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

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