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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

User Karey
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1 Answer

6 votes

Answer:


J=1963W/m^2


q_(rad)=963w/m^2


\triangle T= -0.378k/s

Step-by-step explanation:

From the question we are told that:


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

at Temp=400K


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

Generally the equation for Radiosity is mathematically given by


J=eG+\in E_p


J=(1-\alpha)G+\in \sigma T^4


J=(1-0.14)1000+0.76 (5.67*10^_(8)) (400)^4


J=1963W/m^2

Generally the equation for net radiation heat flux
q_(rad) is mathematically given by


q_(rad)=J-G\\q_(rad)=1963-1000


q_(rad)=963w/m^2

Generally the equation for and the rate of plate temp
\triangleT is mathematically given by


\triangle T= -(q_(con) +q_(rad))/(Ecl)


\triangle T= (45(400)-(30+273+963))/((2702*949*0.005))


\triangle T= -0.378k/s

User Omar Freewan
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