Answer:
The electrical power is 96.5 W/m^2
Explanation:
The energy balance is:
Ein-Eout=0
[tex]qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0[/tex]
if:
Gsky=oTsky^4
Eb=oTs^4
qc=h(Ts-Tα)
[tex]\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }[/tex]
[tex]\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }[/tex]
if Gl≈El(l,5800)
[tex]\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))[/tex]
lt= 2*5800=11600 um-K, at this value, F=0.941
[tex]\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77[/tex]
The hemispherical emissivity is equal to:
[tex]E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))[/tex]
lt=2*333=666 K, at this value, F=0
[tex]E=0+(1-0.7)(1)=0.3[/tex]
The hemispherical absorptivity is equal to:
[tex]qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}[/tex]
