# **Solution to Question 1** Let's break down the problem and solve step by step. --- ## **Given Data** - **Mass flow rate (\(\dot{m}\))** = 4.5 kg/s - **Inlet (Stagnation) Pressure (\(P_\))** = 8.3 bar = 830,000 Pa - **Inlet Temperature (\(T_\))** = 327°C = 600 K - **Exit Pressure (\(P_e\))** = 1.38 bar = 138,000 Pa - **Coefficient of Discharge (\(C_d\))** = .96 - **Nozzle Efficiency (\(\eta\))** = .92 - **Assume Air Properties:** - \(R = 287\) J/kg·K - \(\gamma = 1.4\) --- ## **Step 1: Find Critical Pressure Ratio (for throat, sonic condition)** \[ \left(\frac{P_t}{P_}\right) = \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}} \] \[ \left(\frac{P_t}{P_}\right) = \left(\frac{2}{1.4+1}\right)^{\frac{1.4}{1.4-1}} = (.588)^{3.5} \approx .528 \] \[ P_t = .528 \times 830,000 = 438,240~\text{Pa} \] --- ## **Step 2: Find Throat Temperature (\(T_t\))** \[ \frac{T_t}{T_} = \frac{2}{\gamma+1} = .833 \] \[ T_t = .833 \times 600 = 500~\text{K} \] --- ## **Step 3: Find Throat Density (\(\rho_t\))** \[ \rho_t = \frac{P_t}{R T_t} = \frac{438,240}{287 \times 500} = \frac{438,240}{143,500} \approx 3.055~\text{kg/m}^3 \] --- ## **Step 4: Find Throat Velocity (\(V_t\)), Sonic Condition** \[ V_t = \sqrt{\gamma R T_t} = \sqrt{1.4 \times 287 \times 500} = \sqrt{200,900} \approx 448.2~\text{m/s} \] Apply nozzle efficiency (\(\eta\)): \[ V_{t,\text{actual}} = V_t \sqrt{\eta} = 448.2 \times \sqrt{.92} = 448.2 \times .959 = 429.8~\text{m/s} \] --- ## **Step 5: Find Throat Area (\(A_t\))** \[ \dot{m} = C_d \cdot \rho_t \cdot A_t \cdot V_{t,\text{actual}} \] Solve for \(A_t\): \[ A_t = \frac{\dot{m}}{C_d \cdot \rho_t \cdot V_{t,\text{actual}}} \] \[ A_t = \frac{4.5}{.96 \times 3.055 \times 429.8} \] \[ A_t = \frac{4.5}{126.6} = .00357~\text{m}^2 = 35.7~\text{cm}^2 \] --- ## **Step 6: Find Exit Temperature (\(T_e\)), Pressure (\(P_e\)), and Density (\(\rho_e\))** Isentropic relation: \[ \frac{T_e}{T_} = \left(\frac{P_e}{P_}\right)^{\frac{\gamma-1}{\gamma}} \] \[ \frac{T_e}{600} = \left(\frac{138,000}{830,000}\right)^{.286} \] \[ = (.166)^{.286} \approx .635 \] \[ T_e = .635 \times 600 = 381~\text{K} \] \[ \rho_e = \frac{P_e}{R T_e} = \frac{138,000}{287 \times 381} \approx \frac{138,000}{109,347} \approx 1.262~\text{kg/m}^3 \] --- ## **Step 7: Find Exit Velocity (\(V_e\))** Isentropic enthalpy drop: \[ V_e = \sqrt{2 \eta C_p (T_ - T_e)} \] Where \(C_p = \frac{\gamma R}{\gamma - 1} = \frac{1.4 \times 287}{.4} = 1004.5~\text{J/kg·K}\) \[ V_e = \sqrt{2 \times .92 \times 1004.5 \times (600 - 381)} \] \[ = \sqrt{2 \times .92 \times 1004.5 \times 219} \] \[ = \sqrt{2 \times .92 \times 219,985.5} \] \[ = \sqrt{2 \times 202,387.46} \] \[ = \sqrt{404,774.93} \] \[ = \sqrt{404,774.93} \approx 636~\text{m/s} \] --- ## **Step 8: Find Exit Area (\(A_e\))** \[ \dot{m} = C_d \cdot \rho_e \cdot A_e \cdot V_e \] \[ A_e = \frac{\dot{m}}{C_d \cdot \rho_e \cdot V_e} \] \[ A_e = \frac{4.5}{.96 \times 1.262 \times 636} \] \[ = \frac{4.5}{771.4} = .00583~\text{m}^2 = 58.3~\text{cm}^2 \] --- # **Final Answers** ### **(a) Throat Area (\(A_t\)):** \[ \boxed{.00357~\text{m}^2~\text{(35.7 cm}^2\text{)}} \] ### **(b) Exit Area (\(A_e\)):** \[ \boxed{.00583~\text{m}^2~\text{(58.3 cm}^2\text{)}} \] --- **Assumptions:** - Inlet velocity is negligible - Isentropic flow up to efficiency - Air as ideal gas - Constant specific heats - Nozzle flow is steady --- **Let me know if you want a detailed step-by-step for the next question!**