9. Radiation Heat Transfer - He= at and Mass Transfer, 2nd Edition

This equation is called as r= eciprocity theorem and is a function of the temperature of bodies. For diff= erential surfaces dA₁ and dA2, the shape factor is given by

and

Equations (9.12) and (9.13) = are equal if these are multiplied by dA_{1 and dA₂ on both sides, respectively. =
We get}

A_{1 dF_d_A=
₁_=E2=88=92_d_A2 =3D <=
em>A₂dF_{d_A₂_=E2=88=92_d_A1 =
; (9.14)}}

By int= egrating the elemental view factors dF_d= _A₁_=E2=88=92_d_A2 and dF_d<= sub class=3D"EBsub">A₂_=E2=88=92_d_A1 we get Equati= on (9.11).

9.10 PROPERTIES OF SHAPE FACTOR

It is function of geometrical par= ameters.
If radiation takes place between two s= urfaces, then A₁ F_1=E2=88=922 =3DA₂ = F_2=E2=88=921.
If a surface radiates energy, it falls= on =E2=80=98n=E2=80=99 surfaces then,

F_{i=E2=
=88=921} + F_i=E2=88=922 + F_i3 + =E2=80=A6 + F_i=E2=88=92N =3D 1

The shape factor for convex and plane = surfaces is equal to zero. (No radiation energy is falling on the body.)
For concave surfaces, the shape fa= ctor is not equal.

9.11 SHAPE FACTOR ALGEBRA

The standard shape factors ar= e available in charts for a limited number of simple configurations. For a = complicated system, it may be split into a number of simple configurations = for which shape factors can be determined from the available charts. Then t= he shape factor of the original system is determined by the algebric sum of= the simple dividable configuration shape factors. Such an approach is know= n as shape factor algebra. The shape factor algebra is based on the princip= le of conservation of energy.

Consider a configuration as = shown in Fig. 9.4.

Figure 9.4 configuration for shape Factor Algebra

The area A₂ is divided into two areas A₃ and A₄. The shape factor from= Area 1 to Area 2 can be written as

F_{1=E2=
=88=922} =3D F_1=E2=88=923 + F<= /em>_1=E2=88=92₄.

Multiplying Equation (9.15) by A_=
1 on both sides,

A1 F_1=E2=88=922 = =3D A₁ F_{1=
=E2=88=923} + A₁ F_1=E2=88=924       &n= bsp;  (9.15)

The r= eciprocity theorem gives

A_{1 F_1=E2=88=922 =3D A₂ F_2=E2=88=921}

A_{1 F_1=E2=88=923 =3D A₃ F_3=E2=88=921}

A_{1 F_1=E2=88=924 =3D A₄ F_4=E2=88=921}

Equation (9.15) then becomes

A_{2 F_2=E2=88=921 =3D A₃ F_3=E2=88=921 + A}₄ F_{4=E2=88=
=921}

If the area =E2=80=98A₂=E2=80=99 is divided into m= ore parts, then

Shape factors for various con= figurations are shown in Figs. 9.5<= /a>=E2=80=93 9.14.

Figure 9.5 Two Parallel Plates with Midlines Joined by a Perpendi= cular

Figure 9.6 Iwo Inclined Parallel Plates of Equal width with a com= mon side

Figure 9.7 Two Perpendicular Plates with a Common Side

Figure 9.8 Three Planes of an Enclosure

Figure 9.9 Two Parallel Cylinders with Different Radii

Figure 9.10 A Cylinder and a Parallel Plate kept at Rectangle Po= sition

Figure 9.11 An Infinite Plane with a Row of Cylinders

Figure 9.12 Rectangle Plates Aligned Parallel

Figure 9.13 Two Parallel Disks kept coaxi= al

Figure 9.14 Two Perpendicular Rectangle Plates with a Common Edg= e

9.12 ELECTRICAL ANALOGY FOR THERMAL RADIATION

An electrical analogy method = is also used to determine the radiation heat exchange between the surfaces.= Two terms, radiocity and irradiation, are very important in this analysis.=

Radiocity (J): It is the total energy leaving a surface per unit area per unit time.<= /p>
Irradiation (a<= /em>): It is the total radiation incident upon a surface per unit area per = unit time.

Let us consider a gray surfa= ce as shown in Fig. 9.15(a) ha= ving a total emissive power E, transmissivity (=CF=84), and absorb= icity (=CE=B1). These are constant over the entire surface. Total = incident radiation over the surface is G (in W/m²). Radiocity is the sum of energy emitted and the energy transmit= ted

Figure 9.15 (a) Gray Thermal Surface; (b) Electrical Analogy

then net radiation energy lea= ving the surface is written as

where

The equation is in the form = of an electrical network shown in = Fig. 9.15(b). The term is called th= e surface resistance. If the surface is black (=CE=B5 =3D 1), then Rs =3D 0;

J =3D E_{b =3D}=CF=83T⁴.  &nbs= p;       (9.18)

If a surface is perfectly in= sulated, it re-radiates all the incident energy and is known as re-radiatin= g surface. Such surfaces increase the heat transfer between surfaces enclos= ed by them. Radiating surfaces provide an additional parallel path for radi= ation heat flow, as shown in Fig. = 9.16 for black surfaces.

Figure 9.16 Radiating Surfaces

Let us consider the exchange= of radiation energy between two surfaces A_{1<=
/sub> and A₂ having radiocities J₁ and J₂.=
The amount of energy reaches A₂ from <=
em>A₁ is J_{1A₁F₁₂₌ and the amount of energy reaches A_{1 from A₂ is J₂ A₂ F_2=E2=80=931.}}}

The net interchange energy,<= /p>

where

The term R_Sp is known as space resistance between the bodies. It is= purely based on the location and geometry of the surfaces. It can be prese= nted in electrical analogy as shown in Fig. 9.17.

Figure 9.17 Electrical Analogy

Then net heat exchange betwe= en the two gray surfaces is represented in Fig. 9.18.

Figure 9.18 Net Heat Exchange

The equation can be applied = to parallel systems.

9.12.1= Two Infinite Parallel Plates

In this case (Fig. 9.19), radiation loss from the sides of = the plates is negligible. The radiation energy leaves the surface 1 and rea= ches plate 2 and vice versa. This gives the shape factor as unity.

Figure 9.19 Infinite Parallel Plates

Then,= Equation (9.20) is reduced to

9.12.2= concentric cylinders or spheres

Similar to parallel plates fo= r concentric cylinders or spheres as shown in Fig. 9.20, we have

Figure 9.20 concentric Cylinder

F_{1=E2=80=
=932} =3D 1 and for cylinders=

= for spheres.

Then Equation (9.20) becomes

9.12.3= Small Body Lies Inside a Large Enclosure

Here F_1=E2=88=922 =3D 1; A₁ is v= ery small compare to A₂; therefore, . Then Equation (9.20) becomes

Shape factors for various co= nfigurations of different shapes can also be found from charts shown in Figs. 9.21=E2=80=939.27.

