Let's break down the problem step-by-step. --- # Given Data: - **Flow type**: Constant-area, insulated duct - **Mach number, \( M \)**: .60 - **Static pressure, \( P_ \)**: 150 kPa - **Static temperature, \( T_ \)**: 300 K - **Duct length, \( L \)**: 45 cm = .45 m - **Duct diameter, \( D \)**: 3 cm = .03 m - **Friction coefficient, \( f \)**: .005 --- # Step 1: Find the cross-sectional area \[ A = \frac{\pi}{4} D^2 \] \[ A = \frac{\pi}{4} \times (.03)^2 = \frac{\pi}{4} \times .0009 \approx .0007069 \text{ m}^2 \] --- # Step 2: Find the speed of the flow at the inlet Using Mach number: \[ a = \sqrt{\gamma R T_} \] Where: - \( \gamma \) (ratio of specific heats) for air: 1.4 - \( R \) (specific gas constant for air): 287 J/(kg·K) Calculate the speed of sound: \[ a = \sqrt{1.4 \times 287 \times 300} \\ a = \sqrt{1.4 \times 86100} \\ a = \sqrt{120540} \approx 347.3 \text{ m/s} \] Now, flow velocity: \[ V = M \times a = .60 \times 347.3 \approx 208.4 \text{ m/s} \] --- # Step 3: Find the mass flow rate, \( \dot{m} \) \[ \dot{m} = \rho A V \] Where density \( \rho \) at the inlet: \[ \rho = \frac{P_}{R T_} = \frac{150,000}{287 \times 300} \approx \frac{150,000}{86100} \approx 1.74 \text{ kg/m}^3 \] Therefore, \[ \dot{m} = 1.74 \times .0007069 \times 208.4 \approx .257 \text{ kg/s} \] --- # Step 4: Find the change in total pressure due to friction The pressure loss \( \Delta P \) in a duct with friction is given by Darcy-Weisbach: \[ \Delta P = f \frac{L}{D} \frac{\rho V^2}{2} \] Calculate: \[ \frac{L}{D} = \frac{.45}{.03} = 15 \] Now, \[ \Delta P = .005 \times 15 \times \frac{1.74 \times (208.4)^2}{2} \] Calculate \( V^2 \): \[ (208.4)^2 \approx 43474 \] So, \[ \Delta P = .005 \times 15 \times \frac{1.74 \times 43474}{2} \] \[ = .075 \times \frac{1.74 \times 43474}{2} \] Calculate numerator: \[ 1.74 \times 43474 \approx 75609 \] Then, \[ \Delta P = .075 \times \frac{75609}{2} = .075 \times 37804.5 \approx 2835.3 \text{ Pa} \] --- # **Part (a): Mach number at the outlet** Since the duct is insulated and flow is steady, the change in stagnation pressure is due to friction. For an ideal flow with friction, the Mach number at the outlet can be approximated by considering the pressure loss. The static pressure at the outlet: \[ P_{out} = P_ - \Delta P = 150,000 - 2835.3 \approx 147,164.7 \text{ Pa} \] Using isentropic relations and the fact that the flow is steady and adiabatic, the Mach number can be approximated from the pressure ratio: \[ \frac{P}{P_} = \left(1 + \frac{\gamma - 1}{2} M^2\right)^{-\frac{\gamma}{\gamma - 1}} \] Rearranged as: \[ \left(\frac{P}{P_}\right)^{\frac{\gamma - 1}{\gamma}} = \frac{1}{1 + \frac{\gamma - 1}{2} M^2} \] Calculate the pressure ratio: \[ \frac{P_{out}}{P_} = \frac{147,165}{150,000} \approx .981 \] Now, \[ \left(.981\right)^{\frac{1.4 - 1}{1.4}} = .981^{.2857} \approx .9958 \] Thus, \[ .9958 = \frac{1}{1 + .2 M^2} \] which leads to: \[ 1 + .2 M^2 = \frac{1}{.9958} \approx 1.0042 \] \[ .2 M^2 = .0042 \] \[ M^2 = \frac{.0042}{.2} = .021 \] \[ M_{out} = \sqrt{.021} \approx .145 \] --- # **Final Answers:** ### a. Mach number at the outlet: \[ \boxed{ M_{out} \approx .145 } \] ### b. Static pressure and static temperature at the outlet: - Static pressure (already computed): \[ P_{out} \approx 147.2 \text{ kPa} \] - Static temperature at outlet: Using isentropic relations: \[ \frac{T_{out}}{T_} = \left(1 + \frac{\gamma - 1}{2} M_{out}^2\right)^{-1} \] Calculate: \[ 1 + .2 \times (.145)^2 = 1 + .2 \times .021 = 1.0042 \] Thus, \[ T_{out} = T_ \times 1.0042^{-1} \approx 300 \times .9958 \approx 298.7 \text{ K} \] --- # **Summary:** | Quantity | Approximate Value | |---|---| | Mach number at outlet | **.145** | | Static pressure at outlet | **147.2 kPa** | | Static temperature at outlet | **298.7 K** | --- Let me know if you'd like further clarification!