Steel balls 10 mm in diameter are annealed by heating to 1150 K and then slowly cooling to 450 K in an air environment for which T[infinity] = 325 K and h = 25 W/m2 ·K. Assuming the properties of the steel to be k = 40 W/m·K, rho = 7800 kg/m3, and c = 600 J/kg·K, estimate the time required for the cooling process.

the time required for the cooling process = 0.1635 hours

Explanation:

We are given;

Thermal conductivity; k = 40 W/m·K

Density; ρ = 7800 kg/m³

Specific heat capacity; c = 600 J/kg·K

h = 25 W/m². k

Diameter; D = 10mm = 0.01m

T_ (∞) = 325K

T_i = 1150K

T = 450K

The formula for time required for the cooling process is given by;

t = (ρVc/hA) [In ((T_i - T_ (∞)) / (T - T_ (∞))) ]

Where;

V is volume = πD³/6

A is Area = πD²

Other terms remain are previously stated.

Thus;

t = (ρ (πD³/6) c / (h (πD²)) [In ((T_i - T_ (∞)) / (T - T_ (∞))) ]

This is simplified into;

t = (ρDc / (6h) [In ((T_i - T_ (∞)) / (T - T_ (∞))) ]

Plugging in the relevant values to obtain;

t = (7800*0.01*600 / (6*25) [In ((1150 - 325) / (450 - 325)) ]

t = 312In (825/125)

t = 588.766 seconds = 0.1635 hours