A lead sphere has a diameter that is 0.05% larger ...?

Question

A lead sphere has a diameter that is 0.05% larger than the inner diameter of a steel ring when each has a temperature of 70.0 oC. Thus, the ring will not slip over the sphere. At what common temperature will the ring just slip over the sphere?


Cheers, any help is much appreciated.

Answer 1

.If D is the diameter, after expansion it is

D + Dαө

‘α’ is the linear expansivity and ‘ө’ is the increase in temperature.

Dl + Dl*α*ө = Ds + Ds*β*ө--------------1


α and βfor lead and steel respectively.

Dl -Ds = {Ds*α - Dl*β} ө

ө = (Dl -Ds) / {Ds*α - Dl*β }

Dividing both numerator and denominator of RHS by Ds
And representing ‘Dl/Ds’ by K
ө = (K -1 ) / {α - Kβ }



Dl = Ds + (0.05/100) Ds

Dl / Ds =100.05/100 = 1.0005 = K say.


α = 15x10^(-6) ; β = 29 x10^(-6) .

ө = (K -1) / {α - K*β}

ө = 33.37degre centigrade.

The common temperature is 70 +33.37

= 103.37degree centigrade.

Answer 2

if steel expands faster than lead, it will be at a higher temperature, if it contracts slower than lead, it will be at a lower temperature.

do you have a reference for how much lead and steel expand with temperature? Obviously, you need the amount of expansion for one metal to be 5% larger than that of the other for a particular change in temperature.

try searching for "coefficient of thermal expansion"

hope this gets you off to a good start!

Answer 3

You need to at least do your own grunt work on this by looking up the coefficients of thermal expansion for the two materials.

Sheesh!