Answer:
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
