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

Question 4 (continued)

4.2.2 To draw the shear force diagram, we calculate the shear force at critical points along the beam.

• $F_A = 20 \text{ kN}$ (down) at $x=0$
• $R_B = 31.25 \text{ kN}$ (up) at $x=1 \text{ m}$
• UDL = $5 \text{ kN/m}$ from $x=1 \text{ m}$ to $x=6 \text{ m}$
• $F_C = 30 \text{ kN}$ (down) at $x=6 \text{ m}$
• $R_D = 73.75 \text{ kN}$ (up) at $x=7 \text{ m}$
• $F_E = 30 \text{ kN}$ (down) at $x=8 \text{ m}$

Step 1: Calculate shear forces at key points.

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

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 ($1 Pa = 1 N/m^2$).

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 ($A_r$) = $0.524 m^2$
• Neglect ALL losses.

Step 1: Calculate the area of the plunger. $A_p = 0.25 \times A_r = 0.25 \times 0.524 m^2 = 0.131 m^2$

5.3.1 Step 2: Calculate the volume of liquid displaced by the plunger in 12 pump strokes. Volume per stroke: $V_{stroke} = A_p \times L_p = 0.131 m^2 \times 0.161 m = 0.021091 m^3$ Total volume for 12 strokes: $V_{total} = V_{stroke} \times 12 = 0.021091 m^3 \times 12 = 0.253092 m^3$ The volume of liquid displaced by the plunger in 12 pump strokes is $\boxed{0.253 m^3}$.

5.3.2 Step 1: Apply the principle of conservation of volume. Volume displaced by plunger = Volume moved by ram piston $A_p \times L_p = A_r \times L_r$ $0.021091 m^3 = 0.524 m^2 \times L_r$ Step 2: Calculate the distance moved by the ram piston ($L_r$). $L_r = \frac{0.021091 m^3}{0.524 m^2} \approx 0.04025 m$ Step 3: Convert to mm. $L_r = 0.04025 m \times 1000 mm/m = 40.25 mm$ The distance moved by the ram piston in 1 pump stroke is $\boxed{40.25 mm}$.

5.3.3 Step 1: Calculate the force exerted by the ram piston ($F_r$). According to Pascal's law, the pressure is constant throughout the fluid: $P = \frac{F_p}{A_p} = \frac{F_r}{A_r}$ $F_r = F_p \times \frac{A_r}{A_p} = 560 N \times \frac{0.524 m^2}{0.131 m^2} = 560 N \times 4 = 2240 N$ The force exerted by the ram piston is $\boxed{2240 N}$.

5.3.4 Step 1: Calculate the mechanical advantage (MA). $MA = \frac{Force on ram}{Force on plunger} = \frac{F_r}{F_p} = \frac{2240 N}{560 N} = 4$ The mechanical advantage of the press is $\boxed{4}$.

5.3.5 Step 1: Calculate the fluid pressure in the liquid. $P = \frac{F_p}{A_p} = \frac{560 N}{0.131 m^2} \approx 4274.81 Pa$ The fluid pressure in the liquid is $\boxed{4274.81 Pa}$.

5.4 Given:

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

Step 1: Calculate the area of one plunger. $A = \frac{\pi d^2}{4} = \frac{\pi (0.171 m)^2}{4} \approx 0.02294 m^2$ Step 2: Calculate the theoretical volume displaced per minute for one cylinder. $V_{theoretical, 1cyl} = A \times L \times N = 0.02294 m^2 \times 0.300 m \times 230 r/min \approx 1.58286 m^3/min$ Step 3: Calculate the theoretical volume displaced per minute for 3 cylinders. $V_{theoretical, 3cyl} = 3 \times V_{theoretical, 1cyl} = 3 \times 1.58286 m^3/min \approx 4.74858 m^3/min$ Step 4: Account for slip. Actual volume delivered = Theoretical volume $\times (1 - slip)$ $V_{actual} = 4.74858 m^3/min \times (1 - 0.16) = 4.74858 m^3/min \times 0.84 \approx 3.9888 m^3/min$ Step 5: Convert to litres per minute. $V_{actual} = 3.9888 m^3/min \times 1000 litres/m^3 \approx 3988.8 litres/min$ The volume of water delivered is $\boxed{3988.8 litres/minute}$.

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 $273.15$ ($0 K = -273.15 \text{ °C}$).

6.2 Given:

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

Step 1: Calculate the volume of the bubble at the surface ($V_1$). Radius at surface $r_1 = d_1/2 = 0.042 m / 2 = 0.021 \text{ m}$ $V_1 = \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi (0.021 m)^3 \approx 3.879 \times 10^{-5} m^3$ Step 2: Calculate the volume of the bubble at the bottom ($V_2$). Radius at bottom $r_2 = d_2/2 = 0.016 m / 2 = 0.008 \text{ m}$ $V_2 = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (0.008 m)^3 \approx 2.145 \times 10^{-6} m^3$ Step 3: Apply the combined gas law to find the pressure at the bottom ($P_2$). $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ $P_2 = P_1 \frac{V_1}{V_2} \frac{T_2}{T_1}$ $P_2 = 101325 Pa \times \frac{3.879 \times 10^{-5} m^3}{2.145 \times 10^{-6} m^3} \times \frac{283.15 K}{300.15 K}$ $P_2 \approx 101325 Pa \times 18.0848 \times 0.9433 \approx 1730000 Pa$ Step 4: Calculate the depth of the ocean ($h$) using the hydrostatic pressure formula. $P_2 = P_1 + \rho g h$ $h = \frac{P_2 - P_1}{\rho g}$ $h = \frac{1730000 Pa - 101325 Pa}{1000 kg/m^3 \times 9.8 m/s^2}$ $h = \frac{1628675 Pa}{9800 N/m^3} \approx 166.19 m$ The depth of the ocean is $\boxed{166.19 m}$.