Download - DA-40 Analysis
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PERFORMANCE ANALYSIS OF THE DIAMOND DA-40
Indu Shekhar Kumar
AEROENG 2200
12/02/2013
For
Dr. James W. Gregory
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TABLE OF CONTENTS
1. Summary.....(3)
2. List of Figures.(4)
3. List of Tables..(5)
4. Nomenclature.(6)
5. Introduction.(7)
6. Drag Polar Estimation..(8)
7. Power Required(11)
8. Power Available(13)
9. Climb Performance..(15)
10. Range and Endurance...(20)
11. Gliding Flight...(22)
12. Turning Flight..(25)
13. Takeoff and Landing Performance..(27)
14. Conclusion(34)
15. References(35)
16. Appendix(36)
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SUMMARY
The results obtained in this project by the analysis of the Diamond DA-40 aircraft are
shown in the following paragraph. The drag polar estimation resulted in values parasitic
drag coefficient and the Oswald efficiency factor which were equal to 0.03257 and 0.73
respectively. The maximum lift to drag ratio was found to be 14.3794. The analysis of
the power required curve resulted in values of stall speed, the velocity at minimum
power required and velocity at maximum lift to drag ratio for all three altitudes. These
values are listed on page 11. The maximum power available at 0 ft., 5000 ft. and 10000
ft. is 137.3378 hp. 114.4482 hp. and 91.5586 hp. respectively. The values of maximum
ascent rate and the velocity corresponding to the max ascent rate for all three altitudes
are listed in a table on page 15. The absolute and service ceilings for the aircraft are
17445.92 ft. and 15403.52 ft. respectively. The time required by the aircraft to climb to
10000 ft. from sea level is 17.4051 minutes. The true airspeeds for best climb angle
and best rate of climb are 67.25 knots and 81.763 knots respectively. The range and
endurance of the DA-40 is 803.5 nautical miles and 10.25 hours respectively. The
maximum range for the DA-40 for a descent of 4000 ft. is 9.460 nautical miles and the
indicated airspeed for this range is 78.87 knots. The airspeed for maximum time aloft is
59.57 knots and the time required for descending a distance if 4000 ft. is 8.25 minutes.
The minimum turn radius and maximum turn rate for the DA-40 is 249.67 ft. and 0.7
rad/s. The takeoff ground roll distance at maximum takeoff weight at standard sea level,
5000 ft. and ground roll distance at sea level with 2400 lb. is 3272 ft., 4385 ft. and 2767
ft. respectively. The landing ground roll distance at maximum weight and standard sea
level conditions is 621.35 ft.
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LIST OF FIGURES
1. Drag Polar graph (Task 1)(10)
2. (L/D vs. CL) graph (Task 1)(10)
3. Power required vs. Airspeed graph (Task 2)..(12)
4. Power available vs. Airspeed graph (Task 3).(13)
5. Rate of Climb vs. Airspeed graph (Task 4).(15)
6. Altitude vs. Rate of Climb graph (Task 4)(17)
7. Climb Hodograph (Task 4)..(18)
8. Glide Hodograph (Task 6)(22)
9. V-n Diagram (Task 7)(25)
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LIST OF TABLES
1. Geometric Table (Task 1)...(8)
2. Table (Task 2)(11)
3. Table (Task 3)(14)
4. Table 1 (Task 4)(15)
5. Table 2 (Task 4)(19)
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NOMENCLATURE
Cd0 = Parasitic drag coefficient; CF = Aerodynamic cleanliness factor; Swet= Wetted area
S= Wing Plane-form area; = taper ratio of chord; K=slope of the drag polar;
AR= aspect ratio; e= Oswald efficiency factor; Pr= Power required; Pa= Power
available; = density; V= velocity; W= weight; VPrmin = velocity at min. power required;
Vstall = stall speed; V(L/D)max = speed at max. lift to drag ratio; V= free-stream velocity; SHPa = Shaft horse power; R/C = rate of climb; T= time; h= height; Vmax= velocity at max. Climb angle; Vrcamax = velocity at max. Rate of climb; E= endurance of aircraft;
R= range of aircraft; c= specific fuel consumption; = propeller efficiency; W0 = initial
weight of aircraft; W1 = final weight of aircraft; = angle between free-stream velocity
and the horizontal direction; VH= horizontal component of free-stream velocity; VV=
vertical component of free-stream velocity; n= loading factor; = turn rate; r = turn
radius; g= acceleration due to gravity; = geometric wing constant; VTD= touch-down
velocity; VTO= take-off velocity; Favg= average force acting on the aircraft during take-off
and landing; SG= take-off ground roll distance; SL= landing ground roll distance;
T0.707VTO = thrust required at 0.707 times the take-off velocity; P0.707VTO = power required at 0.707 times the take-off velocity; = coefficient of friction;
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INTRODUCTION
This projected is centered on the performance analysis of the Diamond DA-40 aircraft.
