,, = 0.96 ; aPl,,,= 0.95 Using these values in eqn. (1); 0.96 x 700 = 10 x (T - 298) + 0.95 x 5.67 x lo4 x (T4- 298*) p - 424.787 672 = 10 T - 2980 + 5.3865 x T4 + 18,56,49,308.5 T - 75.6852 x 109= 0 It is a non linear equation and its numerical (Newton Raphson method) solution gives ; T = 337.65 K or 64.65"C. Ans.

12.6.4. Radiation Function and Band Emission Eqn. (12.23) gives the total amount of radiant energy emitted by a blackbody a t temperature T over wavelength 31. = 0 to 31. = m. There are often situations, when it is necessary to evaluate the energy over certain wavelength band, like 0 to h or A1 to 1,. The radiation energy emitted by a blackbody per unit area, over a wavelength band from h = 0 to h (Fig. 12.10) is determined as

The eqn. (12.25) is evaluated numerically by usFig. 12.10. Radiation emission from a ing eqn. (12.19). But the integration does not have sim- blackbody in spectral band of to ple closed form solution and therefore, performing integration is not practical solution. Therefore, a dimensionless quantity fo- ;, called the blackbody radiation funqtien is used, which is defined as


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The function fo - represents fraction of radiation energy emitted from a blackbody at temperature T in the wavelength band from h = 0 to A. A table of computed blackbody radiation function fo - ;, as a function of hT is given in Table 12.2.

TABLE 12.2. Blackbody radiation functions

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THERMAL RADIATION : PROPERTIES AND PROCESSES ENGINEERING HEAT AND MASS TRANSFER

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R,ange O I hS0.4 OShI0.7

The fraction of radiation energy emitted by a blackbody a t a temperature T over a finite wavelength band from h = hl to h = h2 (Fig. 12.11) is evaluated as

In terms of blackbody radiation function

Blackbody radiation function f0-,, = 0.1245 f, - = 0.4914. (Pune Univ., Dec. 1999)

,,

Solution Given : Solar radiation a t T, = 5800 K, and blackbody radiation functions with wavelengths. To find :Amount of solar radiation in visible spectrum. Analysis :The total radiation emitted fram the sun in visible spectrum

A

= 5.67 x lo4 x (5800)~[0.4914 - 0.12451

= 23.54 x lo6 W/m2. Ans.

or

fh,-h,

=f0-h, -f0-h,

Example 12.11.Determine (a) the wavelength at which the spectral emissive power o f a tungsten filament at 1400 K is maximum, (b) the spectral emissive power at that wavelength, and (c) the spectral emissive power at 5 pm.

...(12.28)

where fo - h, and fo - h , are the blackbody radiation functions corresponding to hlT and h, T, respectively.

Solution Given :For a radiating surface T, = 1400 K. To find : (i)A,, corresponds to peak emissive power, (ii)Peak spectral emissive power corresponding to h, (iii)Spectral emissive power a t h = 5 pm. Assumption :Blackbody radiation and o = 5.67 x lo4 W/m2. K4. Analysis :(i)The wavelength corresponds to maximum emissive power. h, T = 2897.6 pm.K

Fig. 12.11. Blackbody radiation in wavelength band 2. = hl to h = h2

Example 12.9.The temperature of a filament of a n incandescent light bulb (a blackbody) is maintained at 2500 K Calculate the fraction of radiant energy emitted by the filament in the visible spectrum. Also calculate the wavelength at which the emission from the filament reache8 a maximum value. Solution Given :Radiation from a filament of an incandescent light bulb in visible range. T = 2500 K h, = 0.4 gm & = 0.76 pm For visible range. To find : ( i ) Fraction of radiation energy emitted in visible range. (ii)Wavelength corresponding to maximum emissive power. Analysis :( i )The blackbody radiation function corresponds to h1T and h, T are fo-1, = 0.000321 hlT = (0.4 pm) x (2500 K ) = 1000 pm.K+ h2T = (0.76 pm) x (2500 K ) = 1900 pm K--+ fo- h , = 0.052111 - fO - hl = 0.052111 - 0.000321 = 0.05179 fhl - h2 = fO I t indicates that only 5.18% of the radiation energy emitted falls in the visible range. Ans. (ii)Wavelength corresponding to maximum emissive power is obtained by using Wien'~ displacement law 2897.8 hm,T = 2897.8 gm.K or h, - -pm = 1.16 pm. Ans. - 2500 Example 12.10. Solar radiation has approximately same spectral distribution as a11 ideal radiating body at temperature of 5800 K. Determine the amount of solar radiation, which is in the visible range of 0.4 pm to 0.7 pm, use following data :

Lax= 2897'6 - 2.07 p.Ans. 1400 --

'

(ii) Spectral emissive power a t h = h, can be obtained from eqn. (12.21) ; = 12.87 x 10-lo T5 Ebhmax = 12.87 x 10-lo x (1400Y = 69.23 x loS W/m2.pn. Ans. (iii) Monochromatic emissive power a t 5 pm. hT = 5 x 1400 = 7000 ym.K +fo - = 0.808144 Ebh= o T4. fo- = 5.67 x lo4 x (1400)4x 0.808144 = 1.76 x 105W/m2.pm. Ans. Example 12.12. A window glass 0.3 cm thick has a monochromatic transmissivity of 0.9 in the range of 0.3 pm to 2.5 pm and nearly zero elsewhere. Estimate the total transmissivity of the window for (a) near black solar radiation at 5800 K, and (b) black room radiation at 300 K.

Solution Given :Transmission through a glass window Z~ = 0.9, hl = 0.3 pm, h2 = 2.5 pm ( a ) T = 5800 K, ( b )T = 300 K. To find : Transmissivity in range of ill = 0.3 pm to A, = 2.5 pm for (i) T = 5800 K, and (ai)T = 300 K.


772

773


Assumptions : ( i ) Blackbody behaviour, ( i i ) Stefan Boltzmann constant, o = 5.67 x 104 W/m2. K4. transmissivity of a surface is defined as Energy transmitted through body Z= Energy incident on the body For a black body

For black surface (sun)

=0.4x0.98+0.8x0.02=0.408. Ans. ( i i ) Source condition, T,,, = 527°C = 800 K h,T = 3 pm x 800 K = 2400 pm.K -+ fo - L , = 0.140266

and

fIl-,

= 1- f 0 - ~ , = 1- 0.140266 = 0.859734

Absorptivity of the surface for source

asource = ahlfO - A,

+ ah2fA , = 0.4 x 0.140266 + 0.8 x 0.859734 = 0.7439.

= 2.5 x 5800 = 14500 ym.K +fo - A , = 0.96597 = 0.9[0.96597 - 0.032851 = 0.834. Ans. T = 300 K h,T = 0.3 x 300 = 90 pm.K -4 fo - A, = 0.000

The radiation emitted by a real body at temperature T is always less than that of black body. Therefore, the blackbody emission is considered as reference. The emissivity is defined as the ratio of the radiation energy emitted by a surface to that emitted by a blackbody at the same temperature. It is a dimensionless quantity, a property of a radiating surface to measure of how closely a surface approximates a black surface for which E = 1. It is designated as E and varies between 0 and 1. The emissivity of real surfaces is not constant. It varies with temperature of surface, a s well as wavelength and direction of emission. Therefore, different emissivities may be defined for a surface, depending upon the effect considered.

Example 12.13. The aluminium paint is used to cover the surface of a body that isl maintained at 27OC. I n one installation this body is irradiated by the sun, i n another by a source at 527OC. Calculate the effective absorptance of the surface for both conditions, assuming - the sun is a blackbody at 5800 K. Take Solution Give

hs.

12.7.1. Hemispherical and Total Emissivity The emissivity of a surface that is averaged over all directions is called the hemispherical emissivity and the emissivity averaged over all wavelengths is called the total emissivity. Thus the total hemispherical emissivity E ( T )of a surface is defined as ratio of the radiation heat flux emitted over all wavelengths into a hemispherical space (all directions) to that which would have been emitted by a blackbody at same temperature. Mathematically

tion on a aluminium painted cover surface. * 27% = 300 K, T, = 5800 K, Tsource= 527°C = 800 K a , . =0.4 for O l h c 3 p m ty of the surface for

( i ) Solar radiation, and ( i i ) Source radiation. Anslysis :t i ) For solar radiafio a

- w = 1- f O - l i = 1-0.98 = 0.02 Thus 98% solar radiation falls below 3 pm and the remaining 2% between h = 3 pm to

h = 00. The effective absorptivity of the surface

3

i

i

E

For a given value of emissivity, the emissive power of a real surface a t a temperature T is determined by ...(12.30) E(T) = E E,(T) = E o T4 The total, normal emissivities for some selected materials are shown in Figs. 12.12 and 12.13. and listed in Table 12.3. The following observations are listed below : 1.The emissivity of the metallic surfaces is very small having the values as low as 0.02 for highly polished gold and silver. 2. The presence of oxide layers may improve the emissivity of metallic surfaces. 3. The non conductors have the large value of emissivity, generally exceeding 0.6.

778



779

Consider the spectral emissivity of a real surface is represented by a wiggly line as shown in Fig. 12.17. It differs from Planck's distribution and it has large variation with wavelength and consists of several peaks and valleys. Further, the spectral emissivity also varies with temperature. Thus, its evaluation to average ernissivity becomes very complicated. A useful simplification is to replace the wiggly line by an average emissivity line as shown by dashed line in Fig. 12.17. It is the gray body approximation. The effect of apprdximation on emissive power is shown in Fig. 12.18. A gray surface should emit same radiation as the real surface. That is

Then the average emissivity for a gray surface can be expressed as

EA

=

i

E!

= constant ; 0 Ih < hl = constant ; hl l h < X z = constant ; h2 I h < =

Fig. 12.19. Approximation of actual variation of spectral emissivity with wavelength by a step function

I

Black body E, = E = 1

Thus a gray surface is considered for which the spectral emissivity ck is independent of the wavelength and thus for a gray surface, the spectral emissivity is equal to total emissivity E c(T T E;(T) ...(12.40) Further, for a gray surface, the absorptivity, reflectivity and transmissivity are also independent of wavelength. For such a surface ; E(T)= a(T) ...(12.41) U

1

2

Wavelength, h, pm Fig. 12.17.iCom~arisonof the emissivit~of a real surfa&, a gray surface and a blackbody a t the same temperature

4

3

5

6

'

Fig. 12.18. Comparison of hemispherical spectral emission for black, gray and real surfaces

If the variation of spectral emissivity is quite large, but it is constant over certain wavelength as shown in Fig.12.19, then E(T)is expressed as step function and eqn. (12.38) is integrated by dividing the spectrum into a number of wavelength bands and assuming the emissivity c(T) remains constant over each wavelength band. Then the average emissivity can be determined as

Example 12.14. In an isothermal enclosure at uniform temperature two small surfaces A and B are placed. The irradiation to the surface by the enclosure is 6200 W l m 2 . The absorption rates by the surfaces A, and B are 5500 W l m 2and 620 Wlm2. When steady state is established, calculate the following :. ( i ) What are the heat fluxes to each surfaces ? What are their temperatures ? (ii)Absorptivity of both surfaces, (iii)Emissive power of each surface, and ( i v )Emissivity of each surface. Solution Given :Two small body A, and B are placed in an enclosure. G = 6200 W/m2 G,, = 5500 W/m2 G,, = 620 W/m2.


780

To find : ( i )Heat flux to each surface and surface temperatures. (ii)aAand aB (iii)EAand EB (iv) and EB. Analysis :( i )At steady state, the net heat flux is zero. The enclosure is considered to be a black body and isothermal, thus T A = TB= T Irradiation to the surface = Emission from the surface 6200 = o T4 = 5.67 x 10-8(T)4 or T = 575 K. Ans. q* = 0, q* = 0. Ans. Gd = 5500 = 0.887. Ans, (ii)Absorptivity of surface, A, a, = G 6200 a , = -G,B - -620 = 0.1. Ans. Absorptivity of surface B, G 6200 (iii)Emissive power of each surface According to Kirchhoff 's law, at thermal equilibrium Energy absorbed = Energy emitted .. E, = aAG = 5500 W/m2. Ans. E, = %G = 620 W/m2. Ans. (iv)Emissivity of each surface According to Kirchhoff's law

EB

= % = 0.1' Ans.

Example 12.15. A solar collector surface has an absorptivity of 0.85 for wavelength 0 I h < 3 pm and a value of 0.15 for wavelength h z 3 pm. (i) Calculate the energy absorbed from the source at 5000 K, (ii) If the flux is 800 W / m 2 ,calculate the energy radiated by the body, if its temperature was 350 K.

ah, =a2=0.15 for


Analysis :( i )At temperature 5000 K, from Table 12.2.

The average absorptivity is ~=a~fo~~l+~2(~-fo~~l)=0.85x0.969+0.15~(1-0.969)=0 For flux of 800 W/m2. The energy absorbed = a G = 0.8283 x 800 = 662.64 W/m2. Ans. (ii)When surface temperature is 350 K

h,T = 3 ym x 350 K = 1050 ym +fo-hl = 0.00057 From Kirchhoff's law a , = E,, a2= E~ Energy radiated = [EI~O-X, + ~2 ( 1- fo+ )I O. T4 = [(0.85 x 0.00057 + 0.15 x ( 1 - 0.00057)] x 5.67 x lop8x (350)4 = 128 W/m2. Ans. Example 12.16. The spectral emissivity as a function of an opaque surface at 800 K is approximated by step function and is given below : E, = 0.3 for O I h I 3 p m E~ = s2 = 0.8 for 3 p n I h 5 7 p m E~ = 0.1 for 7 p n S h S m . Calculate the average emissiuity of the surface and the emissiue power.

Solution Given :Spectral emission from an opaque surface. T = 800 K = 0.3 for O I h I h , = 3 p m E~ = 0.8 for 3 p m I h < h 2 = 7 p m E~ = 0.1 for 7 p m l h S w . To find : ( i )Average emissivity of the surface, and (ii)Emissive power of the surface. Analysis :( i )Average emissivity of the surface can be determined by integrating it into steps as

h>3pm

T, = 5000 K. T, = 350 K , G = q = 800 W/m2 (ii) To find : ( i )Energy absorbed by solar collector surface. (ii)Emissive power, if s Assumptions : ( i )Diffuse, gray surfaces. (ii)Kirchhoff s law holds good.

781

Here Then

h,T = 3 pm x 800 K = 2400 pm K 4 fO+ = 0.140256 fo-h2-0.701046 h2T = 7 pm x 800 K = 5600 pm K+ E

= 0.3 x 0.140256

+ 0.8 x (0.701046 - 0.140256) + 0.1 x ( 1 - 0.701046)

.67 x lo4 x (500)4 = 12.09 x lo3 W/m2. Ans. -


786

Thus

a=& J = &Eb+. (1- E ~ G

The total rate of energy leaving the surface = AJ The total rate of energy incident on the surface = AG Thus the net radiant energy leaving the surface Q=AJ-AG=A(J-G)

...(12.58)

This equation is not valid for black surface for which E = a = 1 ; p = 0 For a black surface, J = Eb ...(12.60) Q = A(Eb- G) Example 12.18. A gray, diffuse opaque surface (a = 0.8) is at 100°C and receives an irradiation 1000 Wlm2. If the surface area is 0.1 m2. Calculate (i) Radiosity of the surface, and (ii) Net radiative heat transfer rate from the surface. (iii) Calculate above quantities, if surface is black.

or

Solution Given :A gray, diffuse opaque surface G = 1000 W/m2 a = 0.8, A, = 0.1 m2. T, = 100°C = 373 K, To find : (i) Radiosity J, (ii) Net heat transfer rate, (iiil For black surface J and Q,,. Analysis :(i) The radiosity of the surface J = EE, + pG Fig. 12.24. Surface energy balance with radiant energy For a gray, diffuse and opaque surface, a = E T=O, a + p = l p=l-a=l-0.8=0.2 J = a o T4+ pG = 0.8 x 5.67 x 10*(373)~+ 0.2 x 1000 = 1078 W/m2. Ans. .. (ii) The net heat transfer rate using eqn. (12.'60) Qnet= A(J - G) = 0.1(1078 - 1000) = 7.8 W/m2. Ans. Alternatively using eqn. (12.59) [5.67 x (373)' - 10781 = 7.81 wlmZ. - = 0.1 x 0.8 Qnet = AF(Eb 1- E 1 - 0.8 (iii) For black surface E = a = 1 (a) Radiosity J = Eb = o T4 = 5.67 x lo* x (373)4= 1097.5 W/m2. Ans. ( b )The heat transfer rate, eqn. (12.60) Q = A(E, - G) = O.l(l097.5 - 1000) = 9.753 W/m2. Ans.

J,


787

Example 12.19. A plane ,gray, diffuse and, opaque surface (absorptivity = 0.7) with a surface area of 0.5 m2, is maintained a t 500°C and receives radiant energy at a rate of 10,000 1 Wlm2. Determine per unit time (ij The energy absorbed. (ii) The radiant energy emitted. (iii) The total energy leaving the surface per unit area. (iv) The radiant energy emitted by the surface in the wave band 0.2 pm to 4 ym. (v) The net radiative heat transfer from the surface. Solution Given :Plane, gray, diffuse, opaque surface a = 0.7 A,(surface area) = 0.5 m2 G = 10000 W/m2 Ts = 500°C. To find : (i) Rate of energy absorbed, (ii) Emitted radiant energy, (iii) Total radiant energy leaving the surface per unit area Fig. 12.25. Schematic for energy transactions on the surface (iv) Emitted radiant energy in the wave band 0.2 pm-4 pm ( v ) Q(net radiative heat transfer rate from the surface). Analysis :(i) Rate of energy absorbed = aAG = 0.7 x 0.5 x 10000 = 3500 W. Ans. (ii) Rate of radiant energy emitted = A E o T4 (For a gray surface, E = a = 0.7) A E o T4 = 0.5 x 0.7 x 5.67 x x (773)4= 7091 W. Ans. (iii) Total energy flux leaving the surface is the sum of the emitted energy flux and the reflected energy flux J = EE, + pG where p = 1- a = 1- 0.7 = 0.3 :. Radiosity, J = 0.7 x 5.67 x x (773)4+ 0.3 x 10000 = 17182 W/m2. Ans. (iv) Rate of radiant energy emitted in the wave band 0.2 pm to 4 ym is given by EL,-A, = d f o - h , - fo-x,]Eb W/m2 At h, and A,, T = 773, from Table 12.2 h2T = 4.0 x 773 = 3092 pm.K +fo-h, = 0.294 Ehl-hZ= 0.7 x [0.294 - 01 x 5.67 x

lo4

x (773)4= 4166.2 W/m2.

