Can someone please help me with this? A steel ball bearing is 5 cm in diameter a
ID: 2252534 • Letter: C
Question
Can someone please help me with this?A steel ball bearing is 5 cm in diameter at 29 degrees Celsius. A bronze plate has a hole that is 4.995 cm. in diameter at 29 degrees Celsius. What common temperature must they have in order for the ball to just squeeze through the hole? Answer in units of degrees Celsius. Can someone please help me with this?
A steel ball bearing is 5 cm in diameter at 29 degrees Celsius. A bronze plate has a hole that is 4.995 cm. in diameter at 29 degrees Celsius. What common temperature must they have in order for the ball to just squeeze through the hole? Answer in units of degrees Celsius.
Explanation / Answer
Alpha(a) for ac cu =17 *e-6 and ab brass 19*e-6
volume expansion r^3=r(cu) ^3 (1+3 ac dT) gamma=3 alpha
are expansion r^2=r(brass)^2 (1+2 ab dT) beta =2 alpha
the final r should be same
substuiting (5/4.995)^6=(1+2 ab dT)^3/(1+3 ac dT)^2 (1+a)^t=1+at for small a
1.006=1+6*ab *dT/1+6*ac*dT
0.006=6(19-17)*dT*E-6
dT=0.006/12 *1000000
dT=500
now 529 degrees celcius