# 5.a) Definitions ### i) Lagrangian vs. Eulerian Flow Descriptions **Step 1:** - **Lagrangian Description:** Follows individual fluid particles as they move, tracking their position and properties over time. - **Eulerian Description:** Observes changes in fluid properties at fixed locations in space as the fluid moves past. **Explanation:** Lagrangian tracks particles; Eulerian fixes attention at spatial points. --- ### ii) One, Two, and Three Dimensional Flows **Step 2:** - **One-Dimensional Flow:** Flow parameters change only along one spatial dimension (e.g., flow in a straight pipe). - **Two-Dimensional Flow:** Parameters vary in two dimensions (e.g., flow over a flat plate). - **Three-Dimensional Flow:** Parameters vary in all three dimensions (e.g., real fluid flows in open space). **Explanation:** Dimensionality refers to how many directions the flow properties change. --- ### iii) Steady vs. Unsteady Flows **Step 3:** - **Steady Flow:** Fluid properties at any given point do not change with time. - **Unsteady Flow:** Fluid properties at points change with time. **Explanation:** Steady = time-invariant; unsteady = time-variant flow characteristics. --- ### iv) Control Volume **Step 4:** - **Control Volume:** A defined region in space (real or imaginary) through which fluid may flow, used for analysis in fluid mechanics. **Explanation:** It helps analyze the behavior of fluids as they enter, exist, or interact within a specific region. --- ### v) Continuity and Linear Momentum Equations **Step 5:** - **Continuity Equation:** Expresses mass conservation; for incompressible flow, the mass entering a control volume equals the mass leaving it. - **Linear Momentum Equation:** Derived from Newton's second law, relates the sum of forces on a control volume to the rate of change of momentum. **Explanation:** Continuity ensures mass conservation; momentum deals with force and motion in fluids. --- --- # 5. (b) Flow Rate Calculations ### **Given Data** - Specific Gravity (SG) = 1.4 - Pipe diameter, \( D = 150\, \text{mm} = .15\, \text{m} \) - Velocity, \( V = .8\, \text{m/s} \) --- ## **Step 1: Calculate Volumetric Flow Rate** \[ Q = A \times V \] \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (.15)^2 = .01767\, \text{m}^2 \] \[ Q = .01767 \times .8 = .01414\, \text{m}^3/\text{s} \] - **In L/s:** \( Q = .01414\, \text{m}^3/\text{s} \times 100 = 14.14\, \text{L/s} \) **Explanation:** Area of pipe is found and multiplied by velocity to get flow rate. --- ## **Step 2: Calculate Mass and Force Flow Rate** - **Density:** \( \rho = \text{SG} \times 100 = 1.4 \times 100 = 140\, \text{kg/m}^3 \) - **Mass flow rate:** \[ \dot{m} = \rho \times Q = 140 \times .01414 = 19.80\, \text{kg/s} \] - **Force flow rate (N/s or kN/s):** (Force = mass × gravity; \(g = 9.81\, \text{m/s}^2\)) \[ \text{Force flow rate} = \dot{m} \times g = 19.80 \times 9.81 = 194.24\, \text{N/s} = .194\, \text{kN/s} \] **Explanation:** Mass flow is density × volumetric flow. Force flow is mass flow × gravity. --- ## **Final Answers** - **Flow rate:** - \( \boxed{14.14\, \text{L/s}} \) - \( \boxed{.01414\, \text{m}^3/\text{s}} \) - \( \boxed{19.80\, \text{kg/s}} \) - \( \boxed{.194\, \text{kN/s}} \)