An aircraft of mass $m = 6000\text{ kg}$ flies in steady level flight at sea level ($\rho = 1.225\text{ kg/m}^3$). Wing area is $S = 30\text{ m}^2$, zero-lift drag coefficient is $C_{D0} = 0.020$, and induced drag factor is $K = 0.050$. What is the minimum thrust required $T_{R,\min}$ in kilonewtons (kN)? (Take $g = 9.81\text{ m/s}^2$, round to two decimal places).

Answer Key:

3.72

### Solution: Weight $W = 6000 \times 9.81 = 58860\text{ N}$. Maximum lift-to-drag ratio $(L/D)_{\max} = \frac{1}{2\sqrt{C_{D0} K}} = \frac{1}{2\sqrt{0.020 \times 0.050}} = \frac{1}{2\sqrt{0.001}} = \frac{1}{2 \times 0.0316227} = \frac{1}{0.063245} \approx 15.811$. Minimum thrust required $T_{R,\min} = \frac{W}{(L/D)_{\max}} = \frac{58860}{15.81139} \approx 3722.6\text{ N} \approx 3.72\text{ kN}$. Correct: 3.72 (range 3.68 to 3.76).

← Prev Question Next Question →