## Step-by-Step Solution Let's solve each part step by step using the **Bernoulli Equation** and the **Continuity Equation**. --- ### **Given Data** - **Pipe diameters:** - Top: \( D_1 = 2\, \text{m} \) - Bottom: \( D_2 = 1\, \text{m} \) - **Heights:** - Top: \( z_1 = 10\, \text{m} \) - Bottom: \( z_2 = 0\, \text{m} \) (choose bottom as reference) - **Pressures:** - Both gages: \( p_1 = p_2 = 50\, \text{kPa} \) - **Fluid:** Water (incompressible, inviscid) - **Find:** Flow rate \( Q \) --- ## **Part 1: Flow Downward** ### 1. **Continuity Equation** \[ Q = A_1 v_1 = A_2 v_2 \] where \( A_1 = \frac{\pi}{4} D_1^2 \), \( A_2 = \frac{\pi}{4} D_2^2 \) \[ A_1 = \frac{\pi}{4} (2^2) = \pi \, \text{m}^2 \] \[ A_2 = \frac{\pi}{4} (1^2) = \frac{\pi}{4} \, \text{m}^2 \] So, \[ v_2 = \frac{A_1}{A_2} v_1 = \frac{\pi}{\pi/4} v_1 = 4v_1 \] --- ### 2. **Bernoulli's Equation (from top to bottom):** \[ \frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2 \] Given \( p_1 = p_2 \), so those terms cancel. \[ \frac{v_1^2}{2g} + z_1 = \frac{v_2^2}{2g} + z_2 \] Since \( z_2 = 0 \): \[ \frac{v_1^2}{2g} + 10 = \frac{v_2^2}{2g} \] Recall \( v_2 = 4v_1 \): \[ \frac{v_1^2}{2g} + 10 = \frac{(4v_1)^2}{2g} \] \[ \frac{v_1^2}{2g} + 10 = \frac{16v_1^2}{2g} \] \[ 10 = \frac{16v_1^2}{2g} - \frac{v_1^2}{2g} \] \[ 10 = \frac{15v_1^2}{2g} \] \[ v_1^2 = \frac{2g \times 10}{15} = \frac{20g}{15} = \frac{4g}{3} \] \[ v_1 = \sqrt{\frac{4g}{3}} \] With \( g = 9.81 \, \text{m/s}^2 \): \[ v_1 = \sqrt{\frac{4 \times 9.81}{3}} = \sqrt{\frac{39.24}{3}} = \sqrt{13.08} \approx 3.62\, \text{m/s} \] So, \[ v_2 = 4v_1 = 4 \times 3.62 \approx 14.48 \, \text{m/s} \] --- ### 3. **Flow Rate:** \[ Q = A_1 v_1 = \pi \times 3.62 \approx 11.4\, \text{m}^3/\text{s} \] --- ### **Final Answer (Part 1):** \[ \boxed{Q \approx 11.4\ \text{m}^3/\text{s}} \] --- ## **Part 2: Flow Upward** Now flow is **upward** (from 1 m section to 2 m section). The **lower gage** (narrow section) is reading \( 50\, \text{kPa} \). We are to find the pressure indicated at the **upper gage**. #### Let: - \( p_2 = 50\, \text{kPa} \) (bottom/narrow/1 m section) - \( v_2 = 14.48\, \text{m/s} \) (from previous calculation) - \( z_2 = 0 \) - \( v_1 = 3.62\, \text{m/s} \) - \( z_1 = 10 \) - \( p_1 = ? \) Apply Bernoulli (from bottom to top): \[ \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2 = \frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 \] Plug in values (\( z_2 = 0 \)): \[ \frac{p_2}{\rho g} + \frac{v_2^2}{2g} = \frac{p_1}{\rho g} + \frac{v_1^2}{2g} + 10 \] \[ \frac{p_1}{\rho g} = \frac{p_2}{\rho g} + \frac{v_2^2 - v_1^2}{2g} - 10 \] \[ p_1 = p_2 + \rho \left( \frac{v_2^2 - v_1^2}{2} - g \times 10 \right) \] Water: \( \rho = 1000\, \text{kg/m}^3 \) - \( p_2 = 50,000\, \text{Pa} \) - \( v_2 = 14.48 \, \text{m/s} \), \( v_1 = 3.62\, \text{m/s} \) Plug in: \[ p_1 = 50,000 + 1000 \left[ \frac{(14.48)^2 - (3.62)^2}{2} - 9.81 \times 10 \right] \] Calculate each term: - \( (14.48)^2 = 209.72 \) - \( (3.62)^2 = 13.10 \) - Difference: \( 209.72 - 13.10 = 196.62 \) - Half: \( 196.62/2 = 98.31 \) - \( 9.81 \times 10 = 98.1 \) - Difference: \( 98.31 - 98.1 = 0.21 \) So, \[ p_1 = 50,000 + 1000 \times 0.21 = 50,000 + 210 = 50,210 \, \text{Pa} \] **So the upper gage will read:** \[ \boxed{50.2\, \text{kPa}} \] --- ## **Summary Table** | Part | Flow Direction | Gage Readings | Flowrate (\(Q\)) / Pressure (\(p_1\)) | |------|-------------------|------------------------------|---------------------------------------| | 1 | Downward | Both 50 kPa | \( Q = 11.4\ \text{m}^3/\text{s} \) | | 2 | Upward | Lower: 50 kPa, Upper: ? | \( p_1 = 50.2\ \text{kPa} \) | --- ### **Let me know if you'd like further clarification!**