The top surface of an L = 5mmthick anodized aluminum plate is irradiated with G = 1000 W/m2 while being simultaneously exposed to convection conditions characterized by h = 50 W/m2 ⋅ K and T[infinity] = 25°C. The bottom surface of the plate is insulated. For a plate temperature of 400 K as well as α = 0.14 and ε = 0.76, determine the radiosity at the top plate surface, the net radiation heat flux at the top surface, and the rate at whic

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okpalawalter8 okpalawalter8 · Answer 1 · 2021-04-27T04:13:01+02:00

Answer:

[tex]J=1963W/m^2[/tex]

[tex]q_{rad}=963w/m^2[/tex]

[tex]\triangle T= -0.378k/s[/tex]

Explanation:

From the question we are told that:

[tex]L=5mm => 5*10^{-3}\\Irradiation G=1000W/m^2\\h=50W/m^2\\T_{infinity} = 25C.\\Plate\ temperature\ T_p= 400 K\\\alpha=0.14\\E=0.76[/tex]

at Temp=400K

[tex]E=2702kg/m^2,c=949J/kgk[/tex]

Generally the equation for Radiosity is mathematically given by

[tex]J=eG+\in E_p[/tex]

[tex]J=(1-\alpha)G+\in \sigma T^4[/tex]

[tex]J=(1-0.14)1000+0.76 (5.67*10^_{8}) (400)^4[/tex]

[tex]J=1963W/m^2[/tex]

Generally the equation for net radiation heat flux [tex]q_{rad}[/tex] is mathematically given by

[tex]q_{rad}=J-G\\q_{rad}=1963-1000[/tex]

[tex]q_{rad}=963w/m^2[/tex]

Generally the equation for and the rate of plate temp [tex]\triangleT[/tex] is mathematically given by

[tex]\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}[/tex]

[tex]\triangle T= \frac{45(400)-(30+273+963)}{(2702*949*0.005)}[/tex]

[tex]\triangle T= -0.378k/s[/tex]