An artisan was instructed to design a hydraulic system that will be used to press bearings - NSC Mechanical Technology Fitting and Machining - Question 11 - 2021 - Paper 1

Using the area of the ram:

$A = \frac{\pi D^2}{4} = \frac{\pi (0.12)^2}{4} = 11.31 \times 10^{-2} \text{ m}^2$

Next, we calculate the force:

$F = P \times A = 25.98 \times 10^6 \times 11.31 \times 10^{-2} = 293.88 \text{ kN}$

A hydraulic reservoir functions as a fluid storage tank, ensuring a supply of hydraulic fluid to the system.

1. Pneumatic tools are environmentally friendly.
2. They are easy to use and require less power.

1. Air blow guns.
2. Nail guns.

Using the equation for rotational speed:

$N_1D_1 = N_2D_2$

Where:

• $D_1 = 0.6 \text{ m}$ (pulley diameter on the washing machine)
• $D_2 = 0.8 \text{ m}$ (diameter of the driving pulley)
• $N_1 = \frac{7}{2} \text{ r/s}$ (input speed)

Thus, $N_2 = \frac{N_1 D_1}{D_2} = \frac{3.5 \times 0.6}{0.8} = 5.4 \text{ r/s}$

The power can be calculated using:

$P = \frac{(T_1 - T_2) \times D N}{60}$

Where:

• $T_1 = 300 \text{ N}$ (tight side tension)
• $T_2 = 120 \text{ N}$ (slack side tension)
• $D = 0.8 \text{ m}$
• $N = 5.4 \text{ r/s}$

Thus:

$P = \frac{(300 - 120) \times 0.8 \times 5.4}{60} = 2.44 \text{ kW}$

Using the gear ratio formula:

$N_{output} = \frac{T_A \times T_C \times N_A}{T_B \times T_D}$

Where:

• $T_A = 30$, $T_B = 40$, $T_C = 20$, $T_D = 60$

Substituting: $N_{output} = \frac{30 \times 20 \times 2300}{40 \times 60} = 575 \text{ r/min}$

The gear ratio can be calculated with:

$\text{Gear ratio} = \frac{\text{Product of teeth on driven gears}}{\text{Product of teeth on driver gears}} = \frac{40 \times 60}{30 \times 20} = 4:1$