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

`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`