Question

(II)What is the lift (in newtons) due to Bernoulli’s principle on a wing of area\(88\;{{\rm{m}}^2}\)if the air passes over the top and bottom surfaces at speeds of 280 m/s and 150 m/s, respectively?

Short answer

The lift force on wing is \(31.97 \times {10^5}\;{\rm{N}}\).

Step-by-step solution

Understanding the concept of Bernoulli’s equation

The lift force on the wing is estimated by utilizing the relation of pressure variation with fluid velocity which is obtained by reducing Bernoulli’s equation.

Given data

The area of wing is \(A = 88\;{{\rm{m}}^2}\).

The speed of air over the top surface of wing is \({v_1} = 280\;{\rm{m/s}}\).

The speed of air over the bottom surface of wing is \({v_2} = 150\;{\rm{m/s}}\).

The standard value for the density of air is \(\rho = 1.3\;{\rm{kg/}}{{\rm{m}}^3}\).

Evaluating the lift force on a wing by using Bernoulli’s equation reduced form

The lift force on a wing is calculated as:

\(\begin{array}{c}{P_2} - {P_1} = \frac{1}{2}\rho \left( {{{\left( {{v_1}} \right)}^2} - {{\left( {{v_2}} \right)}^2}} \right)\\\frac{F}{A} = \frac{1}{2}\rho \left( {{{\left( {{v_1}} \right)}^2} - {{\left( {{v_2}} \right)}^2}} \right)\\F = \frac{1}{2}\rho A\left( {{{\left( {{v_1}} \right)}^2} - {{\left( {{v_2}} \right)}^2}} \right)\end{array}\)

Here, \({P_1}\) and \({P_2}\) are the pressure at bottom and the top surface of the wing and F is the lift force.

Substitute the values in the above equation.

\(\begin{array}{c}F = \frac{1}{2}\left( {1.3\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {88\;{{\rm{m}}^2}} \right)\left( {{{\left( {280\;{\rm{m/s}}} \right)}^2} - {{\left( {150\;{\rm{m/s}}} \right)}^2}} \right)\\F = 31.97 \times {10^5}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2} \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\F = 31.97 \times {10^5}\;{\rm{N}}\end{array}\)

Hence, the lift force on a wing is \(31.97 \times {10^5}\;{\rm{N}}\).