Question

For an electrochemical machining process \[ \frac{dy}{dt}=\frac{\lambda}{y}-f , \] where $y$ is the inter-electrode gap in mm at time $t$ (in minute), and $f$ is the feed of the tool in mm/minute. The value of $\lambda$ is $6\times10^{-3}$ cm$^{2}$/minute. For maintaining a constant inter-electrode gap of 0.1 mm, the feed (in mm/minute) should be \(\underline{\hspace{2cm}}\). [round off to one decimal place]

Hint:

For constant-gap ECM, set $dy/dt = 0$ and match the tool feed exactly to the dissolution rate $\lambda/y$.

Answer 1

Correct Answer: 5.9 - 6.1

To maintain a constant gap, \[ \frac{dy}{dt}=0. \] Thus, \[ 0 = \frac{\lambda}{y} - f \Rightarrow f = \frac{\lambda}{y}. \] Convert \(\lambda\) from cm²/min to mm²/min: \[ 1\ \text{cm}^2 = 100\ \text{mm}^2 \] \[ \lambda = 6\times10^{-3} \times 100 = 0.6\ \text{mm}^2/\text{min} \] Given: \[ y = 0.1\ \text{mm} \] So, \[ f = \frac{0.6}{0.1} = 6\ \text{mm/min} \]