A fully loaded boeing aircraft has a mass of $$5.4 \times 10^5$$ kg. Its total wing area is $$500$$ m$$^2$$. It is in level flight with a speed of $$1080$$ km h$$^{-1}$$. If the density of air $$\rho$$ is $$1.2$$ kg m$$^{-3}$$, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be ($$g = 10$$ m s$$^{-2}$$)

Solution :

For an aircraft in horizontal flight, lift force is equal to the weight of the aircraft.

Using Bernoulli’s principle :

$$\Delta P = \frac{1}{2}\rho(v_u^2-v_l^2)$$

Lift force :

$$\Delta P \times A = mg$$

Given :

$$m = 5.4 \times 10^5\text{ kg}$$

$$A = 500\text{ m}^2$$

$$\rho = 1.2\text{ kg m}^{-3}$$

$$g = 10\text{ m s}^{-2}$$

Speed of aircraft :

$$1080\text{ km h}^{-1}$$

$$= \frac{1080 \times 1000}{3600}$$

$$= 300\text{ m s}^{-1}$$

Therefore,

$$\frac{1}{2}\rho(v_u^2-v_l^2)A = mg$$

Substituting values :

$$\frac{1}{2}(1.2)(v_u^2-v_l^2)(500) = 5.4 \times 10^5 \times 10$$

$$300(v_u^2-v_l^2) = 5.4 \times 10^6$$

$$v_u^2-v_l^2 = 18000$$

Using,

$$v_u^2-v_l^2 = (v_u-v_l)(v_u+v_l)$$

Since difference in speeds is small,

$$v_u+v_l \approx 2v$$

$$= 2 \times 300$$

$$= 600$$

Therefore,

$$v_u-v_l = \frac{18000}{600}$$

$$= 30\text{ m s}^{-1}$$

Fractional increase in speed :

$$\frac{v_u-v_l}{v_l}\times 100$$

$$= \frac{30}{300}\times 100$$

$$= 10\%$$

Final Answer :

$$10\%$$