Subtopic: Minimum Drag & Maximum Lift-to-Drag Ratio
Question Statement:
A conventional low speed aircraft had the following aerodynamic characteristics: wing area $S = 20$ m$^2$, weight $W = 30000$ N, and zero-lift drag coefficient $C_{D0} = 0.02$. At sea level with $\rho = 1.225$ kg/m$^3$, if the thrust required at minimum drag speed is 2400 N, the maximum lift-to-drag ratio $(L/D)_{\max}$ of the aircraft is _________ (round off to 1 decimal place).
Official Answer Key:
12.5
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.40)**
**1. Intuitive Physical Setup & Given Parameters**:
- Aircraft weight: $W = 30000\text{ N}$
- Minimum thrust required for steady level flight: $T_{R,\min} = D_{\min} = 2400\text{ N}$
- Wing reference area: $S = 20\text{ m}^2$
**2. Step-by-Step Flight Mechanics Derivation**:
1. **Steady Level Flight Equilibrium**:
In steady, unaccelerated level flight:
$\text{Lift } L = W = 30000\text{ N}$
$\text{Thrust } T_R = \text{Drag } D$
2. **Maximum Lift-to-Drag Ratio Relation**:
At the condition of minimum thrust required ($T_{R,\min}$), the aerodynamic drag is at its absolute minimum ($D_{\min}$), which corresponds precisely to the condition of maximum lift-to-drag ratio:
$\left( \frac{L}{D} \right)_{\max} = \frac{L}{D_{\min}} = \frac{W}{T_{R,\min}}$
3. **Evaluating $(L/D)_{\max}$**:
$\left( \frac{L}{D} \right)_{\max} = \frac{30000\text{ N}}{2400\text{ N}} = \frac{300}{24} = 12.5$
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Evaluation**: **12.5** (Acceptable Range: [12.0, 13.0]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **12.5**
- **Golden Rule**: $(L/D)_{\max} = \frac{W}{T_{R,\min}}$ for jet aircraft in level flight.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.