Question

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

Answer 1

Answer:

$J=1963W/m^2$

$q_{rad}=963w/m^2$

$\triangle T= -0.378k/s$

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

$\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}$

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

$\triangle T= -0.378k/s$