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 fogether. Suppose an aluminum rod with diameter \(D_{1}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right)\) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right) .\) 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 \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} ?\) 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 of the aluminum rod at 20°C is approximately 10.048 mm.

Step-by-step solution

Calculate the change in diameter for the aluminum rod

First, we need to find the change in diameter for the aluminum rod when cooled to 77 K. The change in diameter is given by: ΔD = αD₀ΔT ΔD = (22 * 10^{-6} °C^{-1})(D_{1})(20°C - 77K) Note that we need to make sure we have the same units for temperature. Since 0°C is equal to 273.15 K, the given temperatures can be converted as: 20°C = 293.15 K Now we can write the change in diameter equation as: ΔD = (22 * 10^{-6} °C^{-1})(D_{1})(293.15K - 77K)

Calculate the change in diameter for the brass hole

Next, we need to find the change in diameter for the brass hole when the plate is heated to 20°C. The change in diameter is given by: ΔD = αD₀ΔT ΔD = (19 * 10^{-6} °C^{-1})(10mm)(293.15K - 293.15K) Since the brass plate is not changing temperature (remaining at 20°C), the change in diameter for the brass hole is 0. D_{2} = D_{2 initial} + ΔD = 10mm + 0 = 10mm Note that D_{2} remains constant at 10mm.

Determine the expanded diameter of the aluminum rod

Now we will find the expanded diameter of the aluminum rod (D_{1 final}) at 20°C. When the rod returns from 77K to 20°C, it will expand back to its original size D_{1}, which should fit into the brass hole. D_{1 final} = D_{1 initial} - ΔD Now, we substitute the expression for ΔD for aluminum from Step 1: D_{1 final} = D_{1} - (22 * 10^{-6} °C^{-1})(D_{1})(293.15K - 77K)

Determine the maximum diameter of the aluminum rod

Finally, we can set the expanded aluminum rod diameter equal to the brass hole diameter, and solve for D_{1}: D_{1} - (22 * 10^{-6} °C^{-1})(D_{1})(293.15K - 77K) = 10mm D_{1} - (22 * 10^{-6})(D_{1})(216.15) = 10mm (1 - 22 * 10^{-6}(216.15))D_{1} = 10mm D_{1} = 10mm / (1 - 22 * 10^{-6}(216.15)) D_{1} ≈ 10.048 mm The largest possible diameter of the aluminum rod at 20°C is approximately 10.048 mm, so it can fit into the brass hole if the rod is cooled to 77 K and the brass plate is left at 20°C.

Key concepts

Linear Expansion Coefficient

When materials are heated or cooled, they expand or contract respectively. This property is described by the linear expansion coefficient, denoted as \( \alpha \). This coefficient indicates how much a material will change in size per degree of temperature change. \( \alpha \) is measured in \( \text{°C}^{-1} \).

• The formula for linear expansion is: \( \Delta D = \alpha D_0 \Delta T \)
• \( \Delta D \) is the change in diameter.
• \( D_0 \) is the initial diameter.
• \( \Delta T \) is the change in temperature.

Each material has its own linear expansion coefficient. For example, in our problem, aluminum has a coefficient of \( 22 \times 10^{-6} \, \text{°C}^{-1} \), whereas brass has \( 19 \times 10^{-6} \, \text{°C}^{-1} \). These coefficients help us predict changes in dimensions due to temperature variations.

Aluminum and Brass

Aluminum and brass are common materials used in construction and manufacturing due to their mechanical and thermal properties.

• Aluminum: Lightweight and with a higher expansion rate, meaning it changes size more with temperature variations. It is more sensitive to thermal expansion compared to brass.
• Brass: Known for its strength and moderate expansion rate. It’s less sensitive to thermal expansion compared to aluminum, providing stability under temperature fluctuations.

Understanding these properties is crucial when fitting parts together like in a mechanical fitting involving an aluminum rod and a brass plate. Their distinct expansion characteristics make them suitable for specific applications but require careful calculation for precise fits.

Temperature Conversion

Temperature conversion is essential in thermal expansion calculations as it ensures consistency in units, typically between Celsius (°C) and Kelvin (K).

• The Kelvin scale is often used in scientific calculations because it starts at absolute zero. \( 0 \, ^{\circ}\text{C} = 273.15 \, \text{K} \)
• Therefore, to convert °C to K, add 273.15 to the Celsius temperature.
• For example, \( 20 \, ^{\circ}\text{C} = 293.15 \, \text{K} \).

This conversion is vital, especially in exercises that involve large temperature changes, as these can significantly affect the results of calculations related to material contraction or expansion.

Mechanical Fitting

Mechanical fitting involves creating a precise fit between two or more components, often made of different materials. This process is critical in engineering and manufacturing.

• In the context of thermal expansion, it's important to consider how temperature changes affect the dimensions of each component.
• Certain techniques, such as cooling or heating parts, are employed to ensure proper fits. Components may initially not fit perfectly due to their sizes, but thermal expansion or contraction can allow them to fit snugly.
• For example, cooling an aluminum rod to fit into a brass hole involves calculating the size changes due to temperature decreases to ensure a secure, precise fit.

This method ensures that parts are joined together safely and efficiently despite differing thermal expansion rates.