Question

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air that is 200C and provides a convection heat transfer coefficient of 10 W/m2K, solar irradiation of 600 W/m2, and an effective sky temperature of -400C. Calculate the electrical power (W/m2) required to maintain the plate surface temperature at Ts

Answer 1

The electrical power is 96.5 W/m^2

Step-by-step explanation:

The energy balance is:

Ein-Eout=0

`qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0`

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

`\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }`

`\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }`

if Gl≈El(l,5800)

`\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))`

lt= 2*5800=11600 um-K, at this value, F=0.941

`\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77`

The hemispherical emissivity is equal to:

`E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))`

lt=2*333=666 K, at this value, F=0

`E=0+(1-0.7)(1)=0.3`

The hemispherical absorptivity is equal to:

`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)`