Let's analyze and solve this problem step-by-step. --- ## **Problem Summary:** - A **variable mesh screen** produces a **linear and axysymetric velocity profile**. - **D = 2 ft** (diameter of the duct). - The duct has **upstream** and **downstream** static pressures: - **p₁ = .2 psi** (upstream) - **p₂ = .15 psi** (downstream) - **V₁ = 100 ft/s** (average velocity upstream). - The **flow cross-sectional area** is **uniform**. - **Question:** Calculate the **screen drag force** (drag exerted by the screen on the air), neglecting the force exerted by the duct wall. --- ## Step 1: Understand the flow and what is asked - The problem involves airflow through a variable mesh screen, which causes a velocity change. - The static pressure drops from **p₁** to **p₂**. - The **drag force** exerted by the screen relates to the momentum change of the air passing through it. - Since the problem states **neglecting the force exerted by the duct wall**, the focus is on the **force exerted by the screen** that causes the velocity change. --- ## Step 2: Key assumptions and concepts - **Flow is steady, incompressible**, and **non-viscous** (since no mention of viscosity, typical in these problems). - **Flow is 1D** in the cross-section. - The **pressure difference** is responsible for accelerating the airflow through the screen. - The **mass flow rate** \( \dot{m} \) is conserved across the section. --- ## Step 3: Calculate the **mass flow rate** \( \dot{m} \) - Cross-sectional area \( A \): \[ A = \frac{\pi D^2}{4} = \frac{\pi \times (2)^2}{4} = \pi \text{ ft}^2 \] - **Upstream velocity** \( V_1 = 100 \, \text{ft/s} \). - **Density** of air \( \rho \): We need to find \( \rho \). Using the ideal gas law: \[ p = \rho R T \] But since the static pressures are given in psi, and typical air density at standard conditions (~14.7 psi, 59°F) is about: \[ \rho \approx .075 \, \text{lb}_m/\text{ft}^3 \] (standard density of air at sea level). --- ## Step 4: Calculate **mass flow rate** \[ \dot{m} = \rho \times A \times V_1 \] \[ \dot{m} = .075 \times \pi \times 100 \approx .075 \times 3.1416 \times 100 \approx 23.56 \, \text{lb}_m/\text{s} \] --- ## Step 5: Determine **velocity downstream** \( V_2 \) - **Static pressure difference**: \[ \Delta p = p_1 - p_2 = .2 - .15 = .05 \, \text{psi} \] - Convert pressure to **lb/ft²**: \[ 1 \, \text{psi} = 144 \, \text{lb/ft}^2 \] \[ \Delta p = .05 \times 144 = 7.2 \, \text{lb/ft}^2 \] - **Bernoulli's equation** (assuming negligible height differences and no losses): \[ p_1 + \frac{1}{2}\rho V_1^2 = p_2 + \frac{1}{2}\rho V_2^2 \] Rearranged: \[ \frac{1}{2}\rho_2^2 = \frac{1}{2}\rho V_1^2 + (p_1 -_2) \] - Solve for \( V_2 \): \[ V_2 = \sqrt{V_1^2 + \frac{2}{\rho} (p_1 - p_2)} \] - Plug in values: \[ V_2 = \sqrt{(100)^2 + \frac{2}{.075} \times 7.2} \] \[ V_2 = \sqrt{10000 + \frac{2 \times 7.2}{.075}} \] \[ V_2 = \sqrt{10000 + \frac{14.4}{.075}} = \sqrt{10000 + 192} \] \[ V_2 = \sqrt{10192} \approx 101 \, \text{ft/s} \] (So, the velocity increases slightly due to the pressure drop). --- ## Step 6: Calculate the **drag force** exerted by the screen The **drag force** \( F_d \) is equal to the **rate of change of momentum** across the screen: \[ F_d = \dot{m} (V_2 - V_1) \] - \( V_2 \approx 101 \text{ ft/s} \), - \( V_1 = 100 \text{ ft/s} \), - \( \dot{m} \approx 23.56 \, \text{lb}_m/\text{s} \). Calculate: \[ F_d = 23.56 \times (101 - 100) = 23.56 \times 1 = \boxed{23.56 \, \text{lb}_f} \] --- ## **Final Answer:** ### The **screen drag force** is approximately **23.56 lb_f**. --- ## **Summary:** - Calculated the **mass flow rate** from upstream conditions. - Used Bernoulli's equation to estimate the **velocity downstream**. - Determined the **drag force** as the **momentum change** across the screen. --- If you need further clarification or details, feel free to ask!