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

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

Let the diameter of the rivet at [tex]\rm T_i=23.0\ ^\circ C[/tex] be [tex]\rm d_o[/tex], such that,

[tex]\rm d_o=d+\Delta d[/tex]

[tex]\rm \Delta d[/tex] 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:

[tex]\rm \dfrac{\Delta d}{d_o}=\alpha \Delta T.[/tex]

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

Also, [tex]\rm \Delta d=d_o-d.[/tex]

Using these values,

[tex]\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.[/tex]

It is the required diameter of the rivet at [tex]-78.0\ ^\circ C[/tex].

Eduard22sly Eduard22sly · Answer 2 · 2022-04-01T13:24:26+02:00

The diameter of the rivet when it is cooled to –78 °C, the temperature of dry ice is 4.49 mm

Data obtained from the question

Original diameter (D₁) = 4.5 mm
Original temperature (T₁) = 23 °C
New temperature (T₂) = –78 °C
Coefficient of expansion (α) = 2.4×10¯⁵ °C¯¹

New diameter (D₂) =?

How to determine the new diameter

α = D₂ – D₁ / D₁(T₂ – T₁)

2.4×10¯⁵ = D₂ – 4.5 / 4.5(–78 – 23)

2.4×10¯⁵ = D₂ – 4.5 / 4.5(–101)

2.4×10¯⁵ = D₂ – 4.5 / –454.5

Cross multiply

D₂ – 4.5 = 2.4×10¯⁵ × –454.5

D₂ – 4.5 = –0.010908

Collect like terms

D₂ = –0.010908 + 4.5

D₂ = 4.49 mm

Complete question

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