Question

Increasing the surface temperature of a ball bearing for a short time without significantly raising the temperature inside the ball bearing produces a hard outer surface desirable in ball bearings. For a steel (r = 7800 kg/m3; Cp = 500 J/kg*K; k = 50 W/m*K) ball bearing, assume that any location within the bearing where the temperature is greater than 1000 K will be hardened. The ball bearing, initially at 300 K, will be placed in a very hot (1300 K) fluid with an extremely high convection coefficient (5,000 W/m2*K). Determine the time required for the outer 1 mm of a 20 mm diameter ball bearing to be hardened.

Answer 1

Answer:

required time is 5.7 seconds

Explanation:

solution

we first find here biot number by given formula that is

Bi = $\frac{hL}{k}$

put here h is coefficient = 5000 and k = 50 W/m K and L is radius = $\frac{0.02}{2}$  = 0.01

so

Bi = $\frac{5000 * 0.01}{50}$

Bi = 1

so for obtaining coefficient by table from term approx for the sphere at biot no 1

$\xi$ = 1.5708 and

C = 1.2732

so

and

center line temp. is here

θ* = $\frac{\theta}{\theta1}$

θ* = $\frac{T - Tx}{T1 - Tx}$

here T is 1000 K and Tx = 1300 K and T1 = 300 K

so

θ* = $\frac{1000 - 1300 }{300-1300}$

θ* = 0.3

so

radius for steel ball bearing ratio is here

r* = $\frac{r}{ro}$

here r is 10 -1 and ro is 10

so

r* = $\frac{10-1}{10}$

r* = 0.9

so that

fourier no for term approx for sphere is here

Fo = $\frac{-1}{\xi^2} ln[ \frac{\theta*}{C} \frac{1}{\xi * r} sin ( \xi * r)]$   ......1

put here value

Fo = $\frac{-1}{1.5708^2} ln[ \frac{0.3}{1.2732} \frac{1}{1.5708*0.9} sin (1.5708*0.9 *\frac{180}{\pi})]$  

Fo = - 0.4053 ln ( 0.16462)

Fo = 0.7312

so time fpr 1 mm to 20 mm dia

Fo = $\frac{\alpha*t}{r^2}$

Fo = $\frac{kt}{\rho C r^2}$

0.7312 = $\frac{50 * t}{7800 * 500*10^{-3} ^2$

t = 5.703 sec

so required time is 5.7 seconds