Answer:
[tex]h=133.333 W/m^2K[/tex]
Step-by-step explanation:
Heat transfer due to conduction:
[tex]Q_c=-KA\frac{T_s-T_b}{L}[/tex]
Where:
K is the thermal conductivity
A is the area of the Plate
T_s is the temperature at the upper surface
T_b is the temperature at the lower surface
L is the length of thickness of the plate
Heat Transfer due to convection:
[tex]Q_convection=hA(T_s-T_a)[/tex]
Where:
h is the convection heat transfer coefficient.
A is the area of the Plate.
T_s is the temperature at the upper surface.
T_a is the air temperature
Heat Transfer due to conduction=Heat Transfer due to convection
[tex]Q_c[/tex]=[tex]Q_convection[/tex]
[tex]-KA\frac{T_s-T_b}{L}[/tex]=[tex]hA(T_s-T_a)[/tex]
[tex]h=-K*\frac{T_s-T_b}{T_s-T_a}* \frac{1}{L} \\h=-80*\frac{50-60}{50-10}* \frac{1}{0.15m}\\h=133.333 W/m^2K[/tex]