Question

To harden a steel sphere having a diameter of 50.8 mm, it is heated to 1033 K and then dunked into a large water bath at 300 K. Determine the time for the center of the sphere to reach 366.5 K, and the temperature at 10 mm from the centre. The surface coefficient can be assumed as 710 W/m^2.K and k=45 W/m.K and alpha=0.0325 m2/h.

Answer 1

diameter of steel d = 50.8 mm x 1m/1000mm = 0.0508 m

Radius r = 0.0508 / 2 = 0.0254 m

Volume of sphere = 4/3 x ?r^3 = 4/3 x 3.14 x (0.0254^3)

= 6.86 x 10^-5 m3

Area = 4?r^2

= 4*3.14*(0.0254^2)

= 0.0081 m2

Bi = hV/Ak

= 710 x 6.86 x 10^-5 / 0.0081* 45

= 0.1

T0 = 1033 K

T1 = 300 K

T = 366.5 K

Surface heat transfer coefficient h = 710 W/m^2.K

Thermal conductivity k = 45 W/m.K

Thermal diffusivity = 0.0325 m2/h x 1h/3600s = k/Cp

Cp = 45*3600/0.0325 = 4984615.38

Unsteady state heat conduction

ln (T - T1) / (T0 - T1) = - hAt/VCp

ln (366.5-300)/(1033-300) = - 710*0.0081*t/4984615.38*6.86 x 10^-5

2.4 = 0.0168 x t

t = 142.69 s x 1h/3600s = 0.04 hr