Let's solve the problem step by step. ### **Given:** - Air in pipeline (\(\gamma_{air} = 0.0763 \text{ lb/ft}^3\)) - Neglect compressibility effects - Air flow rate (\(Q\)) = 75 cfs - Area at section 1 (\(A_1\)) = 1.0 sq. ft - Piezometer at section 2, area (\(A_2\)) = ? - Water should just be drawn up to the piezometer opening (i.e., pressure at the piezometer opening equals the pressure of the water surface). - Piezometer height = 6 in = 0.5 ft - Manometer reading = 1 in Hg = 1/12 ft Hg ### **Step 1: Apply the Continuity Equation** \[ Q = V_1 A_1 = V_2 A_2 \] \[ 75 = V_1 \cdot 1 = V_2 \cdot A_2 \] \[ V_1 = 75 \text{ ft/s} \] \[ V_2 = \frac{75}{A_2} \] --- ### **Step 2: Apply Bernoulli's Equation (Neglecting Losses)** Between section 1 and section 2 (both at the same elevation): \[ \frac{P_1}{\gamma_{air}} + \frac{V_1^2}{2g} = \frac{P_2}{\gamma_{air}} + \frac{V_2^2}{2g} \] --- ### **Step 3: Pressure at Section 2** Since water is just being drawn up to the piezometer opening, the pressure at section 2 is equal to the pressure at the water surface plus the hydrostatic column: \[ P_2 = P_{atm} - \gamma_{water} h \] where \(h = 0.5\) ft and \(\gamma_{water} = 62.4 \text{ lb/ft}^3\). --- ### **Step 4: Manometer Reading** The manometer shows a difference between the two points (1 inch Hg): \[ 1 \text{ in Hg} = 0.0833 \text{ ft Hg} \] \[ \gamma_{Hg} = 849 \text{ lb/ft}^3 \] Pressure difference: \[ \Delta P = \gamma_{Hg} h_{Hg} = 849 \times 0.0833 = 70.74 \text{ lb/ft}^2 \] --- ### **Step 5: Set up Bernoulli's Equation** \[ \frac{P_1}{\gamma_{air}} + \frac{V_1^2}{2g} = \frac{P_2}{\gamma_{air}} + \frac{V_2^2}{2g} \] \[ P_1 - P_2 = \gamma_{air} \left( \frac{V_2^2 - V_1^2}{2g} \right) \] But from the manometer: \[ P_1 - P_2 = 70.74 \text{ lb/ft}^2 \] --- ### **Step 6: Substitute the Velocities** \[ 70.74 = 0.0763 \left( \frac{ (\frac{75}{A_2})^2 - 75^2 }{2 \times 32.2} \right) \] --- ### **Step 7: Solve for \(A_2\)** \[ 70.74 = 0.0763 \left( \frac{ \left(\frac{75^2}{A_2^2}\right) - 75^2 }{64.4} \right) \] \[ 70.74 \times 64.4 = 0.0763 \left( \frac{75^2}{A_2^2} - 75^2 \right) \] \[ 4555.56 = 0.0763 \left( \frac{5625}{A_2^2} - 5625 \right) \] \[ \frac{4555.56}{0.0763} = \frac{5625}{A_2^2} - 5625 \] \[ 59743 = \frac{5625}{A_2^2} - 5625 \] \[ 59743 + 5625 = \frac{5625}{A_2^2} \] \[ 65368 = \frac{5625}{A_2^2} \] \[ A_2^2 = \frac{5625}{65368} \] \[ A_2^2 = 0.0861 \] \[ A_2 = \sqrt{0.0861} = 0.293 \text{ ft}^2 \] --- ### **Final Answer:** **The largest \(A_2\) that will cause water to be drawn up to the piezometer opening is:** \[ \boxed{0.293 \text{ ft}^2} \] --- **If you have any follow-up or need clarification on any step, let me know!**