Question

2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flux of sun q′′ =750W/m2 on the other side. The exposed surface has an absorptivity of 0.87 α sun = for solar radiation; that is, 87% of the sun’s energy are absorbed whereas the other 13% are reflected. Assuming that the energy absorbed by the panel is dissipated by both natural convection and radiation (with a surrounding temperature of Tsur=300K, and a surface emissivity of εpanel = 0.85) into the surrounding quiescent air at atmospheric pressure and T[infinity] = 300 K, determine the surface temperature of the panel at steady state.

Answer 1

Answer: 383.22K

Step-by-step explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.