Complete question is;
An airplane has a mass of 2.5 × 10^(6) kg , and the air flows past the lower surface of the wings at 80 m/s .
If the wings have a surface area of 1600 m² , how fast must the air flow over the upper surface of the wing if the plane is to stay in the air?
Answer:
v_a = 178.74 m/s
Explanation:
We can find the sum of pressure head and velocity head above and below the wing using Bernoulli's theorem given by;
P_a + ½ρ(v_a)² = P_b + ½ρ(v_b)²
Where;
P_a is pressure above wings
P_b is pressure below wings
v_a is speed above wings
v_b is speed below wings
ρ is density of air
We want to find V_a, so let's make V_a the subject;
v_a = √[(2(P_b - P_a)/ρ) + (v_b)²]
Now, we don't know (P_b - P_a)
However, P_b - P_a is the pressure difference and we know pressure is force/area
Thus;
(P_b - P_a) = Force/Area
(P_b - P_a) = mg/Area
We are given m = 2.5 × 10^(6) kg and area = 1600 m²
Thus, (P_b - P_a) = (2.5 × 10^(6) × 9.81)/1600 = 15328.125 N/m²
Density of air will be taken as 1.2 kg/m³
Thus;
v_a = √[(2(15328.125)/1.2) + (80)²]
v_a = √31946.875
v_a = 178.74 m/s
