A horizontal opaque flat plate is well insulated on the edges and the lower surface. The top surface has an area of 8 m2 and it experiences uniform irradiation at a rate of 6000 W. The plate absorbs 5000 W of the irradiation and the surface is losing heat at a rate of 750 W by convection. If the plate maintains a uniform temperature of 350 K, determine the absorptivity, reflectivity, and emissivity of the plate.

a = 0.8333

p = 0.1667

ε = 0.6243

Explanation:

Given:

- Solar Irradiation G = 6000 W

- Irradiation absorbed G_abs = 5000 W

- Heat Loss by convection Q_convec = 750 W

- The temperature of surface remains constant @ 350 K

Find:

Determine the absorptivity, reflectivity, and emissivity of the plate

Solution:

- Absorptivity is the ratio of energy absorbed to the incident energy given as:

a = G_abs / G

- Plug in values: a = 5000 / 6000

a = 0.8333

- Reflectivity is the ratio of energy not absorbed to the incident energy given as:

p = (1 - G_abs) / G

- Plug in values: p = 1000 / 6000

p = 0.1667

- Emissivity is the of ratio amount energy that is radiated from the body to the black body:

- We will use energy balance on the plate surface:

E_in - E_out = - k*A*dT / L

- We know that the plate temperature remains constant, dT = 0:

E_in = E_out

Q_abs = Q_convec + Q_rad

Q_rad = Q_abs - Q_convec

- Plug in values: Q_rad = 5000 - 750 = 4250 W

- The expression for surface radiation is given by:

Q_rad = ε*A*б*T_b^4

- Re-arrange: ε = Q_rad / A*б*T_b^4

- Plug values in: ε = 4250 / 8 * (5.67*10^-8) * (350) ^4

ε = 0.6243