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ME307 1/11
SCHOOL OF ENGINEERING
MODULAR HONOURS DEGREE COURSE
LEVEL 3
SEMESTER 2
2002/03
AIRCRAFT PERFORMANCE AND CONTROL
Examiners: Mr J Hutchinson/Dr D F Pearce
Attempt FOUR questions only
Time allowed: 2 hours
Total number of questions = 6
All questions carry equal marks
The figures in brackets indicate the relative weightings
of parts of a question.
Special Requirements:
Table 1, I.C.A.O standard atmosphere table.
Table 2 of the dynamic pressure ‘Q’ versus Equivalent airspeed in
knots.
Table 3 of the values of the relative air density versus altitude and
ambient sea-level temperature.
Chart 1 of true Mach number versus calibrated airspeed and
altitude.
Chart 2 of compressibility correction to calibrated airspeed versus
calibrated airspeed and altitude.
ME307 2/11
1)
a) Define what the listed airspeeds are and explain the difference between them:-
Indicated airspeed-knots.
Kias
Calibrated airspeed-knots.
Kcas
Equivalent airspeed-knots.
Keas
True airspeed-knots.
Ktas
Mach number.
M
(5)
b) An aircraft is flying at 300 Kcas at 35000 feet. Using the applicable charts calculate
the following parameters:-
Equivalent airspeed-Keas.
True airspeed-Ktas.
Mach number.
At 35000 feet the square root of the ambient density ratio √σ = 0.5567
(7)
c) Explain how a pitot-static system is used to measure airspeed using the
incompressible Bernoulli equation with an annotated sketch of such a system.
(7)
d) With respect to pitot-static systems, explain what position corrections are, why they
are necessary and how they are determined.
(6)
ME307 3/11
2)
a) The ground run distance of an aircraft, XG, can be calculated from the equation
below:-
XG =
W
V2
log e ( 1 +
) metres.
2gK
N
where
W= Aircraft weight-N.
V = Aircraft velocity-ms-1.
g = gravitational constant = 9.81 ms-2
N
(T  W )
K
K  ½ S(C D  CL )
T = Engine thrust-N.
S = Wing area-m2
ρ = Air density- kg.m-3
μ = Rolling coefficient of friction.
and the air distance, XA, from lift-off to the 15 metre screen height from the equation
below:-
0.75V 2
XA =
2g
metres
ME307 4/11
THIS QUESTION CONTINUES ON THE NEXT PAGE
For the following conditions:-
CLMAX = 2.2 in the take-off configuration.
The lift off speed is 1.2 times the stalling speed.
CL
= 0.4 during the ground run.
CD
= 0.05 during the ground run.
T
= 444,822.0 N (constant with forward speed).
W
= 2,224,110.0 N.
S
= 321.0 m2.
ρ
= 1.225 Kg.m-3 air density.
μ
= 0.015 (rolling coefficient of friction).
i) Calculate the takeoff safety speed.
ii) Calculate the ground run distance to lift-off.
(8)
(10)
iii) Calculate the air distance from lift-off to the 15 metre screen height.
(4)
iv) Calculate the total takeoff distance if the takeoff safety factor is 1.15.
(3)
ME307 5/11
3)
a) An aircraft with a wing area of 5000 square feet has for its drag equation:CD = 0.025 + 0.06 CL2
At the start of cruise it weighs 450,000 lb and uses 150,000 lb of fuel during the
cruise at 35000 feet. The specific fuel consumption is 0.7 lb/lb thrust/hour.
If the flight plan is to maximise the range calculate:-
i) The cruise Mach number.
(4)
ii) The cruise speed in knots calibrated airspeed, Kcas.
(3)
iii) Maximum range in nautical miles.
(4)
Note:At 35000 feet the square root of the ambient density ratio √σ = 0.5567
and the square root of the ambient temperature ratio √θ = 0.8714
The speed of sound ‘a0’ = 661 knots at sea-level.
Speed of sound ‘a’ = 576 knots at 35000 feet.
b) The aircrew are required to change the flight plan to a search and rescue mission and
to maximise the flight time at an altitude of 5000 feet where the specific fuel
consumption is 1.0 lb/lb thrust/hour. The weight at the start of the new mission is still
450,000 lb. Neglecting the descent calculate:-
THIS QUESTION CONTINUES ON THE NEXT PAGE
ME307 6/11
i) The new cruise Mach number.
(4)
ii) The new cruise speed in knots calibrated airspeed, Kcas.
(3)
iii) The flight time in hours.
(4)
Note:At 5000 feet the square root of the ambient density ratio √σ = 0.9283
and the speed of sound ‘a’ = 650 knots.
c) If in the drag equation CD0 is a function of Mach number of the form given in the
table below, explain, with annotated diagrams, how the maximum range would be
calculated:-
(3)
Note:- Calculations are not required for part (c).
Mach
CD0
0.2
0.015
0.6
0.016
0.7
0.018
0.75
0.022
0.8
0.04
0.85
0.07
0.9
0.08
ME307 7/11
4)
a) An aircraft weighs 1779288 N and has a wing area ‘S’ of 371.612 m2 . The engines
give a thrust ‘T’ of 355857.6 N at 128.61 ms-1 on a standard day at sea-level. The
aircraft drag equation is:CD = 0.015 + 0.05 CL2
i) If the aircraft is in a steady unaccelerated rate of climb at a true airspeed of 128.61
ms-1 find the rate of climb at 3048 metres if the thrust varies as the ambient
pressure ratio. At 3048 metres the ambient pressure ratio δ = 0.6877
and the ambient density ratio σ = 0.7385.
The sea-level ambient density ρ0 = 1.225 Kg-m-3.
(8)
ii) Find the angle of climb at the conditions in (i) above.
(8)
b) Show that if the aircraft is in an accelerated climb the rate of climb is given by:-
dh

