# Nozzle Problem: Step-by-Step Solution ## Given Data | Property | Inlet (1) | Exit (2) | |----------------------------|---------------------|--------------------| | Pressure, \(P\) | 1.5 atm | 1 atm | | Temperature, \(T\) | 77 °F | \(T_2\) (unknown) | | Velocity, \(V\) | 32 ft/s | 650 ft/s | - **Energy lost**: \( q = -2.33 \) Btu/lb (energy leaves system) - **Find**: 1. Exit temperature (\(T_2\), °F) 2. Area ratio (\(A_2/A_1\)) --- ## Step 1: Apply Steady-Flow Energy Equation (SFEE) For a nozzle (no work, negligible potential energy change): \[ h_1 + \frac{V_1^2}{2g_c} + q = h_2 + \frac{V_2^2}{2g_c} \] Where: - \(h\): specific enthalpy (Btu/lb) - \(V\): velocity (ft/s) - \(g_c\): 32.174 lb·ft/(lbf·s²) (US units) - \(q\): heat transfer per mass (Btu/lb, negative if lost) --- ## Step 2: Convert Temperatures to °R (Rankine) \[ T (\text{°R}) = T (\text{°F}) + 459.67 \] - \(T_1 = 77 + 459.67 = 536.67~\text{°R}\) --- ## Step 3: Express Enthalpy Change Using \(c_p\) For ideal gas (e.g., air), \[ h_2 - h_1 = c_p (T_2 - T_1) \] \[ h_1 = c_p T_1 \] \[ h_2 = c_p T_2 \] For air (approx): - \(c_p = .24~\text{Btu/lb} \cdot \text{°F}\) (Can use °F or °R for differences; 1 °F = 1 °R for differences) --- ## Step 4: Solve for Exit Temperature Plug values into the SFEE: \[ c_p T_1 + \frac{V_1^2}{2g_c} + q = c_p T_2 + \frac{V_2^2}{2g_c} \] Rearrange: \[ c_p (T_2 - T_1) = \left( \frac{V_1^2 - V_2^2}{2g_c} \right) + q \] \[ T_2 = T_1 + \frac{1}{c_p} \left[ \left( \frac{V_1^2 - V_2^2}{2g_c} \right) + q \right] \] ### Plug in Numbers - \(V_1 = 32~\text{ft/s}\) - \(V_2 = 650~\text{ft/s}\) - \(g_c = 32.174~\text{lb} \cdot \text{ft}/(\text{lbf} \cdot \text{s}^2)\) - \(q = -2.33~\text{Btu/lb}\) - \(c_p = .24~\text{Btu/lb} \cdot \text{°F}\) - \(T_1 = 77~\text{°F}\) Calculate kinetic energy change: \[ \frac{V_1^2 - V_2^2}{2g_c} = \frac{32^2 - 650^2}{2 \times 32.174} \] \[ = \frac{1024 - 422500}{64.348} \] \[ = \frac{-421476}{64.348} \] \[ = -6552.5~\text{ft}^2/\text{s}^2 \] Convert kinetic energy unit (since \(1~\text{Btu} = 778.17~\text{ft}\cdot\text{lbf}/\text{lb}\)): \[ -6552.5~\frac{\text{ft}^2}{\text{s}^2} \times \frac{1~\text{lbf}}{32.174~\text{lb}} = -203.8~\frac{\text{ft}\cdot\text{lbf}}{\text{lb}} \] But for energy per mass, in Btu/lb: \[ \frac{-6552.5}{778.17} = -8.42~\text{Btu/lb} \] So, \[ T_2 = 77 + \frac{1}{.24} \left[ -8.42 + (-2.33) \right] \] \[ = 77 + \frac{1}{.24} \left[ -10.75 \right] \] \[ = 77 - \frac{10.75}{.24} \] \[ = 77 - 44.79 \] \[ = 32.21~\text{°F} \] **Final Exit Temperature:** \[ \boxed{T_2 = 32.2~\text{°F}} \] --- ## Step 5: Find Area Ratio (\(A_2/A_1\)) Use continuity (mass flow) equation: \[ \dot{m} = \rho_1 V_1 A_1 = \rho_2 V_2 A_2 \] So, \[ \frac{A_2}{A_1} = \frac{\rho_1 V_1}{\rho_2 V_2} \] For ideal gas: \[ \rho = \frac{P}{R T} \] Where: - \(P\) in lbf/ft² - \(R\) for air = 53.35 ft·lb/(lb·°R) - \(T\) in °R ### Convert Pressures to lbf/ft² - \(P_1 = 1.5~\text{atm} \times 14.696~\text{psia/atm} \times 144~\text{in}^2/\text{ft}^2 = 1.5 \times 14.696 \times 144 = 3176~\text{lbf/ft}^2\) - \(P_2 = 1~\text{atm} \times 14.696 \times 144 = 2117~\text{lbf/ft}^2\) ### Find Densities \[ \rho_1 = \frac{P_1}{R T_1} = \frac{3176}{53.35 \times 536.67} = \frac{3176}{28682.8} = .1107~\text{lb/ft}^3 \] \[ \rho_2 = \frac{P_2}{R T_2} = \frac{2117}{53.35 \times 491.21} = \frac{2117}{26230.9} = .0807~\text{lb/ft}^3 \] (Use \(T_2 = 32.2 + 459.67 = 491.87~\text{°R}\)) ### Area Ratio \[ \frac{A_2}{A_1} = \frac{\rho_1 V_1}{\rho_2 V_2} \] \[ = \frac{.1107 \times 32}{.0807 \times 650} \] \[ = \frac{3.542}{52.455} \] \[ = .0675 \] **Final Area Ratio:** \[ \boxed{\frac{A_2}{A_1} = .0675} \] --- ## **Summary of Answers** 1. **Exit Temperature:** \[ \boxed{32.2~\text{°F}} \] 2. **Area Ratio (\(A_2 / A_1\)):** \[ \boxed{.0675} \] --- ### **Key Steps Recap** 1. Used the energy equation for a nozzle, including heat loss. 2. Converted all units to a consistent system. 3. Found the exit temperature using enthalpy and velocity changes. 4. Used the ideal gas equation and mass conservation to get the area ratio. --- **If you need clarification on any step, just ask!**