I think that you will find that there is an error in your first question.
By what percentage does the lift increase in a steady level turn at 45° angle of bank, compared to straight and level flight?
a. 52%.
b. 41%.
c. 19%.
d. 31%.
In order to carry out a constant altitude turn the aircraft must be banked in the direction of the turn. This tilts the lift force towards the lower wing, so that part of it acts horizontally towards the centre of the turn. It is this horizontal component of lift that accelerates the aircraft in the direction of the intended turn.
But tilting the lift force reduces the vertical component of lift, such that it no longer equals the weight of the aircraft. This means that more lift is required in order to prevent the aircraft from sinking. This increase in lift constitutes an increase in load factor.
The load factor at any bank angle in a constant altitude turn is equal to 1 divided by the cosine of the bank angle. So in a 45 degree banked turn the load factor is 1/Cos 45 which is 1/0.7071, or approximately 1.414.
The load factor is equal to the lift divided by the weight. In straight and level flight the lift is equal to the weight of the aircraft and the load factor is 1. So an increase in load factor from 1 to 1.41 means that lift has increased to 141% of its initial value. This means that the lift has increased by 41% (option b).
But had the question been about the change in stalling speed we would have:
By what approximate percentage will the stall speed increase in a horizontal coordinated turn with a bank angle of 45°?
a. 52%.
b. 19%.
c. 31%.
d. 41%.
In order to carry out a constant altitude turn the aircraft must be banked in the direction of the turn. This tilts the lift force towards the lower wing, so that part of it acts horizontally towards the centre of the turn. It is this horizontal component of lift that accelerates the aircraft in the direction of the intended turn.
But tilting the lift force reduces the vertical component of lift, such that it no longer equals the weight of the aircraft. This means that more lift is required in order to prevent the aircraft from sinking. This increase in lift constitutes an increase in load factor.
The load factor at any bank angle in a constant altitude turn is equal to 1 divided by the cosine of the bank angle. So in a 45 degree banked turn the load factor is 1/Cos 45 which is 1/0.7071, or approximately 1.414.
The stall speed in any manoeuvre can be calculated by using the standard equation:
Stall speed at new load factor = 1g stall speed x Square root of (new load factor).
Inserting the data above gives:
Stall speed in 45 degree turn = 1g stall speed x Square root of (1.414) which = 1g stall speed x 1.189.
This means that the stall speed has increased by 18.9% or approximately
19% (option b).
For your second question
2. Twin engine airplane (59.000 Kg) climbing with all engines operating. Lift-Drag ratio is 12. Each engine producing 60.000 N thrust. The gradient of climb is (assume g = 10 m/sē)?
Correct answer is 12%
If we make the simplifying assumption that lift = weight in a steady climb, then we can calculate climb gradient using the standard equation below.
% Gradient = 100 x ( (thrust/weight) (1/Lift: Drag ratio)).
This can also be expressed as:
% Gradient = 100 x ( (thrust/weight) (Drag/Lift)).
It is essential that both all forces are expressed in the same units.
In this question weight is in kg and thrust is in Newton, so one must be of these must be converted to match the other.
To convert Newton to kg divide by g.
So thrust per engine = 60000 N /10 m/s/s = 6000 kg thrust.
The question asks for the all engines climb gradient so the total thrust is 2 x 6000 = 12000 kg.
Inserting the data into the equation gives:
% Gradient = 100 x ( (thrust/weight) (Drag/Lift) ).
% Gradient = 100 x ( (12000 kg /59000 kg) (1/12)).
% Gradient = 12%
Last edited by keith williams; 31st Jan 2013 at 20:00.