Figure 9.21 shape Factor for Plates Held Parallel

Figure 9.22 Shape Factor for Plates Placed opposite in Rectangle= s

Figure 9.23 Shape Factor for Perpendicular Rectangle Adjacent Pl= ates with a Common Edge

Figure 9.24 Shape Factors for Parallel square, Rectangle, and Di= sk-Type Plates

Figure 9.25 Shape Factor for coaxial Circular Parallel Plates

Figure 9.26 Shape Factor for cylinders of Finite Length (Outer C= ylinder to Itself)

Figure 9.27 shape Factor for cylinders of Finite Length (Outer C= ylinder to Inner Cylinder)

9.13 RADIATION SHIELDS

If a low-emissivity material = (radiation shield) is placed between the two surfaces, the radiation heat t= ransfer can be considerably reduced. The shield increases the thermal resis= tance to radiation heat flow. Normally, thin sheets of aluminum, copper, et= c. are used as shields.

Consider two large radiating= parallel surfaces as shown in Fig= . 9.28(a). Let T_l and T₂ be the temperatures and =CE=B5₁ and =CE=B5₂ be the emissiviti= es of the surfaces. The net heat transfer between the plates, we know that<= /p>

Figure 9.28 Plates with shield Assembly and Networks

Let a radiation shield havin= g emissivity of =CE=B5_s be placed betwe= en the plates as shown in Fig. 9.2= 8(b).

The shield radiates the heat= on both sides to plates 1 and 2. The shape factor F_1=E2=80=93s and F_s=E2=80=932 are equal to 1. The net heat transfer between the plates at steady-state= condition can be written as

If the emissivities of the s= urfaces are equal to =CE=B5, then Equations (9.24) and (9.25) redu= ce to

and

This shows that the shield r= educes the heat transfer between the plates by 50%.

If N shields are u= sed between two surfaces with the emissivities of all surfaces equal, the h= eat transfer is

The ratios of heat transfer = rates for parallel plate systems with N shields and without shield= s can be written as

9.14 GAS RADIATION

In combustion furnaces, radia= tion exchange takes place between gases and the surrounding surfaces. The g= ases such as O₂, N₂, = H₂, and dry air neither emit nor absorb radiatio= n at low temperatures. These gases are transparent to incident thermal radi= ation. The gases H₂O, CO_{2, CO, SO₂, NH₃, a=
nd the hydrocarbons absorb the incident radiation and also emit radiation a=
t specific wavelength intervals called bands.}

In solids, the emission and = absorption of radiation are surface phenomena whereas for a gas layer, the = properties are based on thickness, pressure, and surface area. Beer's law g= ives the relation between the intensity of radiation entering the gas and i= ntensity of radiation leaving the gas medium as

I_=CE=
=BB_L =3D I_=CE=BB_o exp(=E2=88=92K= _=CE=BBL)    &n= bsp;     (9.30)

where

    &nbs= p;I_=CE=BBL =E2=80=93 intensity of radia= tion at =E2=80=98L=E2=80=99

    &nbs= p;I_=CE=BB0 =E2=80=93 intensity of incid= ent radiation

    &nbs= p;L =E2=80=93 thickness of gas medium

    &nbs= p;K_=CE=BB =E2=80=93 monochromatic absor= ption coefficient

Then the transmissivity of t= he gas is

For gases,

=CF=84_{=CE=BB+ =CE=B1_=CE=BB =3D 1; =CF=84_=CE=BB=3D (1 =E2=80=93 =CE=B1_=CE=
=BB)}

=CE=B1_{=CE=BB =3D 1 =E2=80=93 exp (=E2=80=93K_{=CE=BBL) =3D =CE=B5_=CE=BB   =
;  (=E2=88=B4 As per Kirchoffs law).     =
;     (9.31)}}

The total absorptivity or em= issivity is obtained by summing up the narrow wavelength bands. In most of = the engineering problems, radiation heat transfer takes place from combusti= on products to adjoining surfaces. The emission from a mixture per unit are= a of the surface is written as

where =CE=B5_g is the emissivity of the gas. It is determined from th= e charts by knowing the temperature (T_a= ) of the gas, total pressure (P), the partial pressure (P_g) of the radiating species, and characteristic = length (L) of the geometry. The characteristic lengths for differe= nt geometries are available as shown in

Ta= ble 9.4 Approximate values of mean beam length for various gas vol= umes

The emissivity of a gas in a = mixture is obtained by

where

    &nbs= p;=CE=B5_g =E2=80=93 emissivity of a gas= at given condition

    &nbs= p; =E2=80=93 emissivity ot the gas at total mixture pres= sure of 1 atm

    &nbs= p;C_s =E2=80=93 correction factor if the= total pressure is other than 1 atm

For m= ixtures of two gases in a volume, the emissivity of the mixture may be calc= ulated by adding the emissivities of the gases determined separately and su= btracting a factor =CE=94=CE=B5, which accounts for emission in co= mmon wavelength bands as

=CE=B5_g1 + =CE=B5_g2 = =E2=88=92 =CE=94=CE=B5       &n= bsp;  (9.35)

The radiation between a blac= k enclosure and the gas is given by

where A_s is the surface area, T_g= and T_s are gas and surfaces temperatur= es, =CE=B5_g is the emissivity of the ga= s at temperature T_g and =CE=B1_g is the absorptivity of the gas. The value is c= alculated as

=CE=B1_g =3D =CE=B1_l + =CE=B1₂ =E2=88=92 =CE=94=CE=B1     &n= bsp;    (9.37)

and

Here =CE=B5_l and =CE=B5₂ are de= termined at surface temperature T_s for = the valve of PL given by PL (T_{s<=
/sub>/T_g}). The products of combustion are m= ainly water vapor and carbon dioxide. The emissivities of H₂O and CO₂ estimated from charts are = shown in Figs. 9.29=E2=80=939.33.

Figure 9.29 emissivity of CO₂(Total P= ressure =3D 1 atm)

Figure 9.30 correction Factor for the Emissivity of Carbon Dioxi= de at Pressures Other Than 1 atm

Figure 9.31 Emissivity of H₂O vapor (= Total Pressure =3D 1 atm)

Figure 9.32 correction Factor for the Emissivity of water vapor = (Pressures Other Than 1 atm)

Figure 9.33 Factor =CE=94=CE=B5 for Emissivity Correcti= on of a Mixture of Water Vapor and CO₂

9.15 CONCLUSIONS

Energy emitted by the bodies,= due to their temperature, in the form of electromagnetic waves is known as= thermal radiation. Thermal radiation is emitted in wavelengths of 0.1=E2= =80=93100 um. The radiation emitted by the Sun at a temperature of about 57= 60 K and wavelengths between 0.1 and 3 =CE=BCM, is known as solar radiation= .

A body that absorbs all the = incident radiations from all directions at all wavelengths without reflecti= ng, transmitting is considered as a black body.

