Let's **solve step-by-step** and check if the Reynolds number is high enough for the given \( C_d \). --- ## **Step 1: List Given Data** - **Pipe diameter, \( D \) = 10 cm = .10 m** - **Orifice diameter, \( d \) = 4.6 cm = .046 m** - **Manometer reading, \( h_{Hg} \) = 18 cm = .18 m (Hg)** - **Discharge coefficient, \( C_d = .61 \)** - **High Reynolds number (Re > 30,000)** - **Fluid: Water** --- ## **Step 2: Calculate Pressure Difference Across Orifice** A mercury manometer measures the pressure difference as a height of mercury: \[ \Delta P = \rho_{Hg} \cdot g \cdot h_{Hg} \] But the manometer is measuring water, so: \[ \Delta P = (\rho_{Hg} - \rho_{water}) g h_{Hg} \] - \( \rho_{Hg} = 13,600 \) kg/m³ - \( \rho_{water} = 1,000 \) kg/m³ - \( g = 9.81 \) m/s² - \( h_{Hg} = .18 \) m \[ \Delta P = (13600 - 100) \cdot 9.81 \cdot .18 = 12600 \cdot 9.81 \cdot .18 \] \[ = 12600 \cdot 1.7658 \approx 22,265 \text{ Pa} \] --- ## **Step 3: Orifice Flow Equation** The volumetric flow rate through an orifice: \[ Q = C_d \cdot A_2 \cdot \sqrt{\frac{2 \Delta P}{\rho (1-\beta^4)}} \] Where: - \( A_2 = \) area of orifice \( = \frac{\pi}{4} d^2 \) - \( \beta = \frac{d}{D} \) - \( \rho = \rho_{water} = 100 \) kg/m³ Calculate \( \beta \): \[ \beta = \frac{.046}{.10} = .46 \] Calculate \( 1 - \beta^4 \): \[ \beta^4 = (.46)^4 \approx .0447 \] \[ 1 - \beta^4 = 1 - .0447 = .9553 \] Calculate \( A_2 \): \[ A_2 = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.046)^2 \approx .00166 \text{ m}^2 \] --- ## **Step 4: Plug Values into the Orifice Equation** \[ Q = .61 \cdot .00166 \cdot \sqrt{\frac{2 \cdot 22265}{100 \cdot .9553}} \] First, calculate the square root term: \[ \frac{2 \cdot 22265}{100 \cdot .9553} = \frac{44530}{955.3} \approx 46.62 \] \[ \sqrt{46.62} \approx 6.827 \] Now, multiply: \[ Q = .61 \cdot .00166 \cdot 6.827 \] \[ = .61 \cdot .01134 \] \[ = .00692 \text{ m}^3/\text{s} \] --- ## **Step 5: Final Answer** \[ \boxed{Q = .00692 \ \text{m}^3/\text{s} = 6.92 \ \text{L/s}} \] --- ## **Step 6: Check Reynolds Number** Find velocity at orifice: \[ v_2 = \frac{Q}{A_2} = \frac{.00692}{.00166} \approx 4.17 \text{ m/s} \] \[ Re = \frac{v_2 \cdot d}{\nu} \] - Kinematic viscosity of water at room temp, \( \nu \approx 1 \times 10^{-6} \) m²/s \[ Re = \frac{4.17 \cdot .046}{1 \times 10^{-6}} = .1918 \times 10^6 = 191,800 \] **Since \( Re \gg 30,000 \), the flow is highly turbulent and using \( C_d = .61 \) is valid.** --- ## **Summary Table** | Quantity | Value | |---------------------------|--------------| | Volumetric Flow Rate, Q | .00692 m³/s (6.92 L/s) | | Orifice Velocity, \( v_2 \) | 4.17 m/s | | Reynolds Number, Re | 191,800 | | Is \( C_d \) valid? | **YES** | --- ## **Conclusion** **The volumetric flow rate of water is:** \[ \boxed{Q = .00692 \ \text{m}^3/\text{s} = 6.92 \ \text{L/s}} \] And \( C_d = .61 \) is valid since \( Re \gg 30,000 \).