Rate of radiant energy on surface area As = 0.5 m2 is = AEh,

= 0.5 x 4166.2 = 2083.13 W. Ans.

788

789



( i )The emissive power

( v )From an energy balance on the surface Net radiative heat transfer rate from the surface = Total energy leaving the surface - Energy reaching the surface = 0.5 x 17182 - 0.5 ~ 1 0 0 0 0= 3591 W. Ans.

where 1, = 2 ym,

Example 12.20. A specially coated diffuse, opaque surface whose absorptivity is 1 for 0 < 3L < 2 ym and 0.1 for 2 pm < 3L < is exposed to solar radiation i n the outer reach of the atmosphere (Fig. 12.26). The incident solar energy is 1353 W l m 2 . Determine ( i ) The heat flux by radiation from the surface to the surroundings, if the surface is maintained at 100°C by a coolant. ( i i ) The equilibrium temperature of the surface, if the coolant flow stops and the surface is insulated on the side that does not receive solar radiation. (iii)Compare the values i n parts a and b, if the surface is black. Assume the sun behaves as a blackbody at 5760 K. The sky is at 0 K.

E, =

a, = 1.0, and

E,

=

a, = 0.1

00

Emissive power of the surface E = [lx 0 + 0.1(1- O ) ] x o Ts4= 0.1 x 5.67 x 104 x (373)4= 109.75 W/m2 Energy absorbed/m2 by the surface of solar radiation

Coolant

= a1fo-ilG + a, fhl - G = [a,fo-i;, + a2(1- f0-,,)1G = [ l x 0.9392 + 0.1 x ( 1 - 0.9392)l x 1353 = 1278.96 W/m2 The heat flux from surface to surroundings q = 109.75 - 1278.96 = - 1169.21 W/m2 The -ve sign indicates, that the net heat flux is towards the surface. Ans. ( i i )If coolant flow stops then at thermal equilibrium Emissive power = Rate of energy absorbed = 1278.96 W/m2

-

Fig. 12.26. Schematic of a specially coated hurface

Solution. Given : Surface : specially coated O
when coolant stops, q = 0 , and

4

Now fO-k1 and Eb are unknown both depends on temperature of the surface, which is also unknown. Thus using iterative method for its solution. The computed yalues for this problem are given in the following table. The emissive power of the surface is denoted by E. In arriving at the values in the table, the principles of bisection for finding the roots of eqn. ( i i )have been employed. Value of T Too low ' Too high Too low Too high A little too high A little too low

At a temperature of 680 K the emitted energy (1280.3 W/m2)is very nearly equal to the absorbed energy (1279 W/m2).Hence, under equilibrium conditions T = 680 K. Ans. (iii)For a black surface we have qb = Emitted energy - Absorbed energy = 5.67 x x (373.15)4- 1353 = - 253.7 W/m2. Am. Under equilibrium conditions, Eb = G and

790

-


Comments. The special surface is much more effective as a solar collector than a black surface. The special surface transfers 1169.2 W/m2 to the coolant, whereas the black surface transfers only 253.7 W/m2.The reason for this difference is that the major part of the radiation from the sun is in the wave band 0-2 pm (94%),and all of it is absorbed by the special surface.

THERMAL RAD+ATlUN : PROPERTIES AND PROCESSES

T, = 2000 K

79 1

Brick wall

\

Example 12.21. A diffuse fire brick wall at temperature of 500 K has discontinuous spectral emissivity as shown i n Fig. 12.27 (a) and is exposed to coal bed at 2000 K.

Coals

I

1 .o

I I I

Fig. 12.27. ( b ) Schematic

Analysis :( i )The total hemispherical emissivity

Breaking integral into parts

Fig. 12. 27 ( a )

Calculate the total hemispherical emissivity and emissive power of the fire brick wall. What is the total absorptivity of the wall to the irradiation resulting from emission by the coals. Solution Given :Fire brick wall exposed to coal bed T, = 2000 K T s = 500 K, EL,

=

= 0.1

for 0 5 h < 1.5

Ex,

=

= 0.5

for 1.5 C: h < 10

Ex,

=

= 0.8

for 10 l h < w

Hence, E(TJ = 0.1 x 0 + 0.5 x [0.634 - 01 + 0.8[1- 0.6341 = 0.610. Ans. ( i i )Total emissive power E(TJ = E(TJ Eb(Ts)= d T S )x o Ts4 = 0.610 x 5.67 x lo4 x (500)4= 2161 W/m2. Ans. (iii)Total absorptivity of wall

.

To find : ( i )Total hemispherical emissivity of the fire brick wall, ( i i )Total emissive power of the brickwall, (iii)Absorptivity of the wall to irradiation from the coals. Assumptions : ( i )Brick wall opaque and diffuse (2 = 0 ) ( i i ) Spectral distribution of brick wall approximates that due to emission from a blackbody a t 2000 K.

JU

Since surface is diffuse and opaque. eh = a, Thus Moreover, the spectral distribution of the irradiation approximates that due to emission from a blackbody at 2000 K ; G,a Ebh . It follows

792


At Tc = 2000 K frem Table 12.2 hlTc = 1.5 x 2000 = 3000 ym.K-+ Hence,

fO+

= 0.273

h2Tc= 10 x 2000 = 20000 ym.K---+ fo-),, = 0.9856 a(TJ = 0.1 x 0.273 + 0.5[0.9856 - 0.2731 + 0.8[1- 0.98561 = 0.3951. Ans.

12.10. SOLAR RADIATION

The sun is our primary source of energy. The energy Normal coming out the sun is called solar energy and it 4 reaches to earth in the form of electromagnetic Earth's Go = G, cos 0 waves. The sun is considered as a nuclear reactor, atmosphere where the heat being generated due to continuous \ fusion reaction of hydrogen atoms to form helium. The sun experiences very large temperature in its core region, but its temperature drops to approximately 5800 K i n its outer region, due to continuous dissipation of energy by radiation. The sun is nearly spherical body of diameter of 1.392 x lo6 km and a mass of 2 x 1030 kg. - It is located at a mean distance of 1.496 x lo8 km from 12.28. solar radiation reaching the the earth. The earth has its mean diameter of 1.27 x earth's atmosphere, solar constant G,, a n d lo4 km and its surface gets only small fraction of e x t r a t e r r e s t r i a l solar i r r a d i a t i o n Go sun's energy, because, the sun subtmds only an angle of 32 minute at the earth's surdce. The intensity of solar radiation reaches outside the earth's atmosphere is almost constant. The solar constant G, is the rate at which the solar radiation flux is received on a surface normal to the sun rays just outside the earth's atmosphere, when the earth is its mean distance from the sun. The radiation coming from the sun is equivalent to blackbody radiation. Using the Stefan Boltzmann law, the solar constant can be calculated as :

G;,

I-[

793

THERMAL RAMATION : PROPERTIES AND PROCESSES

The spectral distribution of solar radiation on the earth's atmosphere and physical significance of G, and Go are illustrated in Fig. 12.29. The solar radiation travels in atmosphere about 30 km outside the eadh's surface. As s d a r radiation passes through this atmosphere, it is absorbed and scattered by atmospheric material. The absorption occurs mainly due to pres. ozone absorbs complete ence of ozone, water vapours, CO,, NO,, CO, 0, and CH, e t ~The ultraviolet radiation at wavelength below 0.3 ym and considerably in the range of 0.3 to 0.4 ym and some radiation in the visible range. Thus the ozone layer in the upper regions of atmosphere guards biological systems on the earth from the harmful ultraviolet radiation. In turn the ozone layer must be protected from the destructive chemicals commonly used as refrigerants, cleaning agents and propellants in the aerosol cans. The carbon dioxide and water vapour absorb mainly longer wavelength radiation (infrared radiation). As a results of these absorption, the solar energy reaching the earth surface is weakened considerably. Its absorption is large as 950 W/m2 on a clear day and much less on cloudy or smoggy days.

2

YT :

= 1353 W/m2

...(12.61)

rSU,= radius of sun = 6.9598 x lo8 m. rOrbi,= distance between sun and earth = 1.496 x 1Ol1 m. o = 5.67 x lo4 W/m2-K4,Stefan Boltzmann constant. Ts = effective temperature of sun = 5762 K Due to very small eccentricity in the earth, the distance between the earth and sun varies throughout the year. Therefore, solar constant also varies from its maximum value of 1399 W/m2on December 21 to a minimum of 1310 W/m2 on June 21 and on any day of year, it can be calculated as : where,

-

where n is the day of year. The extraterrestrial solar irradiation Go incident normal to the outer surface of the earth's atmosphere is calculated as Go = G, cos 8 (W/m2) ...(12.63)

0

0.5

1.O

1.5 2.0 Wavelength, pm

2.5

3.0

Fig. 12.29. Spectral d i s t r i b u t i o n o f solar r a d i a t i o n

The solar energy reaching the earth's surface is also weakened by scattering or reflection when it passes through the atmosphere. The scattering or reflection occurs due to all gaseous molecules as well as particulate matter in the atmosphere. The radiation at wavelength corresponding to violet and blue colours is scattered most. These scattered radiation is redistributed in all directions and gives bluish colour sky. The same phenomenon is responsible for red sunrise and sunset. Early in the morning and late in evening, the sun rays passes through a larger. thickness of atmosphere. Therefore, the violet and blue colours of light experience a large number of scattering and thus do not reach the earth's surface. While the colours correspond to longer wavelength such as yellow, orange and red reach the earth's


T h e surface shown i n Fig. 12.34 a r e assumed to be opaque. T r a n s p a r e n t or semitransparent material can also be selective with somewhat different designs. Example 12.22.A flat plate solar collector with no cover has a selective absorber surface with E = 0.1 a n d a, = 0.95. At a particular time of the day, the absorber surface temperature T, is 120°C, when the solar irradiation is 750 Wlm2, the effective sky temperature is - 10°C a n d ambient air temperature T_ is 30°C. Assume the natural convection is given by q = 0.22(TS - T_)413Wlm2. K. Calculate the useful heat removal rate (Wlm2) from the collector for these conditions. What is the corresponding efficiency of the collector ?

Solution Given :A flat plate solar collector with its operating conditions. E = 0.1 a, = 0.95 T, = 120°C = 393 K Air T, = 30°C a, = 0.95 T, = 30°C Tsky= - 10°C = 263 K h = 0.22 (T, - T,)~'~ G, = 750 W/m2 and relation for calculation of h. To find : 1 qu,, w/m2 (i) Useful heat removal rate per unit area, W/m2. Fig. 12.35. Schematic (ii) Efficiency of the collector. Assumptions : (i)Steady state conditions. (ii) Bottom of the collector is well insulated. (iii) Diffuse absorber surface. (iv) Sky radiation is in approximately same spectral region that of surface emission i.e., E = asky = 0.1. Analysis :(i) The rate of incoming energy on the absorber surface/m2

--

* \

-

799


798

EL A =

+ %kYG ~ k y = 0.95 x 750 + 0.1 x 5.67 x lo4 x (263j4 = 739.62 W/m2 The energy outgoing the absorber surface per unit area

-%ut- - q,,,, A

qrad + quse= h(Ts - Tm)+ + quse = 0.22(Ts - T_)4/3+ E o Ts4 + quse = 0.22 (120 - 30)*3 + 0.1 x 5.67 x lo4 (393 K)4 + quse = 88.73 + 135.25 + q,,, = 223.98 + qUse In steady state conditions,

EL A

-

+

%Ut

A 739.62 = 223.98 +-q,,,

or quse= 515.65 W/m2. Ans.

(iij The collector efficiency is defined a s the fraction solar radiation extracted as useful energy.

T=-----G, - 515'65 750

= 0.687

68.7%. Ans.

Example 12.23. The white paint on a roof as a selective solar absorber (a, = 0.26). Consider now a bare roof under a sunlit sky. The solar radiation on the plane of the roof is 600 W/m2, the a i r temperatu-re is 35"C, a n d a light breeze produces a convective heat transfer coefficient of h = 8 W/m2.K. The sky temperature is 18°C. Find the temperature of the roof, if it is painted with either white acrylic paint or a non-selective black paint having E = 0.9. Solution Given :A roof surface with operating condition T_ = 35°C = 308 K, h = 8 W/m2.K, a, = 0.26 G, = 600 W/m2, Tsky = 18°C = 291 K, E = 0.9 for non-selective black or white acrylic paint. To find :The surface temperature of roof with (i) White acrylic paint, and (ii) Non-selective black paint. Assumptions : (i) Steady state conditions. (ii) No heat transfer to interior of roof. (iii) For non-selective black paint a, = E. Analysis :In steady state condition, the energy balance yields to Incident (solar + sky radiation) energy on the roof surface = Heat loss by (convection + radiation) from the roof surface For 1m2 surface area asG, + E o Ts;y= h(Ts - T,) + E 0 Ts4 (i) For white acrylic paint a, = 0.26, E = 0.9 (Z9lI4 = 8 (Ts - 308) + 0.9 x 5.67 x lo-' Ts4 0.26 x 600 + 0.9 x 5.67 x 5.103 x 1O4TS4 + 8T, - 2986 = 0 or Ts4+ 156.77 x lo6 T, - 5.851 x 1010 = 0 or It is a transcendental equation and its solution by iterative technique converge to T, = 312 K = 39°C. Ans. (ii) For non-selective black paint, a, = E = 0.9 and then above eqn. leads to Ts4 + 156.77 x lo6 T, - 6.6038 x 101° = 0 Its solution converges to Ts = 338K = 65°C. Ans. Example 12.24. An artificial spherical satellite flies around :he earth. Calculate the temperature of the satellite surface, a s s u m i n g ~ h a tthere is no heat sources a n d surface temperature is uniform all over the surface. The solar radiation reflected from the earth a n d radiation emitted from the earth should also be ignored. (i)Ifa,=0.2 and ~ = 0 . 1 ; (ii) If surface of the satellite is gray ; (iii) Find the ratio a , / & , when the temperature of the satellite surface becomes 30°C. The incident solar radiation is 1500 Wlm2.

800


Solution Given :A satellite around the earth ; Gs = 1500 W/m2, To find :Satellite surface temperature if ( i ) as = 0.2 and E = 0.1 ( i i ) Surface of the satellite is gray, and (iii)a& ratio for T, = 30°C. Analysis :For steady state conditions, the energy balance Heat gain by incident radiation = Heat lost by emission a, P$rojG = EA, CT T,4 where P$roj= projected area of satellite for solar irradiation = nr2 A, = surface area of satellite for emission = 4nr2 Therefore, 0.2 x nr2 x 1500 = 0.1 x 4 x 9 x 5.67 x 104Ts4 (i) or Ts4= 1.3227 x 1010 or T, = 339.13 K = 66.13"C. Ans. ( i i )For gray surface a = E = 0.2 0.2 x nr2 x 1500 = 0.2 x 4nr2 x 5.67 x Ts4 or TS4 = 6.6137 x lo9 or T, = 285.17 K = 12.17"C. Ans. (iii)For given temperature T , = 30°C = 303 K a, x nr2 x 1500 = E x 4nP x 5.67 x x (303)4 or a, / E = 1.27. Ans. Example 12.25. In the process of estimating the emission from the sun, it may be treated as blackbody with the surface temperature of 5800 K a t a mean distance of 15 x 1010m from the earth. The diameter of the sun is 1.4 x lo9 m and that of the earth is 12.8 x 106 m. Estimate the following : ( i ) The total energy emitted by the sun. ( i i ) The emission received per m2 just outside the earth's surface. (iii) The total eiiergy received by the earth, i f no radiation is blocked by the earth's atmosphere. ( i v ) The energy received by a 1.5 m x 1.5 m, solar collector, whose normal is inclined at 45" to the sun. The energy loss throug?z the atmosphere is 40% and the diffused radiation is 20% of the direct radiation. (N.M.U., Nov. 1999) Solution Given :Average solar constant for determination of temperature of sun = 5800 K, DSin = 1.4 x l o 9 m, rorbit TSUn = 15 x 1010 m, Dearth= 12.8 x lo6 m A = 1.5 x 1.5 = 2.@5 m2, I$, = 45", Energy loss through the atmosphere = 40% Diffuse radiation = 20% of direct radiation.

80 1


To find : (i) Total energy emitted by the sun. ( i i )Emission received by the earth per m2,just outside the earth surface. (iii)Total energy received by the earth, if no radiation is blocked by the earth's atmosphere. ( i v ) Energy received by solar collector. Assumptions : ( i )The negligible emissive power of earth in comparison of sun. ( i i )No convection and conduction effects. D ,,

9

= 1.4 x 10 m

D ,,,

6

= 12.8 x 10 m

Fig. 12.36

Analysis : ( i )Total energy emitted by the sun : Surface area of the sun = 4 n r,: = .rr D& = n (1.4 x l o 9 m)2= 6.1575 x 1018 m2 The energy emitted by the sun, Q = OAT,; = 5.67 x lo4 x 6.1575 x 1018 x (5800)4 = 3.95 x W. Ans. ( i i )Emission received by 1 m2,just outside the earth's atmosphere : The sun emits radiation in all direction over a distance of 15 x 1010 m from the earth. The orbit area, just out side the earth's atmosphere : Aorbit= 4n (rorbJ2= 4n (15 x 1010m)2 = 2.8274 x loz3m2 The radiation received per m2 outside the earth's surface

Q Gs = ---Aorbit

3.95 x 2.8274 x

= 1397 W/m2.

Ans.