Total emissive power is defi= ned as the total amount of radiation emitted by a body per unit area and ti= me. Absorptivity is the fraction of incident radiation absorbed by a body. = Reflectivity is the fraction of incident radiation reflected by a body. Tra= nsmitivity is the fraction of incident radiation transmitted through body. = Opaque body has zero transmitivity. White body has zero transmitivity and absorptivity and its reflectivity is 1. Gray body abs= orptivity does not vary with temperature and wavelengths of incident radiat= ion. Colored body absorptivity varies with wavelengths and radiation. The e= missitivity of a surface is equal to its absorptivity. Emissitivity of a su= rface is defined as the ratio between emissive power of a surface at a part= icular temperature to the emissive power of a black surface at the same tem= perature.

Irradiation is the total rad= iation incident upon a surface per unit time per unit area. Radiosity is th= e total radiation leaving a surface per unit time per unit area.

The angle between the reflec= ted beam and normal to the surface equals the angle made by the incident ra= diation with the same normal and is known as regular reflection. In diffuse= reflection, the incident beam is reflected in all the directions from a su= rface.

Stefan-Boltzman law states t= hat the emissive power of a black body is directly proportional to the four= th power of the absolute temperature.

Kirchoff's law of radiation = states that at any temperature, the ratio of total emissive power to the to= tal absorpticity is constant for all substances.

Wien's displacement law rela= tes the temperature of a black body and the wavelength at which maximum val= ue of monochromatic emissive power occurs. Planks law and Wien's displaceme= nt law are used to find the maximum monochromatic emissive power. Lambert's= cosine law states that the total emissive power from a radiating plane sur= face in any direction is directly proportional to the cosine of the angle o= f emission.

Intensity of radiation is de= fined as the rate of energy leaving the emitting surface in a given directi= on per unit solid angle per unit area of the surface and normal to the mean= direction in space.

Plane angle is the ratio of = an element of arc length on a circle to the radius of their circle. Solid a= ngle is the ratio of an element of an area on a sphere to the square of the= radius of sphere.

View factor in radiation is = defined as the fraction of radioactive energy that is diffused from one sur= face element and strikes the other surface directly with no intervening ref= lections.

The shape factor for certain= geometric arrangements for which the equations are not available can be de= rived in terms of known shape factors of other geometries. The inter relati= on between various factors is called shape factor algebra. The radiant ener= gy coming out from one part of concave surface is intercepted by another pa= rt of the same surfaces. Therefore, shape factor with respect to itself of = a concave surface is not zero. Whereas the shape factor with respect to its= elf of a convex or flat surface is zero.

The radiation shield reduces= the radiation heat transfer by effectively increasing the surface resistan= ce without removing any heat from the system. In solids, absorption of radi= ation takes place within a small distance from the surface. Surface radiate= s all wavelengths over the entire spectra. Gases emit radiation only betwee= n narrow ranges of wavelengths. Intensity of radiation through the gaseous = medium decreases with the length of passage. H₂O= , CO₂, SO₂, NH₃, CO, and hydro carbons are radiating gases.

KEY CONCEPTS

Thermal radiation is closely associated to elecromagnetic radi= ation, which can be described as material heat converted into light energy = and conversely.

Light here does not mean necessary visible=E2=80=94thermal radiatio= n also occurs in the infra-red spectrum.

This type of heat transfer will depend on the temperature and t= ype of materials of the various elements considered.

EXAMPLE PROBLEMS

9.1. A black body = emits radiation of maximum intensity at a wavelength (=CE=BB_max) of 0.5 =CE=BCm. Calculate the surface te= mperature and heat flux at the surface of the body.

As per Wien's displaceme= nt law

=CE=BB_mT =3D 2.989=C3=9710^=E2=80=933mK

Surface temperature of t= he body,

Heat flux at the surface= ,

E_b =3D =CF=83T ⁴

      = ;            &n= bsp;            = ;=3D 5.67 =C3=97 l(10^=E2=80=938 =C3=97(5796)4

      = ;            &n= bsp; E_b =3D 6.399 =C3=9710^=E2=80=93⁷ W/m².

9.2. Calculate the= monochromatic emissive power of a black body emitting radiation at a tempe= rature of 2500 K (T) and wavelength of 1.4 =CE=BCm (=CE=BB). Also = determine the total emissive power of the body.

As pe= r Planck's law

Here

C₁ =3D 3.7415=C3=9710^=E2=88=9216 W-m²

C₂ =3D 1.4388 =C3=97 10^=E2=88=922 mK

Monochromatic emissive power= ,

Total emissive power,

E_b=3D =CF=83T⁴

      = ;            &n= bsp;            = ;            &n= bsp;            = ;       =3D 5.67 =C3=97 10^=E2=80=938 =C3=97(2500)⁴ =3D2.215= =C3=9710⁶ W/m².

9.3. What is the t= emperature at which a black body will have an emissive power (E_b_=CE=BB) equal to 2 = =C3=9710³ W/m² at a w= avelength of 4.4 mm (=CE=BB).

As per Planck's law

Here

C_{1=3D 3.7415 =C3=9710^=E2=80=9316 Wm²}

C₂=3D1.4388 =C3=9710^=E2=80=932 mK

Therefore,

Therefore,

Therefore, temperature o= f the black body, T =3D 176.3 K.

9.4. A large spher= ical furnace inside surface temperature is 1000 K (T). If there is a hole o= f diameter 0.5 cm (d), determine the rate of emission of radiation= through the opening.

As per the Stefan-Boltzmann = law

Therefore, rate of emission = of radiation,

E_b =3D 1.963 =C3=97 10^=E2=88=925 =C3=97 5.67 = =C3=97 10^=E2=88=928 =C3=97 (1000)⁴

=3D 1.113 W.

9.5. A filament in= a bulb is heated to 3000 K (T). Find the radiation energy that falls in th= e visible range.

Visible range of wavelen= gth is from =CE=BB₁ =3D 0.4 =CE=BCm to =CE=BB₂ =3D 0.7 =CE=BCm Here

=CE=BB₁<= em>T =3D 0.4=C3=9710^=E2=80=936=C3=973000 = =3D 1200=C3=97l0^=E2=88=926 mK

=CE=BB_2T =3D 0.7 =C3=9710^=E2=88=926=C3=973000 = =3D 2100=C3=9710^=E2=88=926 mK

FromTable 9.3, the radiation functions for =CE=BB₁T and =CE=BB₂T are and . Then . That is, 8.3% of the energy= is in the visible range.

Therefore, radiation ene= rgy

9.6 What is the te= mperature of a black body such that 63.4% of the energy emitted should lie = in the wavelength spectrum =CE=BB =3D 0 =E2=88=9210 =CE=BCm?<= /p>
At =CE=BB₁=3D 0, E_b =CE=BB₁ =3D 0.

Radiation function .

