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One side of metallic plate absorbs a heat flux of 1000 $\frac{W}{m^2}$. Its other side is insulated.The emissivity of surface is 0.8 and convective heat transfer coefficient is 20 $\frac{W}{m^2K}$

Ambient is maintained at 300K. Determine the temperature of plate under steady state conditions σ=5.67×$10^(-8)$

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$q_r$=1000w/$m^2$ ,$h_c$=20W⁄($m^2$ K,ε=0.8,$T_∞$=27℃=300K)

Analysis: The radiant heat flux absorbed by the plate,will be dissipated by convection and radiation.

Thus,

$q_r$=$h_c$ ($T_s-T_∞$ )+εσ($T_s^4-T_∞^4$ )

Using the numerical values,

1000=20($T_s-300$)+0.8×5.67×$10^(-8)$ ($T_s^4-300^4$ )

$T_s$=338.57K

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