Please answer the following questions:

Question

A) The flight condition for which an aircraft has minimum aerodynamic drag is important from an
aircraft performance perspective. Consider a propeller aircraft (the Cessna 210) which has a
constant maximum power available as a function of airspeed.
For which of the following cases should the aircraft be flown in the minimum drag condition in
order to achieve maximum performance? Multiple answers can be selected.
1) Minimum airspeed
2) Maximum airspeed
3) Maximum range
4) Maximum endurance
5) Maximum climb angle
6) Maximum rate of climb
7) Minimum descent angle (in case of no engine thrust)
8) Minimum rate of descent (in case of no engine thrust)
B) The aerodynamics of the Cessna 210 is approximated by a 2-term lift-drag polar:
CD = 0.0123 + 0.00075CL + 0.0553C
2
L
The weight of the aircraft is given as W = 14, 715 N, and the wing surface area is S = 16.23 m2
.
Assume that the aircraft is flying a steady, horizontal and symmetric flight
Calculate the airspeed (in m/s) corresponding to the minimum drag condition at sea-level (ρ =
1.225 kg/m3
).
C) Calculate the minimum drag (in N) at sea-level.
D) For increasing altitude, which of the following statements is true?
1) Airspeed for minimum drag decreases and the corresponding minimum drag decreases as
well.
2) Airspeed for minimum drag increases and the corresponding minimum drag decreases.
3) Airspeed for minimum drag decreases and the corresponding minimum drag increases.
4) Airspeed for minimum drag decreases and the corresponding minimum drag stays constant.
5) Airspeed for minimum drag increases and the corresponding minimum drag stays constant.
6) Airspeed for minimum drag stays constant and the corresponding minimum drag stays
constant as well.
7) Airspeed for minimum drag stays constant and the corresponding minimum drag increases.
8) Airspeed for minimum drag stays constant and the corresponding minimum drag decreases.
9) Airspeed for minimum drag increases and the corresponding minimum drag increases as
well
For now, assume that CLopt = 0.60. Note: This is different than the true answer to the previous
question.
E) Calculate the airspeed (in m/s) corresponding to the minimum drag condition at 3000 m altitude
(ρ = 0.9091 kg/m3
).
F) Calculate the minimum drag (in N) at 3000 m altitude. Again assume that CLopt = 0.60.

Answer 1

Let's go through each of the questions step-by-step.

A) Minimum Drag Condition for Maximum Performance

In general, the minimum drag condition corresponds to best lift-to-drag ratio, which is crucial for various flight performance scenarios:

• 3) Maximum range: True. You want to fly at minimum drag to maximize distance for a given amount of fuel.
• 4) Maximum endurance: True. Minimum drag condition yields the longest time in the air for a given amount of fuel.
• 5) Maximum climb angle: Not directly related to minimum drag. Climb performance often requires more thrust relative to drag, rather than maximizing lift-to-drag ratio.
• 6) Maximum rate of climb: Not directly maximizing drag but rather thrust excess over drag.
• 8) Minimum rate of descent (in case of no engine thrust): True. Similar to maximum endurance, in this condition you would want to minimize drag to maintain the longest flight time before landing, but you are not aiming for minimum drag for performance.

So, the applicable cases for minimum drag are 3, 4, and 8.

B) Calculate the Airspeed Corresponding to Minimum Drag

To find the airspeed for minimum drag, we need to determine the value of \(C_L\) at \(CL_{opt}\):

1. Use the relationship \(W = \frac{1}{2} \rho V^2 S C_L\).
2. Solve for \(V\):

\[ C_L = CL_{opt} = 0.60 \] \[ W = 14715 , N \] \[ S = 16.23 , m^2 \] \[ \rho = 1.225 , kg/m^3 \]

Setting the equations: \[ 14715 = \frac{1}{2} (1.225) V^2 (16.23) (0.60) \]

Rearranging: \[ V^2 = \frac{14715 \cdot 2}{1.225 \cdot 16.23 \cdot 0.60} \]

Calculating: \[ V^2 = \frac{29430}{12.03435} \approx 2443.57 \] \[ V \approx 49.43 , m/s \]

C) Calculate Minimum Drag at Sea-Level

Using the drag polar \(C_D = 0.0123 + 0.00075C_L + 0.0553C_L^2\) to find drag:

For \(C_L = 0.60\): \[ C_D = 0.0123 + 0.00075(0.60) + 0.0553(0.60)^2 \] \[ C_D = 0.0123 + 0.00045 + 0.01998 \approx 0.03273 \]

Calculating the drag force: \[ D = \frac{1}{2} \rho V^2 S C_D \] \[ D = \frac{1}{2} \cdot 1.225 \cdot (49.43)^2 \cdot 16.23 \cdot 0.03273 \] \[ D \approx \frac{1}{2} \cdot 1.225 \cdot 2443.57 \cdot 16.23 \cdot 0.03273 \] \[ D \approx 24.05 , N \]

D) True Statements for Increasing Altitude

As altitude increases, the air density decreases. This affects both the airspeed for minimum drag (which should increase because \(L = D\) becomes more efficient with lower density) and the corresponding minimum drag (which decreases because of weight and lower density).

So, for altitude increase:

• 1) Airspeed for minimum drag increases and the corresponding minimum drag decreases: This is the true statement.

E) Calculate the Airspeed at 3000 m Altitude

At 3000 m: \[ \rho = 0.9091 , kg/m^3 \] Using the same formula:

\[ W = 14715 , N, \] \[ C_L = 0.60 \] So: \[ W = \frac{1}{2} \rho V^2 S C_L \] Rearranging gives:

\[ V^2 = \frac{2W}{\rho S C_L} \] \[ V^2 = \frac{2 \cdot 14715}{0.9091 \cdot 16.23 \cdot 0.60} \]

Calculating: \[ V^2 = \frac{29430}{8.813298} \approx 3332.81 \] \[ V \approx 57.73 , m/s \]

F) Calculate Minimum Drag at 3000 m Altitude

Using: \[ C_D = 0.0123 + 0.00075(0.60) + 0.0553(0.60)^2 \] So, it remains the same \(C_D \approx 0.03273\).

Now calculate drag force: \[ D = \frac{1}{2} \rho V^2 S C_D \] \[ D = \frac{1}{2} \cdot 0.9091 \cdot (57.73)^2 \cdot 16.23 \cdot 0.03273 \]

Calculating gives: \[ D \approx \frac{1}{2} \cdot 0.9091 \cdot 3332.85 \cdot 16.23 \cdot 0.03273 \approx 17.04 , N \]

Summary:

A) 3, 4, 8 B) 49.43 m/s C) 24.05 N D) 1 E) 57.73 m/s F) 17.04 N