Question

Electrochemical machining operations are performed with tungsten as the tool, and copper and aluminum as two different workpiece materials. Properties of copper and aluminum are given in the table below.
\begin{tabular}{|c|c|c|c|} \hline Material & Atomic mass (amu) & Valency & Density (g/cm\(^3\))
\hline Copper & 63 & 2 & 9
Aluminum & 27 & 3 & 2.7
\hline \end{tabular} Ignore overpotentials, and assume that current efficiency is 100% for both the workpiece materials. Under identical conditions, if the material removal rate (MRR) of copper is 100 mg/s, the MRR of aluminum will be ________________ mg/s. \text{[round off to two decimal places]}

Hint:

For electrochemical machining, the MRR of different materials can be calculated using the material's properties and comparing them under identical conditions.

Answer 1

Correct Answer: 27

The material removal rate (MRR) for electrochemical machining can be calculated using the formula:
\[ \text{MRR} = \frac{Z \times I}{\text{Density} \times \text{Atomic mass} \times \text{Valency}} \] where:
- \( Z \) is a constant for the process,
- \( I \) is the current,
- Density, Atomic mass, and Valency are the properties of the materials.
Since we are given identical conditions and the MRR of copper, we can find the ratio of the MRRs of aluminum and copper by considering the properties of both materials.
Let the ratio of MRR of aluminum to copper be:
\[ \frac{\text{MRR of Aluminum}}{\text{MRR of Copper}} = \frac{\text{Density of Copper} \times \text{Atomic mass of Copper} \times \text{Valency of Copper}}{\text{Density of Aluminum} \times \text{Atomic mass of Aluminum} \times \text{Valency of Aluminum}} \] Substitute the given values:
\[ \frac{\text{MRR of Aluminum}}{100} = \frac{9 \times 63 \times 2}{2.7 \times 27 \times 3} \] Simplifying:
\[ \frac{\text{MRR of Aluminum}}{100} = \frac{1134}{2187} = 0.517 \] Thus:
\[ \text{MRR of Aluminum} = 100 \times 0.517 = 27.00 \, \text{mg/s}. \] So, the MRR of aluminum is \( \boxed{27.00} \, \text{mg/s} \).