Thermal expansion copper

Question

In mechanical engineering, various components are often connected with thermal expansion/shrinkage. For this task, for example, a copper ring was pressed onto a steel rod so that the ring has been heated before sliding it onto the rod. At the temperature T = 21 ◦C when the copper ring is pressed on, it has an inner diameter of exactly D= 3.4950 cm, the rod has an outer diameter of DSt = 3.5000 cm outside the point of pressing. The rod is therefore slightly dented at the compression.

a) To what temperature must ring and rod be heated so that the copper ring can be separated from the steel rod ? To solve that a circumferential gap between the inside of the ring and the outside for looseningo f the rod of at least s = 10 µm is required. To simplify, assume that ring and rod have the same temperature. The linear expansion coefficient of copper is αKupfer = 17 · 10^−6/ K, the of the steel is αsteel = 12 10^−6/K.

b) What is the density of the copper ring at T = 450 ◦C? The density of copper at T = 0 ◦C isρ(T = 0 ◦C) = 8.95 103 kg/m3 .

Answer 1

To solve this problem, we'll address both parts (a) and (b):

### Part (a): Heating Temperature for Separation

We need to calculate the temperature to which the copper ring and steel rod must be heated so that the ring expands sufficiently to create a gap of at least 10 µm (0.01 mm) between the ring's inner diameter and the rod's outer diameter.

Given:

- Inner diameter of the copper ring at \(T = 21°C\), \(D_{Cu} = 3.4950\) cm

- Outer diameter of the steel rod, \(D_{St} = 3.5000\) cm

- The required gap, \(s = 10\) µm

- Linear expansion coefficients: \(\alpha_{Cu} = 17 \times 10^{-6}/°C\), \(\alpha_{St} = 12 \times 10^{-6}/°C\)

To create the gap, the copper ring must expand more than the steel rod. The expansion for each material can be calculated using the formula for linear expansion: \(\Delta L = \alpha L \Delta T\), where \(\Delta L\) is the change in length, \(\alpha\) is the linear expansion coefficient, \(L\) is the original length, and \(\Delta T\) is the change in temperature.

For the ring to be removed:

- The new diameter of the copper ring should be at least \(D_{Cu} + s + (D_{St} - D_{Cu})\).

- The change in diameter for copper and steel can be calculated using their respective linear expansion coefficients.

We'll calculate the necessary temperature increase \(\Delta T\) such that \(\Delta D_{Cu} - \Delta D_{St} \geq s\).

### Part (b): Density of the Copper Ring at \(T = 450°C\)

The density of a substance changes with temperature due to thermal expansion. As the copper expands, its volume increases, and its density decreases. We can use the formula:

\[

\rho(T) = \frac{\rho_0}{1 + 3\alpha_{Cu} \Delta T}

\]

Where:

- \(\rho(T)\) is the density at temperature \(T\),

- \(\rho_0\) is the initial density (8.95 g/cm³ at 0°C),

- \(\alpha_{Cu}\) is the linear expansion coefficient for copper,

- \(\Delta T\) is the change in temperature from the reference temperature (0°C to 450°C in this case).

We'll calculate the temperature required to separate the copper ring from the steel rod and the density of copper at 450°C using these formulas.

a) To separate the copper ring from the steel rod, ring and rod must be heated to approximately 365.53°C. At this temperature, the copper ring will have expanded enough to create a circumferential gap of at least 10 µm between the inside of the ring and the outside of the rod.

b) The density of the copper ring at 450°C is approximately 8749.21 kg/m³. This decrease from the original density at 0°C is due to the thermal expansion causing an increase in volume.