Let's break down the given problem and solve it step by step. --- ## **Problem Statement** Air (as an ideal gas) flows through a diverging duct, 15 m long. Given: - **Stagnation temperature, \( T_0 = 500\,K \)** - **Stagnation pressure, \( P_0 = 200\,kPa \)** - **Inlet diameter, \( D_1 = 0.1\,m \)** - **Exit diameter, \( D_2 = 0.15\,m \)** - **Shock at \( x = 10\,m \) from entrance** - **Mass flow rate, \( \dot{m} = 2\,kg/s \)** - **Exit pressure, \( P_2 = 100\,kPa \)** - **Heat transfer to the duct, \( \dot{Q} = 70\,kW \)** - **Coefficient of friction, \( f = 0.02 \)** - **Find:** - Exit Mach number (\( M_2 \)) - Total force acting on the duct --- ## **Step 1: List Important Equations** ### **Continuity Equation** \[ \dot{m} = \rho A V \] ### **Energy Equation (with Heat Transfer)** \[ \dot{Q} = \dot{m}(h_2 - h_1) \] ### **Mach Number** \[ M = \frac{V}{a} \qquad a = \sqrt{\gamma R T} \] ### **Isentropic Relations** \[ T_0 = T \left[ 1 + \frac{\gamma - 1}{2}M^2 \right] \] \[ P_0 = P \left[ 1 + \frac{\gamma - 1}{2}M^2 \right]^{\frac{\gamma}{\gamma-1}} \] > Where for air, \( \gamma = 1.4, R = 287~J/kg\cdot K \) --- ## **Step 2: Find Areas** \[ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} \times (0.1)^2 = 0.00785~m^2 \] \[ A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} \times (0.15)^2 = 0.01767~m^2 \] --- ## **Step 3: Inlet Properties** ### **Find Inlet Mach Number, \( M_1 \)** Use isentropic relations, mass flow, and area: \[ T_0 = 500~K \] \[ P_0 = 200~kPa \] \[ A_1 = 0.00785~m^2 \] \[ \dot{m} = 2~kg/s \] The stagnation density: \[ \rho_0 = \frac{P_0}{R T_0} = \frac{200,000}{287 \times 500} = 1.393~kg/m^3 \] From isentropic relations: \[ \dot{m} = \rho_0 A_1 \sqrt{\gamma R T_0} M_1 \left[1 + \frac{\gamma-1}{2}M_1^2 \right]^{-\frac{\gamma+1}{2(\gamma-1)}} \] Plug in values: \[ 2 = 1.393 \times 0.00785 \times \sqrt{1.4 \times 287 \times 500} \times M_1 \left[1 + 0.2 M_1^2\right]^{-3} \] \[ \sqrt{1.4 \times 287 \times 500} = \sqrt{200,900} \approx 448.21 \] \[ 1.393 \times 0.00785 \times 448.21 \approx 4.899 \] So: \[ 2 = 4.899 \times M_1 \left[1 + 0.2 M_1^2\right]^{-3} \] \[ M_1 \left[1 + 0.2 M_1^2\right]^{-3} = \frac{2}{4.899} = 0.408 \] Let’s solve for \( M_1 \): - Try \( M_1 = 0.6 \): \( [1 + 0.2 \times 0.36]^{-3} = [1.072]^{-3} = 0.773 \), \( 0.6 \times 0.773 = 0.464 \) - Try \( M_1 = 0.5 \): \( [1 + 0.2 \times 0.25]^{-3} = [1.05]^{-3} = 0.863 \), \( 0.5 \times 0.863 = 0.431 \) - Try \( M_1 = 0.45 \): \( [1 + 0.2 \times 0.2025]^{-3} = [1.0405]^{-3} = 0.888 \), \( 0.45 \times 0.888 = 0.4 \) So, \( M_1 \approx 0.45 \) --- ## **Step 4: Exit Mach Number, \( M_2 \)** This is more complex due to: - Heat addition - Friction - Area change - Shock But, you are given **exit pressure** after the shock and heat transfer. ### **Approach:** 1. Find properties before the shock (using Fanno or Rayleigh tables is ideal, but here we use concepts). 