9.1 Calculate the: 9.1.1 Rotation frequency of the driver pulley in r/min (4) 9.1.2 Power transmitted in this system (4) 9.2 FIGURE 9.2 shows a gear-drive system - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2016 - Paper 1

To calculate the power transmitted, we can use the formula:

$P = (T_t - T_s) \times V$

Where:

• $T_t$ is the tension in the tight side (350 N)
• $T_s$ is the tension in the slack side (140 N)
• $V$ is the belt speed (36 m/s)

Based on the tension ratio, we have:

$T_s = \frac{T_t}{2.5} = \frac{350}{2.5} = 140 N$

Now substituting into the power formula:

$P = (350 - 140) \times 36 = 7560~\text{W}~\text{or}~7.56~\text{kW}$

Given the gear system, the rotation frequencies are linked through gear ratios:

Let:

• $N_{OUTPUT} = 160~\text{r/min}$
• $N_{INPUT}$ be the unknown frequency for the motor shaft.

Using the gear ratio relationships:

$N_{INPUT} = \frac{T_A}{T_B} \times \frac{T_C}{T_D} \times \frac{T_E}{T_F} \times N_{OUTPUT}$

Substituting the teeth counts:

$N_{INPUT} = \frac{18}{36} \times \frac{16}{46} \times \frac{40}{60} \times 160$

Calculating:

$N_{INPUT} = 0.5 imes \frac{16}{46} imes \frac{40}{60} imes 160 \approx 1380~\text{r/min}$

To find the velocity ratio (VR), we use:

$VR = \frac{N_{INPUT}}{N_{OUTPUT}}$

Using the previously calculated rotation frequencies:

$VR = \frac{1380}{160} = 8.625\approx 8.63$

To calculate the fluid pressure, we first find the area of piston A:

$A_A = \frac{\pi D^2}{4} = \frac{\pi (0.04)^2}{4} \approx 1.2566 \times 10^{-3}~m^2$

Now apply the pressure formula:

$P_A = \frac{F_A}{A_A} = \frac{275}{1.2566 \times 10^{-3}} \approx 218,884~\text{Pa} = 218.84~\text{kPa}$

For piston B, first calculate its area:

$A_B = \frac{\pi (0.075)^2}{4} \approx 4.42 \times 10^{-3}~m^2$

To find the force on piston B:

$F_B = P_B \times A_B = 218,884 \times 4.42 \times 10^{-3} \approx 967.32~N$

Now converting this to mass:

$\text{Mass} = \frac{F_B}{g} = \frac{967.32}{9.81} \approx 96.73~kg$

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