Respuesta :
Answer:
Q convection = 11.361 W
Q radiation = 21.438 W
Explanation:
Given data:
[tex]v = 1.524 \times 10^{-5}m^2/s[/tex]
Pr = 0.7
Isobaric volume expansion is calculated as
[tex]\beta = \frac{1}{v} {\partial v}{\partial T}[/tex]
[tex]\beta = \frac{1}{T_{ambient}} = \frac{1}{22+273} = 3.39\times 10^{-3} K^{-1}[/tex]
L = 0.5 m Since door is vertical
grasshoff number is
[tex]Gr = \frac{g \beta(T_S - T_A)L^3}{v^2}[/tex]
[tex]Gr = \frac{9.81 \times 3.39\times 10^{-3} (32.22) \times 0.5^3}{(1.524\times ^{-5})^2}[/tex]
Gr = 1.79\times 10^8
Rayleigh number
Ra = GrPr
[tex]Ra = 1.253\times 10^8 \times 0.7 = 1.253\times 10^8< 10^8[/tex]
therefore given flow is laminar
Nusselt number is
[tex]Nu = 0.5 (Ra)^{1/4}[/tex]
[tex]Nu= 0.59(1.253\times 10^8)^{1/4})[/tex]
Nu = 62.4223
Heat transfer coefficient
[tex]h = \frac{NuK_{air}}{L}[/tex]
[tex]h = 62.4223 \times 0.026}{0.5}[/tex]
h = 3.246 W/m^2 K
convection heat transfer
[tex]Q = hA(T_S - T_A)[/tex]
[tex]Q = 3.246\times 0.5\times 0.7(32-22)[/tex]
Q = 11.361 W
Radiation heat transfer
[tex]Q_R = \sigma \epsilon A(T_s^4 - T_A^4}[/tex]
[tex]Q_R = 5.67 \times 10^{-8} \times 1\times 0.5\times 0.7(305^4 - 294^4)[/tex]
[tex]Q_R = 21.438 W[/tex]
