Answer:
[tex]G=6750\ W.m^2[/tex] is the irradiation
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
Explanation:
Given that:
Temperature of air, [tex]T_{\infty}=310\ K[/tex]
temperature of plate, [tex]T_s=360\ K[/tex]
convective heat transfer coefficient, [tex]h=45\ W.m^{-2}.K^{-1}[/tex]
absorptivity of plate, [tex]\alpha=0.4[/tex]
radiosity of the plate, [tex]J=4500\ W.m^{-2}[/tex]
From the energy balance eq. :
[tex]E_{in}=E_{out}[/tex]
since two surfaces are involved
[tex]2G=2J+2q_{conv}[/tex]
[tex]G=J+q_{conv}[/tex]
where
[tex]q_{conv}=[/tex] heat transfer per unit area due to convection
[tex]G=[/tex] irradiation per unit area
[tex]G=J+h.\Delta T[/tex]
[tex]G=4500+45\times (360-310)[/tex]
[tex]G=6750\ W.m^2[/tex] is the irradiation
Since radiosity includes transmissivity, reflectivity and emissivity.
[tex]J=G.\tau+G.\rho+E[/tex]
[tex]J=G.\tau+G.\rho+\epsilon\times \sigma\times T^{4}[/tex]
where:
[tex]\tau=[/tex] transmissivity
[tex]\rho=[/tex] reflectivity
[tex]\epsilon=[/tex] emissivity
Since the absorptivity of the plate is 0.4 and the plate is semitransparent so the transmissivity and reflectivity = 0.3 each.
[tex]4500=0.3\times 6750+0.3\times 6750+\epsilon\times 5.67\times 10^{-8}\times 360^4[/tex]
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
