Let's break down the problem step-by-step. **Given Data:** - Subsonic nozzle, choked at exit. - Exit pressure = Ambient pressure (\(P_e = P_{amb}\)) - Inlet parameters: - Mach number (\(M_1\)) = 0.5 - Total pressure (\(P_{t1}\)) = 1.9 \(P_{amb}\) - Temperature (\(T_1\)) = 325°C = 598 K (since \(T(K) = T(°C) + 273\)) - Find: **Force on the nozzle (\(F\))** ### Step 1: **Calculate Inlet Static Pressure (\(P_1\))** Use the isentropic relation between total and static pressure: \[ \frac{P_{t1}}{P_1} = \left[1+\frac{\gamma-1}{2}M_1^2\right]^{\frac{\gamma}{\gamma-1}} \] Assume air, so \(\gamma = 1.4\). Plugging in the values: \[ \frac{P_{t1}}{P_1} = \left[1+\frac{1.4-1}{2}(0.5)^2\right]^{\frac{1.4}{0.4}} = \left[1+0.2 \times 0.25\right]^{3.5} = \left[1+0.05\right]^{3.5} = (1.05)^{3.5} \] Calculate: \[ (1.05)^{3.5} \approx e^{3.5 \cdot \ln(1.05)} \approx e^{3.5 \cdot 0.04879} \approx e^{0.1708} \approx 1.186 \] So, \[ P_1 = \frac{P_{t1}}{1.186} = \frac{1.9\,P_{amb}}{1.186} \approx 1.602\,P_{amb} \] --- ### Step 2: **Find Inlet Density (\(\rho_1\))** \[ \rho_1 = \frac{P_1}{R T_1} \] For air, \(R = 287 \, \text{J/kg}\cdot\text{K}\). \[ \rho_1 = \frac{1.602\,P_{amb}}{287 \times 598} \] --- ### Step 3: **Find Inlet Velocity (\(V_1\))** \[ V_1 = M_1 \cdot a_1 \] Where speed of sound at inlet: \[ a_1 = \sqrt{\gamma R T_1} = \sqrt{1.4 \times 287 \times 598} = \sqrt{1.4 \times 287 \times 598} \approx \sqrt{239,774.8} \approx 489.7\,\text{m/s} \] \[ V_1 = 0.5 \times 489.7 = 244.85\,\text{m/s} \] --- ### Step 4: **Mass Flow Rate (\(\dot{m}\))** \[ \dot{m} = \rho_1 A_1 V_1 \] But area (\(A_1\)) is not given. But for force calculation, it will cancel out. --- ### Step 5: **Force on Nozzle** The net force on the nozzle is given by the **momentum equation**: \[ F = (\dot{m} V_{exit} + P_{exit} A_{exit}) - (\dot{m} V_{inlet} + P_{inlet} A_{inlet}) \] But, for a nozzle discharging to ambient, the pressure at exit matches ambient, so the exit pressure force cancels out (relative to ambient). So, \[ F = [\dot{m} V_e] - [\dot{m} V_1 + (P_1 - P_{amb}) A_1] \] Since it's **choked at exit**, \(M_e = 1\) and exit pressure = \(P_{amb}\). #### Calculate Choked Exit Properties: \[ \frac{P_{t1}}{P_e} = \left[1 + \frac{\gamma-1}{2} M_e^2\right]^{\frac{\gamma}{\gamma-1}} \] \[ P_e = P_{amb} \implies \frac{P_{t1}}{P_{amb}} = \left[1 + 0.2 \cdot 1^2\right]^{3.5} = (1.2)^{3.5} \approx e^{3.5\cdot 0.182} \approx e^{0.637} \approx 1.89 \] But \(\frac{P_{t1}}{P_{amb}} = 1.9\) (approximately matches), so \(M_e \approx 1\). #### Find Exit Velocity (\(V_e\)): \[ T_e = T_{t1}/[1 + 0.2 M_e^2] = T_{t1}/1.2 \] But \(T_{t1} = T_1 [1 + 0.2 M_1^2] = 598 \times [1+0.05]=627.9\,\text{K}\) So, \[ T_e = 627.9/1.2 = 523.3\,\text{K} \] \[ a_e = \sqrt{1.4 \times 287 \times 523.3} \approx \sqrt{210,069} = 458.5\,\text{m/s} \] \[ V_e = M_e \cdot a_e = 1 \times 458.5 = 458.5\,\text{m/s} \] #### Find Mass Flow Rate in terms of Area: \[ \rho_e = \frac{P_e}{R T_e} = \frac{P_{amb}}{287 \times 523.3} \] \[ \dot{m} = \rho_e A_e V_e \] But, \[ \dot{m} = \rho_1 A_1 V_1 \] --- ### Step 6: **Plug Values Back for Force (per unit area)** \[ F = \dot{m} (V_e - V_1) - (P_1 - P_{amb}) A_1 \] Expressing per unit inlet area (\(A_1\)): - \(\dot{m} = \rho_1 A_1 V_1\) So, \[ F = \rho_1 A_1 V_1 (V_e - V_1) - (P_1 - P_{amb})A_1 \] \[ \frac{F}{A_1} = \rho_1 V_1 (V_e - V_1) - (P_1 - P_{amb}) \] --- ### Step 7: **Calculate All Terms** From earlier: - \(\rho_1 = \frac{1.602 P_{amb}}{287 \times 598} = \frac{1.602 P_{amb}}{171,526} = 9.34 \times 10^{-6} P_{amb}\) - \(V_1 = 244.85\) m/s - \(V_e = 458.5\) m/s - \(P_1 - P_{amb} = 1.602 P_{amb} - P_{amb} = 0.602 P_{amb}\) Plug in values: \[ \frac{F}{A_1} = (9.34 \times 10^{-6} P_{amb}) \times 244.85 \times (458.5 - 244.85) - 0.602 P_{amb} \] Calculate: \[ 458.5 - 244.85 = 213.65 \] \[ 244.85 \times 213.65 = 52,322 \] \[ 9.34 \times 10^{-6} \times 52,322 = 0.4885 \] \[ \frac{F}{A_1} = 0.4885 P_{amb} - 0.602 P_{amb} = -0.1135 P_{amb} \] --- ### **Final Answer** \[ \boxed{ F = -0.1135 \, P_{amb} \, A_1 } \] **Interpretation:** The negative sign indicates the net force acts opposite to the flow direction (i.e., the nozzle is being "pushed" backwards by the flow). --- **If you want a numerical answer, plug in values for \(P_{amb}\) and \(A_1\).** Otherwise, the force on the nozzle is approximately \(\boxed{-0.1135 \, P_{amb} \, A_1}\).