Question

5. Steel balls 50 mm in diameter are annealed by heating to 1200 K and then slowly cooling to 450 K in an air environment for which the ambient temperature is 300 K and h = 20 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 this cooling process.

Answer 1

time required for cooling process = 0.233 hours

Step-by-step explanation:

In Transient heat conduction of a Sphere, the formula for Biot number is;

Bi = hL_c/k

Where L_c = radius/3

We are given;

Diameter = 12mm = 0.012m

Radius = 0.006m

h = 20 W/m²

k = 40 W/m·K

So L_c = 0.006m/3 = 0.002m

So,Bi = 20 x 0.002/40

Bi = 0.001

The formula for time required is given as;

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

Where;

A is Area = πD²

V is volume = πD³/6

So,

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

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

We are given;

T_i = 1200K

T_(∞) = 300K

T = 450K

ρ = 7800 kg/m³

c = 600 J/kg·K

Thus, plugging in relevant values;

t = (7800 x 0.012 x 600/(6 x20) )•In[(1200 - 300)/(450 - 300)]

t = 468•In6

t = 838.54 seconds

Converting to hours,

t = 838.54/3600

t = 0.233 hours