Let's solve **Question 4** step by step as requested. --- # **Part (a): Turbojet—Cruising Speed and Condition for Existence** ### **Step 1: Write Down the Drag Polar** Given: \[ C_D = C_{D} + K C_L^2 \] ### **Step 2: Force Equilibrium** At steady cruise: \[ T = \frac{1}{2} \rho S V^2 C_D \\ W = \frac{1}{2} \rho S V^2 C_L \] Let \( q = \frac{1}{2} \rho V^2 \) (dynamic pressure). So, \[ T = S q C_D \] \[ W = S q C_L \] ### **Step 3: Substitute Drag Polar** \[ T = S q (C_{D} + K C_L^2) \] But \( C_L = \frac{W}{S q} \): \[ T = S q \left[ C_{D} + K \left( \frac{W}{S q} \right)^2 \right] \] \[ T = S q C_{D} + K \frac{W^2}{S q} \] Rearrange to a quadratic in \( q \): \[ S^2 C_{D} q^2 - S T q + K W^2 = \] ### **Step 4: Condition for Real, Positive Solution** Quadratic form: \[ a q^2 + b q + c = \] For real roots, discriminant \( \geq \): \[ (S T)^2 - 4 S^2 C_{D} K W^2 \geq \] \[ T^2 \geq 4 S C_{D} K W^2 \] But let's check the solution for \( q \): \[ q = \frac{S T \pm \sqrt{(S T)^2 - 4 S^2 C_{D} K W^2}}{2 S^2 C_{D}} \] For real solution, discriminant \( \geq \): \[ (ST)^2 - 4 S^2 C_{D} K W^2 \geq \] \[ T^2 \geq 4 S C_{D} K W^2 \] ### **Step 5: Thrust Condition** Given in the question: \[ T \geq \frac{W}{E_{max}} = T_{min} \] where \( E_{max} \) is the maximum endurance (minimum thrust required). ### **Step 6: Optimum Speed for Maximum Endurance** At minimum thrust: \[ T_{min} = \frac{W}{E_{max}} \] The corresponding speed (\( V_1^* \)): \[ V_1^* = \sqrt{\frac{2 W}{\rho S C_L^*}} \] Where \( C_L^* = \sqrt{\frac{C_{D}}{K}} \). --- ## **Part (a) Final Answers** ### **Quadratic Equation for Cruise:** \[ S^2 C_{D} q^2 - S T q + K W^2 = \] ### **Existence Condition:** \[ T \geq T_{min} = \frac{W}{E_{max}} \] ### **Optimum Speed:** \[ V_1^* = \sqrt{\frac{2W}{\rho S C_L^*}} \] where \[ C_L^* = \sqrt{\frac{C_{D}}{K}} \] --- # **Part (b): Piston-Prop—Cruising Speed and Condition for Existence** ### **Step 1: Power Available** Available power: \[ \eta_p P_s = T V \] where \( P_s \) is shaft power, \( \eta_p \) is propeller efficiency. ### **Step 2: Drag and Thrust Balance** Set up the equation for cruise: \[ T = S q C_D \] \[ T V = S q C_D V = S q V (C_{D} + K C_L^2) \] But \( q = \frac{1}{2} \rho V^2 \) and \( C_L = \frac{W}{S q} \): \[ \eta_p P_s = S q C_{D} V + K \frac{W^2}{S q} V \] Substitute \( q \): \[ \eta_p P_s = S \left(\frac{1}{2} \rho V^2\right) C_{D} V + K \frac{W^2}{S \cdot \frac{1}{2} \rho V^2} V \] \[ = \frac{1}{2} \rho S C_{D} V^3 + \frac{2 K W^2}{\rho S V^3} \] Rearranging: \[ \frac{1}{2} \rho S C_{D} V^4 - \eta_p P_s V + \frac{2 K W^2}{\rho S} = \] Or: \[ f(V) = \frac{1}{2} \rho S C_{D} V^4 - \eta_p P_s V + \frac{2 K W^2}{\rho S} = \] The question uses slightly different notation: \[ f(V) = \frac{1}{2} \rho S C_{D} V^4 - \eta_p P_s V + \frac{2 K W^2}{\rho S} = \] or \[ f(V) = \frac{1}{2} \rho S C_{D} V^4 - \eta_p P_s V + \frac{2 K W^2}{\rho S} = \] ### **Step 3: Condition for Real Roots** The minimum value of \( f(V) \) must be less than or equal to zero for real roots to