# Supersonic Nozzle Problem Solution Given: - **Exit speed (\(V_e\))**: 1300 ft/s - **Reservoir temperature (\(T_0\))**: 9000°R - **Reservoir pressure (\(p_0\))**: 6 atm - **Weight flow (\(\dot{W}\))**: 1.4 lbm/s - **Specific heat (\(c_p\))**: 6000 ft·lb/(slug·°R) - **Gas constant for air (\(R\))**: 1716 ft·lb/(slug·°R) - **1 atm** = 2116 lb/ft² We are to find: 1. Temperature at the exit (\(T_e\)) 2. Reservoir density (\(\rho_0\)) 3. Density at the exit (\(\rho_e\)) 4. Nozzle exit area (\(A_e\)) --- ## 1. Temperature at the Exit (\(T_e\)) The total (reservoir) temperature is: \[ T_0 = T_e + \frac{V_e^2}{2 c_p} \] Solve for \(T_e\): \[ T_e = T_0 - \frac{V_e^2}{2 c_p} \] Plug in the values: - \(T_0 = 9000\,^\circ R\) - \(V_e = 1300\) ft/s - \(c_p = 6000\) ft·lb/(slug·°R) \[ T_e = 9000 - \frac{(1300)^2}{2 \times 6000} \] \[ T_e = 9000 - \frac{1,690,000}{12,000} \] \[ T_e = 9000 - 140.83 \] \[ T_e \approx 8859\,^\circ R \] --- ## 2. Reservoir Density (\(\rho_0\)) Use the ideal gas law: \[ \rho_0 = \frac{p_0}{R T_0} \] Where: - \(p_0 = 6 \times 2116 = 12,696\) lb/ft² - \(R = 1716\) ft·lb/(slug·°R) - \(T_0 = 9000\,^\circ R\) \[ \rho_0 = \frac{12,696}{1716 \times 9000} \] \[ = \frac{12,696}{15,444,000} \] \[ \approx 0.000822 \text{ slug/ft}^3 \] --- ## 3. Density at the Exit (\(\rho_e\)) Again, use the ideal gas law at exit: \[ \rho_e = \frac{p_e}{R T_e} \] But \(p_e\) is not given directly. However, for supersonic nozzles, if only the exit velocity is given, we can use the isentropic relation for air (assuming \(\gamma = 1.4\)): The total and static temperature relation is: \[ \frac{T_e}{T_0} = \left(1 + \frac{\gamma-1}{2} M_e^2\right)^{-1} \] First, find the exit Mach number (\(M_e\)): \[ a_e = \sqrt{\gamma R T_e} \] \[ M_e = \frac{V_e}{a_e} \] But \(T_e\) is already found; let's use it: \[ a_e = \sqrt{1.4 \times 1716 \times 8859} \] Calculate inside the square root: - \(1.4 \times 1716 = 2402.4\) - \(2402.4 \times 8859 = 21,285,333.6\) - \(\sqrt{21,285,333.6} \approx 4,614\) ft/s \[ M_e = \frac{1300}{4614} \approx 0.282 \] Wait: this is subsonic! This suggests the temperature drop is small because the initial temperature is extremely high (9000°R), so even at \(1300\) ft/s, the Mach number is low due to a very high speed of sound. Now, to get \(p_e\): \[ \frac{p_e}{p_0} = \left( \frac{T_e}{T_0} \right)^{\frac{\gamma}{\gamma-1}} \] \[ \frac{T_e}{T_0} = \frac{8859}{9000} = 0.9846 \] \[ \frac{\gamma}{\gamma-1} = \frac{1.4}{0.4} = 3.5 \] \[ \left(0.9846\right)^{3.5} \approx e^{3.5 \ln(0.9846)} \approx e^{-0.0545} \approx 0.947 \] \[ p_e = 0.947 \times p_0 = 0.947 \times 12,696 = 12,026 \text{ lb/ft}^2 \] Now, plug into the density formula: \[ \rho_e = \frac{12,026}{1716 \times 8859} \] \[ = \frac{12,026}{15,202,644} \] \[ \approx 0.000791 \text{ slug/ft}^3 \] --- ## 4. Nozzle Exit Area (\(A_e\)) First, convert weight flow rate to mass flow rate: - \(g_c = 32.174\) ft/s² \[ \dot{m} = \frac{\dot{W}}{g_c} = \frac{1.4}{32.174} = 0.0435 \text{ slug/s} \] The mass flow rate formula: \[ \dot{m} = \rho_e A_e V_e \] \[ A_e = \frac{\dot{m}}{\rho_e V_e} \] Plug in the values: - \(\dot{m} = 0.0435\) slug/s - \(\rho_e = 0.000791\) slug/ft³ - \(V_e = 1300\) ft/s \[ A_e = \frac{0.0435}{0.000791 \times 1300} \] \[ = \frac{0.0435}{1.0283} \] \[ \approx 0.0423 \text{ ft}^2 \] --- # **Final Answers Summary** | Quantity | Value | |----------------------------|-----------------------------------| | Temperature at exit (\(T_e\)) | 8859 °R | | Reservoir density (\(\rho_0\)) | 0.000822 slug/ft³ | | Exit density (\(\rho_e\)) | 0.000791 slug/ft³ | | Nozzle exit area (\(A_e\)) | 0.0423 ft² | --- **Notes:** - All calculations assume ideal gas and isentropic expansion. - The Mach number at exit is subsonic due to the extremely high reservoir temperature. - Double-check units for consistency in engineering problems.