Question

A wire of 5 mm diameter is drawn into a wire of 4 mm diameter through a conical die at a constant pulling speed of 5 m/s. Neglecting the coefficient of friction and redundant work, the drawing stress \( \sigma_d \) in MPa for the above process is given by \( \sigma_d = \sigma \ln \left( \frac{1}{1 - r} \right) \), where \( \sigma \) is the mean flow strength of wire material in MPa, and \( r \) is the ratio of decrease in area of cross-section to initial area of cross-section of the wire. If the mean flow strength of wire material is 600 MPa, then the power required in kW in the above wire drawing process is _________.
\text{[round off to 2 decimal places]}

Hint:

To calculate the power required in a wire drawing process, use the drawing stress, cross-sectional area, and velocity.

Answer 1

Correct Answer: 16.8

First, calculate the ratio \( r \) of decrease in area of cross-section: \[ r = \frac{A_0 - A_1}{A_0} = 1 - \left( \frac{d_1}{d_0} \right)^2 = 1 - \left( \frac{4}{5} \right)^2 = 1 - \frac{16}{25} = \frac{9}{25} = 0.36. \] Now, substitute into the formula for drawing stress: \[ \sigma_d = 600 \ln \left( \frac{1}{1 - 0.36} \right) = 600 \ln \left( \frac{1}{0.64} \right) = 600 \ln(1.5625) \approx 600 \times 0.4463 = 267.8 \, \text{MPa}. \] The power required is given by: \[ P = \sigma_d \times \text{cross-sectional area} \times \text{velocity}. \] The cross-sectional area is: \[ A = \frac{\pi d_0^2}{4} = \frac{\pi (5 \, \text{mm})^2}{4} = 19.634 \, \text{mm}^2. \] Convert area to m²: \[ A = 19.634 \times 10^{-6} \, \text{m}^2. \] Now, calculate the power required: \[ P = 267.8 \times 19.634 \times 10^{-6} \times 5 = 16.8 \, \text{kW}. \] Thus, the power required is approximately \( 16.80 \, \text{kW} \).