Question

Air is streaming past a horizontal airplane wing such that its speed is 120 meter per sec over the upper surface and 90 meter per sec at the lower surface. If the density of air is $1.3{\text{ }}kg{\text{ }}per{\text{ }}mete{r^3}$ and the wing is 10 meter long and has an average width of 2 meter, then the difference of the pressure on the two sides of the wing is:
A) 4095.0 Pascal.
B) 409.50 Pascal.
C) 40.950 Pascal.
D) 4.0950 Pascal.

Answer 1

Bernoulli's theorem says that with increase in the pressure there is an increase in velocity of the fluid. Bernoulli's theorem is valid only for incompressible fluids. Bernoulli's theorem is derived by applying conservation of energy.

Formula used:
The formula of the Bernoulli’s theorem is given by,
${p_1} + \dfrac{1}{2}\rho {v_1}^2 = {p_2} + \dfrac{1}{2}\rho {v_2}^2$
Where the pressure at two points is ${p_1}$and${p_2}$, the density of the fluid is $\rho$ also the velocity of the fluid at two points is ${v_1}^2$and${v_2}^2$.
The formula of the pressure is equal to,
$\Rightarrow P = \dfrac{F}{A}$
Where pressure is P the force is F and the area of cross section is A.

Complete Step by step solution:
It is given in the problem that air is streaming past a horizontal airplane wing such that its speed is 120 meter per sec over the upper surface and 90 meter per sec at the lower surface the density of air is $1.3{\text{ }}kg{\text{ }}per{\text{ }}mete{r^3}$ and the wing is 10 meter long and has an average width of 2 meter, then we need to find the difference on the two sides of the wing.
Applying the Bernoulli’s theorem,
The formula of the Bernoulli’s theorem is given by,
$\Rightarrow {p_1} + \dfrac{1}{2}\rho {v_1}^2 = {p_2} + \dfrac{1}{2}\rho {v_2}^2$
Where the pressure at two points is ${p_1}$and${p_2}$, the density of the fluid is $\rho$ also the velocity of the fluid at two points is ${v_1}^2$and${v_2}^2$.
The velocity on the both sides of wings are$120\dfrac{m}{s}$and$90\dfrac{m}{s}$, the density of the air is equal to $1.3{\text{ }}kg{\text{ }}per{\text{ }}mete{r^3}$ therefore we get,
$\Rightarrow {p_1} + \dfrac{1}{2}\rho {v_1}^2 = {p_2} + \dfrac{1}{2}\rho {v_2}^2$
$\Rightarrow {p_2} - {p_1} = \dfrac{1}{2}\rho {v_1}^2 - \dfrac{1}{2}\rho {v_2}^2$
$\Rightarrow {p_2} - {p_1} = \dfrac{1}{2}\rho \left( {{v_1}^2 - {v_2}^2} \right)$
$\Rightarrow {p_2} - {p_1} = \dfrac{1}{2} \times 1 \cdot 3 \times \left( {{{120}^2} - {{90}^2}} \right)$
$\Rightarrow \Delta p = 0 \cdot 65 \times \left( {14400 - 8100} \right)$
$\Rightarrow \Delta p = 4095Pa$.
The difference in the pressure on the wings is equal to $\Delta p = 4095Pa$.

The correct answer for this problem is option A.

Note: The pressure is defined as the ratio of the perpendicular force upon the surface area. The force on a surface can be calculated by simply multiplying the surface area with the pressure on that particular surface area.