An 8cm diameter orange, approximately spherical in shape undergoes ripening process and generates $5000W/m^3$ of energy. If the external surface of the orange is at $6.5^0C$. Calculate the temperature at the center and also find the heat flow from the outer surface.

Take K = 0.22 W/m-K for the orange. Assume steady state heat transfer.

Given,

d = 8cm = 0.08m (Diameter of orange)

qg = 5000W/m3 (Rate of Heat generation/volume)

ts = 6.5℃ = 279.5K (Surface temperature of orange)

K = 0.22W/mK (Thermal conductivity of orange)

Find:

(1) tc (Center temperature of orange)

(2) Q (Heat transfer rate)

Solution:

Heat transfer rate

Q = qg × volume = qg × 43 × π × (d2)3

Q = qg × volume = 5000 × 43 × π × (0.082)3

Q = 1.3404 W

Temperature distribution equation

tc - ts = qgR26K

tc - 279.5 = 5000 × (0.04)26 × 0.22

tc = 285.56K = 12.56℃