(iii)The earth may be assumed as a spherical body and energy received by the earth will be proportional to the projected area i.e., area of the earth ; Energy received by the earth = G, x Aearth= 1397 x ( ~ 1 4(12.8 ) x lo6rnl2 = 1.797 x 1017 W. Ans. (iu)Energy received by a solar collector : The direct energy blocked by atmosphere = 40% Hence, the direct energy reaching the earth surface = 60% = 0.6 G& = 0.6 x 1397 = 838.2 W/m2

802


The diffuse radiation = 0.2 x 838.2 = 167.64 W/m2 The total radiation energy reaching the collector = Projected area of collector (cos Q2) x (838.2 + 197.64) W/m2 = 1.5 x 1.5 cos (45") x (1005.84) = 1600.28 W. Ans.

12.1 I . SUMMARY The radiation refers to t h e energy emitted i n form of electromagnetic waves by the bodies because of their temperature.The radiatiop energy emitted i n wavelength between h = 0.1 and h = 100 p m is refirred th'ermal radiation. The s u n emits thermal radiation at a n effective surface temperature of 5760 K and bulk of this energy lies between h = 0.1 to 31. = 3 pm, therefore this spectrum is referred a s the solar radiation. The radiation emitted by t h e s u n is in r, avelength between h = 0.4 to h = 0.76 pm, is visible to h u m a n eye, therefore, this spectrum is referred as the visible radiation (light). The glossary of radiation terms and their definition a r e given in Table 12.5.

TABLE 12.5. Glossary of the radiation terms Terms

Absorption Absorptivity, a Blackbody Blackbody radiation function Diffuse Directional Emission Emissive power, E Emissivity,

E

Gray surface

Green house effect Hemispherical Irradiation Kirchhoff 's law Planck's law Radiation intensity

_

Reflection Reflectivity, p Semitransparent Solar energy Solar constant Solid angle, w

Spectral Spectral distribution Specular Stefan Boltzmann law

Definition The process of converting the radiation intercepted by the matter to internal thermal energy. Fraction of the incident radiation absorbed by the surface. An ideal body which absorbs all incident radiation and emits maximum energy. Fraction of radiation energy emitted by a blackbody at temperature T in wavelength band h = 0 to h. A surface, whose properties are independent of directions. The property pertains to a particular direction, denoted by 8. The process 'of radiation production by the surface at a finite temperature. The rate of radiant energy emitted by a surface in all direction per unit area of the surface. It is measured in W/m2. Ratio of the emissive power of a surface to the emissive power of the blackbody at the same temperature. A surface for which the spectral absorptivity and emissivity are independent of the wavelength over the spectral region of the surface irradiation and emission. The warming up process due to remission between earth's surface and atmosphere. The quantity pertains to all directions above the surface. The rate at which the radiation is incident on a surface from all direction per unit area of the surface, G(W/m". Relation between emission and absorption properties of a surface at thermal eqbilibrium. It is associated with spectral distribution of emission from a blackbody. The rate of radiation energy propagation in a particular direction, per unit area normal to the direction, per unit solid angle about the direction, I (W/m2.sr).

803


Thermal radiation Total solar radiation Transmission ~ransmissivit~ Wien's displacement law

_

_

I

_

_

_

_

Rate at which energy leaving the surface due to emission and reflection in all directions per unit area per unit time, J (W/m2). The process of redirection of radiation energy incident on a surface. The fraction of incident radiation energy reflected by the surface. It is a medium in which radiation absorption is the volumetric process. It is energy coming out the sun. Rate at which the solar radiation flux is received on a surface normal to sun's rays just outside the earth's atmosphere, G,(W/m2). Ratio of area of spherical surface enclosed by a conical surface with vertex of the cone at the centre of sphere to square of radius of sphere. It is measured in straradian (sr). It refers to a single wavelength (monochromatic) radiation. The quantity is denoted by subscript h. It refers to properties variation with wavelength. It refers to the surface for which the angle of reflected radiation is equal to the angle of incident radiation. The emissive power of the blackbody is directly proportional to fourth power of the absolute temperature ; E6 = o T4, where o = 5.67 x lo4 W/m2.K4,is Stefan Boltzmann constant, and T is an absolute temperature in K. It is the electromagnetic energy emitted by a matter at a finite temperature in the spectral region from approximately from 0.1 to 100 ym. Sum of direct and diffuse solar radiation. It is process of the thermal radiation passing through the maftek. It is the fraction of radiation energy transmitted by the matter. Relation between wavelength,,A and absolute temperature T at which Ebhreaches a maximum ; h,, T = 2897.6 pm.K.

REVIEW QUESTIONS What is an electromagnetic wave ? How does it differs from a sound wave ? What are the ranges of wavelengths of electromagnetic waves covering ultraviolet, visible, infrared and thermal radiation ? What is the speed of energy propagation between two bodies when the space between them is evacuated ? What do you mean by ultraviolet, visible and infrared radiation ? What is a blackbody ? What are its properties ? Why does a cavity with a small hole behave as a blackbody ? Why are microwave oven suitable for cooking ? What are the total and spectral emissive power of a blackbody ? What do you mean by spectral, terms used in thermal radiation ? State Planck's distribution law and list down its features. What is Wien's displacement law ? Derive an expression for its relation. What is a diffuse body ? State and explain Stefan Boltzmann law. Derive an expression for total emissive power of a blackbody. What is radiation intensity ? How do you distinguish between spectral emissive power and spectral radiation intensity ?

_

81 8


819

RADIATION EXCHANGE BETWEEN SURFACES

Consider a geometry as shown in Fig. 13.13. The view factor Fl-2 between surface 1and 2 can be evaluated by following procedure. First identify the end points of the surface as A, B, C and D. Connect them with tightly stretched string, which are indicated by dashed line. Hottel has proved that the view factor F1 can be expressed in terms of lengths of these stretched strings which are straight lines as

-,

Two infinitely long parallel planes

Two curved surfaces

Fig. 13.13 (L, + L6) - (L3 + L4) ...(13.11) 2L 1 where L, + L, is sum of lengths of crossed strings and L3 + L, is sum of lengths of uncrossed strings attached to the end points. Therefore, Hottel cross string method can be expressed as X (crossed strings) - X (uncrossed strings) F.2 -J . = 2 x (string on surface i) Fl-, =

Example 13.1. Calculate the view factor Fl - 2 between a small area dAI and a parallel circular disc A,. The elemental area dA, is located a t the axis of the disc A2, at a distance L. Solution The elemental area of the disc d& = 27t r d r pl = P2 (due to symmetry) Here Using the relation FA,-A, 0rFl-2 cos p1 cos P2 dAl dA2 7c s 2 A1 A l A2 Since the area A, is small, hence assuming it as constant.

='J

alc

Fig. 13.12. The view factor for three very small surfaces "looking at" the large surfaces (Al << AJ 13.1.3. The Cross String Method The view factor between very long surfaces can be determined by a very simple crossed string method developed by H.C. Hottel in the 1950s, the surfaces of the geometry do not need to be flat ; they may be convex, concave, or any irregular shape.

J

L~ s2 = r2 + L2 and cos2 P = L~ + r 2 and limits for & (or r) are from r = 0 to r = R

where

L

a ,A,

Fig. 13.14


820

82 1


,

,

Example 13.4. Calculate the view factor Fl - and F2 - for the following geometries :

Substituting in above relation,

Put r2 + L2 = t ; 2 r dr = dt t = 0 + L2to R2 + L2 and limits : Then the integration :

Example 13.2. Determine the view factor between a small area dAl and a rectangular surface of dimension a and b, where the rectangular surface is i n horizontal plane and the small area dAl lies i n the vertical plane and below one corner of the rectangle at a height H. (P.U., May 2000) Solution The view factor between two surfaces is given by

ea-

.

dA2

cos plcos P2 dA1 dA2

c / 7 dY

st

For element dAl = Al ;

-

cos pl cos P2 dA2 For given geometry

H b p1 = --, cos p2 = - '

S

S

and .

s= Then, F,=

'5(1

1. Sphere of diameter D inside a cubical box of length D. 2. Diagonal partition within a long square duct. 3. End and side of a circular tuba of equal length and diameter. Solution Given : Surface geometries. To find :The view factors F1- and F2Analysis : 1.Sphere within a cube : By inspection I?1_, = 1

.

, dA2 = d'x dy

Fig. 13.15 ~ a b ~ Hb Hab n o o (El2 +a212dx dy = . ( H ~ + a212Sody = n(H2 + a2 12 ' Ans. b

Example - 13.3. A flat surface, 1 is completely enclosed by a second surface, 2 Fig. 13.16. Determine the view factors F1 F2 - and F2 - 2 . Solution -27

.

A1 .D2 F 2 4= F13=6L2 X I = - - 6 Aos. 2. Partition within a square duct : \ From summation rule, F1-1 + F1-2 + F1-, = 1 Where for flat surface, r F1-, = 0 ! BY symmetry, F1-2=F1-3 Hence Fl-, = 0.5 A1 J2 x 0.5 = 0.71. Ans. By reciprocal theorem, F2-, = A, F,-2 = -y3. Circular tube : From Fig. 13.9 with r 3 L = 0.5 and Llrl = 2, Fl = 0.17 From summation rule, F1-1 + q - 2 + Fl-3 = 1 Where for flat surface, F,-1 = 0 Fl-, = 1-F,-, = 1-0.17 ~ 0 . 8 3 A1 By reciprocity rule, F2-1- - F1-2=- nD2'4 .DL x 0.83 = 0.21. h s . - A2 Example 13.5. Calculate the view factor F1 F l - , and F2-, for the followinggeometries :

BY reciprocity rule,

s2

cos

Fig. 13.17

Si;

-,

-.

As no part of surface 1 "sees" itself, all the energy leaving surface 1reaches surface 2, and from the definition of view factor, F,, = 1 To determine F2-,, from eqn. (13.5)

A, F2-1 = A1 Fl-2 F2-, =

AI'FI-2 A2

- -A1 A,

Fig. 13.16. A flat surface, 1,com-

pletely enclosed by a second surface, 2.

From eqn. (13.6)

F2-I + F2-, = 1 and F2, = 1- F2-, = 1- A2 Aos. Fig. 13.18

822


1. A black body inside a black enclosure. 2. A tube whose section is equilateral triangle. 3. Radiation exchange between a hemisphere and a plane surface.

Example 13.7. Calculate the shape factor of a conical hole. Solution Analysis :Considering an imaginary flat plane covers the hole, hence by summation rule F , - , + F l - 2 = 1 and F 2 - 1 + F 2 - 2 = 1 But F2- = 0, hence F2 = 1. By reciprocity relation, AIF1- 2 = A2F2-1 F1-l= 1-Fld2 A2 A2 Therefore, F1- = 1- A1 F2 = 1- A1

Solution Given : Surface geometries. To find :The view factor. Analysis : 1.A black body inside a black enclosure : By inspection, F2-,= 1 By summation rule, F1- + F1- = 1 By reciprocity rule, AzF2 - 1 = AIF1 - 2

and

823


2

T

-,

i

tan a = (Dl2h)

-,

Fig. 13.20

D (where L = Wcos a ) = I - (nI4) D~ = 1- n D L/2 2L n ~ tan2 ' a Fl-, = 1= 1 - sin a. Ans. n~~ sin a

2. A tube of equilateral triangle : From summation rule, F1-1+F1-2 + F 1 - 3 = Where for flat surface, F14= 0, F2-2 = 0, F3-3 = 0 F 2 - , + F 2 - , = 1 or F2-,= 1 - F 2 - , By symmetry, F1-2=F1-3

cos2 a Example 13.8. Calculate the shape factor for cylindrical cavity as shown in Fig. 13.21 with respect to itself: I

Hence

A1~ F l - =0.5 2 Fl-,=0.5 or F 2 - l = 2

Hence F2-,=I-0.5=0.5. Ans. 3. The radiation heat exchange between a hemisphere surface 1 and a flat surface 2. From summation rule, Fl- + Fl - 2 = 1 and F2 + F, =1 But for flat surface, F2 = 0, hence F2- = 1 By reciprocity rule, h F 2 - 1 =A1F1-2

-,

To flat surface hemisphere,

-, -,

A2F2-1 = nR2 = 0.5 F1- = A, 2nR2

Solution Assuming cylindrical surface 1and plane surface 2 From summation rule, F1- + F1- = 1 F1-,= l - F 1 - 2 F2-1 + F2-, = 1 But for flat surface, F2-2 = 0 From surface 2 to surface 1, F 2 4= 1 By reciprocity rule, A2F2-l=AlFl-2

T

,

or and

I

Flat surface of hemisphere,

Example 13.6. Calculate the view factor between two opposite sides of a hollow cube, if view factor between two adjacent sides of it is 0.2. (P. U., 2000)

D Fl-, = 1- -----D+4L

Solution Given :The geometry as shown in Fig. 13.19

I

Fig. 13.21

4L-. -D+4L

hs.

Example 13.9. Consider a triangular duct of length L a s shown in Fig. 13.22. For given dimensions, prove that F2- = 0.75 Surface 3 where, ab = ac = x and bc = x/2.

By summation rule,

Solution Given :The triangular duct.

F,-l+F,-2+F,-,+Fl-4+Fl-,+Fl-6= 1 Flat surface 1, F1-, = 0 0 +0.2 +F1-,+0.2+0.2+0.2 = 1 Fl-, = 1- 0.8 = 0.2. h s . The view factor between two opposite faces = 0.2.

Surface 2

X

A --L, 4=A3=xL l- 2 To prove :View factor F, - = 0.75. Assumption :Length of duct is sufficiently large so negligible energy leaving through open ends.

,

Fig. 13.19

Lrface 1

Fig. 13.22

824

ENGINEERING HEAT AND MASS TRANSFER -

Analysis :From the summation rule for flat surfaces F, - 2 F2-l + F2-,= 1 and F3-, + F3-, = 1 By symmetry F l - 2 = F1-3

-

+ F, -,= 1

--

(

.:

-


,

F, - = 0)

For surface 1 and surface 2 : A, = 5 mm x 20 mm = 100 mm2 A,=7mmx20mm=140mm2 w = 8 m m , L = 7 m m , D=2Omm 3 8 L 7 For ~=-=-=0.4, y=-=-=0.35 D 20 D 20 From Fig. 13.10 ; F2 = 0.3

[

Hence Hence

-,= 1- 0.25 = 0.75.

F2

825

-,

Ans.

Example 13.10. Two rectangular plates are a t right angle in following showngeometries. Calculate the view factor F, for the given configurations.

-,

'

Then

F,

140 -,= x (0.35 - 0.3) = 0.07. 100

Ans.

(c) Assuming surface 5 and surface 6 as A, = A, + A, and A, = 4 + A, By additive rule F1 - = F1- + F1F1-2 =F1-,-F1-4 By reciprocity rule

(d)

Fig. 13.23

Solution

From Fig. 13.10, view factor, F, - = 0.34. Ans. ( b )Assuming surface 4 as combination of surface 1and surface 3 ; A, = A, + A, By additive rule F2- = F2- + F2- or F2- = F2- - F2By reciprocity rule

13 7 = 0.35 F6-, ; X= - = 0.65; y = - = 0.35+F6-, 20 20 8 7 For F6-3;x=-=0.4; y=-=0.35+F6.,=0.3 20 20 13 4 For F 4 - , ; x = % = 0 . 6 5 ; y = - - =0.2 -F4-, =0.367 20 8 4 For F4-, ;x = - = 0.4; y = - = 0.2 +F4-,= 0.358 20 20 Then F,-, = 1.4(0.35 - 0.3) = 0.07 F, = 0.8(0.367 - 0.358) = 0.0072 F, = 0.07 - 0.0072 = 0.063. Ans. (d)Assuming surface 5 and 6 as A, = A, + A3 For

and

-, -,

A6=&2+A4

By additive rule F1- = F1 -

+ F1 -

--$

,

F1 - = F1 - - F1 -

826


827


By reciprocal rule A6 A1

=A6F6-1+F1-6=-

*

F6-1

F1-2=

A6 1

F6-l-F1-4

Again by Additive rule

*

F6-5=F6-1+F6-3~F6-1=F6-5-F6-3

-46

Fl-2=

1

(F6-5-F6-3)-F1-,

By reciprocity :

F6- =

A3 A F36

=

A3

(F3-

6

+ F3-

Reciprocity rule A3 A6

A3F3-, =&3F6-3+F6-3=

*

-F3-6

F1-2=

A1 F, -, , -, = A

Due to symmetry : F3- = F2

2

F3-2 = F1-4

,

BY symmetry rule F~- = F~- = Therefore,

F 1 2=

6 - 6 -

and 3

-

2

F3-2=F1-4

~2

Therefore

,

From Fig.13.8, the view factors F6- 5, and F1 - : L 3 w 5 For F 6 - 5 :x = - = - =0.375, y = - = - = 0.625+ D 8 D 8

F6-5=0.06

-,and Fl -,can be obtained

From Fig. 13.10, the view factor F6- ,, F3

Then,

- 0.103 Fl-2 =

= 0.120. Ans.

Example 13.11. Calculate the view factor Fl - 2 for the following geometry.

Example 13.12. Fig.13.25 depicts window i n the wall of a room. Find the view factor FB- A , where A is the window and B is the floor. Solution

Solution Given :Geometry as shown in Fig. 13.24

l------B

c

m

4

Fig. 13.24

Analysis :Assuming surface 5 and surface 6 as A5 = Al + A3 &=&+A4

(4 Fig. 13.25. Schematic for example 13.12

828


Given :The geometry as shown above. L = 6 m , w=2.5m, D = 5 m Dl = 4 m, D2 = 1m. TO find :View factor FB-A=F5-4 Assumption :Surface 5 as A5 = A, + A2 Surface 6 as AG= A, + A,. Analysis : For convenience, divide the floor into two parts, 1 and 2 as shown in Fig. 13.25 (b).The window is represented by 4 and the remaining part of the wall by 3 and we need F5- ; using reciprocity and additive rule simultaneously.

