Question

Consider a long duct constructed with diffuse and gray walls. The width of each wall is 1 m. Heat transfer is from surface 1
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Answer #1

Solution:

Here,

Assumptions:

Surfaces are diffuse grey

Duct is very long

End effects are negligible

Therefore,

View factor 4 on 4, F2=0.5 For symmetry Fir = Fur = F2 Areas 4 =wl 4 = wl The net radiation per unit length gi 91 - 2 L 0(71*For radiation balance 9x = 9; = 0 F3 - Jy+ F» -J2 = 0 4F31 1 43F32 2E,9 = J, +J Ja+J₂ Es 2 Here, 91 (1-&) Jy = E51 - 4 & (1)From equation (1) En = 36651+23489 2 Ena = 30070 OT;* = 30070 30070 5.67x10-5 T = 853K

Since the third wall is adiabatic or completely insulated so there is no effect of change of emissivity.

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