From Table 9.3, the corresponding value

=CE=BB₂<= em> =3D 5=C3=97l0^=E2=88=923 mK

Therefore, temperature of th= e black body,

9.7. Glass transmi= ts 93% of the incident radiation in the wavelength range between 0.35 (=CE=BB₁) and 3.0 =CE=BCm (=CE=BB₂) and is opaque at longer and shorter wavelengths= . Calculate the percentage of solar radiation which the glass will transmit= . The Sun=E2=80=99s temperature is 5800 K (T). (Assume the Sun to = be a black body.)

Here

=CE=BB₁<= em>T =3D 0.35=C3=975800 =3D 2030=C3=97l0^=E2=88=92=
6 mK

=CE=BB_2T =3D 3.0=C3=975800 =3D 7400=C3=9710^=E2=88=926= mK

FromTable 9.3, the values of and .

The fraction of the tota= l radiant energy incident upon the glass is 0.9094.

Percentage of solar energy transmit= ted =3D 0.9094 =C3=97 transmissivity

      = ;            &n= bsp;            = ;            &n= bsp;            = ;             = =3D 0.9094 =C3=97 0.93 =3D 0.846 =3D 84.6%.

9.8 Tungsten filam= ent is used in a 100 Watt (E) light bulb. The filament temperature (T) is 3= 000 K and emissivity (=CE=B5) is 0.3. Calculate the minimum surface area of= the tungsten filament if the bulb is completely evacuated and is at a stea= dy-state condition.

Energy radiated =3D Power of the bu= lb

E =3D =CE=B5=CF=83AT⁴

100 =3D 0.3(5.67 =C3=97 10^=E2=88=928) =C3=97 A=C3=97 (3000)

Therefore, 100 =3D 0.3 =C3=97 5.67 = =C3=97 10^{=E2=88=92 8} =C3=97 A =C3=97 8.= 1 =C3=97 10¹³

A =3D 7.26 =C3=97 l0^=E2=88=925 m².

9.9. Find the shap= e factor F₁₂ of the following hollow co= nfiguration (d =3D 5 cm) and depth (L) is 1 m as shown in figures (a=E2=80=93c).

Case (a)

F_{1=E2=
=88=921}+F_1=E2=88=922+F13 =3D 1       =    (S1)

Also

F_{2_=E2=88=92₂ + F=
₂_=E2=88=92₁ + F_2=E2=88=923 =
=3D 1          (S2)}

And

F_{3=E2=
=88=923}+ F_3=E2=88=921 + F_3=E2=88=922 =3D l      = ;     (S3)

Here F₁_=E2=88=92₁, F_2=E2=88=922 and F3_=E2=80=93₃ are zeros because the surfaces are plane surfaces.

From (S2), we get

T= herefore

Case (b)
Also F_2=E2=88=922 + F_{2=E2=88=921<=
/sub> =3D 1}

F_{2=E2=
=88=921} =3D 1 {since F_2=E2=88=922= =3D 0}

According to the recipro= city theorem,

A_{1F_1=E2=88=922 =3D A₂F_{2 =E2=88=92 1}}

Therefore,

Case (c)
Here also

9.10. Find the sha= pe factor F_1=E2=88=922 for the configurations gi= ven in figure (a) and (b).<= /p>

This is also a case of n= on-intersecting perpendicular rectangles. We cannot directly use the tables= . In such a case, the factor F_{1=E2=88=922 is calculated by introducing hypothetical areas A₃ and A₄. In this, A₅ =3D A₄ =
+ A₁ and A₆ =3D A₂ + A₃ Then}

For F_5=E2=88=926, F_4=E2=88=923, = F_5=E2=88=923 and F_4=E2=88=926, values are available in the charts (perpendic= ular rectangles).

F= or F_5=E2=88=926, we have to calculate t= he parameters.

Here x₂/l =3D 7/6 =3D 1.17; x_1/l =3D 7/6 =3D 1.17

F_5=E2=88=926 =3D 0.19

Similarly, we can use= the charts and

9.11. Two small su= rfaces having areas 4 cm² (dA₁) and 8 cm² (dA_{2) are separated by a distance of 100 cm (r). The orientations of the su=
rfaces are given in the figure. Find the view factor between the surfaces.<=
/em>}

Here

Both surfaces can be tak= en as differential surfaces. The view factor

Then

9.12. A room 2 m i= n length (l), 1.5 m in width (b) and 3 m in height (h) is heated f= rom the ceiling by maintaining it at a temperature 500 K (T₁). If the side walls and floor are maintained at 300 K (T₂), calculate the rate of heat transfer from = the ceiling to floor and side walls when the emissivity of all surfaces is = 0.6 (=CE=B5).

This problem is equivale= nt to that of radiation between two planes with low emissivity material in = between them (gas radiation). See the figure below for reference.

and

Total resistance,

R =3D R₁+ R_1=E2=80=932 + = R₂

=3D 0.22 + 0.33 + 0.028 =3D 0.578/m= ².

Heat transfer,

9.13. Calculate th= e net heat flux between two parallel, opaque, large plate= s that are kept at 600 K (T₁) and 300 K (T₂), respectively. The emissivities of both th= e plates are the same, that is, 0.8 (=CE=B5).

For infinite parallel= plates problem,

Net heat flux,

9.14. A steam pipe= of outer diameter (d_o) and length 60 cm (l) who= se surface is at 200=C2=B0C (T_P) passes through = a room with a wall at 10=C2=B0C (T_w). Assuming t= hat the emissivity (=CE=B5) of the pipe is 0.8, determine the rate= of the heat loss from the pipe by r adiation.

In this case, the ent= ire surface area of the pipe is enclosed in a room. Therefore, the shape fa= ctor of pipe to room is one.

Heat loss from the pi= pe,

Where

A =3D =CF=80dl

      = ;            &n= bsp;            = ;    =3D =CF=80=C3=970.2=C3=970.6 =3D 0.377 m2

Therefore

Q =3D 0.377=C3=970.8=C3=97= 5.67=C3=9710^=E2=88=928=C3=97[(473)⁴ =E2=80=93 (283)⁴]

=3D 746.3 w.

9.15. A cubical room of side 2 m (A) is shown in the figure and it is = heated from the floor by maintaining it at an uniform temperature of =E2=80= =9323=C2=B0C (T), while the walls and the ceiling are at =E2=80=93= 73=C2=B0C (T₂). Assume that the floor has an emi= ssivity 0.9 (=CE=B5) and the walls and the ceiling have emissivity 0.6 (=CE= =B5 ). Calculate, the rate of heat transfer from the floor.
Let us call the floor= , surface 1 and the walls and ceiling, surface 2. The equivalent electric c= ircuit is,

Rate of heat loss from t= he floor,

9.16. A cubical ro= om of side 3 m (A₁), as shown in the fi= gure below, is heated from the floor by maintaining it at an unifo= rm temperature of 300 K (T₁). Heat transfer take= s place to the ceiling and side walls, which are maintained at 200= K. The floor has an emissivity (=CE=B5₁) of 0.9= and the ceiling has an emissivity (=CE=B5₂) of = 0.6. Determine the heat transfer (i) to the side walls having emissivity (= =CE=B5₃) of 0.5 (ii) to the side walls if they a= re well insulated and (iii) find the radiocities when walls are insulated.<= /em>

Side walls are not insulated
In this case, let us call the= floor, surface 1, the ceiling, surface 2, and the side walls, surface 3. T= he details are shown in the figure with the electrical network.