We can obtain F1 - from Fig. 13.10, now we need F2 F,-6 = F,-3 + F5-4

-,;

1 -{A1F1-3+A2F2-3+A1F1-4+A2F2-41 A5 From law of corresponding corner (symmetry)

Example 13.13. Consider a n enclosure formed by closing one end o f cylinder, (diameter = D, Height = L) by a flat surface and other end by a hemispherical dome. Determine the view factors of all the surfaces of the enclosure, if height is the twice the diameter. Solution Given :A geometry as shown in Fig 13.26. Surface 1 = flat cylinder end Surface 2 = cylindrical surface Surface 3 = hemispherical end For 3 surfaces in the enclosure View factors, = N2 = 9 The view factors to be determine directly = 112 N(N - 1)= 112 x 3(2) = 3 Here Fl- = 0 with help of imaginary surface 4. F4-, + F4-, = 1 F4-3 = 1

,

From Fig. 13.10.

Fl-3

-

-

F5-6=

4F2-4=Al

829


R I -- - --'I2L 2D-4. D R2 ='Z=L 4 For enclosure summation rule For surface 1.

From Fig. 13.9

For surface 3. and

.. 4F2-4= Now from eqn. ( i )

7t

-

D2

Again, F3- + F3- = 1 F3-, = 1 - 0.5 = 0.5 Here it should be noted that fraction of radiation energy emitted by surface and intercepted by hemispherical surface 3 will be exactly same that intercepted by flat imaginary surface 4. Thus For F1-3=F1-4 or F4-3

,-

For surface 2.

Now from eqn. (ii)

Fig. 13.26

(6 x 5) x 0.14 - (6 x 4) x 0.12 - (6 x 1)x 0.06 = 0.48 2

Similarly

F1- 4 = 0.06

.

I


830

831


For surface 3. By reciprocity rule -3 =

Therefore, 0.1175+F2-,+0.1175=1 or F2-,=0.765 Therefore, the view factor matrix :

F3-1 =

and

Example 13.14. Calculate all view factors for conical geometry shown in Fig. 13.27.

-] AI A3

-

900n x 0.883 = 0.526. Ans. F1- 3 = 1509.34~

%F2-3

By summation rule F3-, = 1- F 3 - , - F 3 - 2 = 1-0.526-0.0793 = 0.388. Ans.

Solution Given :A conical geometry r, = 30 mm, r2 = 15 mm, H = 30 mm. To find :The view factors F2-27 F2-37 F1-l, F1-2, F1-3>

Example 13.15. Calculate the view factor between two parallel disc in the form of circular rings as shown in Fig. 13.28 below. These are coaxial, spaced 10 cm apart. The inner and outer radii for lower ring are 8 cm and 20 cm, respectively, while that for upper ring are 5 cm and 10 cm, respectively.

F2-17

F3-19

F3-2? F3-3 '

Analysis : For given geometry A, = nr? = n x (30 mm)2 = 900 n mm2 A,= nr; = TC x (15 mm)2 = 225 n mm2

Fig. 13.27

A3 = a (rl + r2)-/,

Solution Hollow, A, Given :Two parallel, coaxial rings as Lower ring r, = 8 cm r2 = 20 cm I Upper ring r3 = 5 cm L=lOcm Hollow, A, r, = 10 cm / I Spacing L = 10 cm. To find :The view factor F1 - 2. Assumptions : (i) The hollow portions of two rings as A, and A, Fig. 13.28. Schematic of parallel, (ii) Two surface 5 and 6 as coaxial circular rings A, = A, + A, A6 = Al + A2. Analysis :Since these two coaxial circular rings are parallel. We may use Fig. 13.9. for determination of view factor between them. View factor F5 - between two circular disc : .t

For surface 1. (Top surface), For two parallel discs. from Table 13.2

F,_,=:[X-,/X~

11

- 4 ( R 2 / ~ 1 ) 2= - 2.25-

-

d

Alternatively from Fig. 13.9 F, = 0.117 R, = 1, R, = 0.5 Now by summation rule F,-,+F,-,+F,-,= 1.0 Surface 1is flat, thus F1-, = 0 .. F, = 1- F, - = 0.883. For surface 2. From reciprocity rule

-,

103j

2.25, - 4 -

=0.117. Ans.

-,

View factor F3- :

Ans.

-41'3 - 2 = A2F2 - 1 A1 F 2 - I = - Fl-2 = -x 0.117 = 0.468. A2 225n

View factor F, -

Ans.

By summation rule F2-, + F 2 _ , + F 2 - 3 = 1.0 Surface 2 is flat ; F2-2 = 0 F2-,= l - F z - , = 1-0.468=0.532. A ~ s . ..

View factor F3-


832

833


But A,=A4-A, Hence the radiation from surface 1 i.e., surface 3 and 4 will deliver energy to surface 4 and 5 ; =A4F4-2-A3F3-2

Again we have, A2F2- = 4 F3- = A3 [F3-

or AlFl-2 = A 4 ( F 4 - 6 - F 4 - 5 ) - A 3 (F3-6-F3-5) where, 0 . 0) 5=~0.00785 ) m2 the surface area, A3 = n ( ~= ~ ( ~ the surface area, A, = n(R2) = n(0.12)= 0.0314 m2 the surface area of ring 1,A, = (A4-A3) = 0.0236 m2 Using given relation for calculation of F4- : R4 10cm R6 20cm B=--== 1 andC=--- =2 H lOcm H lOcm X = 1+ B2 + C2 = 1+ + (212 = 6

- F3-

,

Therefore,

and

Example 13.16. Find the shape factor between two areas 1 and 2 which are i n the form of circular ring, coaxial and are i n two parallel planes at a distance 10 cm. Area 1 has inner radius of 5 cm and outer radius of 10 cm. Area 2 has inner radius of 8 c m and outer radius of 20 cm. Use following formula for calculating the shape factor between two circular areas located coaxially i n two parallel planes is given by

Using given relation for calculation of F, : R, R4 10cm B=--=-- 1 and C = - = H 10cm H

or or

8cm = 0.8 lOcm

-,

Rl R2 B=,,C= ,andX=(1+B2+C2) where, 11 11 where, RI and R2 are the radii of the circular planes and H is the distance between them. (P.U., Nov. 1992) Solution Given :Two circular parallel surfaces located coaxially ; R3 = 5 cm, R4 = 10 cm, H = 10 cm R 5 = 8 c m , R6=20cm. To find :The shape factor between the two parallel surfaces. 10 cm Assumptions :A4 = A, + A3 A, = A5 + 4 Analysis :Using the property relations for shape factor F3-6=F3

...(A)!

Using given relation for calculation of F3 : R3 5cm R, 20cm B=-== 0.5 and C = - =2 H lOcm H lOcm

Using given relation for calculation of F3- : R, 5cm = 0.5 and C = R5 = 8cm = 0.8 B=-=H lOcm H 10cm

-5+F3-2

Substituting the values in eqn. (A) ; 0.0236 x F, = 0.0314 x (0.764 - 0.27) - 0.00785 x (0.7917 - 0.3553) F, = 0.5121. Ans.

-, -,

F3-2 = F 3 - 6 - F 3 - 5

or

F2-4=F2-3+F2-l

Example 13.17. Two concentric cylinders have inner and outer radii 5 c m and 10 em, respectively and length 20 cm. Calculate the all possible view factors.

F2-1=F2-4-F2-3 F4-6=F4-5

+F4-2

20 cm

Applying the reciprocal theorem A1Fl-2 =A2F2-1=4(F2-4-F2-3)

Fig. 13.29

Solution Given :Two concentric cylinders as shown below : To find :All possible view factors.

834


835


Analysis :For parallel cylinders from Fig. 13.11. L 20 For surface 2. - - r2 - 10 = 2

42

13.2. BLACKBODY RADIATION EXCHANGE

and

-,

F2 = 0.43 and For surface 1. By reciprocity

-~=20cm

+

Fig. 13.30

By summation rule Fz-1+F2-2+F2-3+F2-4=

-,= F2-,

By symmetry

F2

1 (1- 0.43 - 0.34) = 0.115 F2- 3 = F2- = 2 By summation rule

Then,

F 1 4 + Fl-, + F1-3 + F1-4= 1 Convex surface 1, thus Fl = 0 By symmetry F1- = F, -

-,

..

1

0.86+2Fl-,=I

or F1-3=-[1-0.86]=0.07 2 For surface 3. By reciprocity : A3 F3- = AIF1-

For surface 4. Due to symmetry F 3 -3 . = F 4 - j , j = 1 , 2 , 3 , 4 . By summation rule F 3 - 1 + F 3 - 2+ F 3 - 3 + F 3 - 4 = 1

Flat surface 3, thus F3- = 0 :. 0 . 1 8 7 + 0 . 6 1 3 + O + F 3 - 4 = 1 or View factor matrix 1

2

F1-3

Fl-4

F2-1

F2-2

F2-3

F2-4

F3-1 F3-2 F4-1 F4-2

F3-3 F3-4 F4-3 F4-4

-

The radiation may leave a surface due to reflection and emission and on reaching on the second surface, there may be reflection as well as absorption. For a blackbody radiation, there is no reflection (p = 0). Hence energy leaves a surface as a result of emission, while it absorbs all incident energy. Considering the radiation exchange between two blackbodies having surface area A, and 4. The rate of energy leaves the surface 1and absorbed by surface 2 ...(13.11~) Q1-2 = A1 Fl-2 Ebl = Al F1- 0 TI4 Similarly, the energy leaves the surface 2 and reaches the surface 1 ...(13.11b) Q2-1 = 4 F s 1 Eb2 = A2 F2-1 0 Tz4 The net radiation exchange between the two surfaces ...(13.12) - A1 F 1-2 0 (T; - T24)= 4 F2-1 Q (TI4- T24) Qnet Since by reciprocity relation A, F1, = A2 F2-l, Consider an enclosure consists of N black surfaces maintained a t specified temperatures. The radiation heat transfer from any surface i of the enclosure is sum of radiation heat transfers from surface i to each surface of the enclosure. N

N

Example 13.18. Two black discs of diameter 50 cm are placed parallel to each other concentrically a t a distance of 1m. The discs are maintained at 727OC and 227"C, respectively. Calculate the heat transfer between the discs per hour, when no other surface is present except the discs. Solution Given :Two parallel discs of 50 cm diameter and 1m apart. r, = r2 = 25 cm = 0.25 m T, = 727°C = 1000 K T2 = 227°C = 500 K. To find : Heat transfer rate when no other surface is present between the discs. Analysis :When no other surface is present except two parallel discs. Fig. 13.31. The view factor between the two parallel plates, from Fig. 13.9

L

1

-

r2 -=0'25 0.25 ;---t F, = 4.0 and =rl 0.25 L 1

I Fig. 13.31

-,= 0.06

The heat transfer rate, Q=F~-~AO(T,~-T,~) = 0.06 x (d4) x (0.5)~x 5.67 x lo4 (1000~ - 5004) = 626.24 W = 2.254 x lo6 Jlh = 2254.46 kJk. Ans.

Example 13.19. A 5 cm diameter sphere a t 600°C is placed near a n infinite wall a t 100%. Both surface are black. Calculate the net radiant heat transfer between the two bodies.

836


Solution Given :A hot sphere near a wall T, = 600°C = 873 K, T2 = 100°C = 373 K D = 5 cm = 0.05 m To find :Net radiant heat transfer. Analysis :The net radiation heat exchange

ce

+lo

+ r2)d(r2 - r1)2+ H'

m

F2-

0'00785 x (0.804) = 0.065. Ans.

= 0.09715

(ii) Net radiation heat transfer from heater to shield Qnet= A, F1 - a (TI4- T24) = 0.00785 x 0.804 x 5.67 x lo4 x (14734- 3734) = 1678 W. Ans.

Example 13.21. A jet of liquid metal a t 2000°C pours from a crucible. It is 3 mm in diameter. A long cylindrical radiation shield, 5 cm diameter, surrounds the jet through a n angle of 330°7but there is a 30" slit in it. The jet and the shield radiate as black bodies. The slit in a room is a t 30°C, and the shield has a temperature of 700" C. Calculate the net heat transfer :from the jet to the room through the slit ;from the jet to the shield ;and from the inside of the shield to the room. Solution Given :A jet of liquid metal is surrounded by radiation shield. Dl = 3 mm T, = 2000°C = 2273 K, D 2=5cm T, = 30°C = 303 K, T2 = 700°C = 973 K Qskield = 330" Qslit = 30". To find :Net radiation heat transfer from (i) Jet to room Qnet -,, (ii) Jet to shield, %et - 2, (iii) Shield to room, Q,, - , Analysis :The view factors : (a) F, - : From jet to room, by inspection

cm+

Fig. 13.33. Heat transfer from a disc heater to its radiation shield

(b) F1 - : From jet to shield, by inspection

(i) Radiation heat transfer from jet to room

-

Qnet 1- =A1 F1-- a (T; - TW4) = fx x (0.003 m x 1m)l x (0.08333) (5.67 x

lo4

W/m2.K4)

= 1188.2 Wlm. Ans.

=1

-

F1-,g0 For two parallel discs 1and 3, from Fig. 13.9 ; L - -0.2 r2 = 0.1 = 0.5 F1 r, 0.05 = 4.0 and L 0.2 Then, Fl-, = 1- 0.196 = 0.804 F1-2=1-F1-3

A2 = d r ,

_

AlFl-2 =AzF2-1 A1 F2-,= A, Fl-2 - 2 + F1-3

Area

Then

Example 13.20. A heater as shown in Fig. 13.33 radiates heat partially conical shield that surrounds it. (i) Determine view factor from shield to heater. (ii) If heater and shield are black and are at temperatures 1200°C and 10O0C, respectively, what is the net heat transfer rate from heater to shield ?

F,-, + q

A, = nrI2 = n x (0.05)2= 0.00785 m2

@@I 1

-,

and or

Area

= 60.05 + 0.1) x J(0.1- 0.05)~+ (0.2)' = 0.09715 m2

Q1-2=AlFl-2 E b l - A 2 F 2 - 1 Eb2 =A1F1-, (Ebl-Eb2)=AlF1-2d(T14-T24) Since the sphere is half covered by an infinite wall thus half of the radiation emitted by sphere will fall on infinite wall i.e., Fig. 13.32 F, = 0.5 and A, = n D2 = n x 0 . 0 5 ~= 7.853 x m2 x 0.5 x 5.67 x lo4 x (8734- 3734) = 125 W. Ans. Then Q, - 2 = 7.853 x

Solution Given :A heater a t the base of a conical shield as shown in Fig 13.33. r, = 0.05 m, r2 = 0.1 m H = 0.2 m T, = 1200" C = 1473 K T2 = 100" C = 373 K Both surfaces are black. To find : (i) View factor F,I-&, (ii) Radiation heat transfer from heater to shield. Analysis :(i) The view factor, by reciprocity

837


: '

(ii) Radiation heat transfer from jet to shield

-,= 0.196

Qnet 1-2 =A,F,-, a(T;-T;) = (n x 0.003) x (0.9167) x (5.67 x lo4) (2273*- 9734) = 12637 W/m. Ans. (iii) Radiation heat transfer rate from shield to room Qnet2-ee=A2 2' - - a (T,4 - T 2 )

838


Here F2- ,: View factor from shield to room Aslit

'slit

-

839


Thus,

- 1 = A1 1' - w = 16500 W. Ans. (ii) When ceiling height in increased to 4.5 m, neither view factor F, - nor F, will change, thus the heat transfer rate from floor to window and ceiling, walls will remain same.

-,

13.3. RADIATION FROM CAVITIES

(

Qnet 2-- = n X 0.05 X %) 360 x (0.08545) x (5.67 x 10") x (973' = 618 Wlm. Ans.

- 303')

Example 13.22. A n outlet shoe store with a display window i n the front is shown, with dimensions, i n Fig. 13.34. The store is to be heated by making the floor a black radiant heating panel at 45' C. The glass window acts as a black plane at 10' C and the other walls and the ceiling act as black planes at 25' C.Find the net heat given up by the floor. What difference results, i f the ceiling height is raised to 4.5 m, the other dimensions remaining unchanged ? Solution Given :The display window of a shoe store. We designate floor as surface 1,window as surface 2, and other walls and ceiling as surface 3, then data T, = 45 + 273 = 318 K

Reconsider examples 13.7, 13.8 solved for radiation view factor of conical and cylindrical cavities, respectively. The radiative heat transfer from such cavities can be estimated by using view factor Fl - obtained. Now we go for formulation of some standard relation for such cavities as shown in Fig. 13.35. Let us consider cavity as surface 1 and opening as surface 2. For any of following geometries, all the energy emitted from cavity (surface A,) does not coming out the opening. But a part of radiation emitted from the surface falls on its other part, of which a portion is absorbed and remainder is reflected back.

