What is the difference between the Kelvin scale and the Celsius scale?

• At $x=0$ (just after A): $V_A = -20 \text{ kN}$
• At $x=1 \text{ m}$ (just before B): $V_{B, \text{left}} = -20 \text{ kN}$
• At $x=1 \text{ m}$ (just after B): $V_{B, \text{right}} = -20 \text{ kN} + R_B = -20 + 31.25 = 11.25 \text{ kN}$
• At $x=6 \text{ m}$ (just before C, end of UDL): $V_{C, \text{left}} = V_{B, \text{right}} - (5 \text{ kN/m} \times 5 \text{ m}) = 11.25 - 25 = -13.75 \text{ kN}$
• At $x=6 \text{ m}$ (just after C): $V_{C, \text{right}} = V_{C, \text{left}} - F_C = -13.75 - 30 = -43.75 \text{ kN}$
• At $x=7 \text{ m}$ (just before D): $V_{D, \text{left}} = -43.75 \text{ kN}$
• At $x=7 \text{ m}$ (just after D): $V_{D, \text{right}} = V_{D, \text{left}} + R_D = -43.75 + 73.75 = 30 \text{ kN}$
• At $x=8 \text{ m}$ (just before E): $V_{E, \text{left}} = 30 \text{ kN}$
• At $x=8 \text{ m}$ (just after E): $V_{E, \text{right}} = V_{E, \text{left}} - F_E = 30 - 30 = 0 \text{ kN}$

The shear force diagram starts at $-20 \text{ kN}$ at A, remains constant until B. At B, it jumps up to $11.25 \text{ kN}$. From B to C, it decreases linearly to $-13.75 \text{ kN}$ due to the UDL. At C, it drops to $-43.75 \text{ kN}$. It remains constant until D. At D, it jumps up to $30 \text{ kN}$. It remains constant until E, where it drops to $0 \text{ kN}$.

Question 5

5.1
The unit pascal (Pa) is the SI unit of pressure, defined as one newton per square meter ($1 \text{ Pa} = 1 \text{ N/m}^2$).

5.2
Pascal's law states that a pressure change at any point in a confined incompressible fluid is transmitted undiminished throughout the fluid such that the same change occurs everywhere.

5.3
Given:

• Area of plunger ($A_p$) = 25% of ram area ($A_r$)
• Stroke of plunger ($L_p$) = $0.161 \text{ m}$
• Force applied to plunger ($F_p$) = $560 \text{ N}$
• Area of the ram piston ($A_r$) = $0.524 \text{ m}^2$
• Neglect ALL losses.

$$A_p = 0.25 \times A_r = 0.25 \times 0.524 \text{ m}^2 = 0.131 \text{ m}^2$$

5.3.1

Volume per stroke:

$$V_{stroke} = A_p \times L_p = 0.131 \text{ m}^2 \times 0.161 \text{ m} = 0.021091 \text{ m}^3$$

Total volume for 12 strokes:

$$V_{total} = V_{stroke} \times 12 = 0.021091 \text{ m}^3 \times 12 = 0.253092 \text{ m}^3$$

The volume of liquid displaced by the plunger in 12 pump strokes is $\boxed{\text{0.253 m}^3}$.

5.3.2

Volume displaced by plunger = Volume moved by ram piston

$$A_p \times L_p = A_r \times L_r$$ $$0.021091 \text{ m}^3 = 0.524 \text{ m}^2 \times L_r$$

$$L_r = 0.04025 \text{ m} \times 1000 \text{ mm/m} = 40.25 \text{ mm}$$

The distance moved by the ram piston in 1 pump stroke is $\boxed{\text{40.25 mm}}$.

5.3.3

According to Pascal's law, the pressure is constant throughout the fluid:

$$P = \frac{F_p}{A_p} = \frac{F_r}{A_r}$$ $$F_r = F_p \times \frac{A_r}{A_p} = 560 \text{ N} \times \frac{0.524 \text{ m}^2}{0.131 \text{ m}^2} = 560 \text{ N} \times 4 = 2240 \text{ N}$$

The force exerted by the ram piston is $\boxed{\text{2240 N}}$.

