9.1 The mechanical workshop need a hydraulic press - NSC Mechanical Technology Automotive - Question 9 - 2017 - Paper 1

The pressure exerted on Piston A can be calculated using the formula:

$P_A = \frac{F_A}{A_A}$

Where:

• $F_A = 550\ N$
• $A_A$ was calculated previously as $0.00508\ m^2$

Calculating this gives us:

$P_A = \frac{550}{0.00508} \approx 108,268\ Pa = 108.27\ kPa$

Using hydraulic principles, the force exerted on Piston B can be determined as:

$P_A = P_B \Rightarrow F_B = P_B \times A_B$

The area of Piston B is calculated as follows:

$A_B = \frac{\pi}{4} D_B^2 = \frac{\pi}{4} (0.180)^2 \approx 0.0254\ m^2$

Thus, the force on Piston B is:

$F_B = P_A \times A_B = 108268 \times 0.0254 \approx 2.76\ kN$

The bush material is subjected to compressive stress due to the force exerted on it.

The stress in the material can be calculated using:

$\sigma = \frac{F}{A}$

Where:

• $F = 23\ kN = 23000\ N$
• $A$ can be calculated as the difference in areas:

$A = \frac{\pi}{4}(D_{outer}^2 - D_{inner}^2) = \frac{\pi}{4}((0.040)^2 - (0.030)^2) \approx 0.000785 \ m^2$

Now, substituting these values gives:

$\sigma = \frac{23000}{0.000785} \approx 29,305\ kPa = 29.3\ MPa$

The rotation frequency of the electrical motor can be directly calculated from its rotation speed. Given the shaft rotates at 90 revolutions per minute (rpm):

$N_E = 90\ rpm$

This frequency in Hz is:

$N_E = \frac{90}{60} = 1.5\ Hz$

1. No slip occurs
2. It is much stronger
3. More accurate
4. Lasts longer