Answer:
the rate of heat loss by convection across the air space = 82.53 W
Explanation:
The film temperature
[tex]T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C[/tex]
to kelvin = (5 + 273)K = 278 K
From the " thermophysical properties of gases at atmospheric pressure" table; At [tex]T_f[/tex] = 278 K ; by interpolation; we have the following
[tex]\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}[/tex] → v 13.93 (10⁻⁶) m²/s
[tex]\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})}[/tex] → k = 0.0245 W/m.K
[tex]\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})}[/tex] → ∝ = 19.6(10⁻⁶)m²/s
[tex]\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720}[/tex] → Pr = 0.713
[tex]\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}[/tex]
The Rayleigh number for vertical cavity
[tex]Ra_L = \frac{g \beta (T_1-T_2)L^3}{\alpha v}[/tex]
= [tex]\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}[/tex]
= [tex]8.38*10^5[/tex]
[tex]\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24[/tex]
For the rectangular cavity enclosure , the Nusselt number empirical correlation:
[tex]Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}[/tex]
[tex]NU_L= \frac{hL}{k}= 4.878[/tex]
[tex]\frac{hL}{k}= 4.878[/tex]
[tex]\frac{h*0.06}{0.0245}= 4.878[/tex]
[tex]h = \frac{4.878*0.0245}{0.06}[/tex]
h = 1.99 W/m².K
Finally; the rate of heat loss by convection across the air space;
q = hA(T₁ - T₂)
q = 1.99(1.4*0.96)(20-(-10))
q = 82.53 W
