Takeoff

This section computes parameters related to HL devices for the example aiplane in takeoff conditions. The drag polar in takeoff condition is also plotted, and the maximum \(L/D\) is also determined. Based on constraint analysis, the \(C_{L_{max_{TO}}}\) is 1.8 for the example airplane. Note that \(C_{L_{max}}\) is 1.44, computed in aerodynamics section. Using these values and the process outlined earlier, below code block computes various parameters related to HL devices for takeoff conditions.

# Variables
CLmax = 1.44
CLmaxTO = 1.8
Sf = 58.1 # sq ft
Sref = 134 # sq ft
K = 0.93 # from figure 7.4, Roskam Part 2
cf_c = 0.25
delta_alpha_Clzero = -9 # deg, assumed from Raymer
Clalpha = 0.116 # 1/deg for NACA 23018
alpha_CLzero = -1 # from aerodynamics section

# Requried change in Cl
delta_CLmax = 1.05*(CLmaxTO - CLmax)
delta_Clmax = delta_CLmax * Sref / Sf
delta_Cl = delta_Clmax / K

print(f"Required change in airfoil sectional Cl: {delta_Cl:.2f}")

# Computed change in Cl
# Change the value of delta_f and alpha_delta_f until computed value is same as required
delta_f = 16 # deg
alpha_delta_f = 0.5 # fromdelta_f value and figure 7.8, Roskam Part 2
delta_Cl_comp = Clalpha * alpha_delta_f * delta_f

print(f"Computed change in airfoil sectional Cl with flap deflection of {delta_f} deg: {delta_Cl_comp:.2f}")

# change zero lift angle
delta_alpha_CLzero = delta_alpha_Clzero * Sf / Sref

# zero lift angle with TO flaps
alpha_flap_CLzero = alpha_CLzero + delta_alpha_CLzero

print(f"Zero-lift angle of attack with TO flaps: {alpha_flap_CLzero:.1f}  deg")

Required change in airfoil sectional Cl: 0.94
Computed change in airfoil sectional Cl with flap deflection of 16 deg: 0.93
Zero-lift angle of attack with TO flaps: -4.9  deg

Based on above calculation, it is determined that flap deflection of approximately 15 degrees is enough to achieve the maximum \(C_L\) of 1.8 at takeoff conditions. Below code block plots the lift curve in takeoff conditions.

NOTE: The sinlge slotted flap used for example airplane does not change the lift curve slope, while having approximately same \(\alpha\) for maximum \(C_L\).

# Lift curve
import numpy as np
import matplotlib.pyplot as plt

fs = 14 # fontsize
alpha = np.linspace(-7,16,100) * np.pi/180 # alpha values, rad
CLalpha = 5.1
CLmaxTO = 1.8

CL = CLalpha * (alpha - alpha_flap_CLzero * np.pi/180)

# Splitting data based on linear and nonlinear region
CL_linear = CL[CL<=1.6]
alpha_linear = alpha[CL<=1.6]

CL_nonlinear = CL[np.logical_and(CL>1.6,CL<CLmaxTO)]
alpha_nonlinear = alpha[np.logical_and(CL>1.6,CL<CLmaxTO)]

fig, ax = plt.subplots()
ax.plot(alpha_linear*180/np.pi, CL_linear, "k-")
ax.plot(alpha_nonlinear*180/np.pi, CL_nonlinear, "k--")
ax.scatter(18, CLmaxTO)
ax.set_xlabel("Angle of attack (deg)", fontsize=fs)
ax.set_ylabel("Lift coefficient, $C_L$", fontsize=fs)
ax.set_title("Example airplane lift curve (TO configuration)", fontsize=fs)
ax.annotate(r"$C_{L_\alpha}$: " + f"{CLalpha:.2} " + r"$\text{rad}^{-1}$", (10,0.3), fontsize=fs, ha="center", va="center")
ax.annotate("$C_{L_{max}}$", (18,CLmaxTO-0.15), fontsize=fs, ha="center", va="center")
ax.grid()
plt.tight_layout()

Takeoff Drag Polar

Once lift curve is obtained for the takeoff flap setting, one can plot the drag polar as well. As discussed in aerodynamics section, a simple drag polar model is used. With flaps deployed, one has to compute its effect on \(C_{D_0}\) and \(C_{L_i}\). As described in high-lift devices section, the change in \(C_{D_0}\) can be computed based on amount of flap deflection. To account for the change in induced drag, the value of \(e\) is reduced by 0.05 (based on Table 3.6, Roskam Part 1). Since LG is not fully sized at this stage of design, increase in \(C_{D_0}\) due to LG is set to 0.02. One should also account for the ground effect while computing takeoff drag polar. Section 12.6.4 in Raymer provides an equation to update \(K\) to account for the ground effects.

Below code block computes the change in zero-lift drag:

Ff = 0.0074
cf_c = 0.25
Sf = 58.1 # sq ft
Sref = 134 # sq ft
delta_f = 15 # deg

delta_CD0 = Ff * cf_c * Sf / Sref * (delta_f - 10) + 0.02

print(f"Change in zero-lift drag: {delta_CD0:2f}")

Change in zero-lift drag: 0.024011

Below code block plots the drag polar for example airplane in TO conditions:

fs = 14 # fontsize
A = 8 # aspect ratio
CLalpha = 5.1 # 1/rad
alpha_CLzero = -4.9 # deg
e = 1.78*(1 - 0.045*A**0.68)-0.64 - 0.05 # e in TO
K = 1/np.pi/A/e
h = 4 # ft
b = 33 # ft
Keff = K * ( 33 * (h/b)**1.5 / ( 1 + 33 * (h/b)**1.5 ) ) # eq 12.60, Raymer

alpha = np.linspace(-7, 13, 100) # alpha in deg

CL = CLalpha * (alpha - alpha_CLzero) * np.pi / 180

CD0 = 0.03363 + delta_CD0 # zero-lift drag in TO conditions

CD = CD0 + Keff*CL**2 # drag model

fig, ax = plt.subplots()
ax.plot(CD, CL, "k-")
ax.plot([0.0, 0.135], [0.0, 1.62], "k--")
ax.scatter(0.112, 1.35, zorder=10, s=30)
ax.set_xlabel("$C_D$", fontsize=fs)
ax.set_ylabel("$C_L$", fontsize=fs)
ax.set_title("Drag polar for example airplane - TO configuration", fontsize=fs)
ax.annotate("$L/D$ max", (0.112,1.25), fontsize=fs, ha="left", va="center")
ax.grid()
plt.tight_layout()

As indicated in the above plot, the point where line from origin is tangent to drag polar refers to maximum \(L/D\) in TO configuration. The example airplane has a maximum \(L/D\) of 12. This concludes the HL devices sizing for TO condition. In next section, high-lift devices for landing configuration are sized.