A machinist wishes to insert a brass rod with a diameter of 4 mm into a hole with a diameter of 3.993 mm. By how much would the machinist have to lower the temperature (in °C) of rod to fit in hole

Answer 1

It wasn't noted the starting temperature so let's assume it is 20 C

Then,

Initial diameter of the rod D₁ = 4 mm

Final diameter of the rod D₂ = 3.993 (at least)

Initial Temperature of the rod T₁ = 20 C

Final Temperature of the rod T₂ = ?

pi = 3.14159...

Assume the coefficient of expansion (a) for brass is 19 x 10^-6 m/m-C

The formula is

dL/L₁ = a dT

But the L in this case is the circumference of the rod = pi*D

So,

d(pi*D)/(pi*D₁) = a dT or (D₂ - D₁)/D₁ = a(T₂ - T₁)

or (3.993 - 4)/4 = (19 x 10^-6)(T₂ - 20)

-92.1 = T₂ - 20

So, T₂ = -72.1 C (to get to 3.993 mm)