Calculate mass of an aeroplane with the wings of area $55\;m^2$ flying horizontally. The velocity of air above and below the wings is $155\;m/s$ and $140\;m/s$ respectively.

Here,
Area of the wings $(A) = 55 \; m^2$
Velocity of the air above the wings $(V_a) = 155\; m/s $
Velocity of the air below the wings $(V_b) = 140\; m/s $
Mass of the aeroplane $(M) = ?$

Let us assume, the height difference between the top and bottom of the wind is negligible. So, we can ignore the change in gravitational potential energy. Then from the Bernoulli's principle,

$P_b + \frac{1}{2}\; \rho \; V_b^2$    =     $P_a + \frac{1}{2}\; \rho \; V_a^2$ .......... (i)

Since the velocity of the air above the wings is greater than the velocity of the air below the wings. Then the pressure on the bottom will be greater than the pressure on the above.

Now from the equation (i), we get

$P_b - P_a$       =       $ \frac{1}{2} \rho \; v_a^2 - \frac{1}{2} \rho \; v_b^2$      =      $\frac{1}{2}\rho \; (v_a^2 - v_b^2)$

Now force can be found from;

$F = \Delta P \, A = (P_b - P_a)\,A = \frac{1}{2}\rho\; (v_a^2 - v_b^2)\,A$

Finally, we assume lift goes into balancing weight since the flight is level.

$W = M\;g = \frac{1}{2}\rho\; (v_a^2 - v_b^2)\,A$

$M = $ $\frac{1}{2} \frac{\rho\; (v_a^2 - v_b^2)\,A}{g} = \frac{1}{2} \frac{1.29 \; (155^2 - 140^2) \, 55}{10}$ $ = 1.57 \; * 10^4 \;kg$

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