(a) Spherical cavity

(b) Cylindrical cavity

(c) Conical cavity

Fig. 13.35. Radiations from cavities

Fig. 13.34. Geometry for example 13.22

To find : ( i )Net heat given up by floor, (ii)Difference in heat radiation by floor, if ceiling height is increased to 4.5 m. Analysis :( i ) Since all surfaces are black, thus various emissive powers are Ehl = oT14 = 5.67 x lo4 x (318)' = 579.8 W/m2 x (283)4= 363.7 W/m2 Eb2= oTZ4= 5.67 x Eb3= dI'34= 5.67 x lo4 x (298)4= 447.14 W/m2. The view factors :The geometrical arrangement for floor and window is very similar to example 13.12. F, - = 0 and F1 = 0.058 For an enclosure of floor, window, wall and ceiling F l - l + F l - 2 + F 1 - 3 = l ~ F 1 - 3 =1 - F l - 2 = 1 - 0 . 0 5 8 = 0 . 9 4 2 For complete enclosure Heat radiated by floor = Heat absorbed by window and ceiling

,

-,

Rate of emission from a cavity surface = Al o c1T4 ; of this energy, a part falling on Al and absorbed by it = A, o E, T: (a, F1- ,) = Al o E, T4 el F1 where F1 - is view factor of the cavity surface with respect to itself and a, = E, from Kirchhoff 'S law. Now the reflected radiation from the surface A, = (1- cl) A, o el TI4F1- of this reflected radiation, a part again falls on cavity surface and absorbed = (1- E,) A, o el T: F1 - 12 Then again reflected = (1- E,) A, o E, T: F1 - 12 (1- E,) = (1- E,) Al 0 TI4F1 - 13 (1 - el) E, Again absorbed and so on. = (1Again reflected A, o E, T t F1 - 13 (1Net rate of radiation coming out the surface Q = Total emission rate - Total absorption rate

,

840


84 1


It is an infinite series, and it can be arranged as Integrating to obtain the area of sector of sphere

As =

...(13.14) For a conical cavity For a cylindrical cavity For hemispherical cavity

F,-, = 1 - s i n a 4L F ~ -D ~+ =~L

[see example 13.71

30

= 2nR2

[see example 13.81

30

2

0

r

- COS P

1:

n sin ~ p df3 ~ = 2nR2 [ I - 0.86601 = 0.268 nR2 m2 = 0.00842 m2

A, = Area of sphere - Area of sector of sphere = 4nR2 - As = 4nR2- 0.268 nR2 = 3.732 nR2 = 0.1172 m2 Area of opening, A, = nr2 = n x (0.05)2= 0.00785 m2 If we mark cavity as surface 2 and sphere inner surface as cavity 1then Area of cavity

A2 I?, - = 1 - A1

F,-, + F,-2 = 1 F 2 - I + F 2 - 2= 1+F2-,

Example 13.23. A spherical cavity of radius 10 c m is made i n a large flat metal plate. The cavity has circular opening to the atmosphere. The radius of circular opening is 5 cm. The surface of the cavity is at 227" C. There is no heat conduction through the metal plate and there is no convection in the cavity. If the emissivity of the cavity surface is 0.8. Calculate the net radiative heat loss to the surroundings. Assume atmospheric temperature as 27" C. Solution Given :A spherical cavity in a metal plate. T,= 227" C = 500 K Tm= 27" C = 300 K E = 0.8, R = 10 cm opening radius, rl = 5 cm. To find :Net radiative heat loss to surroundings. Analysis :Considering A = area of sphere Fig. 13.36 A.. = area of sector of sphere cut by opening A-B-C. A, = area of cavity A, = area of opening 1. Area of sector of sphere : Consider a circular differential strip of thickness dr at radius r from vertical centre of sphere. It substends an angle of dp at the centre. r = R sin p Here dr = R dp (since chord is very small) Area of element dA=2nrdr=2nRsinb.Rdp ...( i ) = 2nR2 sin P dB Semi (half) angle substended by sector A-B-C

So d A = S

= 0 since flat

F 2 4= 1 By reciprocity, I?, and

-

A2

= - F2

,

A1

-, = 0.00785 0.1172 x 1 = 0.067

F,- = 1- 0.067 = 0.933. The rate of radiant heat loss through cavity

= 0.1172 x 5.67 x

lo4

x 0.8 (500' - 300') x

1- 0.933 o.933 (1- o.8)

= 23.863 W. Ans.

Example 13.24. Fig. 13.37 shows a cavity having surface temperature of 900°C and emissivity as 0.6. Find the rate of emission from the cavity to the surroundings. (P.U., May 1994) Solution )c-5 cm 4

Opening of

Fig. 13.37

843


842


Assumptions : 1. Interior surfaces as blackbodies. 2. Negligible heat convection. 3. No heat exchange to the surroundings from bottom and sides. Analysis :The heat loss to the surroundings ;

Given :A cavity as shown in Fig. 13.38. T, = 900°C = 1173 K. E = 0.6. To find :The heat transfer rate from cavity to surroundings Analysis :The radiation heat transfer rate from a cavity is given by

Here we need view factor F1

Q=Ql-3

+Q2-3=A1Fl-30(T14-T34)+A2F2-3~(T24-T34)

The shape factor can be obtained as From Fig. 13.9, with r 2 / L= 0.37511.5 = 0.25 and Llr, = 1.510.375 = 4, F, From summation rule, F2-l + F 2 4 + F 2 - 3 = 1 where for flat surface, F2-, = 0 F2-, = 1 - F , - , = 1-0.06=0.94

-, (cavity surface to cavity surface)

By reciprocal theorem, where area of opening A2 = xr12 = li x (0.025)2= 0.00196 m2 Area of cavity A, = Area of hemisphere Ash + Area of frustum of the cone Af

Then F, - = 1and

0.00196 = 0.9268 0.0268

Q = 0.0268 x 6.67 x

lo4

[

x 0.6 x (1173)4

I

1- 0.9268 = 200.61 W. Ans. 1- 0.9268 (1- 0.6)

Example 13.25. A furnace cavity, which i n form of a hollow cylinder of 75 cm i n diameter and 150 c m long, is open at one end to the surroundings at 27OC. The side and bottom of the furnace are approximated as blackbodies, and are heated electrically, well insulated and are maintained at temperatures of 1500°C and 1800°C respectively. How much power is required to maintain the furnace conditions ? Solution Given :A furnace cavity in the form of a hollow cylindrical cavity D = 75 cm = 0.75 m, L = 150 cm = 1.5 m T, = 1500°C = 1773 K T, = 1800°C = 2073 K T3 = 27°C = 300 K. TQfind :Power required to maintain the prescribed cbnditions.

7-

i

Side A, at T, = 1500°C

Bottom, A, T, = 1800°C

Fig. 13.38

-

= 0.06

A1 F 2 - l = x~~ 14 x 0.94 = (0.75)~14 x 0.94 = Fl-, = A9 nDL 0.75 x 1.5

By symmetry, F,-, = F l - 3 = 0.118 Hence, Q = (5.67 x 1O4W1m2 . K4) ((71 x 0.75 m x 1.5 m) x (0.118) [(1773)4- (300)4] + W4) x (0.75 m)2 x (0.06)[(2073)*- (300)4]} = 354612.4 W = 354.6 kW. Ans. Alternatively :It can also determined by using eqn. 13.14.

Here el = 1. Therefore, Q = A, o (TI4- T2'9 (1- F1-

, - T24).Ans.

= Al CY F1 - (TI4

13.4. RADIATION HEAT EXCHANGE BETWEEN DIFFUSE, GRAY SURFACES

In the preceding section, the heat exchange between blackbodies has been discussed. The analysis was simpler, because blackbodies do not reflect any amount of incident radiation. But in practical applications, most of enclosures have nonblack surfaces, which allow multiple surface reflections. Therefore, the analysis of radiation heat transfer for such enclosures becomes very complicated. The analysis of radiation exchange in such enclosures may be simplified by making certain assumptions. These assumptions are 1. Each surface in an enclosure is opaque, d i f u s e and gray. 2. Each surface of the enclosure is isothermal. 3. Each surface of the enclosure is characterised by uniform radiosity and irradiation.

13.4.1. The Net Radiation Exchange by a Surface Consider a surface with the following properties during radiation exchange : E = emissivity of the surface. Eb = emissive power of the black surface, W/m2. p = reflectivity of the surface.



848

By reciprocity rule ; A3F3-1 =A1F1-3 and A 3 F 3 - 2 = A 2 F 2 - 3 Therefore, Al Fl - (J1- J3)- A2F2- (J3- J2)= 0 Hence,

JI-J3

1

849

...(13.35) The total resistance of the network

- J3-J2 1

-

Reradiating wall, T,

The net radiation heat exchange ---

. . .

~ u r n i n bfuel bed, TI

(b) Thermal network

(a) A fuel bed, water tubes and refractory walls make an enclosure in a boiler

,

Since Q12 = - Q21 or for a reradiating surface, net heat transfer is zero, therefore, its temperature can be calculated as JR= EbR= 0 TR4 ...(13.41) Example 13.26. A spherical liquid oxygen tank 0.3 m i n diameter is enclosed concentrically i n a spherical container of 0.4 m diameter and the space i n between is evacuated. The tank surface is at - 183OC and has a n emissivity of 0.2. The container surface is at 15OC and has a n emissivity of O.25. Determine the net radiant heat transfer rate and rate o* f evaworation of liquid oxygen i f its latent heat is 220 k J l k g . L

Solution Given :A spherical oxygen tank with el = 0.2, E~ = 0.25, hfg= 220 kJkg. To find :Net radiation heat transfer and rate of evaporation of oxygen.

A typical three-body configuration

3

(c) Schematic Fig. 13.42. Three surface enclosure with one surface reradiating

The equivalent thermal network for three surface enclosure with a reradiating surface is shown in Fig. 13.42(b).It is simple series parallel arrangement and the total resistances can be expressed as : The resistances lIAIFl and 1/A2F2 for the reradiating surface are in series, therefore, its total resistance is

-,

Further, the resistance Rs1 is in parallel with resistance 11AlF,-2, therefore, its equivalent resistance

Evacuated space

-,

Fig. 13.43

Assumptions : 1. Surfaces are opaque, diffuse and gray. 2. Space between two concentric spheres is evacuated. 3. No conduction and convection heat transfer. Analysis :The net radiation heat exchange between two concentric ca

Q=

A, O(T; - T ~ ~ ) -+ El

where,

( ):; --I

Al = .n dl2= .n x (0.3)~= 0.2827 m2

-

expressed

850

and


A~ =

*2

(

=

)0.4

851 .


When the hot plate temperature is raised to 600 K, then

2

= 0.5625

Therefore, The rate of evaporation of oxygen,

Q

,j.$=-- -

hfg

16.33 = 7.423 x 220 x lo3

kg/s = 0.267 kglh. Ans.

Example 13.27. Two parallel, infinite gray surfaces are maintained at temperature of 127" C and 227" C respectively. If the temperature of the hot surface is increased to 327" C. By what factor is the net radiation exchange per unit area increased ? Assume the emissivities of (N.M.U., May 1997) colder and hotter surfaces to be 0.9 and 0.7, respectively. Solution Given :Two parallel infinite surfaces with T, = 327°C = 600 K, E, = 0.9,

E,

= 0.7.

!& = 3830 = 2.82. Q,

1359

Ans.

Example 13.28. A cubical room 4 m by 4 m by 4 m is heated through the ceiling by maintaining it at uniform temperature of 350 K, while walls and the floor are at 300 K. Assuming that the all surfaces have a n emissivity of 0.8, determine the rate of heat loss from ceiling by radiation. Solution Given : A cubical room with sides of 4 m each Floor L=H=w=4m walls T, = 350 K T2 = 300 K, E, = s2 = 0.8. To find :Heat loss from ceiling to room. Assumptions : 1. Surfaces are diffused and gray. Fig. 13.45 2. Heat is transferred by radiation only. Analysis :Considering ceiling as surface 1 and other surfaces of room as surface 2. The surface area, A, = (4 m x 4 m)= 16 m2 The surface area, A2 = 5 surface x (4 m x 4 m) = 80 m2 The net radiation heat exchange between two surfaces can be calculated by electrical analogy ; The calculation of resistances

Fig. 13.44. Two parallel infinite gray surfaces To find :Net radiation heat transfer. Assumptions : 1. Surfaces are diffused and gray. 2. Heat is transferred by radiation only. Analysis :The net radiation heat exchange between two parallel plates can be expressed

Fig. 13.45 (a) Radiation network

as ; Since all energy leaving the ceiling will be absorbed by room walls and floor, Hence, q - 2 =1 and

852


853


The total radiation resistance between Ebl and Eb2is

Example 13.29. A cubical room 4 m by 4 m by 4 m is heated through the floor by maintaining it at uniform temperature of 350 K, while side walls are well insulated. The heat loss takes place through the ceiling at 300 K. Assuming that the all surfaces have a n emissivity of 0.8, determine the rate of heat loss by radiation through the ceiling. Solution Given :A cubical room with sides of 4 m each L=H=w=4m T, = 350 K T2 = 300 K = c2 = 0.8. To find :Heat loss by radiation through ceiling. Assumptions : 1. Surfaces are diffused and gray. 2. Heat is transferred by radiation only. Analysis :Considering floor as surface 1,ceiling as surface 2 and walls of room as surface 3. The surface area, A, = (4 m x 4 m) = 16 m2 The surface area, A2=(4mx4m)=16m2 The surface area,

Example 13.30. Two parallel discs 50 m m i n diameter are spaced 40 cm apart with one disc located directly above the other disc. One disc is maintained at 500°C and other at 227°C. The emissivities of the discs are 0.2 and 0.4, respectively. The disc are located i n a very large room whose walls maintained at 67OC. Determine the rate of heat loss by radiation from the inside surfaces of each disc.

,Ceiling

0 @ Floor L I

4m I

(a) Schematic

The net radiation heat exchange beQ, = 0 tween two surfaces can be calculated by electrical analogy ; ;&,u The calculation of resistances Eb, l - - ~1-0.8 ~ QIR2 = Rl = '32 A, 16~0.8 R1 RI-2 = 0.0156 (b) Radiation network Since all energy leaving the floor Fig. 13.46 will not reach the ceiling and hence the view factor F, and F1 - are to be determined from Fig. 13.8 for shape factor, hence F, = 0.2

&

-,

By summation rule, But Hence and

-, F, ,+ F1 -

-

+ F1-

=1

F1-,=O F,-, = 1 - F l - , = 1 - 0 . 2 = 0 . 8

-

Q2

Solution Given :Two parallel discs spaced at 40 cm apart, located in a large room with Dl = 50 cm = 0.5 m D, = 0.5 m Tl = 500°C = 773 K E, = 0.2, T2 = 227°C = 500 K = 0.4, L = 40 cm = 0.4 m, T3 = 67°C = 340 K To find :The heat loss by radiation from each disc to room 7a11. Assumptions : 1. Steady state conditions. 2. Diffuse and gray surfaces. 3. All surfaces are opaque. 4. Room has reradiating surfaces. Analysis :The thermal network is shown in Fig. 13.47 ( b ) ; T J3

R1

Rl-2

Fig. 13.47 (a) Schematic

= Eb3

'32

Fig. 13.47 ( b ) Radiation network

Area of each disc, Al = A2 = (d4) Dl2 = ( d 4 ) x (0.5 m)2 = 0.1963 m2

854


approximates a blackbody and is maintained at a temperature of 400 K. Calculate the net rate of radiation heat transfer at each surface during steady state conditions.

The shape factor from Fig. 13.9 Llr, = 40125 = 1.6 and r,/L = 25/40 = 0.625 FlV2= F2-I = 0.24 Hence, By summation rule, F1- = 1- Fl - = 1- 0.24 = 0.76 By summation rule, F2 = 1- F2 = 1- 0.24 = 0.76 The various resistances ;

-,

1 Rl-3=

-

All?,-,

855


Solution Given :A furnace as three surface enclosure ro=H=lm = 0.4, Tl = 700 K T2 = 500 K E~ = 0.8, T3 = 400 K. E~ = 1, To find :Net rate of heat transfer at each surface i.e., Q1, Q2 and Q3. Schematic :

-,

1 = 6.703 m-2 0.1963~016

R2-3 =R1-3

Ebl = CT T14 = 5.67 x lo-" (773)4= 20244.22 W/m2 x (500)4= 3543.75 W/m2 Eb2= CT T24 = 5.67 x = B T = ~ 5.67 ~ x lo-" (340)4 = 757.7 W/m2 J3 = Eb3 Applying Kirchhoff's law of electrical current at each node i.e., summation of incoming currents to each node is equal to zero. The emissive powers,

I

Fig. 13.48

C

Node, 1, for J1 :

Node 2, for J2:

Ehl-J1 R1

Eb2-J2

+

J2-J1

+

%-2

Eh3-J1

=0

R1-3

iEb3-J2

J1-J2

=O

tbl

R1

R2 Rl- 2 R2 - 3 Eb, 3543.75 - J 2 J1- J2 757.7 - J2 + -------=0 7.64 21.22 6.703 Solution, leads to simultaneous equations as JI - 0.2454 J, + 0.0471 J2+ 1106.82 = 0 Rl-2 '32 0.0471 J, - 0.3272 J2+ 576.88 = 0 Fig. 13.47 ( c ) The solution to these equations are Jl = 4986.426 W/m2,J2= 2480.87 W/m2 The heat loss from hot disc at 500°C Ehl - J1 20244.22 - 4986.426 = 749 W. Ans. -

J2

Assumptions : ( i )The surface 1 and 2 are opaque, diffuse and gray. (ii) Steady state conditions without convection effects. Analysis :The radiosity for surface 1 and surface 2 can be determined by writing the energy equation on node 1and 2 and setting it to zero.

+

Q1 =

R1

Eb2

20.37

x 7004 = 13614 W/m2 Ebl = B T14 = 5.67 x x 5004 = 3544 W/m2 Eb2= B T24= 5.67 x x 4004 = 1452 W/m2 Eb3= J3= 0 T34 = 5.67 x Al=A2=n:ro2=n:~12=3.141m2 The view factor F1 - can be obtained from Fig. 13.9 as Fl = 0.38 The view factor F1 - can be obtained from summation rule Fl~l+Fl~2+Fl~3=1+Fl~3=l-O-0.38=0.62 Since the surface 1is flat and thus F1 - = 0.

rhere,

-,

Example 13.31. Consider a cylindrical furnace, whose radius is 1 m and equal to its height. The base and top surface of the furnace have emissivities 0.4, and 0.8, respectively, and are maintained at uniform temperatures of 700 K and 500 K. The curved cylindrical surface

856


857


The radiation resistances

To find : ( i )Qnet - if open (ii)Qnet 1- 2, if connected by an insulated dif-

Case (a) Both sides are open to black surroundings

fuse reflector Case (b) A reflecting shield is Yo (iii) Temperature of reflector. Yood placed on both side o m Il Assumptions : II I-" + N 1. Each surface has uniform radiosity and the enclosure can be treated as a three surface enclosure. 2. Diffuse, gray and opaque surfaces. Fig. 13.49 3. Medium separating the surface does not participate in radiation. 4. Negligible convection inside the enclosure. Analysis : ( i )When two infinite long parallel plates are spaced at a distance L and are open. The view factor from Fig. 13.8

I 1

a

Substituting the numerical values in eqns. (i)and ( i i )

Solving these two equations with two unknowns, we get The net rate of radiation heat transfer for top surface 1 and base surface 2 are In addition, surface 3 may represent reflector or surroundings, then for an enclosure F 1 4 +F1-,+ F1-3= 1 Fl-, = 1- Fl-,= 1- 0.2 = 0.8 F1-,= 01 By symmetry F2-, = Fl-, = 0.8 The heat transfer is towards bottom surface side. The net rate of heat transfer at cylindrical surface

Q3

J3- Jl +--J3- J 2 - 1452 - 11418 = R2 - 3 0.5137 = - 25455 W. Ans.

