Many situations are encountered where it is desired to reduce the overall heat transfer between two radiating surfaces. The task is accomplished by placing radiation shields between the emitting surfaces.
The shields are thin opaque partitions arranged in the direction perpendicular to the propagation of radiated heat, and made of materials of very low absorptivity and high reflectivity (thin sheets of aluminium, copper etc.) The shields introduce a sort of additional resistance in the heat flow path and accordingly the net heat flux is reduced.
(i) With no radiation shields, the net heat exchange between the infinite parallel planes is given by-
Q12 = (Fg)12 A1 σb (T14 – T24)
ADVERTISEMENTS:
For the given configuration of parallel planes which see each other and nothing else,
(ii) The placement of a radiation shield between these two planes would neither remove nor add any heat to the system. Under steady state conditions, the screens attain a uniform temperature of t3, and heat transfer between plane 1 and the shield (Q13)net .
Both faces of the radiation shield have been assumed to have the same emissivity.
ADVERTISEMENTS:
Simplification of expression 8.29 yields:
Each side of expression 8.29 represents the heat flow through the system. Substituting the value of I3 in left hand side of expression 8.29, we obtain –
The ratio of radiant energy transfer with one shield to that without any shield is obtained from expressions 8.28 and 8.31. That gives,
Radiant energy transfer –
When ϵ1 = ϵ2 = ϵ3, the above fraction takes the value 1/2. Thus by inserting one shield, the heat transfer rate is reduced to one-half of the original value.
ADVERTISEMENTS:
The corresponding temperature T3 of the shield attains the value:
Heat exchange without any shield
Comparison of expressions (a) and (b) also shows that the heat flow ratio with a radiation shield becomes just half of what it would have been without the radiation shield.
ADVERTISEMENTS:
If n-radiation shields are inserted between the two planes, then –
(i) There will be two surface resistances for each radiation shield, and one for each radiating plane. When the emissivity of all the surfaces are equal, then all the (2n + 2) surface resistances will have the same value (1 – ϵ)/ϵ.
(ii) There would be (n + 1) space resistance and the configuration factor for each will be unity.
Obviously the total resistance of the physical system will be:
A comparison of expression (a) and (c) does indicate that the presence of n-shields reduces the radiant heat transfer by a factor of (n + 1).
Example:
Two parallel square plates, each 4 m2 area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and a surface emissivity of 0.9. Find the net energy exchange by radiation between the plates.
If a thin polished metal sheet of surface emissivity 0.1 on both sides is now located centrally between the two plates, what will be its steady state temperature? How the heat transfer would be altered? Neglect the convection and edge effects if any. Comment upon the significance of this exercise.
Solution:
The rate of heat interchange between the two plates is given by:
Therefore equilibrium temperature T3 of the shield is 671.65°K
Any one of the expressions (a) and (b) can now be used to work-out the heat interchange between the plates-
... Heat flow through the system,
= 0.09374 × 4 × (5.67 × 10-8) × (8004 – 671.654)
= 4381.66 Watts or 4.382 kW
The placement of a radiation shield reduces the radiant heat transfer by a factor of 51.176/4.382 = 11.678 times. Screens are thus placed between surfaces to cut down the radiation loss in heat insulation structure members. For example, the bulb of a thermometer or a thermocouple junction is often shielded in order to reduce radiation effects to a minimum in the act of temperature measurement of a fluid.