# Problem Breakdown We are asked to check if the **net natural draught** is sufficient for a boiler chimney, and to find the **exit velocity** and **diameter** of the chimney given: - **Mass flow rate of gases:** \( \dot{m} = 3 \) kg/s - **Chimney height:** \( h = 19 \) m - **Friction loss:** \( \Delta P_f = 40 \) kPa - **Exit velocity (given):** \( V = 4 \) m/s - **Air density:** \( \rho_{air} = 1.29 \) kg/m³ - **Flue gas density:** \( \rho_{gas} = 1.28 \) kg/m³ - **Ambient temperature:** \( T_a = 20^\circ \)C - **Average gas temperature:** \( T_g = 120^\circ \)C We need to: 1. Calculate the **net draught** (pressure difference produced by the chimney). 2. Check if it's sufficient (i.e., covers friction + exit velocity losses). 3. Find the **chimney diameter**. --- # Step 1: Draught Produced by Chimney The **static draught** (pressure difference) produced by the chimney is: \[ \Delta P_{chimney} = g h (\rho_{air} - \rho_{gas}) \] where - \( g = 9.81 \) m/s² \[ \Delta P_{chimney} = 9.81 \times 19 \times (1.29 - 1.28) \] \[ = 9.81 \times 19 \times .01 \] \[ = 9.81 \times .19 \] \[ = 1.8639 \ \text{Pa} \] --- # Step 2: Draught Required The **required draught** must cover both the **friction loss** and the **velocity head** at the exit. ## (a) Friction Loss Given: \( \Delta P_f = 40 \) kPa = \( 40,000 \) Pa ## (b) Velocity Head \[ \Delta P_{velocity} = \frac{1}{2} \rho_{gas} V^2 \] \[ = \frac{1}{2} \times 1.28 \times (4)^2 \] \[ = .64 \times 16 = 10.24 \ \text{Pa} \] ## (c) Total Required Draught \[ \Delta P_{required} = \Delta P_f + \Delta P_{velocity} = 40,000 + 10.24 = 40,010.24 \ \text{Pa} \] --- # Step 3: Sufficiency Check Compare **draught produced** vs **draught required**: - Draught produced: \( 1.86 \) Pa - Draught required: \( 40,010 \) Pa \[ \boxed{ \text{The net natural draught produced is \textbf{not sufficient}.} } \] --- # Step 4: Chimney Diameter Calculation Given \( \dot{m} = 3 \) kg/s, \( \rho_{gas} = 1.28 \) kg/m³, \( V = 4 \) m/s The volumetric flow rate (\( Q \)): \[ Q = \frac{\dot{m}}{\rho_{gas}} = \frac{3}{1.28} = 2.34375 \ \text{m}^3/\text{s} \] Chimney cross-sectional area (\( A \)): \[ A = \frac{Q}{V} = \frac{2.34375}{4} = .58594 \ \text{m}^2 \] Chimney diameter (\( D \)), since \( A = \frac{\pi}{4} D^2 \): \[ D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times .58594}{3.1416}} = \sqrt{\frac{2.34375}{3.1416}} = \sqrt{.7465} = .864 \ \text{m} \] --- # Step 5: Exit Velocity Confirmation Already given as \( 4 \) m/s; calculation above confirms this is consistent with mass flow and area. --- # Final Answers - **Net natural draught produced:** \( \boxed{1.86 \ \text{Pa}} \) - **Required draught:** \( \boxed{40,010 \ \text{Pa}} \) - **Is draught sufficient?** **No**, it is not sufficient. - **Chimney exit velocity:** \( \boxed{4 \ \text{m/s}} \) (given & confirmed) - **Chimney diameter:** \( \boxed{.864 \ \text{m}} \) --- ## Summary Table | Quantity | Value | |-------------------------|----------------------| | Draught produced | 1.86 Pa | | Draught required | 40,010 Pa | | Draught Sufficient? | No | | Exit Velocity | 4 m/s | | Chimney Diameter | .864 m | --- **Image alt text:** Diagram illustrating a vertical chimney of height 19 m, showing gas flow, pressure drop due to friction and velocity, with notation for air and gas densities.