Question

Aluminum rivets used in airplane construction are made slightly larger than the rivet holes and cooled by "dry ice" (solid CO2) before being driven.
If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 *C, if its diameter is to equal that of the hole when the rivet is cooled to - 78.0 *C, the temperature of dry ice? Assume that the expansion coefficient of aluminum is 2.4

Answer 1

Answer:

$\rm 4.554\ m m.$

Explanation:

Given:

• Initial temperature, $\rm T_i=23.0\ ^\circ C=23 + 273.15 = 296.15\ K.$
• Final temperature, $\rm T_f = -78.0\ ^\circ C = -78+273.15=195.15\ K.$
• Diameter of the hole, $\rm d = 4.500\ m = 4.500\times 10^{-3}\ m.$
• Expansion coefficient of the Aluminum, $\rm \alpha = 2.4\times 10^{-5}\ K^{-1}.$

Let the diameter of the rivet at $\rm T_i=23.0\ ^\circ C$ be $\rm d_o$, such that,

$\rm d_o=d+\Delta d$

$\rm \Delta d$ is the elongation in the diameter of the rivet.

We know, The change in the diameter of the rivet is related with the change in temperature as:

$\rm \dfrac{\Delta d}{d_o}=\alpha \Delta T.$

where, $\rm \Delta T$ is the change in temperature = $\rm T_f-T_i=-195.15-296.15=-491.30\ K.$

Also, $\rm \Delta d=d_o-d.$

Using these values,

$\rm \dfrac{d_o-d}{d_o}=\alpha \Delta T\\1-\dfrac{d}{d_o}=\alpha \Delta T\\\dfrac{d}{d_o}=1+\alpha \Delta T\\\Rightarrow d_o = \dfrac{d}{1+\alpha \Delta T}\\=\dfrac{4.500\times 10^{-3}}{1+2.4\times10^{-5}\times (-491.30)}\\=4.554\times 10^{-3}\ m.\\=4.554\ m m.$

It is the required diameter of the rivet at $-78.0\ ^\circ C$.