Let's solve the problem step-by-step: **Given:** - Inlet Pressure, \( P_1 = 200 \) kPa - Inlet Temperature, \( T_1 = 950 \) K - Exit Pressure, \( P_2 = 70 \) kPa - \( c_p = 1.099 \) kJ/kg·K = **1099 J/kg·K** - \( k = 1.354 \) - The nozzle is insulated (adiabatic process). We are to **find the maximum velocity at the nozzle exit**. --- ### **Step 1: Apply the Steady Flow Energy Equation (SFEE)** For an insulated nozzle, the SFEE simplifies to: \[ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} \] Assume inlet velocity \( V_1 \) is negligible, so: \[ h_1 = h_2 + \frac{V_2^2}{2} \] \[ \Rightarrow V_2 = \sqrt{2(h_1 - h_2)} \] --- ### **Step 2: Calculate \( h_1 \) and \( h_2 \)** For an ideal gas, \[ h = c_p T \] So, \[ h_1 = c_p T_1 \] To find \( h_2 = c_p T_2 \), we need \( T_2 \). --- ### **Step 3: Find \( T_2 \) using Isentropic Relations** For an isentropic process (adiabatic and reversible): \[ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} \] Plug in the values: \[ \frac{T_2}{950} = \left(\frac{70}{200}\right)^{\frac{1.354-1}{1.354}} \] First, calculate the exponent: \[ \frac{k-1}{k} = \frac{1.354-1}{1.354} = \frac{0.354}{1.354} \approx 0.2614 \] Now, \[ \left(\frac{70}{200}\right)^{0.2614} = (0.35)^{0.2614} \] Calculate \( (0.35)^{0.2614} \): \[ \ln(0.35) = -1.0498 \] \[ -1.0498 \times 0.2614 = -0.2743 \] \[ e^{-0.2743} \approx 0.760 \] So, \[ T_2 = 950 \times 0.760 = 722 \text{ K} \] --- ### **Step 4: Calculate \( h_1 \) and \( h_2 \)** \[ h_1 = 1099 \times 950 = 1,044,050 \text{ J/kg} \] \[ h_2 = 1099 \times 722 = 793,478 \text{ J/kg} \] --- ### **Step 5: Find \( V_2 \)** \[ V_2 = \sqrt{2 (h_1 - h_2)} \] \[ = \sqrt{2 \times (1,044,050 - 793,478)} \] \[ = \sqrt{2 \times 250,572} \] \[ = \sqrt{501,144} \] \[ \approx 708 \text{ m/s} \] --- ### **Final Answer:** The **maximum velocity at the nozzle exit is**: \[ \boxed{708\ \text{m/s}} \] So the correct option is **708 m/s**.