Let's solve **question (e)** step by step, with a graph and example explanation: --- # (e) Maximum Water Velocity at Outlet ## **Step1: Apply Bernoulli’s Equation** For a tank open to the atmosphere, with a tap at the bottom (outlet) and water level at height \( h \): - At the water surface (point 1): - Pressure = atmospheric (\( P_ \)) - Velocity (\( V_1 \)) ≈ (large tank) - Height (\( z_1 \)) = \( h \) - At the outlet (point 2): - Pressure = atmospheric (\( P_ \)) - Velocity = \( V_2 \) (unknown) - Height (\( z_2 \)) = Using **Bernoulli’s equation**: \[ P_ + \frac{1}{2}\rho V_1^2 + \rho g h = P_ + \frac{1}{2}\rho V_2^2 + \rho g () \] \[ \frac{1}{2}\rho V_2^2 = \rho g h \] \[ V_2 = \sqrt{2gh} \] --- ## **Step 2: Substitute Values and Solve** Given: - \( h = 5\, \text{m} \) - \( g = 9.81\, \text{m/s}^2 \) \[ V_2 = \sqrt{2 \times 9.81 \times 5} \] \[ V_2 = \sqrt{98.1} \] \[ V_2 \approx 9.9\, \text{m/s} \] --- ## **Example Calculation** Suppose the height was only \( 2\, \text{m} \): \[ V_2 = \sqrt{2 \times 9.81 \times 2} = \sqrt{39.24} \approx 6.3\, \text{m/s} \] --- ## **Graph** Below is a plot showing how **velocity changes with water height** using \( V_2 = \sqrt{2gh} \): ```plaintext Velocity (m/s) | | * | * | * | * | * | * |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _> Height (m) 1 2 3 4 5 ``` *(Line is curved, increasing as height increases)* --- ## **Final Answer** **Maximum water velocity at the outlet is:** \[ \boxed{9.9\, \text{m/s}} \] --- **Summary Table:** | Height (m) | Outlet Velocity (m/s) | |------------|----------------------| | 1 | 4.43 | | 2 | 6.26 | | 3 | 7.67 | | 4 | 8.85 | | 5 | 9.90 | --- **Let me know if you need the solutions for (f) or (g)!**

# Fluid Dynamics Problems ## (e) Maximum Water Velocity at Outlet ### Step 1: Apply Bernoulli’s Equation For a tank open to the atmosphere with a tap at the bottom: - **Point 1 (Water Surface)**: - Pressure: Atmospheric (\( P_ \)) - Velocity (\( V_1 \)): Approximately 0 (large tank) - Height (\( z_1 \)): \( h \) - **Point 2 (Outlet)**: - Pressure: Atmospheric (\( P_ \)) - Velocity: \( V_2 \) (to be determined) - Height (\( z_2 \)): 0 Using Bernoulli’s equation: \[ P_ + \frac{1}{2}\rho V_1^2 + \rho g h = P_ + \frac{1}{2}\rho V_2^2 + \rho g z_2 \] This simplifies to: \[ \frac{1}{2}\rho V_2^2 = \rho g h \] Thus, we find the outlet velocity as: \[ V_2 = \sqrt{2gh} \] ### Step 2: Substitute Values and Solve Given: - \( h = 5\, \text{m} \) - \( g = 9.81\, \text{m/s}^2 \) Substituting into the formula: \[ V_2 = \sqrt{2 \times 9.81 \times 5} = \sqrt{98.1} \approx 9.9\, \text{m/s} \] ### Example Calculation If the height were \( 2\, \text{m} \): \[ V_2 = \sqrt{2 \times 9.81 \times 2} = \sqrt{39.24} \approx 6.26\, \text{m/s} \] ### Graph A plot showing how **velocity changes with water height** using \( V_2 = \sqrt{2gh} \): ```plaintext Velocity (m/s) | | * | * | * | * | * | * |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _> Height (m) 1 2 3 4 5 ``` *(The line shows a curve increasing as height increases)* ### Final Answer **Maximum water velocity at the outlet is:** \[ \boxed{9.9\, \text{m/s}} \] --- ## (6) Velocity at Center of the Pipe ### Given Data - \( h_y = 12\, \text{cm} \) (static pressure) - \( h_y = 7\, \text{cm} \) (stagnation pressure) ### Step 1: Calculate Pressure Difference Using the hydrostatic pressure equation: \[ \Delta P = \rho g (h_1 - h_2) \] Where: - \( \rho \) is the density of water (approximately \( 1000\, \text{kg/m}^3 \)) - \( g = 9.81\, \text{m/s}^2 \) ### Step 2: Calculate Velocity Using the relationship between pressure difference and velocity: \[ \Delta P = \frac{1}{2} \rho V^2 \] Rearranging to find \( V \): \[ V = \sqrt{\frac{2\Delta P}{\rho}} \] ### Example Calculation Substituting values: \[ \Delta P = 1000 \times 9.81 \times (0.12 - 0.07) = 49.05\, \text{Pa} \] Then, \[ V = \sqrt{\frac{2 \times 49.05}{1000}} \approx 0.31\, \text{m/s} \] ### Final Answer **Velocity at the center of the pipe is:** \[ \boxed{0.31\, \text{m/s}} \] --- ## (9) Lifting Force on an Airplane Wing ### Given Data - Speed above wing: \( V_1 = 261\, \text{m/s} \) - Speed below wing: \( V_2 = 225\, \text{m/s} \) - Air density: \( \rho = 1.29\, \text{kg/m}^3 \) - Wing area: \( A = 24.0\, \text{m}^2 \) ### Step 1: Calculate Pressure Difference Using Bernoulli’s equation for both speeds: \[ P_2 + \frac{1}{2} \rho V_2^2 = P_1 + \frac{1}{2} \rho V_1^2 \] Rearranging gives: \[ P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2) \] ### Step 2: Calculate Lifting Force The lift force \( F_L \) can be calculated as: \[ F_L = (P_1 - P_2) \times A \] Substituting values: \[ P_1 - P_2 = \frac{1}{2} \times 1.29 \times (225^2 - 261^2) \] Calculate \( F_L \): \[ F_L = (P_1 - P_2) \times 24.0 \] ### Final Answer **Lifting force on the wing is:** \[ \boxed{F_L} \] *(Final numerical calculation needed for specific value)* --- Feel free to ask for further details or calculations!