8. BASIC CONCEPTS OF RADIATION HEAT TRANSFER
8.3 R ADIATION EXCHANGE BETWEEN SURFACES
8.3.1 View factors
Radiation heat flux from one surface to another is often of interest when radiation heat transfer in a heat treatment machine is outlined. The heat flux depends on the shapes and sizes of surfaces and distances between them. In other words, it depends on how surfaces see each other. View factors are used to take this view into account. The view factor is the fraction of radiant energy leaving one surface that is incident upon another surface. Heat flux from a black surface 1 to a black surface 2 can be written as
1 4 1 2 1 2
1
F T A
Q
(8.13)where F1-2 is the view factor from a surface 1 to a surface 2. Next, two basic rules are useful, when view factors between surfaces are solved
j i j i j
i A F A
F (8.14)
1
1
n
j j
Fi (8.15)
Let us consider radiation heat transfer in a tempering furnace. In Figure 8.3 a glass is under heating resistors (see Figure 5.1 b) in a tempering furnace. If the length and width of the furnace and glass are large, then the effect of sides can be ignored.
Figure 8.3. Schematic of glass under resistors in tempering furnace.
The view factor from glass to resistors F2-1 is 1, because the glass surface cannot see itself (F2-2 = 0).
Accordingly, Eqs. (8.14) and (8.15) give F1-2 = F2-1(A2/A1) and F1-1 = 1-F1-2.
Figure 8.4 shows a glass on rollers in a tempering furnace. The view factor F2-1 from rollers to glass is the difference between the view factor from the roller to an infinite glass plate and screening caused by other rollers. Screening is a half of the view from a roller to other rollers. Then, the view factor F2-1 is [39]
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X
X X
F 1
sin 1 1
2
1 2 0,5 1
1
2 (8.16)
In Eq. (8.16) X = 1+B/(2R).
Figure 8.4. Glass on rollers in tempering furnace.
In the cases above the geometry was assumed to be infinite at least in one direction, which makes the solution of radiation heat transfer between surfaces easier. A lot of similar view factors can be written for the surfaces in a tempering furnace. These kinds of view factors are suitable only for solving the mean radiation between the bodies with uniform surface temperatures.
8.3.2 Radiation heat flux from tubular resistor to glass
A typical heating resistor consists of a resistance wire, which is spooled around a ceramic tube. Such a shape can be simplified as a long cylindrical tube shown in Figure 8.5.
Figure 8.5. Angles γ and β between resistor and point p in plane.
By using the angles and coordinates defined in Figure 8.5 the projected area of a resistor viewed from point p can be written as
R d Rd
dA
s cos
(8.17)and then Eq. (8.1) gives solid angle
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d ddcos (8.18)
According to [62] the polar angle θ defined in Figure 8.1 can be written by means of angles and as
cos cos
cos
(8.19)The radiation emitted by the resistor in Figure 8.5 to a point p on the plane surface can be solved by inserting Eqs. (8.18) and (8.19) in Eq. (8.5). Thus, the incoming radiative heat flux density in point p is
12
1
1
cos cos
cos
2
d d
i
q
p r (8.20)In Eq. (8.20)
i
r is the emitted intensity of the resistor and α is the angle between normal of the plane and line S1 in point p as shown in Figure 8.6. Line S1 is the distance between resistor mid-point and point p viewed from z-direction. In Eq. (8.20) cosα takes into account that the projection of the elemental area dA1 on plane surface toward the resistor is cosα -times smaller than the elemental area itself and thereby the incoming flux is equally smaller. The integration limits of angles and are shown in Figure 8.7 and Figure 8.8. The vertical distance S2 between midline of the resistor and its projection to point p is defined in Figure 8.8.Figure 8.6. Angle between normal of plane and line S1 in point p.
Figure 8.7. Integration limits of angle .
50 Figure 8.8. Integration limits of angle .
The integration of Eq. (8.20) yields to
1
Eq. (8.21) is ready to be used for practical problems because it includes three typical dimensions for a resistor above the plane geometry: resistor diameter d, resistor length L and vertical distance between the resistor and plane Yv. Coordinates x and z define the point on the plane surface, where the radiative heat flux density is to be solved.
The intensity
i
r needed in Eq. (8.21) iswhen a resistor is approximated as a blackbody. In Eq. (8.22) qr is outgoing heat flux density from the resistor. Usually the temperature Tr of a resistor is not known, but the electric power Qr,e of a resistor is. In thermal balance the total outgoing heat flux Qr from a resistor is the sum of the electric power of the resistor Qr,e and incoming heat flux to resistor from the surroundings.
dL
In the case above only one half of the heating power of a resistor hits the plane (glass) under it, while another half is radiated in an opposite direction and hits a furnace ceiling. The ceiling is made from insulation material, which radiates incoming radiation back to a resistor and plane under it.
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Figure 8.9. Cross section of resistors built in groove-shaped shade. [109]
Figure 8.9 shows a geometry in which resistors are located in groove-shaped shades. Such resistors are used when more focused radiation is needed, for instance in windscreen bending furnaces. The shade cuts the visibility from the plane to the resistor. The equations above are suitable for the case if the shading is considered in integration limits of angle . The emission upwards from the resistor hits the shade, which reflects and emits radiation back to the resistor and directly to the plane under it. Thus, similar equations as above for the resistor tube must be written for the shade also. Again, the temperatures of the surfaces are not known, and difficulties arise when they or outgoing heat fluxes from the resistor and shade are to be solved. The exact solution to such a multi-phased radiation heat transfer problem is complex. It can be solved by using the following assumptions and methods: the crosscut of the shade is semicircle-shaped, the surfaces are diffuse and grey, the temperatures of the resistor and shade are constant over the surface, and the radiative equilibrium temperatures for both surfaces are solved by using the net-radiation method presented in [61] page 391. Some results are shown later in Sec. 14.1.
Direct radiation heat flux from a tubular resistor to a plane underneath is a clear problem to be solved with self-written equations, but non-direct radiation via the ceiling or a groove-shaped shade is a much more complicated problem. A competent way to solve the problem above is to use commercial numerical modelling codes.
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