The emissivity of the opaque horizontal plate = 0.34
The absorptivity of the opaque horizontal plate = 0.8
The radiosity of the opaque plate = 1700 w/m²
Net heat transfer rate = - 700 w/m²
a) Determine the emissivity of the Opaque horizontal plate
E = ξ * σ * Ts⁴
∴ ξ = E / ( σ * Ts⁴ )
= 1200 / ( 5.67 * 10⁻⁸ * ( 227 + 273 ) )
= 0.34
b) Determine the absorptivity of the plate
Given that total absorptivity is reflected to reflectivity
P = Gref / G
= ( 500 w/m² ) / ( 2500 w/m² )
= 0.2
∴ The absorptivity = 1 - P = 1 - 0.2 = 0.8
C) Determine the radiosity of the plate
J = pG + ξ Eb
= Gref + Eb
= 500 + 1200 = 1700 W/m²
D) Determine the net heat transfer rate per unit area
q[tex]_{net}[/tex]" = q"[tex]_{in}[/tex] - q"[tex]_{out}[/tex]
= G - Gref - E - q"[tex]_{conv}[/tex]
= ( 2500 - 500 - 1200 ) - hΔT
= 800 - 15 * ( 227 - 127 )
∴ q[tex]_{net}[/tex]" = - 700 w/m²
Hence we can conclude that the emissivity, absorptivity and radiosity of the plate and the net heat transfer is as listed above.
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