## Airplane Wings

What is the lift due to Bernoulli's principle on a wing of area 44.5 m2 if the air passes over the top and bottom surfaces at speeds of 355 m/s and 274 m/s, respectively? The density of air is 1.29 kg/m3.

**1.46×106 N**## **Approach**

Wings are designed so that the wind speed is higher (lower pressure) over the top of the wing than the bottom. The difference in pressure between the top and bottom of the wing creates lift.

## **Solution**

Given quantities:

Pb + 1/2pvb^2 + pghb = Pt + 1/2pvt^2 + pght

Rearranges to give...

Pb-Pt = 1/2p(Vt^2-Vb^20+ pg(ht-hb)

Now, ht - hb is the thickness of the wing, which is of the order of 1 m. The term ρ g (ht - hb) is insignificant compared to the speed term so can be neglected. The lift is

(Pb-Pt)=1/2p(Vt^2-Vb^2)

Putting in the numbers, the lift is 1.46×106 N.

- density of air, ρair = 1.29 kg/m3
- top wind speed, vt = 355 m/s
- bottom wind speed, vb = 274 m/s
- area of wing, A = 44.5 m2

Pb + 1/2pvb^2 + pghb = Pt + 1/2pvt^2 + pght

Rearranges to give...

Pb-Pt = 1/2p(Vt^2-Vb^20+ pg(ht-hb)

Now, ht - hb is the thickness of the wing, which is of the order of 1 m. The term ρ g (ht - hb) is insignificant compared to the speed term so can be neglected. The lift is

(Pb-Pt)=1/2p(Vt^2-Vb^2)

Putting in the numbers, the lift is 1.46×106 N.