Question

In order to create a tight fit between two metal parts, machinists sometimes make the interior part larger than the hole into which it will fit and then either cool the interior part or heat the exterior part until they fit together. Suppose an aluminum rod with diameter \(D_{1}\) (at \(20 .{ }^{\circ} \mathrm{C}\) ) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(20 .{ }^{\circ} \mathrm{C}\) ). The machinists can cool the rod to \(77.0 \mathrm{~K}\) by immersing it in liquid nitrogen. What is the largest possible diameter that the rod can have at \(20 .{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(20 .{ }^{\circ} \mathrm{C} ?\) The linear expansion coefficients for aluminum and brass are \(22 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and \(19 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\), respectively.

Short answer

Answer: The largest possible diameter that the aluminum rod can have at 20°C and just fit into the hole if the rod is cooled to 77.0 K is 10.048 mm.

Step-by-step solution

Write down the linear expansion formula

The linear expansion formula can be expressed as: \(\Delta L = L_{0} \alpha \Delta T\) Where \(\Delta L\) is the change in length/diameter, \(L_{0}\) is the initial length/diameter, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature.

Find the change in temperature for the aluminum rod

To find the change in temperature for the aluminum rod, subtract the initial temperature from the final temperature: \(\Delta T = T_{final} - T_{initial} = 77.0 \mathrm{~K} - 20 .{ }^{\circ} \mathrm{C}\) Note: Temperature in Kelvin and temperature in Celsius can be used interchangeably when calculating a change in temperature because the difference in the initial and final temperatures is the same: \(\Delta T = 77.0 \mathrm{~K} - 293 .{ }^{\circ} \mathrm{K} = -216 .{ }^{\circ} \mathrm{C}\)

Calculate the change in diameter for the aluminum rod

Using the linear expansion formula, calculate the change in diameter of the aluminum rod: \(\Delta D_{1} = D_{1} \alpha_{Al} \Delta T = D_{1} (22 \cdot 10^{-6} .{ }^{\circ} \mathrm{C}^{-1}) (-216 .{ }^{\circ} \mathrm{C})\) \(\Delta D_{1} = - D_{1} (4.752\cdot 10^{-3})\)

Calculate the final diameter of the aluminum rod

Now, calculate the final diameter of the aluminum rod after it has been cooled: \(D_{1_{final}} = D_{1} + \Delta D_{1} = D_{1} - D_{1}(4.752\cdot 10^{-3})\) \(D_{1_{final}} = D_{1} (1- 4.752\cdot 10^{-3})\)

Set the final diameter of the aluminum rod equal to the diameter of the hole and solve for \(D_{1}\)

Since the aluminum rod has to just fit into the hole, the diameter of the cooled aluminum rod should be equal to the diameter of the hole in the brass plate: \(D_{1_{final}} = D_{2}\) Substitute the expression for \(D_{1_{final}}\) from Step 4: \(D_{1} (1 - 4.752\cdot 10^{-3}) = 10.000 \mathrm{~mm}\) Now, solve for \(D_{1}\), the largest possible diameter of the aluminum rod at \(20 .{ }^{\circ} \mathrm{C}\): \(D_{1} = \frac{10.000 \mathrm{~mm}}{(1 - 4.752\cdot 10^{-3})} = 10.048 \mathrm{~mm}\) The largest possible diameter that the aluminum rod can have at \(20 .{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) is \(10.048 \mathrm{~mm}\).