5.3.4

$$MA = \frac{\text{Force on ram}}{\text{Force on plunger}} = \frac{F_r}{F_p} = \frac{2240 \text{ N}}{560 \text{ N}} = 4$$

The mechanical advantage of the press is $\boxed{4}$.

5.3.5

$$P = \frac{F_p}{A_p} = \frac{560 \text{ N}}{0.131 \text{ m}^2} \approx 4274.81 \text{ Pa}$$

The fluid pressure in the liquid is $\boxed{\text{4274.81 Pa}}$.

5.4
Given:

• Number of cylinders = 3
• Plunger diameter $d = 171 \text{ mm} = 0.171 \text{ m}$
• Stroke length $L = 300 \text{ mm} = 0.300 \text{ m}$
• Speed $N = 230 \text{ r/min}$
• Slip = 16%

Actual volume delivered = Theoretical volume $\times (1 - \text{slip})$

$$V_{actual} = 4.74858 \text{ m}^3/\text{min} \times (1 - 0.16) = 4.74858 \text{ m}^3/\text{min} \times 0.84 \approx 3.9888 \text{ m}^3/\text{min}$$

$$V_{actual} = 3.9888 \text{ m}^3/\text{min} \times 1000 \text{ litres/m}^3 \approx 3988.8 \text{ litres/min}$$

The volume of water delivered is $\boxed{\text{3988.8 litres/minute}}$.

Question 6

6.1
The Kelvin scale is an absolute temperature scale where 0 K represents absolute zero, the theoretical point of no molecular motion. The Celsius scale is a relative temperature scale where 0 °C is the freezing point of water and 100 °C is its boiling point. A 1-degree change in Celsius is equivalent to a 1-Kelvin change, but their zero points differ by 273.15 ($0 \text{ K} = -273.15 \text{ °C}$).

6.2
Given:

• Surface temperature $T_1 = 27 \text{ °C} = 300.15 \text{ K}$
• Surface pressure $P_1 = 101.325 \text{ kPa} = 101325 \text{ Pa}$
• Bottom temperature $T_2 = 10 \text{ °C} = 283.15 \text{ K}$
• Diameter of bubble at bottom $d_2 = 16 \text{ mm} = 0.016 \text{ m}$
• Diameter of bubble at surface $d_1 = 42 \text{ mm} = 0.042 \text{ m}$
• Density of water $\rho = 1000 \text{ kg/m}^3$

Radius at surface $r_1 = d_1/2 = 0.042 \text{ m} / 2 = 0.021 \text{ m}$

$$V_1 = \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi (0.021 \text{ m})^3 \approx 3.879 \times 10^{-5} \text{ m}^3$$

Radius at bottom $r_2 = d_2/2 = 0.016 \text{ m} / 2 = 0.008 \text{ m}$

$$V_2 = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (0.008 \text{ m})^3 \approx 2.145 \times 10^{-6} \text{ m}^3$$

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ $$P_2 = P_1 \frac{V_1}{V_2} \frac{T_2}{T_1}$$ $$P_2 = 101325 \text{ Pa} \times \frac{3.879 \times 10^{-5} \text{ m}^3}{2.145 \times 10^{-6} \text{ m}^3} \times \frac{283.15 \text{ K}}{300.15 \text{ K}}$$ $$P_2 \approx 101325 \text{ Pa} \times 18.0848 \times 0.9433 \approx 1730000 \text{ Pa}$$

$$P_2 = P_1 + \rho g h$$ $$h = \frac{P_2 - P_1}{\rho g}$$ $$h = \frac{1730000 \text{ Pa} - 101325 \text{ Pa}}{1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2}$$ $$h = \frac{1628675 \text{ Pa}}{9800 \text{ N/m}^3} \approx 166.19 \text{ m}$$

The depth of the ocean is $\boxed{\text{166.19 m}}$.