R1-a

+

1452 - 4562 0.5137

Example 13.32. Two very long strips 1 m wide and 2.40 m apart face each other, as shown i n Fig. 13.49 (a) Find Q,,, - 2 w l m ) if the surroundings are black and at 250 K. (b) Find &,, I - (W/ m), if they are connected by a n insulated diffuse reflector between the edges on both sides. Also evaluate the temperature of the reflector i n part (b). Solution Given :Two very long strips parallel to each other

Fig. 13.49 (a)Radiation n e t w o r k for case (i)

The nodal equations for node 1and 2 are : node 1 node 2 Where for 1m2 area

858


& ~ 3

Fig. 13.49 ( 6 ) Radiation network for case

(id

In this case, the surface 3 is adiabatic, thus all the heat leaving the surface 1 will be transferred to surface 2.

Qnet 1-2 -

Thus

Jl-J2

JI-J3

Rl-2

Rl-3

Since the surface 3 is reradiating, thus we can also obtain Qnet Simplification leads to Jl - 0.14 J2- 0.56 x 221.5 = 435.6 - J, + 10.0 J2- 4.0 x 221.5 = 2296.5 Jl - 0.14 J2= 559.6 - Jl + 10.0J2= 3182.5 Its solution leads to Jl = 612.1 W/m2, J2= 379.5 W/m2 Thus the net heat flow from plane 1to plane 2 is J1- J2 _ 612.1- 379.5 = 46.52 Wlrn. Ans. Qnet 1- 2 Rl-2 5 (ii) When two parallel strips are connected by an insulated diffuse reflector (a radiation shield) between the edges of both sides. Q3 = 0 Then and node1 equations node 1 node 2 node 3

-,

The resistances R,, R2, Rl -,, R, and R2- remain unchanged, because the surroundings is replaced by radiation shield, The solution of these simultaneous equation leads to Jl = 987.7 W/m2, J2= 657.4 W!m2, J3= 822.6 W/m2

by relation

A ~ ( T ; -'I'Z4) Qnet 1-2 = 1- E

1--

1

1-E +2

1x 5.67 x lo-' x [4004 - 30o4I - = 198.4 Wlm. Ans. 1- 0.5 1 -1-- 0.3 0.3

+

-I

+d.5

125 125 (iii)Temperature of reflector or reradiating surface Thus

Q3 = 0 J3= Eb3 5.67 x 104 Tz4= 822.6 T, = 347 K. Ans.

Example 13.33.A paint baking oven consists of a long triangular duct i n which a heated surface is maintained at 1200 K and another surface is insulated. Painted panels, which are maintained at 500 K, occupy the third surface. The triangle of width of 1 m on a side, and heated and insulated surfaces have a n emissivity of 0.8. The emissivity of the panels is 0.4. During steady state operation, at what rate must energy be supplied to the heated side per unit length of the duct to maintained its temperature at 1200 K ? What is the temperature of insu(N.M.U., Nov. 1994) lated surface ? Solution Given :A paint baking oven with T2 = 500 K, Tl = 1200 K, = 0.8, E~ = 0.4. To find : 1. The heat supplied rate to duct. 2. Temperature of the insulated surface.

w=lm,


860

Assumptions : 1. Steady state conditions. 2. All surface are opaque, diffuse and gray. 3. Convection effects are negligible. Analysis :The schematic and radiation network are shown below :

2. The temperature of insulated surface may be calculated by obtaining J3. Q1-2 -L

Insulated

-

Q2

Q1-

R2 = 0.4,

T, = 500 K (a) Schematic

861


(b) Radiation network

Fig. 13.50

1. The shape factor : From symmetry, F,, = F1-3 = F2-3 = 0.5 The area of wall : A, = 4 = w . L, where L is the duct length. The various resistances :

Ebl - J 1 R1

For reradiating surface J 1 - J3 -J3 - J 3 -

R2-3 108327.2 - J3 - J3- 59018 2 2 J3= 83683 W/m2 = 5.67 x T3 = 1102.2 K. Ans. 1

3

x T34

Example 13.34.A heater of 1 m diameter is covered by a hemisphere of 4 m diameter. The surface of hemisphere is maintained at 400 K. The ernissivity of the surface is 0.8. The heater surface is maintained at 1000 K. The remaining base area is open to surroundings at 300 K. The surroundings may be considered black. The emissivity of heater surface is also 0.8. Determine the heat exchange from heater to the hemisphere and to the surroundings. Solution Given :A heater (1)is covered by a hemisphere (2) Dl = 1 m T, = 1000 K, E, = 0.8, D2 = 4 m T2 = 400 K, c2 = 0.8, E~ = 1. T, = 300 K, To find : ( i )Heat exchange with hemisphere, and ( i i )Heat exchange with surroundings.

R2-3 = R1-3 = 5.67 x 104 x (1200)4= 117573 W/m2 The emissive powers, Ebl = 0T14 Eb2= 0T24= 5.67 x 104 x (500)4= 3543.75 W/m2 The resistance R2-3and RIS are in series, thus its total series resistance is Rsl = 2 + 2 = 4 m-l This total series resistance Rsl and R,-, are in parallel to each other, thus

(a)

Re, = 1.333 m-l The heat transfer rate from heated surface to painted panel

= 36982.87 W/m = 36.982 kW/m. Ans.

(b)

Fig. 13.51. Schematic and radiation network

Assumptions : (1) Heater and hemispherical surfaces are opaque, diffuse and gray. (2) Steady state conditions. (3) Assuming circular disc heater, facing towards hemisphere.


863


862

1451.51- J2 J1- J2 459.27 - J2 +- 1.273 =0 0.0099 0.085 Simplification leads to 226764.37 - 4.99 J, + J2= 0 163.45 - J, + 0.7884 = 0 Its solution leads to J, = 5.39 x lo4 W/m2,J2= 42556 W/m2 ( i )Now net rate of radiation from heater to hemisphere +

Analysis :The thermal network is shown in Fig. 13.51 ( b ) : The areas are : n n Area of circular heater, A, = = 0.7853 mZ 4 D12 = x n n A - - DZ2= - x (4)2= 25.132 m2 Area of hemisphere, 2- 2 2 7c n 4 (42- 12)= 11.78 m2 Area of opening of hemisphere, A3 = (D: - Dl2) = -

Ebl - J1 - 56700 - 5.39 x lo4 = 8796.7 W. A n s . R1 0.3183 (ii)Net rate of heat radiation from heater to surroundings

View factors. If circular heater faces hemisphere, then F1-,= 1, F1-l=O, F1...3=0 F,-1 + Fz-2 + F2-3 = 1 Y

There will be negligible radiation from surroundings to heater surface, thus

Comment. Heater does not radiate heat directly (F13 = 01, but the heat transfer to hemisphere is transferred to surroundings.

Applying Network theorem (Kirchhoff 's law) at each node (1)and (2)

Example 13.35. A short cylinder enclosure is formed with three surfaces, a circular plane surface 1 of radius 20 cm maintained at 2000 K and having emissivity of 0.8, another circular plane surface 2 of same size as surface 1 having emissivity of 0.5 and maintained at 500 K. The surface 1 and 2 are parallel to each other and the distance between them is 5 cm. The third surface is reradiating, which forms an enclosure. Draw an equivalent circuit and compute all resistances. Also find, (i) temperature attained by reradiating surface a&.(ii) net heat * transfer rate between surface 1 and 2 due to radiation. Use the following expression for finding the shape factor between two circular discs, coaxial and parallel areas :

where

Ebl = o T," = 5.67 x 10" x (1000)4 = 56700 W/m2 Eb2= o T; = 5.67 x 10' x (400)4 = 1451.51 W/m2 Eb3= o T34= 5.67 x 10" x (300)4 = 459.27 W/m2

Using value in eqn. ( i )and (ii)

B = -r1, C = - r2 a n d X = ( l + B 2 + C 2 ) H H where, r, and r2 are the radii of the circular planes and H is the distance between them. (P.U., Nov. 1992) Solution Given :A short circular cylinder consists of a two parallel plane and a reradiating lateral surface with rl = r2 = 20 cm = 0.2 m, H = 5 cm = 0.05 m TI = 2000 K, T2 = 500 K &I = 0.8, e2 = 0.5. To find : 1. To draw an equivalent electric circuit. 2. Net heat transfer rate between two parallel surfaces. 3. Temperature, T3 attained by reradiating surface. where,


864

Assumptions : 1.Steady state conditions. 2. All surfaces are opaque, diffuse and gray. 3. Convection effects are negligible. Analysis :1.The schematic and thermal network are shown below :


'

865

The resistance R14 and R2-3are in series, thus its total series, Rsl = 36.17+ 36.17= 72.34m-2 This total series resistance Rsl and Rld2are in parallel to each other, thus

1

Re, = 8.94 m-2 The heat transfer rate between two surfaces, = 907200 - 3543'75 = 47849 w = 47.849 kw. Ans. R1 + Re, + R2 1.989+ 8.94+ 7.957 3. The temperature of reradiating surface may be calculated by obtaining J,. Q1-2

=

Q1-2

=

E b l -Eb2

Ebl - J l R1

(b) Radiation network

(a) Schematic Fig. 13.52

2.The shape factor :

or

J2= (47849W) x (7.957m-2) + (3543.75W/m2)= 384278.24W/m2 For reradiating surface

Jl-J , J -J2L 3

Using given relation :

%-3

F1, = F2-, = 0.78 From symmetry, Fl-2 + F14 = 1 or F19 = 1 - 0.78 = 0.22 and by summation rule, F,, = FIS = 0.22 BY symmetry, = 0.12566 The area of surfaces : A, = A2 = n r12= (n)x (0.2)~ The varidus resistances :

R23

The emissive powers,

or or

R 2 -3

812028.34- J3 - J3- 384278.24 36.17 36.17 J3= 598153.3W/m2 = 5.67 x lo4 x TS4 T3 = 1802.2 K. Ans.

13.6. RADIATION HEAT TRANSFER I N THREE SURFACE ENCLOSURE

Now we consider an enclosure consisting of three opaque, diffuse and gray surfaces as shown ia Fig. 13.53

= R13

Ebl = o T," = 5.67 x lo4 x (2000)*= 907200 W/m2 *b2

= 0T,' = 5.67 x lo4 x (500)' = 3543.75w/m2 Fig. 13.53. Three surface enclosure and equivalent radiation network


866

Surface 1 , 2 and 3 have surface area A,, A2 and A3, emissivities E ~ , and e3, and uniform temperatures TI, T2 and T,, respectively. The triangular circuit for three bodies radiation network is not so easy to analysis as in line circuit for two surface radiation problem. The basic approach is to apply energy conservation on each radiosity node in the circuit. The three equations for determination of three unknowns J,, J, and J3are obtained and net heat transfer at each node is set to zero. That is -0

J2-J1 J3-J1 AtnodeJ,: Ebl-Jl 1 1 - - +

ElA

At node J, :

J1-J2

+

+

A1F1-3

J3-J2

+Ebz-J, 1- c2

1 AlFl- 2

~

-3

For surface 2

Eb2-J2

For surface 3

Eb3-J3

2

=O

q,

2

J,

Example 13.36. A long square duct has its three surfaces 1, 2, and 3 maintained a t uniform temperatures of 400 K, 500 K and 600 K, respectively, their respective emissivities are 0.9. 1.0 and 0.1. The surface 4 is subjected to a uniform heat flux of 5000 W l m2 and emissivity of 0.8. Determine the net radiative heat fluxes from surfaces 1, 2, and 3 and temperature of surface 4. Assume all surfaces are gray and diffuse.

-

P - B L . ' a " , w 4 z y l.

+

~ A3F3-1 3

J1-J4

+

J2-J4 1

+

A3F3-2 +

A3F3-4

J3-J4

1

=o

A5 F5-I = & x 0.5 = F1-5 = f1i = 0.7071 A1 F,-, + F,-, + F,, = 1 +F,, = 1- F,, = 0.2928 F,, = F,-, = 0.2928 A1 F,-, = 0.2928, F2-, = 0.2928 (BYsymmetry) A2 FZ4 = F13 = 1- F1-2 - F1, = 1- 0.2928 - 0.2928 = 0.4142 A, = A , = A -, = A , = A Areas : Solution of above simultaneous equations gives J, = 1843.1 W/m2, J3= 4555.2 W/m2, J4= 8342.1 W/m2 J, = Eb2= 3543.75 W/m2 (since black = 1) The heat fluxes : =

(1451.2 - 1843.1) = - 3527.1 W/m2. Ans. 1- 0.9 Since surface 2 is black, its surface resistance will be zero, the emissive power equals the heat flux. q2 = F2-, (Eb2- J1) + FZ3 (Eb2- J3)+ F2, (Eb2- 5 4 ) = 0.2928 (3543.75 - 1843.1) + 0.2928 (3543.75 - 4555.2) + 0.4142 [13543.75 - 8342.11 = 1785.6 W/m2.Ans.

+' y3

-

n2-

For surface 1 :

1

--

=OF, -

fin&? .4 (i) Radiation heat f+ux$:, ?!d I (ii) Radiation heam@ q2,1fS;P" A= (iii) Radiation heat flux q$' (iv) Temperature of surface$.= Analysis : The energy bala&e TO

J ~ - J = O~

= 0.9

6%

+-q;% ii L F P o 0

+

+

1

Ebl = a TI4 = 5.67 x lo4 x (400)4= 1451.52 W/m2 Eb2= o TZ4= 5.67 x lo4 x (500)4= 3543.75 W/m2 x (600)4= 7348.32 W/m2 Eb3= o T34= 5.67 x The view factors We assume diagonally imaginary surface 5 as shown in Fig. 13.54 = 0.5 F5-, =F5,=F5,=F5-,

F,

? ;

3

J ~ - J 2

J2-J3 J4-J3 =0 1 1 1 - - -

+Jl-J3

1~

+

1

where, ~

instead of the temperature, then the term (Eb l --E should be replaced by Q.

E,

J1-J2

...[l3.42 (b)]

These three equations are solved simultaneously for determination of J1, J2 and J3. . Then net rate of heat transfer at each surfacb can be determined from eqn. (13.25). The solution of set of equations can be simplified, if one or more surfaces are special in some way. For an example, for a black or reradiating surface J = Eb = a Pand Q = 0 for such a surface at thermal equilibrium. If net rate of radiation heat transfer is specified at a surface

Solution Given :A long square duct T, = 400 K T2 = 500 K T e U 6 0 0K

+

1-E2

...L13.42 (a)] For surface 4

A1F1-2

867


E ~ , - ~ ; JA 2-Jl + J3-Jl I - F , A ~+-r- 1

+ J ~ - J I

1

=0

868


869


The radiation network is shown below :

E,, = o T,4 9592.1 T44= + T, = 641.3 K. Ans. 5.67x lo-' Example 13.37. Two sides of a long triangular duct, as shown i n Fig. 13.55 (a), are made of stainless steel (E = Q.5) and are maintained at 500°C. The third side is of copper (E = 0.15) and has a uniform temperature of 100°C. Calculate the rate of heat transfer to the copper base per metre of length of the duct. Further

Solution Given :A triangular duct as shown in Fig. 13.55 ( a )

4>7'+, s.Lq

' v

$0

//

kv

, To find :Net rate of heat transfer to copper surface 1 per metre length of duct Assumptions : ( i )The duct surfaces are opaque, diffuse and gray. (ii)Negligible convection from surfaces. (iii)Steady state conditions. Analysis :The view factors for the enclosure surfaces. From Table 13.1 for kiangular duct

P

9s

Copper, el = 0.15, T, = ioo0c

I

Fig. 13.55. ( a )Illustration for example 13.37

Fig. 13.55 ( b ) Radiation network

The node1 equations at three nodes are Node 1 : Node 2 : Node 3 : where J,, J2and J3are unknowns and for per metre depth of the duct

By reciprocity

By summation Fl-I + F,,

+ F1, = 1 jF1, = 1 - Fl-, = 0.6

Thus,

1097.5- J1 J2 - J1 + +- 5.0 11.33

J3

- J1 -

3.33

870


-

Simplification leads to three simultaneous equations 6.67 J, - 2.268 J2- 3.395 J, = 1097.5 - 2.0 J, + 56 J2- J, = 60732.65 - 3.0 J, - J2+ 8.0 J, = 80976.8 Solving these equations J, = 15767.3 W/m2 J2= 18434.4 W/m2 J, = 18340 W/m2 - Ebl - J1 -- 1097.5 - 15767.3 qnet - 1

-

R1

11.33

871


&1

&3,2

'3,l

'2

The temperature of radiation shield T,- is unknown. For instance, if all emissivities are equal : or = e2 = c3, = e3, ( ~ , -4 T ~ ~ ) Then T34= 2 The heat transfer rate between the plates with a shield becomes Shield

= - 1295.1 Wlm. Ans.

13.7. RADIATION SHIELDS

The radiation heat transfer rate between the two surfaces can be reduced significantly by placing a thin sheet of high reflectivity (very low emissivity) material, between them Fig. 13.56. Such highly reflective thin sheets are called the radiation shields. The radiation shields increase the thermal resistance in path of radiation heat transfer and hence reduce the heat flow rate. Multilayer radiation shields constructed of about 20 sheets per cm thickness separated by evacuated space are commonly used in cryogenic and space applications. Consider a radiation shield is placed between two large plates as shown in Fig. 13.57 (a). Let the emissivities of the radiation shield facing plate 1and 2 be E,, and E ~2,, respectively. Since, the radiation shield does not deliver or remove any amount of heat from the system. Thus the heat transfer rate between d a t e 1 and radiation shield must be equal to the heat transfer rate between the shield and plate 2.

