Let's solve the questions step by step: --- ## (a) **Define pressure and list 3 different pressure units that are frequently used.** ### **Definition of Pressure:** Pressure is defined as the amount of force exerted per unit area of a surface. \[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \] ### **Three Common Pressure Units:** 1. **Pascal (Pa)** — SI unit, \(1 \text{ Pa} = 1 \text{ N/m}^2\) 2. **Bar** — \(1 \text{ bar} = 100,000 \text{ Pa}\) 3. **Atmosphere (atm)** — \(1 \text{ atm} = 101,325 \text{ Pa}\) --- ## (b) **Absolute Pressure at B Calculation** ### **Given Data:** - Pressure at A, \(P_A = 100\, \text{kN/m}^2 = 100,000\, \text{N/m}^2\) - Water column: \(h_1 = 650\, \text{mm} = .65\, \text{m}\) - Oil column (\(\text{SG}=.85\)): \(h_2 = 250\, \text{mm} = .25\, \text{m}\) - Mercury (\(\text{SG}=13.6\)): \(h_3 = 150\, \text{mm} = .15\, \text{m}\) - Gravitational acceleration, \(g = 9.81\, \text{m/s}^2\) - Densities: - Water, \(\rho_{water} = 100\, \text{kg/m}^3\) - Oil, \(\rho_{oil} = .85 \times 100 = 850\, \text{kg/m}^3\) - Mercury, \(\rho_{Hg} = 13.6 \times 100 = 13600\, \text{kg/m}^3\) --- ### **Step-by-Step Solution:** #### **1. Pressure at point X from A's side:** Starting from A and moving down: - Down .65 m of water, **pressure increases**: \[ P_X (\text{left}) = P_A + \rho_{water} g h_1 \] #### **2. Pressure at point X from B's side:** From B, move: - Down .25 m of oil (**increase**) - Down .15 m of mercury (**increase**) - Up .15 m of water (**decrease**; since the water column is above the X datum on left side) So, \[ P_X (\text{right}) = P_B + \rho_{oil} g h_2 + \rho_{Hg} g h_3 - \rho_{water} g h_3 \] --- #### **3. Equate Pressures at X:** \[ P_X (\text{left}) = P_X (\text{right}) \] \[ P_A + \rho_{water} g h_1 = P_B + \rho_{oil} g h_2 + \rho_{Hg} g h_3 - \rho_{water} g h_3 \] --- #### **4. Solve for \( P_B \):** \[ P_B = P_A + \rho_{water} g h_1 - \rho_{oil} g h_2 - (\rho_{Hg} - \rho_{water}) g h_3 \] --- #### **5. Plug in the values:** ##### Constants: - \(P_A = 100,000\, \text{N/m}^2\) - \(\rho_{water} = 100\, \text{kg/m}^3\) - \(h_1 = .65\, \text{m}\) - \(\rho_{oil} = 850\, \text{kg/m}^3\) - \(h_2 = .25\, \text{m}\) - \(\rho_{Hg} = 13,600\, \text{kg/m}^3\) - \(h_3 = .15\, \text{m}\) - \(g = 9.81\, \text{m/s}^2\) ##### Now calculate each term: - \(\rho_{water} g h_1 = 100 \times 9.81 \times .65 = 6,376.5\, \text{N/m}^2\) - \(\rho_{oil} g h_2 = 850 \times 9.81 \times .25 = 2,085.6\, \text{N/m}^2\) - \((\rho_{Hg} - \rho_{water}) g h_3 = (13,600 - 1,000) \times 9.81 \times .15 = 12,600 \times 9.81 \times .15 = 18,540.9\, \text{N/m}^2\) ##### Substitute into the equation: \[ P_B = 100,000 + 6,376.5 - 2,085.6 - 18,540.9 \] \[ P_B = 100,000 + 6,376.5 - 2,085.6 - 18,540.9 = 100,000 + 6,376.5 - 20,626.5 \] \[ P_B = 106,376.5 - 20,626.5 = 85,750\, \text{N/m}^2 \] --- ## **Final Answers** ### (a) - **Definition:** Pressure is force per unit area. - **Three units:** Pascal (Pa), bar, atmosphere (atm). ### (b) - **Absolute pressure at B:** \[ \boxed{85,750\, \text{N/m}^2\ (\text{or}\ 85.75\, \text{kPa})} \]