Let's break down the problem step by step: --- ## **Given Information** - **Nitrogen (N₂) flows as a real gas** - **Mass flow rate (\(\dot{m}\))** = .3 kg/s - **Enters well-insulated horizontal nozzle** - **Inlet velocity (\(V_1\))** = 70 m/s - **Inlet temperature (\(T_1\))** = 500 K - **Inlet pressure (\(P_1\))** = 400 kPa - **Exit velocity (\(V_2\))** = 300 m/s - **Exit pressure (\(P_2\))** = 100 kPa **To determine:** a) The exit temperature (\(T_2\)) in K b) The exit area (\(A_2\)) in m² --- ## **Step 1: Energy Balance (Steady-flow, Adiabatic Nozzle)** The nozzle is adiabatic and horizontal, so we can use the steady-flow energy equation (no heat, no shaft work, neglecting potential energy): \[ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} \] --- ## **Step 2: Find Exit Temperature (\(T_2\))** We need to use **enthalpy values for real gases** (Nitrogen), but if not given tables, we can make an ideal gas assumption as an estimate. Let's proceed with **ideal gas** as a first step: \[ h = c_p T \] For Nitrogen, \(c_p \approx 1.039\) kJ/kg·K (at moderate T). So, \[ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2} \] Convert velocities to kJ/kg: \[ \frac{V^2}{2} \text{ (in J/kg)} = \frac{V^2}{2} \times \frac{1}{100} \text{ (in kJ/kg)} \] Plug in values: - \(T_1 = 500~K\) - \(V_1 = 70~m/s\) - \(V_2 = 300~m/s\) - \(c_p = 1.039~kJ/kg\cdot K\) \[ 1.039 \times 500 + \frac{70^2}{2 \times 100} = 1.039 \times T_2 + \frac{300^2}{2 \times 100} \] \[ 519.5 + \frac{490}{200} = 1.039 T_2 + \frac{90000}{200} \] \[ 519.5 + 2.45 = 1.039 T_2 + 45 \] \[ 521.95 = 1.039 T_2 + 45 \] \[ 521.95 - 45 = 1.039 T_2 \] \[ 476.95 = 1.039 T_2 \] \[ T_2 = \frac{476.95}{1.039} \] \[ T_2 \approx 459~K \] --- ### **Final Answer for (a):** \[ \boxed{T_2 \approx 459~K} \] --- ## **Step 3: Find Exit Area (\(A_2\))** The mass flow rate is: \[ \dot{m} = \rho_2 V_2 A_2 \] \[ \Rightarrow A_2 = \frac{\dot{m}}{\rho_2 V_2} \] First, find \(\rho_2\) (density at exit): **Ideal Gas Law:** \[ \rho_2 = \frac{P_2}{R T_2} \] For Nitrogen, \(R = .2968~kJ/kg\cdot K = 296.8~J/kg\cdot K\) - \(P_2 = 100~kPa = 100,000~Pa\) - \(T_2 = 459~K\) \[ \rho_2 = \frac{100,000}{296.8 \times 459} \] \[ \rho_2 = \frac{100,000}{136,303.2} \] \[ \rho_2 \approx .733~kg/m^3 \] Now solve for \(A_2\): \[ A_2 = \frac{.3}{.733 \times 300} \] \[ A_2 = \frac{.3}{219.9} \] \[ A_2 \approx .00136~m^2 \] --- ### **Final Answer for (b):** \[ \boxed{A_2 \approx .00136~m^2} \] --- ## **Summary Table** | Part | Quantity | Value | |------|-----------------------|--------------------| | a | Exit Temperature (\(T_2\)) | \(459~K\) | | b | Exit Area (\(A_2\)) | \(.00136~m^2\) | --- **If you need more accurate results, use nitrogen property tables for real gas enthalpy at given pressures and temperatures.**