# Step-by-Step Solution: Force Required to Hold Stationary Aircraft with Turbojet Let's solve the problem step by step. --- ## **Given Data** - **Pressure ratio across compressor (\( r_p \))**: 9 - **Ambient air temperature (\( T_ \))**: \( 17^\circ\text{C} = 290.15\,K \) - **Ambient air pressure (\( P_ \))**: \( 95\,kPa \) - **Air mass flow rate (\( \dot{m}_a \))**: \( 15\,kg/s \) - **Fuel flow rate (\( \dot{m}_f \))**: \( .5\,kg/s \) - **Fuel heating value (\( h_{PR} \))**: \( 42,700\,kJ/kg \) - **Assumptions**: - Neglect diffuser effect - Ignore mass increase at exit (i.e., \( \dot{m}_{exit} = \dot{m}_a \)) - Ideal air, constant \( c_p \) - Engine component efficiencies = 100% - No motion (\( V_ = \)), so all exit velocity is thrust --- ## **Step 1: Principle Used** The **net force** (thrust) generated by the engine is the force that the brakes must counteract. For a stationary aircraft: \[ F_{brakes} = F_{thrust} \] The **thrust** (\( F \)) for a stationary turbojet is: \[ F = \dot{m}_{exit} V_{exit} - \dot{m}_a V_ \] Since \( V_ = \), and neglecting the small mass increase at the exit, use: \[ F = \dot{m}_a V_{exit} \] --- ## **Step 2: Find Exit Velocity (\( V_{exit} \))** Apply the energy balance across the engine (neglecting losses): \[ \dot{m}_a \left( h_{exit} + \frac{V_{exit}^2}{2} \right) = \dot{m}_a h_ + \dot{m}_f h_{PR} \] where: - \( h_ \) = enthalpy of inlet air - \( h_{exit} \) = enthalpy of exit air - \( h_{PR} \) = fuel heating value Assume all fuel energy goes into increasing exit enthalpy and kinetic energy. **Rearrange:** \[ h_{exit} + \frac{V_{exit}^2}{2} = h_ + \frac{\dot{m}_f}{\dot{m}_a} h_{PR} \] \[ \frac{V_{exit}^2}{2} = \left( h_ + \frac{\dot{m}_f}{\dot{m}_a} h_{PR} \right) - h_{exit} \] --- ## **Step 3: Get \( h_ \) and \( h_{exit} \) (Use Ideal Gas Tables)** - For **air at \( 17^\circ\text{C} \):** \( h_ = c_p T_ \) - For **exit:** Assume combustion raises temperature to \( T_{exit} \). ### **a. Find \( T_{exit} \) and \( h_{exit} \)** **Pressure Ratio:** \[ P_{exit} = P_ \times r_p = 95\,kPa \times 9 = 855\,kPa \] But for a stationary engine, the exhaust is at approximately ambient pressure (since the exhaust discharges to the atmosphere). So, \( P_{exit} \approx P_ = 95\,kPa \). **Enthalpy values:** - Use \( c_p \approx 1.005\,kJ/(kg \cdot K) \) for air. \[ h_ = c_p T_ = 1.005 \times 290.15 \approx 291.6\,kJ/kg \] We need \( h_{exit} \), but unless given an exit temperature, we can write the final equation in terms of \( h_{exit} \) and solve for kinetic energy (since there's no nozzle loss, assume all remaining energy is kinetic). --- ## **Step 4: Substitute Numbers** \[ h_ = 291.6\,kJ/kg \] \[ \frac{\dot{m}_f}{\dot{m}_a} = \frac{.5}{15} = .0333 \] \[ h_{PR} = 42,700\,kJ/kg \] \[ h_ + \frac{\dot{m}_f}{\dot{m}_a} h_{PR} = 291.6 + .0333 \times 42,700 = 291.6 + 1,423.91 = 1,715.5\,kJ/kg \] Assume the exhaust is expanded to ambient pressure, and all energy after combustion (except \( h_ \)) is converted to kinetic energy: \[ \frac{V_{exit}^2}{2} = 1,715.5 - h_{exit} \] But if the exit is at ambient pressure and temperature, the minimum \( h_{exit} = h_ \), but that would mean no kinetic energy. **However**, in reality, the exhaust is much hotter due to combustion. For a turbojet, we assume all fuel energy (minus the initial air enthalpy) goes into exhaust kinetic energy: \[ \frac{V_{exit}^2}{2} = 1,423.9\,kJ/kg = 1,423,900\,J/kg \] \[ V_{exit} = \sqrt{2 \times 1,423,900} \] \[ V_{exit} = \sqrt{2,847,800} \approx 1,688\,m/s \] --- ## **Step 5: Calculate Thrust (Force on Brakes)** \[ F = \dot{m}_a V_{exit} = 15\,kg/s \times 1,688\,m/s = 25,320\,N \] --- ## **Step 6: Final Answer** \[ \boxed{ \text{The force that must be applied on the brakes to hold the plane stationary is } 25,300\,N } \] --- ## **Summary Table** | Parameter | Value | |-------------------|------------------| | Air flow rate | \( 15\,kg/s \) | | Fuel flow rate | \( .5\,kg/s \) | | Heating value | \( 42,700\,kJ/kg \) | | Inlet temp | \( 17^\circ C \) | | Exit velocity | \( 1,688\,m/s \) | | Force on brakes | \( 25,300\,N \) | --- ## **Key Steps Illustrated** 1. **Energy balance**: Fuel energy goes into kinetic energy of exhaust. 2. **Thrust calculation**: \( F = \dot{m}_a V_{exit} \) for stationary engine. 3. **Assumptions**: Perfect efficiency, ideal gas, air properties constant. --- **Note:** If you use more precise air property tables or account for actual engine parameters, the answer may vary slightly, but this is the standard textbook approach. If you have those tables, you can look up \( h_ \) and \( h_{exit} \) for more accuracy. --- **If you need a diagram or further clarification on any step, let me know!**