Question

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).

Answer 1

Answer:

q = 3.181 w

Explanation:

$T_w = 21.1$degree celcius

$T_b = 4.4$degree celcius

D = 0.102 m

Radius = 0.051 m

$\Delta T = T_w - T_b= 21.4 - 4.4 = 16.7$ degree celcius

$L^3 \Delta t = 0.051^3\times 16.7 = 2.22\times 10^{-3}$

$h = 1.37 (\frac{\Delta}{L})^{1/4}$

  $= 1.37\times (\frac{16.7}{0.051})^{1/4}$

$h = 5.828 w/m^2 K$

$A =4\pi r^2$

$A = 4\times \pi 0.05^2 = 0.03268 m^2$

$q = hA\Delta t$

h =5.828 (0.03268) 16.7

q = 3.181 w