Question

A small cylindrical filament is heated in a large combination convection-radiation oven. The filament has a diameter \(\mathrm{D}=50 \mathrm{~mm}\) and length \(\mathrm{L}=1 \mathrm{~m}\). During the process, the oven wall temperature is fixed at \(\mathrm{T}_{\text {sur }}=800 \mathrm{~K}\), and the hot air is blowing with a temperature \(\mathrm{T}_{\infty}=500 \mathrm{~K}\) and an average heat transfer coefficient \(\bar{h}=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Assuming the filament surface is opaque and diffuse, and it has a spectral emissivity as shown below: (25 pts)

The oven walls behave like a black surface, and the filament end effects are negligible. If initially the surface temperature of the filament is \(\mathrm{T}_{\mathrm{s}}=300 \mathrm{~K}\),

(a) Determine the total, hemispherical emissivity \(\varepsilon\) and absorptivity α of the surface of filament.

(b) Use energy balance on the filament and determine the net rate of heat transfer to the filament.

Answer 1

given data: oven temp, Tswr=800k fliament Dia, D=50mm Dc0-05 Au temp, Too = 500k heat transfer coetticient, 50을 mok filament length, L: Surface temp TS = 300 K. Assumption: Cylindrical compared with walls Ουιρ verly large CO) Oven filament (11) Cylider i opaque and diffuse depends on temperature. caii) Spectral emissivity Solution: -) Schematic Diagram: don Tour & Eb & conv Ufo To n a) Total hemispherical emissivity: The total hemispherical emissivity of the diffuse, Spectral Surface • given by, Oo EXEX,Ts) Exib r Ist AT= 4 x 300 AT: 1200 um-k O &- & F60AM) +62 (1- Frozum)

For black body radiation functions, Heat and mass tranfer" by Refer textbook Incopora table :12.1 ( Sixth edition) page Number : 740, FC0-1200'4m-le) = 0.002 & 0.8 %0.002 + 0.2 (1-0.002) & = 0.201 =) Absorptivity (X) which, cradiation from oven wall The absoptivity for behave as Isothermal Duro they large and because at 800k. black Surface BOOK 4 XT=0xx Crys Exib c d 1000k) XT= 2000 lm-k dx = {x Fro2000 Umk) = 0.0667 d=& Fco->4km) +&2 C1- F10 44m) -0.8 x 0.0667 + 0-2 (1-0.0667) x=0-24 Net heat transfer: 9 net = (x + ñ (Too-Ts) – 8.0.134) x Area Goo Tscount Area, A TTDL TX 0.05 X 1 A = 0.157m2

-8 x 3004) 40.157 Inet = ((0.24 x 5* 0.24 x 5067X17*x 8004) + 50x ( 500-300) -(0-20185:67x17 ton")]xo - [5573.83 92.313 :5573.836 + 10,000 + Prikritos X0:157 qnet 24 30.59 watts Result a) = 0 201 و d こ 0.24 b) & net 2430.59 watts