uppose that the lift-drag polar of the previous problem belongs to a 32,000 kilogram aircraft with an average wing chord of 3.5 metres and a wingspan of 30 metres. At a certain moment this aircraft flies in cruise at a speed of 300 ft/s, with an air density of 1.0 kilogram per cubic metre. For this flight condition, determine the aircraft drag (in Newton).

Answer 1

To calculate the drag force acting on the aircraft, we can use the drag equation:

\[ D = \frac{1}{2} C_D \cdot \rho \cdot V^2 \cdot S \]

where:

• \( D \) is the drag force,
• \( C_D \) is the drag coefficient,
• \( \rho \) is the air density,
• \( V \) is the velocity of the aircraft,
• \( S \) is the reference area (wing area).

Step 1: Calculate the Wing Area \( S \)

The wing area \( S \) can be calculated using the wingspan and the average wing chord:

\[ S = b \cdot c \]

where:

• \( b \) is the wingspan,
• \( c \) is the average wing chord.

Given:

• \( b = 30 \) m,
• \( c = 3.5 \) m,

\[ S = 30 , \text{m} \times 3.5 , \text{m} = 105 , \text{m}^2 \]

Step 2: Convert Velocity to Appropriate Units

The velocity given is \( V = 300 , \text{ft/s} \). To use SI units in our calculations, we need to convert feet per second to meters per second:

\[ 1 , \text{ft} = 0.3048 , \text{m} \]

Thus,

\[ V = 300 , \text{ft/s} \times 0.3048 , \text{m/ft} \approx 91.44 , \text{m/s} \]

Step 3: Air Density

The air density \( \rho \) is given as:

\[ \rho = 1.0 , \text{kg/m}^3 \]

Step 4: Drag Coefficient Assumption

Since the drag coefficient \( C_D \) is not provided in the problem, we need to assume a value. For cruise conditions of many aircraft, \( C_D \) can vary but is typically around \( 0.025 \) to \( 0.05 \). Let's take \( C_D = 0.03 \) for this calculation.

Step 5: Calculate Drag

Now that we have all the values, we can calculate the drag:

\[ D = \frac{1}{2} \cdot C_D \cdot \rho \cdot V^2 \cdot S \]

Substituting the values:

\[ D = \frac{1}{2} \cdot 0.03 \cdot 1.0 , \text{kg/m}^3 \cdot (91.44 , \text{m/s})^2 \cdot 105 , \text{m}^2 \]

Calculating \( V^2 \):

\[ V^2 \approx (91.44)^2 \approx 8367.91 , \text{m}^2/\text{s}^2 \]

Now substitute this into the drag equation:

\[ D = \frac{1}{2} \cdot 0.03 \cdot 1.0 \cdot 8367.91 \cdot 105 \] \[ D = 0.015 \cdot 8367.91 \cdot 105 \] \[ D = 0.015 \cdot 878630.55 \approx 13179.46 , \text{N} \]

Final Answer

Thus, the drag force acting on the aircraft is approximately:

\[ \boxed{13179.46 , \text{N}} \]