Question

Ice balls of 10 mm diameter at – 30°C are exposed to an air current at 15°C with a convection heat transfer coefficient of 200 W/m2 -K. Determine the time when the surface layer will begin to melt. Also determine the center temperature. Use the following property values: density = 920 kg/m3 , specific heat 2040 J/kg-K, thermal conductivity = 2.00 W/m-K.

Answer 1

Ans

Given data

density ρ = 920 kg/m3

specific heat Cp = 2040 J/kg-K

thermal conductivity k = 2.00 W/m-K

Dia D = 10 mm x 1m/1000mm = 0.010 m

Radius R = 0.01/2 = 0.005 m

Initial temperature Ti = - 30°C

Air temperature Ta = 15°C

heat transfer coefficient h = 200 W/m2 -K

Melting temperature T = 0°C

Biot number Bi = hR/k

= 200 W/m2-K x 0.005 m / 2.00 W/m-K

= 0.5

r/R = 1

From transient conduction and Heisler charts we get

(T - Ta) / (To - Ta) = 0.78

We can write above equation

(T - Ta) / (Ti - Ta) = (To - Ta) (T - Ta) / (Ti - Ta) (To - Ta)

(0 - 15)/(-30 - 15) = (To - Ta)*0.78 / (Ti - Ta)

(To - Ta) / (Ti - Ta) = 0.4274

1/Bi = 2

From Heisler charts

Fourier number = 0.78

ατ/R² = 0.78

τ = 0.78 x R² / α

Thermal diffusivity α = k/(Cp x ρ)

= 0.78 x R² x Cp x ρ/ k

= 0.78 x 0.005² m² x 2040 J/kg-K x 920 kg/m³ / 2 W/m-K

= 18.30 s

For center temperature

(To - Ta) / (Ti - Ta) = 0.4274

(To - 15) / (-30 - 15) = 0.4274

To = - 19.233 + 15

Center temperature To = - 4.23°C