9.1 A driver gear on the shaft of an electrical motor has 30 teeth and meshes with a gear on a countershaft which has 80 teeth - NSC Mechanical Technology Automotive - Question 9 - 2016 - Paper 1

The speed ratio of the gear train can be given as:

$\text{Speed ratio} = \frac{\text{Input}}{\text{Output}} = \frac{T_A}{T_D} = \frac{30}{63} = \frac{2}{4.2}$ That equals: $\text{Speed ratio} = 4.2 : 1$

Let:

• $N_1 =$ Rotation frequency of the pulley on the washing machine.
• $D_1 =$ Diameter of the driven pulley = 800 mm = 0.8 m.
• $D_2 =$ Diameter of the driving pulley = 600 mm = 0.6 m.
• $N_2 =$ Rotation frequency of the driving pulley = 7.2 r.s.

Using the diameter relationship: $N_1 \times D_1 = N_2 \times D_2$ Substituting values: $N_1 \times 0.8 = 7.2 \times 0.6 \Rightarrow N_1 = \frac{7.2 \times 0.6}{0.8} = 5.4 \ r.s.$

The rotational frequency of the pulley on the washing machine is $5.4 \ r.s.$

To calculate the power transmitted we use:

Let:

• $T_1 =$Tensile force in the tight side = 300 N.
• $T_2 =$Tensile force in the slack side = $2.5 \times 300 = 120 N.$
• $N_1 = 5.4 \ r.s.$

The power is calculated as: $P = (T_1 - T_2) \times N_1$ Substituting the values: $P = (300 - 120) \times 5.4 = 2,424.9 \ W = 2.42 \ kW$

The power that can be transmitted is $2.42 \ kW$.

The volume of a certain mass of gas can be changed by altering:

• its pressure,
• its temperature,
• both its pressure and temperature.

Boyle's Law states that the volume of a given mass of gas is inversely proportional to the pressure exerted on it, provided the temperature remains constant. This means that as pressure increases, volume decreases, and vice versa.

To calculate fluid pressure, we use the formula:

Let:

• $A_A = \frac{\pi d_A^2}{4} = \frac{\pi (0.04)^2}{4} = \frac{0.0050265}{4} = 0.0012566 m^2$
• $P_A = \frac{F_A}{A_A} = \frac{80}{0.0012566} \approx 63691.98 \ Pa$

Thus, the fluid pressure in the system is $63.66 \ kPa$.

We use the formula for pressure:

Let:

• $P_B = P_A$
• $F_B = 320 N$
• $A_B = \frac{F_B}{P_A}$
• $A_B = \frac{320}{63691.98}$
• $A_B \approx 0.00504 m^2$

Now, calculating the diameter: $d_B = 2 \times \sqrt{\frac{A_B}{\pi}} = 2 \times \sqrt{\frac{0.00504}{\pi}} \approx 0.08 m = 80 mm$