R_1=E2=88=922 =3D 1/A₁F<= /em>_1=E2=80=932, where F_1=E2=88=922 is 0.2 (from the graphs for parallel rectangular co= nfiguration in opposite locations)

F₁_=E2=88=92_{1 + F_1=E2=88=922 + F_1=E2=88=923 =3D 1, therefore F_1=E2=88=923 =3D 1 =E2=88=92 0.2 =3D 0.8}

Heat transfer to ceiling= ,

Heat transfer to side walls,<= /p>

The side walls are insulated
Insulated walls imply th= at the surfaces are re-radiating surfaces, E_<=
em>b3 =3D J₃, R3 =3D 0. The net electrical work is given in the fig= ure

The equivalent circuit f= or the one given above is shown in the figure below:

E= quivalent resistance,

Heat transfer to the cei= ling,

Applying Kirchoff's law to nodes <= em>J₁ and J₂=

For node

Therefore,

For node J₂,

Solving (S5) and (S6)

1876.95=E2=88=92 (22.49) =C3=97 (50= .02J₁ =E2=88=92 21040.7) + 1.8J ₁=3D 0

1876.95=E2=80=931124.95J1 + 473205.34 + 1.8J_=
1 =3D 0

J_{1 =3D 423 W/m²}

J_{2 =3D 117.31 W/m²}

Checking for energy bala= nce,

9.17. Find the red= uction in radiation heat transfer between two parallel plates when three an= d two shields are placed between them, with all emissivities assum= ed to be equal.

For N shields b= etween plates having same emissivities,

When N =3D 0

When N =3D 3

When N =3D 2

Percentage reduction in = heat transfer rate when N =3D 3 is

Percentage reduction in = heat transfer rate when N =3D 2 is

9.18. Two large pa= rallel plates, as shown in the figure below, with emissiv= ities 0.4 (=CE=B5₁) and 0.8 (=CE=B5) exchange he= at by radiation. If a polished aluminium radiation shield (=CE=B5₃=3D 0.05) is placed between them, calculate the = percentage reduction in heat transfer.

Without shield

Heat transfer,

With shield

Here

Percentage reduction in = heat transfer

9.19. Two large pa= rallel planes at temperatures 1000 K (T₁) and 80= 0 K (T) have emissivities =CE=B5₁ =3D 0.5 and = =CE=B5₂ =3D 0.8, respectively. A radiat= ion shield having an emissivity 0.5 (=CE=B5_s1) o= n one side (facing hot plate) and an emissivity 0.04 (=CE=B5_s2) on the other is placed between the planes. Determine the he= at transfer rate by radiation with and without the radiation shield.

Without shield
Heat transfer rate

With shield

9.20. A spherical = tank with radius 30 cm (R₁) filled with a fluid = at temperature of 100 K (T₁) is placed inside a = spherical container of radius 50 cm (R₂) and is = maintained at temperature 300 K (T₂) The emissiv= ities of the inner and outer tanks are 0.15 (=CE=B5_{1) and 0.25 (=CE=B5₂)}, respectively.= A spherical shield of radius 40 cm (R_s) and hav= ing an emissivity 0.04 (=CE=B5_s) on both sides i= s placed between the spheres. Determine the rate of heat loss without and w= ith shield.

Without shield
Heat loss

With shield

9.21. Determine the = emissivity of CO ₂ in a gas body of spherical sh= ape of diameter 1.5 m (d) at a temperature of 1500 K (T_g) in the following pressures. The gas consists 20% of CO₂ and the rest are non-radiating gases. The total pressure= (P) of the gas is (i) 1 atm, (ii) 4 atm and (iii) <= em>0.4 atm.

Total pressure P is 1 at= m
The characteristic lengt= h of a sphere,

L=3D0.7 D (fromTable 9.4)

L =3D 0.7=C3=971.5 =3D 1.0= 5 m

P_=
co₂ L =3D 0.2 =C3=97 1.05 = =3D 0.21 atm m

From the chart, at T= =3D 1500 K and P_CO₂ L =3D 0.21,

Total pressure, P_T, is 4 atm

P_CO2 =3D 0.2 =C3=97 total pressure =3D 0.2 =C3= =97 4 =3D 0.8 atm

P_CO₂ L =3D 0.8 =C3=97 1.05 =3D 0.84 atm m

From the chart, = =C9=9B_CO₂=3D 0.2= 7.

Then correction facto= r C_CO₂ at be= ing P_g =3D 4 atm and P_CO₂ =3D 0.84 atm m is given b= y

C_CO2 =3D 1.2

Therefore, =C9=9BCO₂ =3D =C9=9B=E2= =80=B2_CO₂ = =E2=88=92 C_CO₂ = =3D 0.27 =C3=97 1.2 =3D 0.324

The total pressure, P_g, is 0.4 atm

Here P_CO₂ =3D0.2=C3=97 P_g =3D 0.2 =C3=97 0.4 =3D 0.08 atm

P_CO2 L =3D 0.08 =C3=97 1.05 =3D 0.084 = atm m

From the chart, = =C9=9B=E2=80=B2_{CO₂ =3D 0.08.}

Then, correction fact= or C_CO₂ at <= em>P_T =3D 0.4 atm and P_CO₂L =3D 0.084 atm m i= s C_CO₂ =3D 0= .8. Therefore,

=C9=9B_CO₂ =3D =C9=9B=E2=80=B2_CO₂ =C3=97 C_CO₂ =3D0.08=C3=97 0.8 =3D 0.06= 4.

9.22. A cubical furnace 0.6 m on a side (A) has gas at temperature 100= 0 K (T). The gas contains 20% H₂o vapour by volu= me at a total pressure of 0.8 atm (P_g). Find the= emissive power of the gas.

Here, characteristic len= gth,

L =3D 0.7 A

=3D 0.7 =C3=97 0.6 =3D 0.42 m

Partial pressure of H₂O =3D 0.2 =C3=97 0.8

From the chart at T =3D 10= 00 K and P_{H₂OL =3D 0.0672 atm m,}

=C9=9B=E2=80=B2H₂O =3D 0.09

To find the correction f= actor,

From the chart, the c= orrection factor C_{H_2O =3D 1.}

Therefore, =CE=B5= =3D =C9=9B=E2=80=B2_H₂OC_{H_2O =3D 0.09=C3=971 =3D 0.09}

The emissive power of= the gas

9.23. A furnace in= the form of a cube of side 3 m (A) has gas at temperature 1250 K (T). The = partial pressure of CO₂(P_CO₂) is 0.2 atm and partia= l pressure (P_{H_2O) of H₂O is 0.15 atm. The total pre=
ssure (Pg) of the gas 2 atm. Determine the emissivity of the gas.}

Characteristic length,

The emissivities of CO2 and H₂O at 1250 K and= for the above PL values (from the chart)

=CE=B5=E2=80=B2_CO₂ =3D0.18; =CE=B5=E2= =80=B2_H₂O =3D 0.19

The correction factor C_CO₂ at P<= sub class=3D"EBsub">T and P_CO₂ L is C C_CO₂ =3D 1.1.=

From the chart, C_{H_2O =3D 1.42.}

Common correction fac= tor =CE=94E, the following parameters are calculated:

=CE=94 =3D 0.048 for = the above values, at T_g.

