Question 258: Properties of Matter

Question

A plane is in level flight at constant speed and each of its two wings has an area of $40 \mathrm{~m}^2$. If the speed of the air is $180 \mathrm{~km} / \mathrm{h}$ over the lower wing surface and $252 \mathrm{~km} / \mathrm{h}$ over the upper wing surface, the mass of the plane is ___________ kg.
(Take air density to be $1 \mathrm{~kg} \mathrm{~m}^{-3}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

Accepted Answer

To solve this problem, we need to employ Bernoulli's equation, which is applied to describe the behavior of fluid flow. For a fluid in steady flow, the principle states that the sum of the pressure potential energy density, kinetic energy density, and the gravitational potential energy density has the same value at all points along a streamline. Since the plane is in level flight, we can ignore changes in gravitational potential energy. Bernoulli's equation can be written as:
$$ P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2, $$
where,
$P_1$ and $P_2$ are the static pressures on the lower and upper surfaces of the wing respectively,
$v_1$ is the airspeed over the lower surface of the wing,
$v_2$ is the airspeed over the upper surface of the wing,
$ho$ is the density of air,
$P_2 > P_1$ because $v_2 > v_1$.
First, let's convert the airspeeds to $ \mathrm{m/s} $:
$$ v_1 = 180 \frac{\mathrm{km}}{\mathrm{h}} 	imes \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 50 \mathrm{m/s}, $$
$$ v_2 = 252 \frac{\mathrm{km}}{\mathrm{h}} 	imes \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 70 \mathrm{m/s}. $$
The pressure difference between the lower and upper wing surfaces can thus be calculated using Bernoulli's equation:
$$ \Delta P = P_1 - P_2 = \frac{1}{2} ho v_2^2 - \frac{1}{2} ho v_1^2. $$
Substitute the values (with $ho = 1 \mathrm{~kg/m}^3$):
$$ \Delta P = \frac{1}{2} (1 \mathrm{~kg/m}^3) (70 \mathrm{m/s})^2 - \frac{1}{2} (1 \mathrm{~kg/m}^3) (50 \mathrm{m/s})^2, $$
$$ \Delta P = \frac{1}{2} (4900 \mathrm{~kg/m \cdot s}^2) - \frac{1}{2} (2500 \mathrm{~kg/m \cdot s}^2), $$
$$ \Delta P = \frac{1}{2} (2400 \mathrm{~kg/m \cdot s}^2), $$
$$ \Delta P = 1200 \mathrm{~N/m}^2. $$
The lift force generated by the pressure difference over one wing is $ \Delta P 	imes 	ext{wing area} $, and since there are two wings, we must double the lift force generated by one wing to find the total lift force sustaining the plane. The weight of the plane is effectively the lift force when in level flight at constant speed. So:
$$ 	ext{Lift} = 2 	imes \Delta P 	imes 	ext{wing area}, $$
$$ 	ext{Lift} = 2 	imes 1200 \mathrm{~N/m}^2 	imes 40 \mathrm{~m}^2, $$
$$ 	ext{Lift} = 2 	imes 48000 \mathrm{N}, $$
$$ 	ext{Lift} = 96000 \mathrm{N}. $$
To find the mass $m$ of the plane, we use Newton's second law, where lift force is equal to the weight ($ mg $, where $g$ is the acceleration due to gravity):
$$ m g = 	ext{Lift}, $$
$$ m = \frac{	ext{Lift}}{g}, $$
$$ m = \frac{96000 \mathrm{N}}{10 \mathrm{m/s}^2}, $$
$$ m = 9600 \mathrm{kg}. $$
Therefore, the mass of the plane is 9600 kg.

Properties of Matter

Integer / Numeric Answer Type