Chapter 10, Problem 44P

To determine

To Calculate: The lift (in newtons) due to Bernoulli’s principle on the wing.

Answer to Problem 44P

  $\text{2}{\text{.3×10}}^{\text{9}}\text{N}$

Explanation of Solution

Given info:

Area, $A=78{\text{m}}^{\text{2}}$

The speeds at the top and bottom surfaces are $\text{260}\frac{\text{m}}{\text{s}}\text{ and }\text{150}\frac{\text{m}}{\text{s}}$ respectively.

Formula used:

Bernoulli’s equation is,

  $\begin{array}{l}{P}_1+\frac{1}{2}\rho v_1{}^2=P_2+\frac{1}{2}\rho v_2{}^2\\ P_1-P_2=\frac{1}{2}\rho v_2{}^2-\frac{1}{2}\rho v_1{}^2\\ \Delta P=\frac{1}{2}\rho \left(v_2{}^2-v_1{}^2\right)\end{array}$

The force on the roof is,

  $\text{F}\text{=}\text{Pressure}\text{×Area}$

Calculation:

The lift force must be equal to the pressure difference between the wing’s surfaces, multiply by the wing’s surface area. Now, assume the two surfaces of the wings to be at the equal height, where subscript 1 represents the bottom surface and subscript 2 represents the top surface. The Bernoulli’s equation becomes for the equal height is given as,

  $\begin{array}{l}{P}_1+\frac{1}{2}\rho v_1{}^2=P_2+\frac{1}{2}\rho v_2{}^2\\ P_1-P_2=\frac{1}{2}\rho v_2{}^2-\frac{1}{2}\rho v_1{}^2\\ \Delta P=\frac{1}{2}\rho \left(v_2{}^2-v_1{}^2\right)\end{array}$

Now substitute the value in this equation,

  $\begin{array}{c}\Delta P=\frac{1}{2}\rho \left(v_2{}^2-v_1{}^2\right)\\ =\frac{1}{2}\times \left(1.29\times {10}^3\right)\times \left({\left(260\right)}^2-{\left(150\right)}^2\right)\\ =\frac{1}{2}\times \left(1.29\times {10}^3\right)\times 45100\\ \Delta P=29089\times {10}^3\frac{\text{N}}{{\text{m}}^{\text{2}}}\end{array}$

Now the net force is,

  $\begin{array}{c}{F}_{net}=\Delta P\times {\text{78 m}}^{\text{2}}\\ {\text{=29089×10}}^{\text{3}}\frac{\text{N}}{{\text{m}}^{\text{2}}}\text{×}{\text{78 m}}^{\text{2}}\\ \text{=}\text{2}{\text{.3×10}}^{\text{9}}\text{N}\end{array}$

Conclusion:

The force is $\text{2}{\text{.3×10}}^{\text{9}}\text{N}$ .