Nozzle Flow Analysis

Cheatsheet Content

Thermodynamics Cheatsheet: Nozzle Flow This cheatsheet provides key formulas and steps for analyzing steady-flow devices like nozzles, focusing on an adiabatic nozzle example. 1. Fundamental Equations for Steady-Flow Devices Mass Conservation: For steady flow, mass flow rate is constant: $$ \dot{m} = \rho_1 A_1 V_1 = \rho_2 A_2 V_2 = \frac{A_1 V_1}{v_1} = \frac{A_2 V_2}{v_2} $$ where $\rho$ is density, $A$ is cross-sectional area, $V$ is velocity, and $v$ is specific volume. Energy Conservation (First Law of Thermodynamics): For a steady-flow system with one inlet and one outlet: $$ \dot{Q} - \dot{W} = \dot{m} \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1) \right] $$ For a nozzle, assume: Adiabatic ($\dot{Q} = 0$) No work done ($\dot{W} = 0$) Negligible change in potential energy ($g(z_2 - z_1) \approx 0$) Simplified energy equation for an adiabatic nozzle: $$ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} $$ For ideal gases, $h = c_p T$, so: $$ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2} $$ 2. Properties of Air (Ideal Gas) Gas Constant for Air: $R = 0.287 \, \text{kJ/kg} \cdot \text{K}$ Specific Heat at Constant Pressure for Air: $c_p = 1.005 \, \text{kJ/kg} \cdot \text{K}$ (at room temperature, can vary with T) Ideal Gas Equation of State: $P v = R T$ or $P = \rho R T$ Specific Volume: $v = \frac{R T}{P}$ 3. Example Problem: Adiabatic Nozzle Given: Inlet Pressure $P_1 = 300 \, \text{kPa}$ Inlet Temperature $T_1 = 200^\circ\text{C} = 473.15 \, \text{K}$ Inlet Velocity $V_1 = 30 \, \text{m/s}$ Inlet Area $A_1 = 80 \, \text{cm}^2 = 0.008 \, \text{m}^2$ Exit Pressure $P_2 = 100 \, \text{kPa}$ Exit Velocity $V_2 = 180 \, \text{m/s}$ (a) Mass Flow Rate ($\dot{m}$) Calculate specific volume at inlet ($v_1$): $$ v_1 = \frac{R T_1}{P_1} = \frac{(0.287 \, \text{kJ/kg} \cdot \text{K}) (473.15 \, \text{K})}{300 \, \text{kPa}} $$ Convert units: $1 \, \text{kJ} = 1 \, \text{kPa} \cdot \text{m}^3$. So, $v_1 = \frac{(0.287)(473.15)}{300} \, \text{m}^3/\text{kg}$ $$ v_1 = 0.4526 \, \text{m}^3/\text{kg} $$ Calculate mass flow rate ($\dot{m}$): $$ \dot{m} = \frac{A_1 V_1}{v_1} = \frac{(0.008 \, \text{m}^2)(30 \, \text{m/s})}{0.4526 \, \text{m}^3/\text{kg}} $$ $$ \dot{m} = 0.5303 \, \text{kg/s} $$ (b) Exit Temperature ($T_2$) Use the energy equation for an adiabatic nozzle: $$ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2} $$ $$ T_2 = T_1 + \frac{V_1^2 - V_2^2}{2 c_p} $$ Substitute values (ensure consistent units, e.g., convert $c_p$ to $\text{J/kg} \cdot \text{K}$): $c_p = 1.005 \, \text{kJ/kg} \cdot \text{K} = 1005 \, \text{J/kg} \cdot \text{K}$ $$ T_2 = 473.15 \, \text{K} + \frac{(30 \, \text{m/s})^2 - (180 \, \text{m/s})^2}{2 \times 1005 \, \text{J/kg} \cdot \text{K}} $$ $$ T_2 = 473.15 \, \text{K} + \frac{900 - 32400}{2010} \, \text{K} $$ $$ T_2 = 473.15 \, \text{K} - 15.67 \, \text{K} $$ $$ T_2 = 457.48 \, \text{K} $$ Convert to Celsius: $$ T_2 = 457.48 - 273.15 = 184.33^\circ\text{C} $$ (c) Exit Area ($A_2$) Calculate specific volume at exit ($v_2$): $$ v_2 = \frac{R T_2}{P_2} = \frac{(0.287 \, \text{kJ/kg} \cdot \text{K}) (457.48 \, \text{K})}{100 \, \text{kPa}} $$ $$ v_2 = 1.3129 \, \text{m}^3/\text{kg} $$ Use mass conservation equation: $$ \dot{m} = \frac{A_2 V_2}{v_2} \implies A_2 = \frac{\dot{m} v_2}{V_2} $$ $$ A_2 = \frac{(0.5303 \, \text{kg/s})(1.3129 \, \text{m}^3/\text{kg})}{180 \, \text{m/s}} $$ $$ A_2 = 0.00387 \, \text{m}^2 $$ Convert to $\text{cm}^2$: $$ A_2 = 0.00387 \, \text{m}^2 \times \left( \frac{100 \, \text{cm}}{1 \, \text{m}} \right)^2 = 38.7 \, \text{cm}^2 $$