Let's solve the problem step by step: ### **Given Data** - **Diameter of pipe (\(d_1\))**: 4 in - **Diameter of orifice (\(d_2\))**: 1.8 in - **Height difference in manometer (\(h\))**: 8 in - **Density of water (\(\rho\))**: 62.36 lbm/ft³ - **Density of mercury (\(\rho_{Hg}\))**: 847 lbm/ft³ - **Dynamic viscosity of water (\(\mu\))**: 7.536 × 10⁻⁴ lbm/(ft·s) - **Temperature**: 60°F --- ## **Step 1: Convert dimensions to feet** - \( d_1 = 4~\text{in} = \frac{4}{12} = 0.333~\text{ft} \) - \( d_2 = 1.8~\text{in} = \frac{1.8}{12} = 0.15~\text{ft} \) - \( h = 8~\text{in} = \frac{8}{12} = 0.667~\text{ft} \) --- ## **Step 2: Calculate Area** - \( A_1 = \frac{\pi}{4} d_1^2 = \frac{\pi}{4} (0.333)^2 = 0.0871~\text{ft}^2 \) - \( A_2 = \frac{\pi}{4} d_2^2 = \frac{\pi}{4} (0.15)^2 = 0.0177~\text{ft}^2 \) --- ## **Step 3: Pressure Difference from Manometer** The pressure difference (\(\Delta P\)) given by a mercury manometer is: \[ \Delta P = (\rho_{Hg} - \rho_{water}) g h \] But since we use lbm/ft³, and USCS units, divide by g_c for force units: \[ \Delta P = (\rho_{Hg} - \rho_{water}) h \] \[ \Delta P = (847 - 62.36) \times 0.667 = 784.64 \times 0.667 = 523.12~\text{lb/ft}^2 \] --- ## **Step 4: Apply Bernoulli and Continuity (Orifice Equation)** **Continuity:** \[ Q = A_1 V_1 = A_2 V_2 \] \[ V_2 = \frac{A_1}{A_2} V_1 \] **Bernoulli:** \[ \Delta P = \frac{1}{2} \rho (V_2^2 - V_1^2) \] \[ 523.12 = \frac{1}{2} \cdot 62.36 \cdot (V_2^2 - V_1^2) \] Substitute \(V_2 = \frac{A_1}{A_2}V_1\): \[ V_2 = \frac{0.0871}{0.0177} V_1 = 4.92 V_1 \] Plug into Bernoulli: \[ 523.12 = \frac{1}{2} \cdot 62.36 \cdot [ (4.92 V_1)^2 - V_1^2 ] \] \[ 523.12 = 31.18 \cdot [24.2 V_1^2 - V_1^2 ] \] \[ 523.12 = 31.18 \cdot 23.2 V_1^2 \] \[ 523.12 = 724.98 V_1^2 \] \[ V_1^2 = \frac{523.12}{724.98} = 0.7216 \] \[ V_1 = \sqrt{0.7216} = 0.849~\text{ft/s} \] --- ### **Step 5: Volume Flow Rate** \[ Q = A_1 V_1 = 0.0871 \times 0.849 = 0.074~\text{ft}^3/\text{s} \] --- ### **Step 6: Average Velocity** \[ \text{Average velocity} = V_1 = 0.849~\text{ft/s} \] --- ### **Step 7: Head Loss Caused by the Orifice Meter** Using energy equation: \[ h_L = \frac{\Delta P}{\gamma} - \frac{V_2^2 - V_1^2}{2g} \] Where \(\gamma = \rho g\), and for water \(\gamma = 62.36~\text{lb/ft}^3\). Calculate velocity terms: \[ V_2 = 4.92 \times 0.849 = 4.18~\text{ft/s} \] \[ V_2^2 - V_1^2 = (4.18)^2 - (0.849)^2 = 17.47 - 0.721 = 16.75 \] \[ \frac{V_2^2 - V_1^2}{2g} = \frac{16.75}{2 \times 32.174} = \frac{16.75}{64.348} = 0.260~\text{ft} \] \[ \frac{\Delta P}{\gamma} = \frac{523.12}{62.36} = 8.39~\text{ft} \] \[ h_L = 8.39 - 0.26 = 8.13~\text{ft H}_2\text{O} \] --- ## **Final Answers** **Volume flow rate of water through the pipe:** \[ \boxed{0.074~\text{ft}^3/\text{s}} \] **Average velocity:** \[ \boxed{0.85~\text{ft/s}} \] **Head loss caused by the orifice meter:** \[ \boxed{8.13~\text{ft H}_2\text{O}} \] --- If you need units in other forms or more explanation, let me know!