11.1 State TWO advantages of a belt drive system compared to a chain drive system - NSC Mechanical Technology Fitting and Machining - Question 11 - 2018 - Paper 1

To calculate the fluid pressure, we can use the formula:

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

Where:

• $P$ is the pressure,
• $F$ is the force exerted (120 N),
• $A$ is the cross-sectional area of the plunger, calculated using:

$A = \frac{\pi d^2}{4}$

With a diameter of 32 mm ($0.032 m$):

$A = \frac{\pi (0.032)^2}{4} = 0.000804\ m^2$

Thus, the fluid pressure is:

$P = \frac{120}{0.000804} \approx 149,200\ Pa\ or\ 0.15\ MPa$

Using the same principles for fluid pressure, we again apply:

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

Where $F = 18,000 N$.

Using the previously calculated area, we find:

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

Rearranging gives:

$A = \frac{18,000}{120} = 0.15 m^2$

Now, using the area to find the diameter:

$A = \frac{\pi d^2}{4} \Rightarrow d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4(0.15)}{\pi}} \approx 0.39088\ m\ or\ 390.88\ mm$

The symbol for a one-way spring-loaded valve can be represented as follows:

 ____
|    |
|    |
|____|
   |  
   |  \
   |   \
   |_____

This represents the flow direction and spring action in a system.

To find the rotation frequency of the driver pulley:

Using the formula for the rotation frequency:

$N_{DR} = \frac{N_{DN} \times D_{DN}}{D_{DR}}$

Where:

• $N_{DR}$ is the rotation frequency of the driver pulley (unknown),
• $N_{DN}$ is the rotation frequency of the driven pulley (80 r/min),
• $D_{DN}$ is the diameter of the driven pulley (240 mm),
• $D_{DR}$ is the diameter of the driver pulley (75 mm).

Calculating gives:

$N_{DR} = \frac{80 \times 240}{75} = 256\ r/min$

Calculate the rotation frequency of the output shaft:

Using the gear relationship:

• Given: $T_A = 20$, $T_B = 35$, shaft speed = 3000 r/min.

Applying the formula:

$N_E = \frac{T_B}{T_A} \times N_1$

This gives:

$N_E = \frac{35}{20} \times 3000 = 5250\ r/min$

To convert to revolutions per second:

$N_E = \frac{5250}{60} \approx 87.5\ r/s$

The gear ratio can be determined by:

$\text{Gear Ratio} = \frac{T_B}{T_A} = \frac{35}{20} = 1.75:1$

Work done can be calculated by:

$W = F \cdot d$

Where:

• $F = 250 N$,
• $d = 15 m$.

Thus, the work done is:

$W = 250 \times 15 = 3750\ Joules\ or\ N.m$