A jet transport aircraft flies at a constant cruise altitude and airspeed where cruise velocity $V = 240\text{ m/s}$ and aerodynamic lift-to-drag ratio $L/D = 18.0$. The thrust-specific fuel consumption of the turbofan engines is $c_t = 0.60\text{ N of fuel / (N of thrust}\cdot\text{hour)}$. If the aircraft burns $20\%$) of its initial gross mass during the cruise leg (mass ratio $m_0 / m_1 = 1 / 0.80 = 1.25$), taking $g = 9.81\text{ m/s}^2$, the Breguet range $R$ (in $\text{km}$) is closest to:

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### 1. Strategy & Theoretical Principles From the Breguet Range equation for a jet-propelled aircraft: $$R = \frac{V}{c_{tsfc}} \left(\frac{L}{D}\right) \ln\left(\frac{W_0}{W_1}\right)$$ where $c_{tsfc}$ must be expressed in consistent SI units ($\text{s}^{-1}$ or $\text{kg/(N}\cdot\text{s)}$): $$c_{tsfc} = \frac{0.60\text{ N/(N}\cdot\text{hr)}}{3600\text{ s/hr}} = \frac{1}{6000}\text{ s}^{-1} \approx 1.6667 \times 10^{-4}\text{ s}^{-1}$$ ### 2. Analytical Formulation & Calculation Given: - $V = 240\text{ m/s}$ - $L/D = 18.0$ - $W_0 / W_1 = 1.25 \implies \ln(1.25) \approx 0.2231435$ - $c_{tsfc} = \frac{0.60}{3600}\text{ s}^{-1}$ ### 3. Step-by-Step Calculation $$R = \frac{240}{\frac{0.60}{3600}} \times 18.0 \times \ln(1.25)$$ $$R = \left(240 \times 6000\right) \times 18.0 \times 0.2231435$$ $$R = 1{,}440{,}000 \times 18.0 \times 0.2231435$$ $$R = 25{,}920{,}000 \times 0.2231435 \approx 5{,}783{,}880\text{ m} \approx 5{,}784\text{ km}$$ ### 4. Distractor Forensics - Forgetting to convert hours to seconds (3600) yields $R = 1.6\text{ km}$. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always convert TSFC from $\text{hr}^{-1}$ to $\text{s}^{-1}$ when multiplying by velocity in $\text{m/s}$. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6 (Breguet Range Equation).

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