Answer:
(a) [tex]T_{s,o}[/tex] = 33.8°C , q = 797 W
(b) [tex]T_{s,o}[/tex] = 27.1°C , q = 675.3 W
(c) [tex]T_{s,o}[/tex] = -9.9°C , q =90,630 W
Comment: Use the special surface finish reduces the solar input, while increasing radiation emission from the surface.
Explanation:
(a)
From an energy balance from the outer surface
Where, [tex]R"_p = (t_{1}/k_{p}) =[/tex] 2.78 x 10⁻⁵ m².k/w
[tex]R"_i = (t_{2}/k_{i}) =[/tex] 1.923 m².k/w
[tex]R_{eL} = U_{w}(L/V) =[/tex] [2.92 m/s x 10 m] / [15.89 x 10⁻⁶ m²/s]
= 1.84 x 10⁴
|h| = (K/L)0.037 [tex]R_{eL}^{4/5}. P_\alpha ^{1/3}[/tex]
substituting parameters,
|h| = 0.0263/10*(0.037)*[(184x10⁷)^(4/5)]*(0.707)^1/3
|h| =562Wm⁻²K⁻¹
Hence,
0.5(950) + 56.2(305-[tex]T_{s,o}[/tex])-0.5*5.67x10⁻⁸ = ([tex]T_{s,o}[/tex] - 268) / (5.56x10⁻⁵ + 1.923)
Solving, we obtain
[tex]T_{s,o}[/tex] = 306.8K
= 33.8°C
Hence, the heat load is
q = (W.L)q" = (3.5 x 10) [tex]\frac{378 + 10}{1.923}[/tex]
q = 797 W
(b) With the special surface finish
[tex]T_{s,o}[/tex] = 301.1K = 27.1°C
Similarly, q = 675.3 W
(c) Without the insulation (t₂ = 0) and with (∝s = E = 0.5)
[tex]T_{s,o}[/tex] = 263.1k = -9.9°C
Similarly, q = 90,630 W