2. Use conservation of mass, energy, and momentum across the duct (with heat and shock). #### **A. Use Energy Equation with Heat** \[ \dot{Q} = \dot{m} (h_2 - h_1) \] \[ h = c_p T \] For air, \( c_p = 1005~J/kg\cdot K \) \[ T_2 = T_1 + \frac{\dot{Q}}{\dot{m} c_p} \] \[ T_0 = T_1 (1 + 0.2 M_1^2) \implies T_1 = \frac{T_0}{1 + 0.2 M_1^2} = \frac{500}{1 + 0.2 \times (0.45)^2} = \frac{500}{1.0405} = 480.5~K \] \[ T_2 = 480.5 + \frac{70,000}{2 \times 1005} = 480.5 + 34.8 = 515.3~K \] #### **B. Use Isentropic Relation at Exit** Given \( T_2 \) and \( T_{0,2} \): \[ T_{0,2} = T_2 (1 + 0.2 M_2^2) \] But since there is heat transfer and friction, \( T_{0,2} \neq T_{0,1} \). #### **C. Use Mass Flow at Exit** \[ \dot{m} = \frac{P_2 A_2}{R T_2} V_2 \] \[ V_2 = M_2 \sqrt{\gamma R T_2} \] \[ 2 = \frac{100,000 \times 0.01767}{287 \times 515.3} \times M_2 \times \sqrt{1.4 \times 287 \times 515.3} \] Calculate terms: \[ \frac{100,000 \times 0.01767}{287 \times 515.3} = \frac{1767}{147,890} = 0.01195 \] \[ \sqrt{1.4 \times 287 \times 515.3} = \sqrt{207,438} = 455.4 \] \[ 2 = 0.01195 \times M_2 \times 455.4 \] \[ 2 = 5.44 \times M_2 \] \[ M_2 = \frac{2}{5.44} = 0.368 \] --- ## **Step 5: Force Acting on the Duct** \[ F = (\dot{m} V_1 + P_1 A_1) - (\dot{m} V_2 + P_2 A_2) \] We need to find \( V_1, V_2, P_1 \): - \( V_1 = M_1 \sqrt{\gamma R T_1} = 0.45 \times \sqrt{1.4 \times 287 \times 480.5} = 0.45 \times 438.6 = 197.4~m/s \) - \( V_2 = M_2 \sqrt{\gamma R T_2} = 0.368 \times \sqrt{1.4 \times 287 \times 515.3} = 0.368 \times 455.4 = 167.6~m/s \) Now, find \( P_1 \) using isentropic relation: \[ P_1 = \frac{P_0}{\left(1 + 0.2 M_1^2\right)^{\frac{1.4}{0.4}}} \] \[ \left(1 + 0.2 \times 0.2025\right)^{3.5} = 1.0405^{3.5} = 1.148 \] \[ P_1 = \frac{200,000}{1.148} = 174,240~Pa \] Now plug in values: \[ F = [2 \times 197.4 + 174,240 \times 0.00785] - [2 \times 167.6 + 100,000 \times 0.01767] \] \[ = [394.8 + 1,367] - [335.2 + 1,767] \] \[ = 1,761.8 - 2,102.2 = -340.4~N \] (The negative sign means the force is directed opposite to the flow, i.e., the duct experiences a net force in the direction of the flow.) --- ## **Final Answers** ### **1. Exit Mach Number** \[ \boxed{M_2 \approx 0.37} \] ### **2. Total Force Acting on the Duct** \[ \boxed{F \approx -340~N} \] --- #### **Note:** - These calculations are based on the information given and standard assumptions for air. The exact answer may vary depending on the treatment of friction and heat addition (Fanno/Rayleigh flow), but these steps cover the main approach for such a problem. - The answer given in the book is \( (3,369.5~N, 576.4~N) \), so if you need to match exactly, check if heat and friction are to be considered in a more detailed manner (using Fanno/Rayleigh tables). --- **If you want, let me know which part you'd like more detailed or a deeper explanation using Fanno/Rayleigh tables!**