Then the emissivity o= f the gas,

9.24. A furnace of= total surface area 25 m² (A) and volume 10 m3 (v) is kept at a constant temperature 500 K (Tw). The furnace contains 20% of CO₂(P_{CO₂) and 10% of water vapour (P_H₂O) by volume. The gas temperature (=
T_g) is 1250 K and total pressure (P)_g is 2 atm, determine the heat exchange from the =
gases to the furnace wall and radiation heat transfer coefficient.}

The characteristic lengt= h

Then

P_CO2 =3D 0.2=C3=97 P_{g=3D0.2=C3=972 =3D 0.4 atm}

P_H2O =3D 0.l=C3=97P_{g =3D0.l=C3=97}2 =3D 0.2 atm

therefore, P_CO₂ L =3D 0.4 =C3=97 1.44 = =3D 0.576 atm m

and P_H₂O L =3D 0.2 =C3= =97 1.44 =3D 0.288 atm m

The emissivities, fro= m the chart, are =CE=B5=E2=80=B2_CO₂ =3D 0.18 and =CE=B5=E2=80=B2_H₂O =3D0.19.

The corrections facto= rs from the charts, are C_CO₂ =3D 1.1 and C_H₂O =3D 1.45.

The common correction factor= , =CE=94=CE=B5, at P_{CO_2<=
/sub> + P_H₂O=L =3D 0.6 atm m and (P_H₂O/P_H₂O+P_CO₂) =3D 0.2/0.2}

+ 0.4 =3D 0.33 is =CE=94= =CE=B5 =3D 0.045.

Therefore, at T_g the emissivity of the gas,

To find the absorptivity= , =CE=B1_g, of the gas at wall temperatu= re, T_w, we have to calculate PL or wall temperatures for CO₂ and H₂O

and

From the chart, the values a= t T_w and (PL)_{T_{_w} are}

=CE=B5=E2=80=B2CO₂ =3D 0.14 and =CE=B5=E2=80=B2 H_{2O =3D 0.2=E2=80=B2.}

Then the correction fact= ors at 2 atm pressure, the PL)_{T_{_w}} values ar= e

C_CO2 =3D 1.1 and C_H₂O =3D 1.46

The common correction fa= ctor =CE=94=CE=B5, from the chart,

The radiation heat trans= fer coefficient, h_r, can be calculated = from the following equation:

9.25. In a semi-ci= rcular tube, air flows at the rate of (m) 36 kg/h and its temperat= ure (T_a) is 500 K at atmospheric pressure. The p= lane surface is maintained (T₁at 1200 K and othe= r surface is insulated and both surfaces have an emissivity (=CE=B5) of 0.8= 5 and the tube radius (R) is 25 mm. Find the heat transfer rate per unit le= ngth that must be supplied to maintain the plane surface at the same temper= ature as the temperature of the insulated surface.

Radiation heat transfer,=

Here,= F₁₂ (shape factor) =3D 1 and {A₁ =3D D; A₂ =3D =CF=80D/2} for unit length of tube

Convection heat transfer,

Q_c =3D hA₂(T₂ =E2=88=92 T_a)

Air properties at (500 K) ar= e k =3D 41 =C3=9710^=E2=80=933 W/mK, =CE=BC =3D 26.7 =C3=97 10^=E2=80=936 N-s/m<= sup class=3D"EBsup">2; Pr =3D 0.68 Reynolds number,

where

For energy balance, = Q_r =3D Q_c

By trial and error, = T₂ =3D 634 K

Total heat transfer per = unit length from plane surface =3D Q_c += convection heat transfer from plane surface (Q=E2=80=B2_c)

9.26. Refrigerant = liquid flows through a tube of 3 cm diameter (D₁= ), which is having outer surface emissivity (=CE=B5₁) 0.03 and temperature (T₁) 80 K. T= he tube is concentric with a larger tube of 6 cm diameter (D₂) and inner surface is having emissivity (=CE=B5₂) of 0.06 and temperature of (T₂) = 350 K. Find the reduction in radiation heat transfer if a thin radiation sh= ield of 45 mm diameter (D₃) and emissivity (=CE= =B5₃) of 0.012 on both sides is inserted between= the tubes. Assume the space between the surface as vacuum.

Radiation heat transfer = per unit length without shield

Radiation heat transfer per = unit length with shield,

and

Heat transfer,

Reduction in heat transf= er from outer cylinder to inside cylinder is (1.95 =E2=80=93 0.528) =3D 1.4= 2 W/m.

9.27. A solar flat plate collector inclined at (=CE=B8) 45=C2=B0 to th= e horizontal and facing south. The air temperature (T_g=) is 295 K and the glass and absorber plate temperatures are (T_g) 325 K and (T_g)345 K= , respectively. A 6 cm thick insulation cover is provided at the bottom= of the system. The wind velocity (v) is 5 m/s and air space between the pl= ate and glass cover is (L) 8 cm. The glass and absorber plate emissivities = are (=CE=B5_g) 0.88 and (=CE=B5_p) 0.95, respectively. Find the overall heat transfer coe= fficient between the flat plate and ambient air.

Grashof number,

Rayleigh number,

R_a=3D Gr. Pr =3D 838280=C3=970.695 =3D 5.8 =C3=9710⁵

Nusselt number,

The convection heat tran= sfer coefficient between plate and cover,

The heat transfer coeffi= cient between cover and atmospheric air,

h_ca<= /em> =3D 2.8 + 3.0(v)

      = ;            &n= bsp;    =3D 2.8 + 3.0=C3=975 =3D 17.8 W/m²K

The effective emissivity= between plate and glass,

The radiation heat trans= fer coefficient between plate and cover,

T= he radiative heat transfer coefficient between glass cover and sky,

where T_s is the sky temperature =3D T_a= =E2=80=93 6 =3D 295 =E2=80=93 6 =3D 289 K.

The total heat transfer = coefficient between plate and cover,

h_{1 =3D h_pc + h _rpc =3D 2+7}

    =3D 9 w/m2k

The total heat transfer = coefficient between cover and ambient air,

h₂ =3D h_ca+ h_rgs=

      = ;   =3D 17.8 + 6.82

      = ;      =3D 24.62 W/(m²k= )

The overall heat transfe= r coefficient between plate and air,

9.28. Molten metal= droplets are ejected in air to cool down during arc welding of metal. Drop= let of 2 mm diameter (D), initial temperature (T_w) of 1100=C2=B0C, and an initial velocity (U) of 50 cm/s= , is cooled in air at 30=C2=B0C (T_{=CE=B1). Find its initial rate of cooling in air}, assuming =CF=81<= em>_mettal =3D 2700 kg/m^{3 and C =3D 1.1 kJ/kg K}, =CE=B5 =3D 0.4, and k =3D 200 = W/mK.

