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ENGINEERING, AE 2024 HSC 22b

The mass of a single-engine propeller aeroplane, without pilot or passengers, is 906 kg . It has a maximum lift to drag ratio of 10.5. When fully loaded, the plane needs to produce 1050 N of thrust to maintain cruising speed and altitude.

What is the maximum total allowable mass for the pilot and passengers in kilograms?   (3 marks)

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\(\text{Max allowable mass (pilot and passengers)}\ =196.5\ \text{kg}\)

Show Worked Solution

\(mg\ \text{(empty plane)}\ = 906 \times 10=9060\ \text{N}\)

\(\text{When plane cruising}\ \ \Rightarrow \ \ \text{Thrust = Drag = 1050 N}\)

\(\text{Max Lift to Drag ratio = 10.5}\)

Mean mark 54%.
\(\dfrac{\text{Max Lift}}{\text{Drag}} \) \(=10.5\)  
\(\text{Max Lift}\) \(=1050\ \text{N}\ \times 10.5\)  
  \(=11\ 025\ \text{N}\)  

 
\(\text{Max}\ mg\ \text{(pilot and passengers)}\ = 11\ 025-9060=1965\ \text{N}\)

\(\therefore\ \text{Max allowable mass (pilot and passengers)}\ =196.5\ \text{kg}\)

ENGINEERING, AE 2017 HSC 6 MC

The diagram shows an aerofoil.
 

Which condition needs to be achieved for lift to occur?

  1. Pressure 1<Pressure 2
  2. Pressure 1= Pressure 2
  3. Pressure 1> Pressure 2
  4. Pressure 1+ Pressure 2=0
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`A`

Show Worked Solution

→ For a net upwards force, the force from pressure below (pushing up) must exceed the force from pressure above (pushing down).

→ Therefore Pressure 1<Pressure 2.

`=A>`

ENGINEERING, AE 2016 HSC 25ai

An image of a small drone is shown.
 


  

Explain how this drone achieves flight.   (3 marks)

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→ The weight of the drone is counteracted by rotating propellers that produce vertical thrust (lift).

→ The drone will achieve flight if the vertical thrust exceeds the weight.

→ Controlled climb or descent is achieved by varying the speed of the motor.

→ The direction of flight or thrust can be affected by a change in the rotational speed of the propellers.

→ Power for flight is supplied by battery systems.

Show Worked Solution

→ The weight of the drone is counteracted by rotating propellers that produce vertical thrust (lift).

→ The drone will achieve flight if the vertical thrust exceeds the weight.

→ Controlled climb or descent is achieved by varying the speed of the motor.

→ The direction of flight or thrust can be affected by a change in the rotational speed of the propellers.

→ Power for flight is supplied by battery systems.


Mean mark 54%.

ENGINEERING, AE 2018 HSC 25c

 A two-engine aircraft with mass 330 tonnes is climbing at 15°. Each engine produces 510 kN of thrust. The aircraft maintains a constant velocity.

  1. Complete the space diagram by indicating the four key forces of flight for this situation.   (2 marks)
     
     

  1. Determine the induced drag force for this situation.   (3 marks)
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i.

ii.   `166\ text{kN}`

Show Worked Solution

i.

ii.  `text{Find Induced drag} (D):`

`text{Weight}\ (W)` `=text{m} xx text{g} `  
  `=330 xx 10^3 xx 10`  
  `=33 xx 10^5\ text {N}`  

 

`text{Thrust}\ (T)` `=510 xx 10^3 xx 2`  
  `=102 xx 10^4` N  

 

`T` `= D + W\ sin\ gamma`  
`D` `= T-W\ sin\ gamma`  
  `=102 xx 10^4  –  33 xx 10^5\ sin15^\circ`  
  `=165\ 897.1512`  
  `=166 xx 10^3`  
  `=166\ text{kN}`  

♦♦ Mean mark (ii) 34%.

ENGINEERING, AE 2022 HSC 24c

An image of a glider is shown.

The glider is currently on a descent at an angle of 19 degrees. The total lift force is 6250 N.

  1. Draw a free-body diagram, indicating all forces acting on the glider.   (1 mark)   
  1. If the mass of the pilot is 95 kg, calculate the mass of the glider.   (3 marks)
  1. Calculate the lift-to-drag ratio.   (2 marks)
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i.   Free-body diagram

ii.  Mass of the glider

`text{cos}19°` `=6250/W`  
`W` `=6250/(\text{cos}19°)`  
  `=6610.13N`  

 
`m=6610.13/10=661.01\ text{kg}`

`:. m_text{glider} =661.01-95=566.01\ text{kg}`
 

iii.    `(text{Lift})/(text{Drag})` `=(Wcos19°)/(Wsin19°)`
    `=1/(tan19°)`

 
`:.\ text{Lift : Drag}\ = 2.9 : 1`

Show Worked Solution

i.   Free-body diagram


♦ Mean mark (i) 49%.

ii.  Mass of the glider

`text{cos}19°` `=6250/W`  
`W` `=6250/(\text{cos}19°)`  
  `=6610.13N`  

 
`m=6610.13/10=661.01\ text{kg}`

`:. m_text{glider} =661.01-95=566.01\ text{kg}`


♦ Mean mark (ii) 46%.
iii.    `(text{Lift})/(text{Drag})` `=(Wcos19°)/(Wsin19°)`
    `=1/(tan19°)`

 
`:.\ text{Lift : Drag}\ = 2.9 : 1`


♦♦ Mean mark (iii) 37%.

ENGINEERING, AE 2021 HSC 23c

A 3000 kg aircraft is flying in level flight at constant speed.

Using a free-body diagram, calculate the thrust on the plane if the lift to drag ratio of the aircraft is 11:1.   (3 marks)

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`2.73\ text{kN}`

Show Worked Solution

`text{Lift} = 30\ 000\ text{N}↑`

`text{Find drag}\ (D):`

`L/D` `= 11/1\ \ text{(given)}`  
`(30\ 000)/D` `= 11/1\ text{N}`  
`D` `= (30\ 000)/11`  
  `= 2730\ text(N)`  

 
`:.\ text{Plane’s thrust} = 2.73\ text(kN)`


Mean mark 56%.