A,, TI,

,

I

--

J1

J3,1

I

-

I

a

I

a

I

I

I

Ebl

Eb3

J3,2

J2

Eb2

(b)

Fig. 13.57. Radiation shield placed between two large parallel plates and radiation network

Q1-2, one shield

- (11 2) CY A(T,~1 1 -

)

-+--I El

E2

Thus by inserting a shield, the radiation heat flow rate becomes just half of that would be experienced without radiation shield. The radiation network with one radiation shield is shown in Fig. 13.57 ( b ) .All the resistances are in series, and thus the radiation heat transfer rate is E b l - Eb2 ...(13.46) 1 1 l & 31 - 1-&32 1-&2 I ' A ' A 1 &lAl ' A1F1-3 ' &3,1A3 ' ~ 3 , 2 ~' A3F3-2 3 ' ~ 2 ~ 2 For large parallel plates, A, = A2 = A, = A and F,, = F3-2 = 1,then eqn. (13.46) simplifies

Q1-2, one shield - 1- &

(a) A plane surface 1, close to a parallel surface 2

(b) Radiation shield between

the two surfaces

Fig. 13.56. Radiation shield between two parallel surfaces to reduce radiative heat transfer



872

Solution Given :Two parallel infinite surfaces with and without radiation shield T, = 1000 K, T2 = 600 K, el = 0.5, E, = 0.8, &a, I = 0.1, c3, = 0.05. To find : 1. The radiation heat transfer rate without radiation shield. 2. The radiation heat transfer rate with radiation shield. Assumptions : 1. Surfaces are diffused and gray. 2. Heat is transferred by radiation only. 3. The conduction resistance of radiation shield is negligible. - Analysis :The net radiation heat exchange between two parallel plates without radiation shield can be expressed as ;

where the terms in the second set of parentheses in the denominator represent the additional resistance to radiation heat transfer introduced by shield. If N radiation shields are inserted between two large parallel plates, then eqn. (13.47) becomes G A ( T , ~- T , ~ )

QI-2, N, shield =

1

{+;1,;}+t[;

1 +G-l)+

1 ......+{-+--I] EN,^

1

EN,^

If the emissivities of all the surfaces are equal then above expression reduces to o A ( T , ~- T ~ ~ ) QI-2. N. shield = (9 1 (Nil)\:-1)

873

4

A radiation shield may have same emissivity on its two faces or the emissivity of one face of shield may differ from that associated with opposite side. If the emissivities associated t with two sides of shield are very low, the heat flow can be reduced drastically.

Example 13.38. Two parallel plates have emissivity of 0.8and 0.5.A radiation shield having same emissivity on both sides is placed between them. Calculate the emissivity of the shield i n order to reduce the radiation losses from the system to one-tenth of that of without shield. Solution Given :Two parallel infinite surfaces with and without radiation shield

-QI- A

5.67 x lo-' (10004 - 6 0 0 ~ ) = 21934 W/m2. Ans. -+ - 1

)

;5 8; (

When a radiation shield is placed between the parallel plates, then the radiation heat transfer can be calculated as Q2 =

Ebl

- Ebz

ZRth

To find :The emissivity of radiation shield. Assumptions : 1.Surfaces are opaque, diffused and gray. 2. Heat is transferred by radiation only. 3. The conduction resistance of radiation shield is negligible. Analysis :The radiation heat exchange ratio between two parallel plates with and without radiation shield can be expressed as ;

For 1 m2 area of the plates where

The heat transfer gain with radiation shield

-L+fl-li -

El

E3

E2

E3

Substituting the numerical values ;

The solution gives, E, = 0.094.' Ans.

Example 13.39.Two large parallel plates at temperature I000K and 600K have emissivity of 0.5and 0.8respectively. A radiation shield having emissivity 0.1 on one side and 0.05 on the other side is placed between the plates. Calculate the heat transfer rate by radiation per square metre with and without radiation shield.

J I

-

The presence of radiation shield reduces the heat transfer rate 21934 - 1579.25 = 92.8% (reduction). 21934 Example 13.40. Two large parallel planes with emissivity 0.6are at 900K a n d 300 K. A radiation shield with one side polished and having emissivity of 0.05,while the emissivity of other side is 0.4 is proposed to be used. Which side of the shield to face the hotter plane, if the temperature of shield is to be kept minimum ? Justify your answer. (P.U., May 2001) Solution Given :A radiation shield between two large parallel planes. = E~ = 0.6 T, = 900 K, T2 = 300 K



874

875

300 K. The space between the surfaces is evacuated. CalcuLate the heat gain by cryogenic fluid per unit length of tubes. If a thin radiation shield of 35 m m diameter kg= 0.02) both sides is inserted midway between the inner and outer surfaces, calculate the percentage change in heat (P.U., Dec. 2001 ; N.M.U., May 1998) gain per unit length of the tube.

- 0.05 = polished &4 = &unpolished= O" T3 = Temperature of shield.

3'

To find :Which side of the shield should face the hotter plane. Assumptions : 1. Each surface has a uniform radiosity ; the configuration can be considered as a n enclosure with two surfaces. 2. The surfaces are gray, diffuse and opaque. 3. The medium between the surfaces does not participate in radiation. Anal.ysis :Arrangement 1 : Let polished surface of the shield face hotter plane, the ensrgy balance on the radiation shield :

Solution Given : Concentric tube arrangement with diffuse and gnay surfaces.

Hot -

Fig. 13.59. Schematic

To find : 1. Heat gain by cryogenic fluid passing through the inner tube. 2. Percentage change in heat gain with radiation shield inserted midway between inner and outer pipe. Assumptions : 1. Space between tubes is evacuated. 2. No conduction and convection involve. 3. Infinite long concentric tubes. Analysis :1. The radiation exchange without radiation shield can be calculated as

6.561 x 1011- T: = 6.526 T34- 5.286 x 10" Fig. 13.58.Schematic 7.526 T34= 7.0896 x 10" 3 3 T3 = 554 K Arrangement 2 : Let the polished side of shield face colder side : Energy balance on shield. 3

where 6.561 x 10" - TS4 - T~~- 8.1 x lo9 3.167 20.67 4.2486 x 1012- 6.526 TS4= T34- 8.1 x lo9

or or 7.526 TS4= 4.2567 x 1012 or T3 = 867.2 K which is greater than the temperature of shield, when it faces the hotter plane. Therefore, the radiation shield will be effective, when its polished side will face the hotter plane.

Example 13.41. A cryogenic fluid flows through a long tube of 20 m m diameter, the = 0.02) at 77 K. This tube is concentric with a outer surface of which is diffuse and gray larger tube of 50 m m diameter, the inner surface of which is diffuse and gray ( E ~= 0.05) and at

Q1 =

i

n x 0.02 x 1 x 5.67 x lo-' (774 - 3004) = - 50 Wlm. Ans. 1 x 0.4

2. When radiation shield is placed in midway between two tubes,

876



Analysis :( i ) Heat-gain rate b y liquid nitrogen line per metre length ( T t-TI4) Qgain= h D 1 ) (2304- 804)= 0.624 Wlm. Ans. = ~ ( 0 . 0 0 6 3 5(0.2) ) (5.67 x ( i i )Heat transfer rate with shield

For 1 m length of tube

+

ER,, = 779.9 + 15.9 + 891.3 121.0 = 1817 m-2 T h e heat gain with radiation shield 101 1

T h e percentage reduction i n heat gain (Q1- Q 2 ) -- (- 0.50 + 0.25) Q1 - 50

= 50%.

Ans.

Since A ,

>7

Al, and

E, =

1, t h e n

Example 13.42. Aphysics experiment uses liquid nitrogen as a coolant. Saturated liquid nitrogen at 80 K flows through 6.35m m O.D. stainless steel line (E, = 0.2)inside a vacuum chamber. The chamber walls are at T,=230 K and are at some distance from the line. Determine the heat gain of the line per unit length. I f a second stainless steel tube, 12.7mm in diameter, is placed around the line to act as radiation shield, to what rate is the heat gain reduced ? Find the temperature of the shield. Solution T h e radiation shield would cut t h e heat gain by 47%. Ans. (iii) The temperature of t h e shield, T , : Qgain,shield = Qnhie~d-charnber= ( x ~ s' s)o 0.328 W / m = ~ ( 0 . 0 1 2 7(0.2) ) (5.67 x Solving, we find T , = 213 K. Ans.

Arrangement-1

Arrangement-2 Fig. 13.60

Given : Flow of liquid nitrogen through a pipe Dl = 6.35 m m Nitrogen ( 1 ) &I = 0.2 Tl =80 K Chamber (c) : T , = 230 K D2 = 12.7 mm Shield ( s ) : T,, E, = 0.2 To find : ( i )Heat gain by liquid nitrogen line per metre length. (ii)Reduction i n heat gain with shield. (iii)Temperature of shield. Assumptions : ( i )The tube surface of nitrogen line at 80 K. (ii)Without shield nitrogen line as small object i n a large enclosure. F1-, = 1, F,, = 1 and >>'

&be'

(iii)Opaque, diffuse and gray surfaces.

(TP - Tt)

(2304- T:)

Example 13.43. A pipe carrying steam having an outside diameter of 20 cm runs in a large room and is exposed to air at a temperature of 30°C.The pipe surface temperature is 400°C.Calculate the heat loss to the surroundings per metre length of pipe due to thermal radiation. The emissivity of the pipe surface is 0.8. 1. What would be the loss of heat due to radiation, i f the pipe is enclosed in a 50 cm diameter brick conduit of emissivity of 0.9 ? 2. What would be the radiation heat transfer from the pipe, if it is enclosed within a (Anna Univ., May 2001) square conduit of 0.5 m side of emissivity of 0.9 ? Solution Given :A pipe carrying steam with

T, = 30°C = 303 K

Fig. 13.61


878

879


D2 = 50 cm = 0.5 m, E, = 0.9 1. Brick conduit, E~ = 0.9. 2. Square conduit of side w = 0.5 m, To find : 1.Net radiation heat transfer from pipe surface. 2. The radiation heat exchange when pipe is enclosed within a 50 cm diameter. brick conduit. 3. The radiation heat exchange, when pipe is enclosed within a square conduit. Assumptions : 1. Surfaces are opaque, diffuse and gray. 2. Space between two concentric pipes is evacuated. 3. No conduction and convection heat transfer. Analysis :1.The net radiation heat exchange from pipe surface to room can be expressed as : QiL = E (n Dl) 0 (Ts4- TW4) = 0.8 x x x 0.2 x 5.67 x lo4 (6734- 3034) = 5606.5 Wlm. Ans. 2. The radiation heat exchange between pipe and a conduit can be calculated as ;

Q=

A, o (TS4- T _ ~ )

Dl = 50 mm Tl = 1000°C = 1273 K c2 = 0.6 T, = 15°C = 288 K If reflector is placed D, = 450 mm. To find :Energy supplied to rod per metre, if 1. Rod is exposed to room, with negligible convection effects, and 2. Rod is covered by half circular reflector. Assumptions . 1. Rod and room surfaces are opaque, diffuse and gray. Q 2. Steady state conditions. 3. Room air does not participate in R1 radiation. Analysis : 1. When cylindrical rod is exposed to room

Fig. 13.62

k

v

Rl-2

R2

Fig. 13.62(a)

When pipe is enclosed within brick conduit : where

The reduction in heat radiation = 5606.5 - 5414 = 192.5 Wlrn. Ans. When pipe is enclosed within a square conduit :

R, = surface resistance of rod

I R1-2 = A (F,-, = 1.0. All heat transfer to room) = 6.366 x 0.05 x 1 n A1F1-2 R2 = 0, negligible transmission to rod from room.

Then

5.67 x Q1 =

(1273~- 288') = 16,330 W. Ans. 2.728 + 6.366 + 0

2. When half circular reflector is placed around the rod.

The reduction in heat radiation = 5606.5 - 5454.2 = 152.3 Wlrn. Ans.

Example 13.44. A cylindrical rod (E = 0.7) of 50 m m diameter is maintained at 1000°C by an electric resistance heating and is kept in a room, the walls (E = 0.6) of which are at 15OC. Determine the energy which must be supplied per metre length of the rod. If an insulated half circular reflector of 0.45 m diameter. is placed around the rod, determine the energy supplied to the rod per metre length. Solution Given :A cylindrical rod (1)exposed in a room (2).

QL0

Fig. 13.62(b)

Rl = 2.728 R2 = 0.0 Since reflector is half circular so half of the energy radiated by rod falls on it

880


Resistance to radiation from half reflector to room 1 1 1 = R2-3 =-= 1.591 m-, 71 5.r A2F2-3 - (D2 - Dl)L F2-3 - (0.45 - 0.05) x 1x 1.0 2 2 Half of the radiation from the rod reaches the room directly 1 1 / R1-2 =--A,F1-2 - x x 0.05 x 0.5 =&.732 m-2 Total of series resistances R1-3 and R2-3 RS1=12.732 + 1.591 = 14.323 m-, The resistance RSlacts parallel to resistance Rl-2, thus

881


Assumptions : 1. One-dimensional temperature distribution in the cylindrical part of the insulation ; i.e., the two-dimensional effects at the corners are negligible 2. Negligible thermal resistance between the water and the inner surface of the insulation, and in the metial cladding. 3. Steady state conditions. Analysis : In steady state, an energy balance yields to Heat transfer rate across the insulation = Heat transfer rate from the cladding to the surroundings The heat transfer from the cladding is by convection to the surrounding air and by radiation to the surrounding surfaces. 2n k~ 1n(rz T1 -IT2 rl) = (h, + hR)A,(T, - T,)

or

Re, = 6.74 m-2 Total thermal resistances CRth = R, + Re, + R, = 2.73 + 6.74 0 = 9.47 m-, The radiation heat transfer rate

In eqn. (I), the left side represents the conduction heat transfer rate across the insulation, and the right side represents the sum of the heat transfer rates by convection and radiation from the metal cladding to the surroundings. With A, = 27cr2L,solving for T2, we obtain

+

Q=

o ( T , ~- T , ~ )- 5.67 x

(1273~- 2 ~ 8 ~ )

= 15,681 W. Ans. CRth 9.47 Example 13.45. Fig. 13.63 shows a hot water heater, 60 cm outside diameter and 1.8 m high, which is insulated with 25 mm thick fiberglass insulation. The insulation is covered with a thin sheet of metal cladding. The water in the boiler is maintained at 80°C. The temperature of the surrounding air and surfaces is 10°C. The painted metal cladding has an emissivity of 0.95. To reduce the heat transfer to the surroundings, it is proposed to replace the cladding with apolished one having an emissivity of 0.1. If the convective heat transfer coefficientis 8 W/m2. "C, determine the reduction in the heat transfer rate from the cylindrical surface by replacing the cladding. Solution

where the temperature T, is calculated iteratively. As the cladding is completely enclosed by second, much larger surface, q, = radiation heat transfer rate from the cladding = A, E~ o (TZ4- T34) and To estimate the average value of hR, treated as constant in eqn. (2) an upper limit of T, (= T, ,,) is found by setting hR = 0, i.e., neglecting radiative heat transfer. The lower limit of T, is T3.

h~ max = o (T2, max + T32)(T2 max + T3) E2

Hot water heater Fig. 13.63. A hot water boiler with fiberglass insulation and a metal cladding

Given : r, = 30 cm, r, = 32. , cm, H = 1.8 m, h,,,, = 8 W/m2. "C T, = 10°C, E, = 0.95. T, = 80°C, To find :The reduction in heat transfer rate, if a radiation shield, E = 0.1 replaces the cladding

$ :

h~ min

= o (T2,

+ T3?(T2

+ TJ

€2

= 5.67 x

(283.15,

+ 283.152)(283.15+ 283.15) = 5.15 W/m2. K

883


882


As the maximum variation of hR is within 3% of the mean value, the effect of the variation of h, with T2 can be neglected. With E, = 0.95, hR = hR,av = 5.295 x 0.95 = 5.03 W/m2. K From eqn. (2)

With

q2 = (hc + hR)A2(T2 - T,) = (8 + 5.03) x 2 x n x 0.325 x 1.825 x (289.9 - 283.15) = 327.8 W. Ans. E, = 0.1, h, = 5.295 x 0.1 = 0.5295 W/m2. K

The last term in eqn. 13.49 is due to the radiation effect and represents the radiation correction. When the convection coefficient is small, then radiation correction term becomes most important, if Tm>> T,. The large error in temperature measurement can be reduced significantly by using 1. The low emissivity thermocouple junction. The special coating of low emissivity metal like aluminium, zinc, chromium etc. can be used. 2. Placing the sensor in a radiation shield without interfering the fluid flow.

-------------- ---Tube wall

q2 = (8 + 0.5295) x 2 x n x 0.325 x 1.825 x (292.9 - 283.15) = 309.9 W. Ans. The heat transfer rate is reduced from 327.8 W to 309.9 W, a reduction of 5.5%.

COMBINED CONVECTIVE AND RADIATION HEAT TRANSFER 13.8. TEMPERATURE MEASUREMENT OF A GAS BY THERMOCOUPLE

The temperature of the flowing fluid through a duct or a pipe is measured by thermocouple as shown in Fig. 13.64. The thermocouple bead is placed in direct contact of gas, the heat is convected from gas to thermocouple sensor tries to gain steady state. Like use of thermometer, the temperature measured by thermocouple is less than the true gas temperature, because a part of heat gain by thermocouple sensor is emitted to wall at low temperature. In absence of conduction, the energy balance on the thermocouple bead yields to h, (Tm- T,) = o E~ (Tc4- TU4)

where,

hc = convective heat transfer coefficient, W/m2.K T, = temperature recorded by thermocouple, K T, = wall temperature, K T_ = gas temperature, K a = 5.67 x W/m2 .K4, the Stefan Boltzmann constant E~ = emissivity of thermocouple sensor.