exist. - At \( V = \), \( f() > \) - For large \( V \), \( f(V) \to +\infty \) - \( f(V) \) is a quartic with positive leading coefficient, so it opens upwards. So, must find the minimum of \( f(V) \): Set derivative to zero: \[ \frac{df}{dV} = 2 \rho S C_{D} V^3 - \eta_p P_s = \\ V_m^3 = \frac{\eta_p P_s}{2 \rho S C_{D}} \] \[ V_m = \left( \frac{\eta_p P_s}{2 \rho S C_{D}}\right)^{1/3} \] Plug \( V_m \) back into \( f(V) \): \[ f(V_m) = \frac{1}{2} \rho S C_{D} V_m^4 - \eta_p P_s V_m + \frac{2 K W^2}{\rho S} \] \[ = \frac{1}{2} \rho S C_{D} V_m^4 - \eta_p P_s V_m + \frac{2 K W^2}{\rho S} \] But \[ \eta_p P_s = 2 \rho S C_{D} V_m^3 \] So, \[ f(V_m) = \frac{1}{2} \rho S C_{D} V_m^4 - 2 \rho S C_{D} V_m^3 V_m + \frac{2 K W^2}{\rho S} = \frac{1}{2} \rho S C_{D} V_m^4 - 2 \rho S C_{D} V_m^4 + \frac{2 K W^2}{\rho S} = -\frac{3}{2} \rho S C_{D} V_m^4 + \frac{2 K W^2}{\rho S} \] Set \( f(V_m) \leq \): \[ -\frac{3}{2} \rho S C_{D} V_m^4 + \frac{2 K W^2}{\rho S} \leq \] \[ \frac{2 K W^2}{\rho S} \leq \frac{3}{2} \rho S C_{D} V_m^4 \] \[ \frac{4 K W^2}{3 (\rho S)^2 C_{D}} \leq V_m^4 \] Recall \[ V_m^3 = \frac{\eta_p P_s}{2 \rho S C_{D}} \] So \[ V_m^4 = \left(\frac{\eta_p P_s}{2 \rho S C_{D}}\right)^{4/3} \] Substitute: \[ \frac{4 K W^2}{3 (\rho S)^2 C_{D}} \leq \left( \frac{\eta_p P_s}{2 \rho S C_{D}}\right)^{4/3} \] Or: \[ \eta_p P_s \geq 2 \left[ \frac{4 K W^2}{3 (\rho S)^2 C_{D}} \right]^{3/4} (\rho S C_{D})^{1} \] But to match the form given in the question, recall: \[ C_L^* = \sqrt{\frac{C_{D}}{K}} \] So, \[ K = \frac{C_{D}}{(C_L^*)^2} \] Substitute this into above and simplify: The question gives the existence condition as: \[ \eta_p P_s \geq 2 \frac{3^{3/4}}{4} E_{max} \sqrt{\frac{2 W^3}{\rho S C_L^*}} \] --- ### **Step 4: Optimum Speed at Minimum Power** At the minimum power (\( \eta_p P_s \) at its minimum value), the corresponding optimum speed (\( V_2^* \)) is: \[ V_2^* = \left( \frac{2 \eta_p P_s}{\rho S C_{D}} \right)^{1/3} \] --- ### **Step 5: Comparison** Compare \( V_2^* \) (optimum for prop) and \( V_1^* \) (optimum for turbojet): - \( V_2^* \) is generally **less than** \( V_1^* \), because the minimum power for prop occurs at a lower speed than the minimum thrust for jet. --- # **Summary Table** | Aircraft Type | Condition for Cruise | Optimum Speed | |---------------|----------------------|------------------------------| | Turbojet | \( T \geq \frac{W}{E_{max}} \) | \( V_1^* = \sqrt{\frac{2W}{\rho S C_L^*}} \) | | Piston-Prop | \( \eta_p P_s \geq 2 \frac{3^{3/4}}{4} E_{max} \sqrt{\frac{2 W^3}{\rho S C_L^*}} \) | \( V_2^* = \left( \frac{2 \eta_p P_s}{\rho S C_{D}} \right)^{1/3} \) | --- ## **Discussion:** - **Existence of solution** depends on whether available thrust/power meets the minimum required for cruise. - **Optimum speed for prop (\( V_2^* \)) is less than that for turbojet (\( V_1^* \))**, reflecting different performance requirements for minimum power vs. minimum thrust. --- **If you need detailed algebraic manipulation for one of the steps, let me know!**