Tham khảo tài liệu 'a heat transfer textbook - third edition episode 3 part 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Radiant heat exchange between two finite black bodies 539 Of course Qneti-2 _Qnet2-i- It follows that A1F1-2 a j4 - t2 -A2F2-1 a T4 - T14 or A1F1-2 A2F2-1 This result called view factor reciprocity is very useful in calculations. Example A jet of liquid metal at 2000 C pours from a crucible. It is 3 mm in diameter. A long cylindrical radiation shield 5 cm diameter surrounds the jet through an angle of 330 but there is a 30 slit in it. The jet and the shield radiate as blackbodies. They sit in a room at 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. By inspection we see that Fjet-room 30 360 and Fjet-shield 330 360 . Thus Qnetjet-room AjetFjet-room T4. Tr4oom n m2 m length X 10-8 22734 - 3034 1 188 W m Likewise Qnetjet-shield AjetFjet-shield Ơ j - Tshield 21 X 10-8 Í22734 - 9734 m length t 12 637 W m The heat absorbed by the shield leaves it by radiation and convection to the room. A balance of these effects can be used to calculate the shield temperature given here. To find the radiation from the inside of the shield to the room we need Fshield-room. Since any radiation passing out of the slit goes to the 540 Radiative heat transfer room we can find this view factor equating view factors to the room with view factors to the slit. The slit sareais Aslit n 30 360 m2 m length. Hence using our reciprocity and summation rules eqns. and Flit-jet 4 Fjet-room n 3 0-0833 Aglit Fslit-shield 1 - Fslit-jet - Flit-slit 1 - - 0 S0 A slit Fshield-room . Fslit-shield Ashield n 330 360 Hence for heat transfer from the inside of the shield only Qnetshield-room AshieldFshield-room Tshield Troom n 330 _ 360 _ X 10-8 9734 - 3034 619 W m Both the jet and the .