Air properties at 30=C2= =B0C are:

k =3D 26.75 =C3= =97 10^=E2=80=933 W/mK =CE=B3 =3D 16.0 =C3=9710=E2=88=926 m²/s; Pr= =3D 0.70; =CE=BC_=CE=B1=3D 18.6 = =C3=97 10^{=E2=88=92=E2=80=936} NS/m²; =CE=BC_m =3D 53.45 =C3=97= 10^=E2=80=936 NS/ m² = at temperature 1100=C2=B0C

Reynolds number,

Nusselt number,

Heat transfer by convection,=

qc =3D h(Tw =E2=80=93 T_=CE=
=B1) =3D 197.13(1100 =E2=80=93 30) =3D 210937 W/m²

Heat transfer by radiation,<= /p>

Biot number,

Using lumped thermal cap= acity mode (as Bi < 0.1),

REVIEW QUESTIONS

Distinguish thermal radiation fro= m other types of radiation.

Explain the physical mechanism of radi= ation heat transfer.

Explain the radiation properties.

Define Stefan=E2=80=93Boltzmann law.
Define radiation intensity.

What is the difference between a black= body and a grey body?

What is radiation shape factor

Discuss the radiation heat transfer be= tween non-black bodies.

Derive the expression for radiation he= at transfer for infinite parallel surfaces.

What is a radiation shield?

Discuss gas radiation.

Explain the radiation exchange wit= h transmitting, reflecting and absorbing media.

OBJECTIVE QUESTIONS

Radiation heat transfer takes pla= ce

through electromagnetic waves
in vacuum

when there is a temperature difference=

all of the above

The speed of electromagnetic waves is = the product of their wavelength and

frequency

emissivity

distance

intensity

The value of the Stefan=E2=80=93Boltzm= ann's constant is =C3=97 10^=E2=80=938

6.599

5.669

5.996

5.966

The ratio of the emissive power of a b= ody to the emissive power of a black body at the same temperature is to the= absorptivity of the body

not related

not proportional

equal

lower

Two parallel black plates of size 5 m = =C3=971 m are maintained at 1000=C2=B0C and 500=C2=B0C with a shape factor = of 0.285. The net heat exchange between them is

13.83

18.88

18.18

18.33

Radiation shields are used to

reduce radiation heat transfer be= tween two surfaces

increase heat transfer

maintain the same heat transfer

all of the above

In radiation exchange between a gas an= d a surface

mostly gases are transparent to r= adiation

gases absorb and emit radiation in nar= row wavelength bands

a more complex heat transfer situation= is seen

all of the above

The total emissive power (kW/m²) of a black body at 100 K is

5669

56.99

569.9

5.699

The intensity of solar radiation is st= rongly dependent on

the atmospheric condition

the time of the year

the angle of incidence

all of the above

According to Wien's displacement law, = the equivalent black body temperature for solar radiation at a wavelength o= f 0.5 =CE=BCm is

5795 K

5975 K

5579 K

5579 K

Answers:

d

a

b

c

d

a

d

d

d

a

EXERCISE PROBLEMS

Temperature of a surface is maint= ained at 632.5=C2=B0C. What is the wavelength at which the maximum radiatio= n will be emitted from the surface? (Ans. 3.19 m)

A body receives total incident radiati= on of 5 kW/m². If the body absorbs 3 kW/m² and reflects 1kW/m², find = the transmissivity. (Ans. 0.2)

If a black body is maintained at 1000= =C2=B0C, what is its total emissive power? (Ans. 148.5 kW/= m²)

What is the maximum monochromatic emis= sive power and the wavelength at which it occurs for a black body maintaine= d at 110 K. (Ans. 112.5 W/m², 7= .49 m)

Two parallel infinite black plates are= maintained at 200=C2=B0C and 300=C2=B0C, what will be the net heat transfe= r rate between the plates per unit area. (Ans. 3.29 kW)
If two parallel plates having equal em= issivity of 0.295 are separated by insulation material that is transparent = to radiation, find the heat transfer between them for unit area. They are m= aintained at 230=C2=B0C and 28=C2=B0C. (Ans. 560 kW)

A circular plate of diameter 2.98 m is= exposed to radiation escaping from a furnace through an opening of area 0.= 099 m². The plate is kept parallel and coaxial w= ith the opening and is placed 4.99 m away from it an= d the opening radiates as a black body at 1815 K. Find the rate of energy t= ransfer from the opening to the plate. (Ans. 5 kW)

A heater of overall dimensions 1.95 m = =C3=97 1 m =C3=97 0.198 m is used to heat the air in a room maintained at 2= 90 K. The surface temperature of the heater is at 335 K and its surface is = can be assumed to be black. The actual surface area of the radiator is 2.5 = times the area of its envelope and the convection heat transfer coefficient= is h =3D 0.00130(=CE=B8/[k])^1/3kW/m2K. Find the rate of heat transfer from the radiato= r to the room air. (Ans. 3.99 kW)

A 100 W electric bulb's filament is as= sumed to be a black body which is maintained at 3275 K. The length of the f= ilament is 249.5 mm. Find the efficiency of the bulb if visible radiation l= ies in the range of wavelengths from 0.5 u to 0.8 =CE=BC (Ans. 16%)

If the Sun is assumed to be a black bo= dy with a surface temperature of 5525=C2=B0C, calculate the amount of radia= nt energy emitted by it per unit time if its diameter can be assumed to be = 1.4 =C3=97 10⁹m. (Ans. 4.4 =C3= =97 10²⁶ W)

The diameter and length of a hot water= tank are 49.5 cm and 99.5 cm, respectively. It is exposed to ambient air a= t 298 K. If the emissivity of the surface is 0.795, find the heat loss from= the tank by radiation. Also find the reduction in heat loss if a coating o= f aluminium paint having an emissivity of 0.295 is given to the tank. (Ans. 690 W, 430 W)

A cylindrical furnace with inner diame= ter 15.95 cm and length 1 m is used to heat a steel rod of length 1 m and d= iameter 19.5 mm from 700 K to 810 K. The inner surface of the furnace is at= a temperature of 1365 K and has an emissivity of 0.849. If the surface of = the rod has an emissivity of 0.599, find the time required for the heating = operation. (Ans. 4 s)

The size of room is 1.495 m =C3=97 1.6= 95 m =C3=97 1.995 m and the walls and ceiling are maintained at 525 K and t= he floor at 405 K. The emissivity of walls and ceiling are 0.819 and that o= f the floor is 0.695. Find the net radiation to the floor. (Ans. 4.9 kW)