dt
VP T  D 
 V
W 1   P
  g
 dVP

 dh



where VP is the flight path velocity
W is the weight
T is the thrust
D is the drag
h is the altitude
t is the time
g is acceleration due to gravity
(9)
ME307 8/11
5)
a) Define what is meant by equilibrium and stability.
b) An aircraft is trimmed in straight and level flight with zero stick force at the following
condition:-
Weight
44482.0 N
Altitude
Sea-level.
Airspeed
83.2 ms-1.
Wing area
23.2 m2.
Centre of gravity xcg
25% mac.
c
mean aerodynamic chord
The aerodynamic characteristics are:-
Vbar is 1.0
The downwash is 0.4 α degrees where α is the angle of attack.
The incidence iw of the wing to the fuselage is zero.
The lift curve slope of the wing ‘aw’ is 0.1 per degree.
The lift curve slope of the tailplane ‘a1’ is 0.06 per degree.
The angle of attack at zero lift is –2.0 degrees.
The tail-off pitching moment coefficient, Cmto, referenced to
the quarter chord point is:Cmto = –0.05 + 0.05 CL
Given that the pitching moment coefficient about the centre of gravity is:Cmcg 
1
xcg  0.25c C L  Cmto  VBAR a1  w  iw  it     a2 e 
c
(6)
ME307 9/11
THIS QUESTION CONTINUES ON THE NEXT PAGE
i) Calculate the incidence, it, of the tailplane for trim if the elevator deflection e is
zero.
ii) Calculate the neutral point.
(9)
(6)
iii) How does VBAR affect the neutral point, eg, if VBAR were decreased from 1.0 to
0.6?
(4)
ME307 10/11
6)
a) Explain the characteristics of the Phugoid motion. What is the governing aerodynamic
derivative which determines the convergence or otherwise of the motion?
(8)
b) Given that the Phugoid motion can be approximated to by simple harmonic motion,
determine the Periodic time and Amplitude for an aircraft flying at a true air speed V
of 259 ms-1at 10668 metres if it encounters a vertical gust which causes the airspeed
to vary by ‘v’ ±5 ms-1.
(10)
The equations for the period and amplitude are given below.
Periodic time T =
Amplitude
A=
where ‘v’ is the airspeed variation.
2π
2g 2
V2
V*v
g
seconds.
metres.
ME307 11/11
THIS QUESTION CONTINUES ON THE NEXT PAGE
c) The same aircraft has to manoeuvre to avoid an oncoming aircraft. What is the
maximum roll rate in degrees per second the aircraft can achieve if the wheel is put
hard over to maximum but due to blowback at this speed the ailerons can only deflect
to 80% of maximum deflection? Maximum aileron deflection is forty (40) degrees at
low speed. The roll rate is such that no sideslip is generated before the aircraft reaches
maximum roll rate. The following conditions and aileron power apply:-
Airspeed
= 250
ms-1.
Altitude
= 10668
metres.
CLLp
 pb 
= – 0.5 

 2V 
Roll damping derivative.
CLLDAIB
= 0.001
Inboard aileron roll power
per degree aileron.
CLLDAOB
= 0.002
Outboard aileron roll power
per degree aileron.
span ‘b’
= 48.768
metres.
(7)