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Natural Convection Cooling of an Orange. An orange 102 mm in diameter having a surface temperature of 21.1°C is placed on an open shelf in a refrigerator held at 4.4°C. Calculate the heat loss by natural convection, neglecting radiation. As an approximation, the simplified equation for vertical planes can be used with L replaced by the radius of the sphere (M1). For a more accurate correlation, see (S2).

Respuesta :

Answer:

q = 3.181 w

Explanation:

[tex]T_w =  21.1 [/tex]degree celcius

[tex]T_b = 4.4 [/tex]degree celcius

D = 0.102 m

Radius = 0.051 m

[tex]\Delta T = T_w - T_b= 21.4 - 4.4 = 16.7[/tex] degree celcius

[tex]L^3 \Delta t = 0.051^3\times 16.7 = 2.22\times 10^{-3}[/tex]

[tex]h = 1.37 (\frac{\Delta}{L})^{1/4}[/tex]

  [tex]= 1.37\times (\frac{16.7}{0.051})^{1/4}[/tex]

[tex]h = 5.828 w/m^2 K[/tex]

[tex]A =4\pi r^2[/tex]

[tex]A = 4\times \pi 0.05^2 = 0.03268 m^2[/tex]

[tex]q = hA\Delta t[/tex]

h =5.828 (0.03268) 16.7

q = 3.181 w

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