Question

A fully loaded Boeing aircraft has mass of 3.3 into 10 ki power 5 kg total wing area is 500 square it is in level flight with a speed of 60 km per hour instrument pressure difference between the lower and upper surface of wing estimate the fractional increase in the speed of air on the upper surface of wing related to the lower surface

Answer 1

Answer:

fractional increase in the speed of air on the upper surface of the wing related to lower surface is 19.4

Explanation:

As we know that Boeing air craft is in level flight

So here we can say that net force in vertical direction must be balanced

So we will have

$mg = \Delta P A$

so we have

$\Delta P = \frac{mg}{A}$

$\Delta P = \frac{3.3 \times 10^5 \times 9.8}{500}$

$\Delta P = 6468 Pa$

now by Bernoulli's principle we can say

$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_1^2$

$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$

$6468 = \frac{1}{2}(1.2)(v_2 - v_1)(v_2 + v_1)$

here we know that

$v_2 + v_2 = 2v$

here we know that

$v = 60 \times \frac{5}{18}$

$v = 16.67$

$v_2 + v_1 = 2(60 \times \frac{5}{18})$

$v_2 + v_1 = 33.3$

so we will have

$v_2 - v_1 = \frac{2 \times 6468}{1.2 \times 33.3}$

$v_2 - v_1 = 323.7 m/s$

So here fractional increase is given as

$\frac{v_2 - v_1}{v} = \frac{323.7}{16.67}$

$\frac{v_2 - v_1}{v} = 19.4$

#Learn

Topic : Bernoulli's Theorem

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