Let's solve the problem **step by step**: --- ## **Given Data** - Volumetric flow rate, \( Q = 0.02 \, \mathrm{m^3/s} \) - Diameter at A, \( D_A = 100 \, \mathrm{mm} = 0.1 \, \mathrm{m} \) - Diameter at B, \( D_B = 165 \, \mathrm{mm} = 0.165 \, \mathrm{m} \) - The Pitot tube at B measures the **stagnation (total) pressure**. - The static tap at A measures the **static pressure**. --- ## **Step 1: Find Velocities at A and B** ### **Area Calculation** \[ A = \frac{\pi}{4} D^2 \] - At A: \[ A_A = \frac{\pi}{4} (0.1^2) = 0.00785 \, \mathrm{m^2} \] - At B: \[ A_B = \frac{\pi}{4} (0.165^2) = 0.0214 \, \mathrm{m^2} \] ### **Velocity Calculation** \[ V = \frac{Q}{A} \] - At A: \[ V_A = \frac{0.02}{0.00785} = 2.548 \, \mathrm{m/s} \] - At B: \[ V_B = \frac{0.02}{0.0214} = 0.935 \, \mathrm{m/s} \] --- ## **Step 2: Apply Bernoulli's Equation** Bernoulli's Equation between point A (static tap) and point B (Pitot tube): \[ P_A + \frac{1}{2}\rho V_A^2 + \rho g z_A = P_{B,\text{static}} + \frac{1}{2}\rho V_B^2 + \rho g z_B \] **But:** - The Pitot tube at B measures **stagnation pressure** (\(P_{B,\text{stag}}\)): \[ P_{B,\text{stag}} = P_{B,\text{static}} + \frac{1}{2} \rho V_B^2 \] So, Bernoulli's equation becomes: \[ P_A + \frac{1}{2}\rho V_A^2 + \rho g z_A = P_{B,\text{stag}} + \rho g z_B \] Assume \(z_A = 0\), \(z_B = 0.4\,\mathrm{m}\) (since B is 400 mm above A): \[ P_A + \frac{1}{2}\rho V_A^2 = P_{B,\text{stag}} + \rho g (0.4) \] --- ## **Step 3: Find the Height h in the Pitot Tube** The **height \(h\)** is the difference in pressure head between the stagnation pressure at B and the static pressure at A, as read by the Pitot tube: \[ h = \frac{P_{B,\text{stag}} - P_A}{\rho g} \] From previous step: \[ P_A + \frac{1}{2}\rho V_A^2 = P_{B,\text{stag}} + \rho g (0.4) \] \[ P_{B,\text{stag}} - P_A = \frac{1}{2}\rho V_A^2 - \rho g (0.4) \] \[ h = \frac{\frac{1}{2}\rho V_A^2 - \rho g (0.4)}{\rho g} \] \[ h = \frac{V_A^2}{2g} - 0.4 \] --- ## **Step 4: Plug in the Values** - \(V_A = 2.548\,\mathrm{m/s}\) - \(g = 9.81\,\mathrm{m/s^2}\) \[ h = \frac{(2.548)^2}{2 \times 9.81} - 0.4 \] \[ h = \frac{6.494}{19.62} - 0.4 \] \[ h = 0.331 - 0.4 = -0.069\,\mathrm{m} \] --- ## **Step 5: Interpret the Result** A negative value means the Pitot tube at B will be **lower** than the static tap at A by 0.069 m (6.9 cm). This makes sense because the static pressure at A is higher than the stagnation pressure at B plus the elevation difference. --- ## **Final Answer** \[ \boxed{ h = -0.069 \text{ m} \approx -6.9 \text{ cm} } \] **The water in the Pitot tube at B will rise to a level 6.9 cm below the static pressure tap at A.** --- ### **Summary of Steps** 1. Calculated areas and velocities at A and B. 2. Applied Bernoulli’s equation, accounting for elevation difference. 3. Related the pressure difference to the height in the Pitot tube. 4. Substituted values and solved for \( h \).