Two parallel planes of very large size= with emissivities 0.295 and 0.795 exchange heat. Find the percentage reduc= tion in heat transfer if a shield with emissivity 0.0395 is placed between = them. (Ans. 94%)

If the temperature on the inner surfac= e of a hemispherical cavity of diameter 49.5 cm is 675 K which has an emiss= ivity of 0.595, find the rate of heat loss from the cavity. (Ans. 1.4 kW)

The diameter of a combustion chamber i= s 34.95 cm and its inner wall temperature is maintained at 775 K. The gas i= nside is at 1275 K at a pressure of 1 atm and contains 11.95% CO₂ and 9.95% H₂O by volume. Find= the net heat transfer per unit surface area. (Ans. 15 kW)=

A furnace flame is assumed to be a pla= ne rectangular black surface, 2.99 m by 1.99 m, and is kept at 1775 K, is e= xposed to an opaque bank of black tubes of the same projected area. These a= re at a surface temperature of 575 K and the shape factor between them is 0= .495. Find the net heat flow to the tube bank. (Ans. 1660 = kW)

Estimate the shape factor of a cylindr= ical cavity of diameter 19.95 cm and depth 49.5 cm with respect to itself. = If its temperature is 325=C2=B0C and emissivity is 0.795, find the rate at = which heat comes out from the cavity. (Ans. 0.9, 18 W)
Find the net rate of heat exchange fro= m at 9.99-cm diameter disc at 799 K, having an emissivity of 0.8, placed at= a distance of 1.99 m from a disc with emissivity 0.695 and 1.99 m diameter= maintained at 299 K. (Ans. 34 W)

A cylindrical rod (=CE=B5 =3D 0.7)= of 50 mm diameter is maintained at 1000=C2=B0C by electrical resistance he= ating and kept in a room, the walls (=CE=B5 =3D 0.6) of which are at 15=C2= =B0C. Determine the energy which must be supplied per metre length of the r= od. If an insulated half-circular reflector of 0.45 m is placed around the = rod, estimate the energy supplied to the rod per metre length. (Ans= . 15 kW)

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Surfaces	Temperature (k)	Emissivity ()
Highly polished aluminum	500=E2=80=93850	0.04=E2=80=930.06
Highly polished brass	1160=E2=80=931525	0.027=E2=80=930.031
Polished copper	350	0.02
Steel (highly polished)	450=E2=80=93500	0.05=E2=80=930.075
Cast iron (polished)	475	0.20
Polished steel casting	1040=E2=80=931310	0.50=E2=80=930.56
cast iron (finish)	1150=E2=80=931260	0.60=E2=80=930.70
Iron plate (rusted red)	295	0.61
Molten cast iron	1575=E2=80=931675	0.29
Molten mild steel	1875=E2=80=932075	0.28
Lead	400=E2=80=93500	0.056=E2=80=930.075
Nickel (polished)	295	0.045
Nickel alloys	325=E2=80=931310	0.65=E2=80=930.75
Platinum (polished)	500=E2=80=93900	0.055=E2=80=930.105
Silver (polished)	500=E2=80=93900	0.02=E2=80=930.032
Tin (bright)	300	0.040=E2=80=930.065
Tungsten	300=E2=80=933590	0.030=E2=80=930.35
Filament	3590	0.395
Zinc	500=E2=80=93600	0.045=E2=80=930.055

SurfacesAsbestos	Temperature (k) 300	Emissivity () 0.965
Asbestos	300	0.935
Brick	295	0.935
Enamel (white)	290	0.895
Oil layers on polished nickel	295	0.045
Oil layers on aluminum foil	375	0.085
white enamel varnish on rough iron plate=	295	0.91
Black sprayed on iron	300	0.875
Oil paints	375	0.92=E2=80=930.96
paper	290	0.92
porcelain	295	0.92
Refractory materials	875=E2=80=931275	0.65=E2=80=930.75
Rubber	295	0.94
water	275=E2=80=93375	0.95=E2=80=930.96

=CE=BBT (mk =C3=97 10³)	f_0-_=CE=BB
1.0	0.321 =C3=97 10^-3
2.0	0.667 =C3=97 10^-1
3.0	0.273
4.0	0.480
5.0	0.634
6.0	0.738
7.0	0.808
8.0	0.856
9.0	0.890
10.0	0.914
15.0	0.969
20.0	0.986
40.0	0.998
50.0	0.999
100.0	1.000

9.10 PROPERTIES OF SHAPE FACTOR

9.11 SHAPE FACTOR ALGEBRA

9.12 ELECTRICAL ANALOGY FOR THERMAL RADIATION

9.12.1= Two Infinite Parallel Plates

9.12.2= concentric cylinders or spheres

9.12.3= Small Body Lies Inside a Large Enclosure

9.13 RADIATION SHIELDS

9.14 GAS RADIATION

9.15 CONCLUSIONS

KEY CONCEPTS

EXAMPLE PROBLEMS

REVIEW QUESTIONS

OBJECTIVE QUESTIONS

Answers:

EXERCISE PROBLEMS

Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20

9

Radiation Heat Transfer=

KEY CONCEPTS

= 9.1 EMISSIVE POWER

= 9.2PLANCK'S LAW

KEY CONCEPTS

= 9.3 WIEN'S DISPLACEMENT LAW

= 9.4 STEFAN-BOLTZMAN LAW

= 9.5 KIRCHHOFF'S LAW

= 9.6 BLACK BODY RADIATION FUNCTION

= 9.7 INTENSITY OF RADIATION (I)

KEY CONCEPTS

= 9.8 LAMBERT'S COSINE LAW

= 9.9 RADIATION SHAPE FACTOR (F)

Table= of Contents for =20 Heat and Mass Transfer, 2nd Edition =20

9

Radiation Heat Transfer=

KEY CONCEPTS

= 9.1 EMISSIVE POWER

= 9.2PLANCK'S LAW

KEY CONCEPTS

= 9.3 WIEN'S DISPLACEMENT LAW

= 9.4 STEFAN-BOLTZMAN LAW

= 9.5 KIRCHHOFF'S LAW

= 9.6 BLACK BODY RADIATION FUNCTION

= 9.7 INTENSITY OF RADIATION (I)

KEY CONCEPTS

= 9.8 LAMBERT'S COSINE LAW

= 9.9 RADIATION SHAPE FACTOR (F)

9.10 PROPERTIES OF SHAPE FACTOR

9.11 SHAPE FACTOR ALGEBRA

9.12 ELECTRICAL ANALOGY FOR THERMAL RADIATION

9.12.1= Two Infinite Parallel Plates

9.12.2= concentric cylinders or spheres

9.12.3= Small Body Lies Inside a Large Enclosure

9.13 RADIATION SHIELDS

9.14 GAS RADIATION

9.15 CONCLUSIONS

KEY CONCEPTS

EXAMPLE PROBLEMS

REVIEW QUESTIONS

OBJECTIVE QUESTIONS

Answers:

EXERCISE PROBLEMS

Highlight

Highlight

Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20