Let's solve the problem step by step, as requested. --- ### **Given Data:** - Pressure in tank, \( P_0 = 196.32 \) kPa - Density in tank, \( \rho_0 = 1.9 \) kg/m³ - At throat: Mach number, \( M^* = 1 \), Area \( A^* = 0.11 \) m² - At exit: Mach number, \( M_e = 1.5 \), Pressure is atmospheric (\( P_e = 101.3 \) kPa, assuming standard atm) - Gas: Air (\( R = 287 \) J/kg·K, \( \gamma = 1.4 \)) --- ## **Step 1: Temperature and Speed of Sound at the Tank** ### **a) Temperature at the Tank (\(T_0\))** Using the ideal gas law: \[ P_0 = \rho_0 R T_0 \implies T_0 = \frac{P_0}{\rho_0 R} \] \[ T_0 = \frac{196320}{1.9 \times 287} = \frac{196320}{545.3} \approx 360 \text{ K} \] **Explanation:** We used the ideal gas law to find the temperature in the tank. ### **b) Speed of Sound at the Tank (\(a_0\))** \[ a_0 = \sqrt{\gamma R T_0} \] \[ a_0 = \sqrt{1.4 \times 287 \times 360} = \sqrt{1.4 \times 103320} = \sqrt{144648} \approx 380 \text{ m/s} \] **Explanation:** We used the formula for the speed of sound in a perfect gas. --- ## **Step 2: Properties at the Throat (\(M^* = 1\))** Use isentropic relations for \(M^* = 1\): #### **a) Temperature at the Throat (\(T^*\))** \[ \frac{T^*}{T_0} = \frac{1}{1 + \frac{\gamma-1}{2} M^{*2}} \] \[ = \frac{1}{1 + 0.2 \times 1^2} = \frac{1}{1.2} = 0.833 \] \[ T^* = 0.833 \times 360 = 300 \text{ K} \] #### **b) Pressure at the Throat (\(P^*\))** \[ \frac{P^*}{P_0} = \left( \frac{T^*}{T_0} \right)^{\frac{\gamma}{\gamma-1}} \] \[ = (0.833)^{3.5} \approx 0.529 \] \[ P^* = 0.529 \times 196.32 = 103.8 \text{ kPa} \] #### **c) Density at the Throat (\(\rho^*\))** \[ \frac{\rho^*}{\rho_0} = \left( \frac{T^*}{T_0} \right)^{\frac{1}{\gamma-1}} \] \[ = (0.833)^{2.5} \approx 0.624 \] \[ \rho^* = 0.624 \times 1.9 = 1.19 \text{ kg/m}^3 \] #### **d) Speed of Sound at the Throat (\(a^*\))** \[ a^* = \sqrt{\gamma R T^*} = \sqrt{1.4 \times 287 \times 300} = \sqrt{1.4 \times 86100} = \sqrt{120540} \approx 347.3 \text{ m/s} \] **Explanation:** We used isentropic flow relations for Mach 1 at the throat. --- ## **Step 3: Mass Flow at the Exit (\(M_e = 1.5\))** First, let's check the properties at the exit. ### **a) Isentropic Relations at the Exit (\(M_e = 1.5\))** \[ \frac{T_e}{T_0} = \frac{1}{1 + \frac{\gamma-1}{2} M_e^2} = \frac{1}{1 + 0.2 \times 2.25} = \frac{1}{1.45} = 0.690 \] \[ T_e = 0.69 \times 360 = 248.4 \text{ K} \] \[ \frac{P_e}{P_0} = \left( \frac{T_e}{T_0} \right)^{\frac{\gamma}{\gamma-1}} = (0.69)^{3.5} \approx 0.273 \] \[ P_e = 0.273 \times 196.32 = 53.6 \text{ kPa} \] \[ \frac{\rho_e}{\rho_0} = (0.69)^{2.5} \approx 0.379 \] \[ \rho_e = 0.379 \times 1.9 = 0.72 \text{ kg/m}^3 \] \[ a_e = \sqrt{\gamma R T_e} = \sqrt{1.4 \times 287 \times 248.4} \approx \sqrt{1.4 \times 71291} \approx \sqrt{99808} \approx 316 \text{ m/s} \] \[ u_e = M_e \cdot a_e = 1.5 \times 316 = 474 \text{ m/s} \] #### **Mass Flow Rate at the Throat (Choked Condition)** \[ \dot{m} = \rho^* u^* A^* \] At throat, \( u^* = a^* = 347.3 \) m/s \[ \dot{m} = 1.19 \times 347.3 \times 0.11 \approx 45.5 \text{ kg/s} \] **Explanation:** We used isentropic relations and continuity to calculate exit and throat properties. --- ## **Final Answers (Summary):** ### **Step 1:** - **Tank Temperature:** **360 K** - **Tank Speed of Sound:** **380 m/s** ### **Step 2:** - **Throat Pressure:** **103.8 kPa** - **Throat Density:** **1.19 kg/m³** - **Throat Temperature:** **300 K** - **Throat Speed of Sound:** **347.3 m/s** ### **Step 3:** - **Exit Mass Flow:** **45.5 kg/s** - **Exit Pressure:** **53.6 kPa** - **Exit Density:** **0.72 kg/m³** - **Exit Temperature:** **248.4 K** - **Exit Speed of Sound:** **316 m/s** - **Exit Velocity:** **474 m/s** Let me know if you want the full calculation for the exit area (using mass flow continuity)!