View factors numericals

Basics for view factor calculation

1. View factor numericals By; k.Vinitha M.Tech 2018694619

2. View factor can be estimated by Inspection Reciprocating relation Summation rule T he energy leaving surface 1 and arriving at surface 2 is E b1 A 1 F 12  and the energy leaving surface 2 and arriving at surface 1 is E b2 A 2 F 21 . All the incident radiation will be absorbed by the blackbody and the net energy exchange will be, Q = E b1 A 1 F 12  - E b2 A 2 F 21 At thermal equilibrium between the surfaces Q 12  = 0 and E b1  = E b2 , thus A 1 F 12  - A 2 F 21 A i F ij  - A j F ji

3. , consider a flat plate (for eg. ) which is emitting the radiation, it can be understood that the radiation of the flat plat cannot fall on its own surface (partly or fully). Such kind or surfaces are termed as “not able to see itself”. F 11  = F 22  = F 33  = F 44  = 0 if the surface can see itself like concave curved surfaces, which may thus see themselves and then the shape factor will not be zero in those cases. The property of the shape factor is that when the surface is enclosed, then the following relation holds

4. 1)Determine the view factors F 12 and F 21 for the following geometries : F 12 = 1 Partition within a square duct: From summation rule, F 11 + F 12 + F 13 = 1, where F 11 = 0 By symmetry F 12 = F 13 F13=0.17 From summation rule, F 11 + F 12 + F 13 = 1 with F 11 = 0, F 12 = 1 - F 13 = 0.83

5. 2)Determine the view factors associated with an enclosure formed by two spheres N= 2 , four view factors , which are F 11, F 12, F 21, and F 22 .   F 11 = 0, since no radiation leaving surface 1 strikes itself F 12 =1, since all radiation leaving surface 1 strikes surface 2 by applying the reciprocity relation to surfaces 1 and 2: the view factor F 22 is determined by applying the summation rule to surface 2:

6. The Superposition Rule: The view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j.

7. 3 ) Determine the fraction of the radiation leaving the base of the cylindrical enclosure that escapes through a coaxial ring opening at its top surface. The radius and the length of the enclosure are r 1 10 cm and L 10 cm, while the inner and outer radii of the ring are r 2 5 cm and r 3 8 cm, respectively. Using the superposition rule, the view factor from surface 1 to surface 3 can be expressed as The view factors F 1 - 2 and F 1 - 3 are determined from the chart

8. 4) Find the view factor from the base of a pyramid to each of its four sides. The base is a square and its side surfaces are isosceles triangles. From symmetry rule, the summation rule yields: Two (or more) surfaces that possess symmetry about a third surface will have identical view factors from that surface. The Symmetry Rule

9. View Factors Associated with a Triangular Duct 5) Determine the view factor from any one side to any other side of the infinitely long triangular duct The duct is infinitely long, the fraction of radiation leaving any surface that escapes through the ends of the duct is negligible. Therefore, the infinitely long duct can be considered to be a three-surface enclosure, N = 3. = =9   The remaining six view factors can be determined by the application of the summation and reciprocity rules.

10. The Crossed-Strings Method Geometries such as channels and ducts that are very long in one direction can be considered two-dimensional (since radiation through end surfaces can be neglected). The Crossed-Strings Method

11. 6) Two infinitely long parallel plates of widths a 12 cm and b 5 cm are located a distance c 6 cm apart,. ( a ) Determine the view factor F 1 - 2 from surface 1 to surface 2 by using the crossed-strings method. ( b ) Derive the crossed-strings formula by forming triangles on the given geometry. Then applying the summation rule to surface 1 yields

12. RADIATION HEAT TRANSFER BLACK SURFACES: The net rate of radiation heat transfer from surface 1 to surface 2 can be expressed as 7) Consider the 5-m 5-m 5-m cubical furnace, whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and 500 K, respectively. Determine ( a ) the net rate of radiation heat transfer between the base and the side surfaces, ( b ) the net rate of radiation heat transfer between the base and the top surface, and ( c ) the net radiation heat transfer from the base surface.   The net rate of radiation heat transfer Q 1 - 3 from surface 1 to surface 3 can be determined

13. Radiation Heat Transfer between Parallel Plates : 8) Two very large parallel plates are maintained at uniform temperatures T 1 = 800 K and T 2 = 500 K and have emissivities 0.2 and 0.7, respectively,. Determine the net rate of radiation heat transfer between the two surfaces per unit surface area of the plates.

14. Radiation Heat Transfer in a Cylindrical Furnace: 9) Consider a cylindrical furnace with r = H = 1 m, The top (surface 1) and the base (surface 2) of the furnace has emissivities 0.8 and 2 0.4, respectively, and are maintained at uniform temperatures T 1 700 K and T 2 500 K. The side surface closely approximates a blackbody and is maintained at a temperature of T 3 400 K. Determine the net rate of radiation heat transfer at each surface during steady operation and explain how these surfaces can be maintained at specified temperatures.

15. To maintain the surfaces at the specified temperatures, we must supply heat to the top surface continuously at a rate of 27.6 kW while removing 2.13 kW from the base and 25.5 kW from the side surfaces.