Tube wall

'

Fig. 13.64. Thermocouple in a gas stream

/

\

Radiation shield

Fig. 13.65. Thermocouple w i t h r a d i a t i o n shield

If thermocouple sensor is surrounded by a radiation shield as shown in Fig. 13.65. The radiation shield receives heat by convection on its two sides and it reradiates. Thus the energy balance for the radiation shield (a small body in compare to enclosure) is ...(13.50) 2hc (T- - Ts) = a E, (T,4 - Ts4) E, = emissivity of the radiation shield. where, Ts = temperature of the radiation shield. The energy balance on the thermocouple bead now yields ...(13.51) ): h, (Tm- T,) = a E, (Ts4- T where T, = temperature recorded by thermocouple. Example 13.46. A thermocouple is used to measure the temperature of a hot gas flowing in a tube maintained a t 100°C. The thermocouple indicates a temperature of 500°C. If the emissivity of thermocouple junction is 0.5 and the convective heat transfer coefficient is 1250 WIm2.K, determine the actual temperature of the gas. Solution T, = 500°C = 773 K Tw= 100°C = 373 K, Given : h, = 250 W/m2.K. E, = 0.5, To find :The true gas temperature. Assumptions : 1. Steady state conditions. 2. Junction surface is gray and diffuse. 3. Constant properties. Analysis : The energy balance on thermocouple bead is hc(Tm- T,) = a E, (T: - Tw4) Using the numerical values,

" P

884

or


(250 W/m2.K)(Tm - 500)("C) = (5.67 x lo4 W/m2.K4)x 0.5 x (7734- 3734) Tm= 500 + 38 = 538°C. Ans.

Example 13.47.A thermocouple (E = 0.6) is used to measure the temperature of exhaust gas in a large duct. The temperature of the duct wall is at 20°C and temperature measured by thermocouple is 500°C. Calculate the true temperature of the gas, if the convection coefficient between gas and thermocouple bead is 200 WIm2.K. To measure the temperature of the gas more correctly, it is enveloped by a thin radiation shield (E = 0.3). Estimate the error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement. Solution Given :Thermocouple without radiation shield : Tw= 20°C = 293 K, T, = 500°C = 773 K E, = 0.6, h, = 200 W/m2.K Thermocouple with radiation shield : Tw= 20°C = 293 K, T, = 500°C = 773 K E, = 0.3, h, = 200 W/m2.K. To find :The true gas temperature. Assumptions : 1. Steady state conditions. 2. Junction surface is gray and diffuse. 3. Constant properties. Analysis :The energy balance on thermocouple sensor without radiation shield is h,(T_ - T,) = o ec (Tc4- Tw4) Using the numerical values, (200 W/m2.K)(Tm- 773)(K) = (5.67 x lo4 W/m2.K4)x 0.6 x (7734 - 2934)(K4) T_ = 773 + 56.67 = 829.67 K. Ans. The energy balance on thermocouple sensor with radiation shield is - Ts4) 2hc (T_ - Ts) = CT E, (TW4 2 x (200 W/m2.K)(829.67- Ts)(K)= (5.67 x lo4 W/m2.K4)x 0.3 x (2934- T,4) 331856 - 400 Ts = 125.36 - 1.701 x lo4 Ts4 Ts4- 23515579070 Ts + 1.95 x 1013= 0 T, = 829.3 K Using the shield temperature to calculate the temperature measured by thermocouple. h, (T_ - T,) = CT E, (Ts4- Tc4) (200 W/m2.K)(829.67- Tc)(K)= (5.67 x lo4 W/m2.K4)x 0.6 x (829.34- )T: It yields Tc = 827 K The error between thermocouple temperature and gas temperature is of only 2°C. Ans.


Example 13.48.A thermocouple is used to measure the temperature ofgas flowing through a duct, records 280°C. I f the emissivity of the junction is 0.4 and convection coeficient is 150 WIm2.K. Find the true gas temperature. The duct wall temperature is 140°C. What should be the emissivity of the junction i n order to reduce the error by 30%. ? Solution Given :Measurement of temperature by a thermocouple (i) T, = 280°C = 553 K, E, = 0.4 h, = 150 W/m2.K, Tw= 140°C = 413 K ( i i )Error to be reduced by 30%. To find : ( i )True temperature of the gas, and (ii)Emissivity of the junction in order to reduce the error by 30%. Analysis :(i) The true gas temperature is determined by eqn. (13.49)

The true temperature of gas is 289.74"C (562.74 K). Ans. (ii)The error in temperature measurement = 289.74 - 280 = 9.74"C Error to be reduced by 30%, thus the remaining error is 70% of 9.74"C = 0.7 x 9.74 = 6.81B°C Thus T, = 286.818"C = 559.818 K Tw= 413 K T_ = 562.74 K, Then, or

562.74 = 559.818 + E,

5.67 x

lo-'

E,

(559.818~- 4 1 3 ~ ) 150

= 0.11. Ans.

13.9. GAS RADIATION

Many gases such as 02,N2, H2 and dry air have a symmetrical molecular structure and they neither emit nor absorb any appreciable amount of thermal radiation, unless they are heated to very high temperature for their ionization. These gases are non polar and may be considered as transparent to thermal radiation. On the other hand, some polar gases and vapours such as CO,, CO, H20, SO,, NH,, and hydrocarbons etc. can emit electromagnetic waves and can also absorb appreciable thermal radiation. The H20 and CO, are most common gases present in the atmosphere as well as in furnaces, the gas radiation can become an important part of heat exchange - .Drocess.

886


Unlike solids, the gases can emit radiation within a short wavelength band only. Hence the intensity of gas radiation at any temperature is less than the blackbody radiation. The gases absorb the incident radiation slowly, because emission and absorption depend on gas layer thickness, pressure, shape and surface area. Moreover, gaseous radiation is not a surface phenomenon but it is a volumetric phenomenon.

,

13.9.1. Absorptivity of the Gases Consider the absorption of thermal radiation by a gas layer as shown in Fig. 13.66, IF is radiation intensity a t the left face and I,.-. propagates in a gas layer is proportional to its thickness dx, thus dI,(x) = - m, Ih(x)dx ...(13.52) where the proportionality constant mh is spectral absorption coefflcient of gas. Integrating both sides, we get ;

elemental surfacedA, located at the centre of the hemisphere's base as shown in Fig. 13.67 where L, is mean beam length, which can be obtained from Table 13.4. For an appropriate calculation, the mean beam length for other shapes can be determined as Volume of gas ...(13.57) L, = 3.6 x Surface area of gas -

TABLE 13.4. Mean beam length L, for various gas geometries Sr. No

(IlLla = IhO- IhL= ILO[(I - exp(- mhL)l Hence, the spectral absorptivity of gases aG, can be expressed as,

,

a,,

,= - 1- exp Lo

(- m,L)

The quantity [I- exp (- m,L)1 represents spectral absorptivity of the gas aG, or accord= EG,,. It depends ing to Kirchhoff 's law, the emissivity at the same wavelength is eG,, or aGL on characteristic coeff~cientm, as well as on thickness dx of the gas layer. For non reflecting gases (p = 0) or

m,x) , ...(13.55) Average or effective absorptivity aGover entire wavelength spectrum can be expressed

as

aG = 1- exp (- mL) where, m =

,..(13.56)

[mh d i .

13.9.2. Gaseous Emission and Absorption Sometimes, it is required to determine the radiant heat flux from a gas to an adjoining surface. Despite the complications of spectral and directional effects,Hottel made the procedure quite simplified. He evaluated the emissivities of various gases a t different pressures and temperatures and plotted his results graphically in Figs. 13.68 to 13.71. The Hottel method involves the determination of radiation emission from a hemispherical gas volume at T, to an

Geometry

Characteristic length

Beam length L,

Sphere (radiation to surface) Infinite circular cylinder (radiation to curved surface) Semi infinite circular cylinder (radiation to base) Circular cylinder of equal height and diameter (radiation to entire surface) Infinite parallel planes (radiation to planes) Cube (radiation to any surface) Arbitrary shape of volume V (radiation to surface area A).

Diameter (Dl Diameter (D)

0.667 D 0.95 D

Diameter (D)

0.65 D

Diameter (D)

0.60 D

Spacing between the planes (L) side (L)

1.80 L 0.66 L 3.6 VIA

The emission from a gas per unit surface area is determined as gas emissivity, can be obtained from Hottel's graphs. The results for emissivity of water vapour and carbon dioxide are plotted in Figs. 13.68 and 13.69 as a function of gas temperature at a total pressure of 1atm and for different values of product of vapour partial pressurep and mean beam length L,. If the total pressure of water vapour or carbon dioxide is other than 1atm, then the emissivities from Figs. 13.68 and 13.69 must be multiplied by correction factor CHzoor CCOzfrom Figs. 13.70 and 13.71.

where

TG,, + ~ G , L = O , T = 1- aG, = exp (-

dA

Fig. 13.67. Gas volume of Hottel's graph.

The radiation intensity decreases exponentially with thickI . C - - L ~ ness of gas layer L. This equat~onis called the Beer's law. Fig. 13.66. Absorption in a gas layer The intensity of radiation absorbed from x = 0 to x = L ;

(ILL)a

887


E,

When both carbon dioxide and water vapour are present in a gas mixture, the effective emissivity of the mixture can be approximated by adding the emissivities of gas constituents as However, the resulting gas mixture emissivity E, is on higher side, because some of emission bands of water vapour and carbon dioxide overlap. In particular, the effective gas emissivity cg can be obtained by subtracting a factor A&.

890


891


respectively. When both water vapour and carbon dioxide are present in tLe mixture, then effective absorptivity of gas mixture can be obtained as T = 400 K

pcL + Lp,

= 1.52atrn rn

Fig. 13.72. Factor A& to correct the emissivity of a mixture of water vapour and C02

I

0

I

I

0.2

0.4

I 0.6

ag = ~ H , O+ aco, where Aa = Ae, and may be obtained from Fig. 13.72.

I 0.8

1.O

1.2

P H ~+OPT atrn 2

Fig. 13.70. Correction factor for the emissivity of water vapour a t pressures other than 1 atm

...(13.65)

Example 13.49. Determine the emissivity of a gas mixture consisting of N,, H20, and CO, at a temperature of 800 K. The gas mixture is i n a sphere with diameter of 0.4 m, and the partial pressures of the gases are piv, = 1 atm, PH,O = 0.4 atm, and Pco, = 0.6 atm.

Solution Given :A gas mixture of N,, H,O, CO, in a sphere D = 0.4 m T, = 800 K, PN, = 1atm,

PH,O = 0.4 atrn

Pco, = 0.6 atm. To find :Emissivity of the gas mixture. Analysis :The mean beam length for a spherical mass of gas is obtained from Table 13.4. Le = (0.667)D = 0.27 m Appropriate values for parameter to be used for Figs. 13.68 and 13.69 T = 800 K PH,OL,= 0.108 atrn Pco, Le = 0.160 atrn Then emissivities at 1 atrn are Fig. 13.71. Correction factor for the emissivity of carbon dioxide at pressures other than 1 atm

Nz does not radiate appreciably at 800 K. Since the total pressure is 2 atm, thus correction factor CH,O and Cco, from Figs. 13.70 and 13.71 are

892


and value of A& to correct the emission in overlapping wavelength bands, from Fig. 13.72 AE = 0.014 The effective emissivity of the mixture ~ , ~ ~ = 1 . 6 2 ~ 0 . 1 5 + 1 . 1 2 ~ 0 . 1 2 5 - 0 . 0 1 4 = 0 .Ans. 369.

Example 13.50.A long cylindrical combustor 40 cm in diameter contains a gas at 1200' ( ' consisting of 0.8 atrn N, and 0.2 atrn CO,. What is the net heat radiated to the walls if they crrv at 300°C ? Solution Given : Gas radiation in combustor D = 40 cm, T,- = 1200°C = 1473 K pNz = 0.8 atrn p = 1 atrn pco, = 0.2 atrn T, = 300°C = 573 K. To find :Net gas radiation to walls of combustor. Analysis :For CO, at 0.2 atm, from Fig. 13.69, Eco, = 0.098 The correction factor for CO, for total pressure pT = 1 atrn

893


F1-2 = F1-3 = "".. The rate of radiation heat transfer between two black surfaces is expressed as = A, F1, o (TI4- TZ4)(W) The net radiation heat transfer rate from a surface i of a black enclosure is sum of radiation heat transfers from surface i to each of the surface of enclosure :

The radiation heat transfer from a cavity surface 1is given by

The net rate of radiation heat transfer from a surface is expressed as E, - J Q = -- (W, R 1-E where R = - is the surface resistance, to radiation. The net rate of radiation heat transfer &4 from surface 1to surface 2 can be expressed as

Ccoz = 1 Thus E~ = Ecoz x CCO2= 0.098 From eqn. (13.62) to obtain ag at T, = 573 K

Now

Qnet

= A, okgTg4- agTs4) = n: x (0.4) x (5.67 x lo-') x [(0.098) x (14.7314- (0.1367) x (573)''I = 31.82 x lo3 Wlm = 31.82 kW/m. Ans.

13.10. SUMMARY

The view factor from surface 1to surface 2 is designated as Fl-, and is defined as fraction of radiation leaving the surface 1 and that strikes surface 2, directly. The view factor FlW1repro. sents the fraction of radiation leaving surface 1and that strikes itself directly. F,-, = 0 for fltrt or convex surfaces while F1-l # 0 for concave surfaces. For view factor the reciprocity rule I* expressed as A, Fl-, = A, F,-, For an enclosure, the sum of view factors from surface i to all surfaces of enclo~uruc

1 where Rl-, = -is the space resistance to radiation. The network method is used to solve A 1Fl-2 the radiation network. The radiation heat transfer between any two opaque, gray, and diffused surfaces is given by

+-+-----"

~ 1 A l AlF1-2 ~ 2 4 The radiation heat transfer rate between two surfaces can be reduced drastically by placing thin, high reflectivity (low emissivity) material sheets between these two surfaces, called radiation shield. The radiation heat transfer rate between two large parallel planes separated by N radiation shield is

-

Q1-2, N shield -

A o ( T , ~- T , ~ )

The radiation effect in temperature measurement can be properly accounted by the relation

N

including itself must be equal to unity i.e.,

F, = 1 1=1

FL-l + F,-, + FLF3+ ...... + F,-N = 1 It is known as summation rule. The superposition or additive rule states that the vlaw factor Fl-, is equal to the sum of view factors from surface 1 to the parts of surface 2. The symmetry rule states that if two or more surfaces are symmetric about the surface 1, then or

where T_ is actual temperature of fluid, T, is temperature measure by thermocouple, and T, is the temperature of surrounding walls. Many gases are transparent to radiation, but water vapour CO,, H20, CO, SO2,NH3 etc. can absorb and emit radiation. The gas radiation is not a surface phenomenon, but is a volumetric phenomenon.

892


and value of AE to correct the emission in overlapping wavelength bands, from Fig. 13.72 A& = 0.014 The effective emissivity of the mixture ~ , ~ ~ = 1 . 6 2 ~ 0 . 1 5 + 1 . 1 2 ~ 0 . 1 2 5 - 0 . 0 1 4 = 0 . Ans. 369.

Example 13.50.A long cylindrical combustor 40 cm in diameter contains a gas at 1200' ( ' consisting of 0.8 atrn N, and 0.2 atrn CO,. What is the net heat radiated to the walls if the.y rtrv at 300°C ? Solution Given : Gas radiation in combustor D = 40 cm, T,- = 1200°C = 1473 K pN2= 0.8 atrn p = 1 atm pco, = 0.2 atrn T, = 300°C = 573 K. To find : Net gas radiation to walls of combustor. Analysis :For CO, at 0.2 atm, from Fig. 13.69, Eco, = 0.098 The correction factor for C0, for total pressure pT = 1 atrn

893


F1-2 = F 1 3 = ...... The rate of radiation heat transfer between two black surfaces is expressed as Q1-2 = A, Fl-, o (TI4- TZ4)(W) The net radiation heat transfer rate from a surface i of a black enclosure is sum of radiation heat transfers from surface i to each of the surface of enclosure :

The radiation heat transfer from a cavity surface 1is given by

The net rate of radiation heat transfer from a surface is expressed as E, - J Q = R(W) 1-E where R = -is the surface resistance, to radiation. The net rate of radiation heat transfer &A from surface 1 to surface 2 can be expressed as

ccoz = 1 Thus E~ = Eco2 x CCO2= 0.098 From eqn. (13.62) to obtain ag at T, = 573 K

Now

Qnet

= As okgTg4- agTs4) = 7c x (0.4) x (5.67 x x L(0.098) x (14.73)4- (0.1367) x (579)'l = 31.82 x lo3 W/m = 31.82 kW/m. Ans.

13.10. SUMMARY

The view factor from surface 1to surface 2 is designated as Fl-, and is defined as fraction of radiation leaving the surface 1 and that strikes surface 2, directly. The view factor F1-l reprub sents the fraction of radiation leaving surface 1and that strikes itself directly. F,-, = 0 for fltrt or convex surfaces while F1-l + 0 for concave surfaces. For view factor the reciprocity rulc* tr expressed as -4, Fl-, = A, F2-, For an enclosure, the sum of view factors from surface i to all surfaces of enclonttm

1 where Rl-, = -is the space resistance to radiation. The network method is used to solve A 1Fl-2 the radiation network. The radiation heat transfer between any two opaque, gray, and diffused surfaces is given by

~ 1 A l' 4F1-2 ' &,A2 The radiation heat transfer rate between two surfaces can be reduced drastically by placing thin, high reflectivity (low emissivity) material sheets between these two surfaces, called radiation shield. The radiation heat transfer rate between two large parallel planes separated by N radiation shield is Q1-2, N shield

--

A o ( T , ~- T , ~ ) 1

1

1

The radiation effect in temperature measurement can be properly accounted by the relation

N

including itself must be equal to unity i.e.,

F,, = 1 1 = l

F,-1+F,-2+F1-3+...... +F1-,+ It is known as summation rule. The superposition or additive rule states that the view factor F1-, is equal to the sum of view factors from surface 1 to the parts of surface 2. The symmetry rule states that if two or more surfaces are symmetric about the surface 1, then or

where T_ is actual temperature of fluid, T, is temperature measure by thermocouple, and Twis the temperature of surrounding walls. Many gases are transparent to radiation, but water vapour C02,H20,CO, SO,, NH, etc. can absorb and emit radiation. The gas radiation is not a surface phenomenon, but is a volumetric phenomenon.

Radiation- R.K.Rajput

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