Joined: 11 Dec 2005
|Posted: Tue Nov 01, 2011 6:52 pm Post subject: Questions on Climb and Descent
Maximum ANGLE OF CLIMB will be achieved when the excess THRUST is greatest and the aircraft is lightest.
The balance of forces in a steady climb show thrust is acting upwards and an element of weight is adding to the drag
As the thrust assists the lift, the lift required is less than in level flight. Verify mathematically by the formula Lift = W.cos gamma
For a steady speed to be maintained the thrust and the two retarding effects of aerodynamic drag and the weight element must be equal.
If Thrust = T, Drag = D and Weight = W, then as a formula it can be written as:
T = D + W sin gamma
Sin gamma = T - D / W
It means that climb angle (Sin gamma) depends on the excess thrust (i.e. thrust less drag) and the weight.
Tan gamma = Opposite / Adjacent
Opposite = Height gained
Adjacent = Distance covered on ground.
Height gained against distance covered on ground is the climb gradient.
Thus Tan gamma = climb gradient
For small angles Adjacent is nearly the same as Hypotenuse and Opposite/Adjacent = Sin relation.
So we get an approximate formula for climb gradient, which is:
Sin gamma = T – D / Weight
Climb gradient = T – D / Weight
Meaning that the greatest climb gradient is obtained when a greatest difference exists between thrust and drag and the weight is least.
In theoretical terms, for a piston engine aircraft this will occur at the minimum power speed (Vmp) and for a jet engine aircraft at the minimum drag speed (Vmd). Vmp is however close to the stall speed, and normally relates to 1.1Vs.
Maximum RATE OF CLIMB will be achieved when the excess POWER is greatest and the aircraft is lightest.
In terms of rate of climb:
Sin gamma = ROC/TAS = (T - D)/W
Now Power = TAS x Thrust
It follows that the ROC = TAS x (T - D)/W
This equates to ROC = (Power available - Power required)/W
or excess power/W, so the maximum rate of climb will be achieved when the excess power is greatest and the aircraft is lightest. This relates to Vy.
Excess power relates to the greatest difference between power available and power required to maintain level flight.
Note that in the power graph that the power available from a jet engine aircraft is a straight line. This is because the thrust is constant and the power varies according to the TAS.
For a piston engine aircraft the best rate of climb occurs when the aircraft is flown at Vmd and for a jet engine aircraft at a speed greater than Vmd.
Note that the speeds relate to theory and are what are expected in the examination. In practice these speeds may however vary depending on the airframe engine configuration.
The lift of an aeroplane of weight W in a constant linear climb with a climb angle (gamma) is approximately:
L = W x cos gamma
In a steady climb: thrust equals drag plus the weight component along the flight path and lift equals the weight component perpendicular to the flight path.
Power Available = Thrust X TAS and Power Required = Drag X TAS
For a jet aircraft the best rate of climb is achieved when excess power available is at a maximum.
In order to achieve the maximum rate of climb, aircraft should be flown at the indicated airspeed that gives maximum excess power.
During a straight steady climb, Lift is less than weight and Load factor is less than 1. Work it out yourself from L = W.Cos gamma
For shallow climb angles the following formula can be used: Sin (gamma) = T/W - CD/CL
From the climb diagram above:
Sin Gamma = Thrust - Drag / Weight
Sin Gamma = Thrust / Weigth - Drag / Weight
For shallow climb angles lift and weight are practically the same. Thus the formula can be written as:
Sin Gamma = Thrust / Weigth - Drag / Lift
Sin Gamma = Thrust / Weigth - CD / CL
Given an Aeroplane mass of 50,000 kg, Lift/Drag ratio 10, Thrust per engine 60,000N, assumed g=10 m/s2. For a straight, steady, wings level climb of a twin engine aeroplane, the all engines climb gradient will be:
Sin gamma = Thrust / Weight - Drag / Weight (as mentioned above)
Thrust = 60,000 per engine. For 2 engines its 2 x 60,000 = 120,000
Weight = mass x g. i.e. 50,000 x 10 = 500,000
L/D = 10 so Drag = Lift/10
Lift = W x cos gamma
Lift decreases as the angle of climb increases. In this questin we cannot calculate lift since we dont know the climb angle. For small angles of climb (up to about 15) lift is almost equal to weight so we can take the weight value as lift (error is very small).
So Lift = 500,000
Thus Drag = 500,000 / 10 = 50,000
Now putting the values in the formula:
Sin (gamma) = T/W - D/W
Sin (gamma) = 120,000/500,000 - 50,000/500,000
Sin (gamma) = 0.14
In percentage: Sin (gamma) = 0.14 x 100 = 14%
For calculating angle:
Sin gamma = 0.14
gamma = Inv Sin of 0.14 = 8 degrees.
An aeroplane performs a continuous descent with 160 kts IAS and 1000 feet/min vertical speed. In this condition weight is greater than lift.
What is the correct relationship between the true airspeed for (i) minimum sink rate and (ii) minimum glide angle, at a given altitude?
(i) is less than (ii)
Minimum sink rate speeed is VMP and minimum glide angle speed is VMD. Thus VMP is lower. At an altitude, VMP TAS will be lower than VMD TAS.
When an aeroplane performs a straight steady climb with a 20% climb gradient, the load factor is equal to:
20% climb gradient (in a right angled triangle) means:
20 up (opposite side) for 100 horizontal (adjacent side)
opposite/adjacent = tan (angle)
20/100 = tan (angle)
tan angle = 0.2
climb angle = Inv tan of 0.2 = 11.3
So gamma = 11.3
Lift in a climb is:
L = W x cos gamma
So load factor in climb is cos of climb angle
cos of 11.3 = 0.98 (Answer)
To maintain equilibrium when an aircraft is climbing, thrust equals the sum of drag and the weight component along the flight path whereas lift equals the weight component perpendicular to the flight path.
When climbing into a headwind, comparedto still air, the climb gradient relative to the ground will be steeper, and the rate of climb unchanged.
In a steady climb at a constant speed the power available must exceed the power required.
Regarding the forces acting on an aircraft in a steady-state descent, the sum of all forward forces is equal to the sum of all rearward forces.
Last edited by K.Haroon on Thu Apr 05, 2012 11:57 am; edited 4 times in total
Joined: 21 Feb 2012
|Posted: Wed Feb 22, 2012 12:54 pm Post subject:
| this make very good reading. i have the book.
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