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

q = 3.181 w

Step-by-step 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* 16.7 = 2.22* 10^(-3)`

`h = 1.37 ((\Delta)/(L))^(1/4)`

`= 1.37* ((16.7)/(0.051))^(1/4)`

`h = 5.828 w/m^2 K`

`A =4\pi r^2`

`A = 4* \pi 0.05^2 = 0.03268 m^2`

`q = hA\Delta t`

h =5.828 (0.03268) 16.7

q = 3.181 w