Suppose an airliner is planning to fly an Airbus A319 very low over a water surface, thereby increasing its lift coefficient by 40%. However, the airliner does not need 40% extra lift, instead it decides to reduce the wing area such that the aircraft generates the same lift as a normal A319. What would the wing area (in square metres) of this modified A319 have to be?
To find the wing area of the modified A319, we first need to calculate the original wing area.
The standard wing area of an Airbus A319 is approximately 122 square meters.
Given that the lift coefficient is increased by 40%, we can express this as a lift coefficient of 1.4 times the original value.
Since lift coefficient is directly proportional to lift, we can assume that the lift of the modified A319 remains the same as the original A319.
Hence, the modified lift coefficient (C_L_modified) is equal to original lift coefficient (C_L_original) multiplied by 1.4.
Let the modified wing area be represented by A_modified.
Since lift (L) is directly proportional to wing area (A) and the lift coefficient (C_L), we can write:
L_modified = C_L_modified * A_modified
L_original = C_L_original * A_original
Since the lift remains the same for both the original and modified A319:
L_modified = L_original
=> C_L_modified * A_modified = C_L_original * A_original
Since C_L_modified = 1.4 * C_L_original, we can substitute it in the equation:
1.4 * C_L_original * A_modified = C_L_original * A_original
Simplifying the equation, we get:
1.4 * A_modified = A_original
A_modified = A_original / 1.4
Plugging in the value of A_original (122 square meters), we can calculate A_modified:
A_modified = 122 / 1.4
A_modified ≈ 87.14 square meters
Therefore, the wing area of the modified A319 would have to be approximately 87.14 square meters.
To find the necessary wing area for the modified A319, we can use the lift equation. The lift equation is as follows:
Lift = (1/2) * ρ * V^2 * A * Cl
Where:
- Lift is the force perpendicular to the direction of motion (N)
- ρ is the air density (kg/m^3)
- V is the velocity of the aircraft (m/s)
- A is the wing area (m^2)
- Cl is the lift coefficient
For the modified A319, the lift coefficient (Cl) is increased by 40%, which means it becomes 1.4 times the normal Cl value. However, the lift (Lift) should remain the same as that of a normal A319.
Now, let's assume the normal A319 wing area is denoted as A_normal. The modified A319's wing area can be represented as A_modified.
Since the lift should be the same for both aircraft:
Lift (modified A319) = Lift (normal A319)
(1/2) * ρ * V^2 * A_modified * 1.4 * Cl (modified A319) = (1/2) * ρ * V^2 * A_normal * Cl (normal A319)
Cancelling out similar terms:
A_modified * 1.4 = A_normal
Dividing both sides by 1.4:
A_modified = A_normal / 1.4
So, the wing area of the modified A319 would need to be A_normal divided by 1.4 square meters.
To find the wing area of the modified A319, we need to understand how lift coefficient, lift, and wing area are related. The lift coefficient (CL) is a dimensionless value that represents the lift generated by an aircraft's wings. It is directly related to the wing area (A) and the lift force (L) by the formula:
CL = L / (0.5 * ρ * V^2 * A)
Where:
- CL is the lift coefficient
- L is the lift force
- ρ (rho) is the air density
- V is the airspeed
- A is the wing area
In this scenario, the airliner is planning to increase the lift coefficient by 40%, but it wants to maintain the same lift as a normal A319. So, the lift force remains constant for both cases.
To determine the wing area of the modified A319, we can set up an equation of the lift generated for the normal A319 and the modified A319:
For the normal A319:
CL_Normal = L / (0.5 * ρ * V^2 * A_Normal)
For the modified A319:
CL_Modified = L / (0.5 * ρ * V^2 * A_Modified)
Since the lift force is the same for both cases, we can equate the two equations:
CL_Normal = CL_Modified
We know that the modified A319 has an increased lift coefficient of 40%, which can be expressed as:
CL_Modified = 1.4 * CL_Normal
Substituting this into the equation, we have:
1.4 * CL_Normal = CL_Normal
Now, we can solve for the wing area (A_Modified):
A_Modified = A_Normal / 1.4
Therefore, the wing area of the modified A319 would be 1.4 times smaller than that of the normal A319.