Qn 9'14

1
Qn 9.14

• Ball bearings (12mm diameter spheres) are
  austenitized at 1145 and then quenched into a
  large tank of oil at 310K. Calculate
• The time to cool the center of a bearing to 480
  K.
• The surface temperature when the center is at 480
  K.
• The space-mean temperature when the center is at
  480 K.
• If 10,000 balls are quenched per hour, calculate
  the rate of heat removal from the oil that is
  needed to maintain its temperature at 310K.
• Data for ball bearing h 1700Wm-2K-1 ?
  7210kgm-3
• Cp 630 JkgK-1 k 43 Wm-1K-1

2
Qn 9.14
Data for ball bearings L 0.006m h1700Wm-2K-1 ?
7210kgm-3 Cp 630 Jkg-1K-1 k 43 Wm-1K-1
3
Qn 9.14 (a)

• At the center of ball bearing
• Tintial 1145K, Tfinal 310K, T 480K
• Bi hL/k (1700)(0.006)/43 0.24
• Tcenter- Tfinal / Tintial- Tfinal
  (480-310)/(1145-310) 0.20
• From Figure 9.10 (a) for r/R 0, Fo 4.5
• Since Fo ?t/L2 kt/?CpL2 where ?k//?Cp
• ?t Fo ?CpL2/k (2.5)(7210)(630)(0.006)2/43
• 17.1secs

4
Qn 9.14 (b)

• Using Figure 9.10 (d) for r/R 1, Fo 4.5, Bi
  0.24
• Tsurface- Tfinal / Tintial- Tfinal 0.17
• ? Tsurface 0.17(1145-310) 310 451.95K

5
Qn 9.14 (c)
Space mean temperature temperature averaged
over the volume of the ball bearing
There are two methods to find T(r).
6
Qn 9.14 (c)

• Method 1 With the values of Fo 4.5, Bi 0.24,
  find the value of ? for r/R 0.4 and r/R 0.8
  from Figure 9.10 (b) and (c) and evaluate Tr/R
  0.4 and Tr/R 0.8.
• ? Tr/R (k)
• r/R0 0.20 480
• r/R0.4 0.19 468.65
• r/R0.8 0.18 460.3
• r/R1 0.17 451.95

7
Qn 9.14 (c)
The volume of the ball bearing is divided into
four regions as shown in the figure. Each region
is divided equally between two points where the
temperatures based on the r/R value are known.
The color of the regions represents its
temperature.
8
Qn 9.14 (c)
Based on this method, the space mean temperature
is calculated by (Tr/R0 x volumegreen region
Tr/R0.4 x volumeorange region Tr/R0.8 x
volumeblue region Tr/R1 x volumegrey region) /
volume of ball bearing
9
Qn 9.14 (c)
Method 2 Plot a graph of T(r) against r and
determine a equation that best fits this
function. In this a linear trendline with a
equation fits the curve as shown in the graph
T(r) -4476r 480
10
Qn 9.14 (d)
Assuming that the balls are quenched for 17.1
secs to the space-mean temperature of
459.86K, Total heat given out by 10,000 balls is
be given by 10000 x mass of one ball x Cp
x(Tinitial Tfinal) 10 000 x 7210 x 4/3?(0.006)3
x 630 x (1145-459.86) 28158kJ Rate of heat
removal 28158kJ/hour 7822kJ/sec