Wing Area 31.83

Mass 5156 kg
Engines 2x 11kN Turbofan
Drag Polar Cdo + KCL^2
Cdo 0.025
k 0.0445
Maximum speed
0/1 point (graded)
The aircraft is flying at sealevel in straight and level flight
What is the maximum speed of the aircraft in kts?

Give the climb angle in degrees corresponding to this speed and rate of climb

Give the climb angle in degrees corresponding to this speed and rate of climb
where possible use python or numerical methods

1 answer

To calculate the maximum speed of the aircraft in straight and level flight, we can use the following principles and formulas.

Given Data

  • Wing Area (S): 31.83 m²
  • Mass (m): 5156 kg
  • Thrust (T): 2 × 11 kN = 22 kN = 22000 N
  • Cdo: 0.025
  • k: 0.0445

Step 1: Calculate Lift at Sea Level

In straight and level flight, lift equals weight:

\[ L = W = m \cdot g \]

Where \( g \) (acceleration due to gravity) is approximately \( 9.81 , \text{m/s}^2 \).

\[ L = 5156 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 50553.96 , \text{N} \]

Step 2: Calculate the maximum speed using the lift equation

Lift for a given speed can be represented as:

\[ L = \frac{1}{2} \cdot \rho \cdot V^2 \cdot S \cdot C_L \]

Where:

  • \( \rho \) (air density at sea level) is approximately \( 1.225 , \text{kg/m}^3 \).
  • \( V \) is the speed in m/s.
  • \( C_L \) (Lift Coefficient) varies with angle of attack and is typically taken as \( 2\pi \cdot \alpha \) for small angles or can be assumed along with drag in some cases.

However, we also have to account for drag. The drag D can be expressed as:

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

Where \( C_D = C_{D0} + k \cdot C_L^2 \).

Step 3: Rearranging for Maximum Speed

The condition for maximum speed occurs when thrust equals drag:

\[ T = D \]

Thus,

\[ T = \left( C_{D0} + k \cdot \frac{2L}{\rho \cdot S \cdot V^2}^2 \right) \cdot \frac{1}{2} \cdot \rho \cdot V^2 \cdot S \]

Step 4: Calculation

We can use numerical methods to solve for V using Python. Here's the code:

import numpy as np
from scipy.optimize import fsolve

# Constants
S = 31.83  # Wing Area in m^2
m = 5156  # mass in kg
g = 9.81  # gravity in m/s^2
rho = 1.225  # air density in kg/m^3
Cdo = 0.025  # zero-lift drag coefficient
k = 0.0445  # coefficient for drag polar
T = 22000  # thrust in N

# Weight of the aircraft
W = m * g

# Climb angle calculation
def equations(V):
    CL = (2 * W) / (rho * S * V**2)  # Lift coefficient
    CD = Cdo + k * CL**2  # Drag coefficient
    drag = CD * (0.5 * rho * V**2 * S)  # Drag force
    return T - drag  # Return thrust - drag

# Estimate max speed
initial_guess = 100  # Initial speed guess in m/s
max_speed_mps = fsolve(equations, initial_guess)[0]

# Convert speed from m/s to knots
max_speed_kts = max_speed_mps * 1.94384  # 1 m/s = 1.94384 knots
max_speed_kts

Step 5: Calculate The Climb Angle

The climb angle (γ) can be calculated by relating the rate of climb to the weight and thrust:

\[ \text{Rate of Climb} = (T - D) \cdot V / W \]

The angle can be calculated as:

\[ \gamma = \arctan\left( \frac{\text{Rate of Climb}}{V} \right) \]

Example Implementation

You may modify the earlier code to include the calculation of climb angle after finding \( V \).

Conclusion

Use the provided code to calculate maximum speed in knots and the associated climb angle derived from the rate of climb under given thrust conditions. The output will give you the required values.

If you need further assistance with running the calculation or anything else, feel free to ask!