# Solution to Question 2 Given: - **Inlet pressure, \( P_ \) = 6.89 bar = 689,000 Pa** - **Inlet temperature, \( T_ \) = 427°C = 700 K** - **Exit pressure, \( P_e \) = 1 bar = 100,000 Pa** - **Throat area, \( A_t \) = 650~\text{mm}^2 = 6.5 \times 10^{-4}~\text{m}^2** - **Exit area, \( A_e \) = 975~\text{mm}^2 = 9.75 \times 10^{-4}~\text{m}^2** - **Exit velocity, \( V_e \) = 680 m/s** - **Inlet velocity is negligible** - **Assume air as perfect gas: \( \gamma = 1.4,~ R = 287~\text{J/kg·K} \)** --- ## 1. **Mass Flow Through the Nozzle** (and under/overexpanding) ### **Step 1: Throat Conditions (Isentropic)** At the throat (sonic conditions, \( M_t = 1 \)): \[ \frac{P_t}{P_} = \left( 1 + \frac{\gamma-1}{2} \right)^{-\gamma/(\gamma-1)} \] \[ \frac{P_t}{P_} = (1.2)^{-3.5} \approx .528 \] \[ P_t = .528 \times 689,000 \approx 363,792~\text{Pa} \] \[ \frac{T_t}{T_} = \left( 1 + \frac{\gamma-1}{2} \right)^{-1} \] \[ \frac{T_t}{T_} = 1.2^{-1} = .833 \] \[ T_t = .833 \times 700 = 583~\text{K} \] ### **Step 2: Density at Throat, \( \rho_t \)** \[ \rho_t = \frac{P_t}{R T_t} = \frac{363,792}{287 \times 583} \approx 2.19~\text{kg/m}^3 \] ### **Step 3: Sonic Velocity at Throat, \( V_t \)** \[ V_t = \sqrt{\gamma R T_t} = \sqrt{1.4 \times 287 \times 583} \approx 487~\text{m/s} \] ### **Step 4: Mass Flow at Throat** \[ \dot{m} = \rho_t A_t V_t = 2.19 \times 6.5 \times 10^{-4} \times 487 \approx .694~\text{kg/s} \] --- ### **Is the Nozzle Underexpanding or Overexpanding?** - **Design exit pressure (isentropic expansion):** \[ \frac{P_e^\text{(ideal)}}{P_} = \left[1 + \frac{\gamma-1}{2} M_e^2 \right]^{-\gamma/(\gamma-1)} \] But with actual exit pressure = 1 bar, check the pressure at the exit for isentropic expansion. #### **Find Ideal Exit Mach Number** Area ratio: \[ \frac{A_e}{A_t} = \frac{9.75 \times 10^{-4}}{6.5 \times 10^{-4}} \approx 1.5 \] For isentropic flow: \[ \frac{A_e}{A_t} = \frac{1}{M_e} \left[\frac{2}{\gamma+1}(1+\frac{\gamma-1}{2} M_e^2)\right]^{(\gamma+1)/2(\gamma-1)} \] Solving numerically for \( M_e \) with \( \gamma = 1.4 \), \( \frac{A_e}{A_t} = 1.5 \): - \( M_e \approx 1.5 \) (from standard tables or iterative calculation) Now, isentropic exit pressure: \[ \frac{P_e^\text{(ideal)}}{P_} = \left[1 + \frac{\gamma-1}{2} M_e^2 \right]^{-\gamma/(\gamma-1)} \] \[ = \left[1 + .2 \times 1.5^2 \right]^{-3.5} = (1 + .45)^{-3.5} = (1.45)^{-3.5} \approx .314 \] \[ P_e^\text{(ideal)} = .314 \times 689,000 \approx 216,000~\text{Pa} = 2.16~\text{bar} \] **Actual exit pressure is 1 bar < ideal 2.16 bar — so the nozzle is *underexpanding* (exit pressure lower than design pressure).** --- ## 2. **Nozzle Efficiency and Coefficient of Velocity** ### **Step 1: Isentropic Exit Velocity (\( V_{e,\text{isen}} \))** \[ V_{e,isen} = \sqrt{2 C_p (T_ - T_e)} \] Where, \( C_p = \frac{\gamma R}{\gamma-1} = \frac{1.4 \times 287}{.4} = 1004.5~\text{J/kg·K} \) First, find \( T_e \) (isentropic): \[ \frac{T_e}{T_} = \left(\frac{P_e}{P_}\right)^{(\gamma-1)/\gamma} = \left(\frac{1}{6.89}\right)^{.286} \approx (.145)^{.286} \approx .603 \] \[ T_e = .603 \times 700 = 422.1~\text{K} \] \[ V_{e,isen} = \sqrt{2 \times 1004.5 \times (700 - 422.1)} = \sqrt{2 \times 1004.5 \times 277.9} = \sqrt{558,763} \approx 748~\text{m/s} \] ### **Step 2: Nozzle Efficiency (\( \eta \))** \[ \eta = \frac{\text{Actual Kinetic Energy}}{\text{Isentropic Kinetic Energy}} = \frac{(V_e)^2}{(V_{e,isen})^2} = \left(\frac{680}{748}\right)^2 \approx .827 \] ### **Step 3: Coefficient of Velocity (\( C_v \))** \[ C_v = \frac{V_e}{V_{e,isen}} = \frac{680}{748} \approx .91 \] --- ## **Summary Table** | Quantity | Value | Units | |-------------------------|-------------|---------------| | Mass flow, \( \dot{m} \)| .694 | kg/s | | Nozzle type | Underexpanding | | | Nozzle efficiency, \( \eta \) | .83 | (dimensionless)| | Coefficient of velocity, \( C_v \) | .91 | (dimensionless)| --- ### **Final Answers** 1. **Mass flow through the nozzle is approximately \( .694~\text{kg/s} \). The nozzle is underexpanding.** 2. **Nozzle efficiency is \( .83 \) and coefficient of velocity is \( .91 \).**