Question

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:

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