The Diamond DA-40 is four-seat, single propeller aircraft. The whole project is divided
into different sections, each focusing on a particular performance aspect. Graphs,
figures and tables have been included to facilitate the understanding of the results. The
required have been put in boxes and the font has been highlighted. The analysis was
done purely on the geometric, aerodynamic and propulsion characteristics of the
aircraft. The results have also been listed in the summary of the project. The equations
used have been highlighted. The heading of each section represents the content and
the task no. to which it corresponds to in the hand-out.
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TASK 1: DRAG POLAR ESTIMATION
Discussion:
1. This section includes the estimation of the Drag Polar for the DA-40. It is done in
three steps. Firstly, the value of Aerodynamic Cleanliness factor (CF) is estimated
from the historical data. After this the engineering drawings of the aircraft is
deconstructed to find the total wetted area (Swet) of the aircraft. Lastly, the
following equation is used to calculate the parasitic drag coefficient.
Cd0 = CF*Swet/S
Aircraft Component Plane Form Area (ft2) Wetted Area (ft2)
Fuselage 11.746 137.957
Horizontal Tail 20.39 40.78
Vertical Tail 14.646 29.291
Wing 145.7 291.4
Swet = 137.957+40.78+29.291+291.4
Cd0 = 0.0095*499.5/145.7
This section also includes the Oswald efficiency factor for the aircraft. This is also done
in three steps. First, the slope of the drag polar (K) is determined. Following this the
Cd0 = 0.03257
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value of delta () is estimated from the aspect ratio and the taper ratio (). Lastly the
values are plugged in the following equation
= 1/( + 1 + ) = tip chord / root chord
AR = (wing span) 2 / wing area
= 2.85 ft. / 5.2125 ft. = 0.547
AR = 39.172 / 145.7 = 10.53
From historical data
= 0.03 e = 1 / ((0.0105* * 10.53) +1+0.03)
2. This section also includes the plotting of Drag Polar. It is done by calculating the
drag coefficient (CD) as a function of function of lift coefficient (CL). The equation
used is as follows
Cd = Cd0 + (CL2/ *AR*e)
e = 0.73
Cd0 = 0.03
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3. The graph for Lift to Drag ratio vs. lift coefficient was plotted using the ratio of CL
and CD
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2Drag Polar
Lift Cofficient
Drag
Coe
ffici
ent
Cd vs ClCd0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
5
10
15
Lift Cofficient
Lift
to D
rag
Ratio
L/D vs Cl(L/D)max
(L/D) max = 14.3794
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TASK 2: POWER REQUIRED
Discussion:
This section includes the power required for the DA-40 is plotted as a function of
airspeed. The equation used for the plot is as follows.
= + ( ) ( ))
The graph of power required on airspeed was plotted at three different altitudes. The
values of velocity at minimum power required, the stall speed and the velocity at
maximum lift to drag ratio were marked on the graph for all three altitudes. The stall
speed was found using the following equation
= ( ) ( ) Calculation for stall speed at sea level:
= ( ) (. . .)
Altitude 0 ft. 5000 ft. 10000 ft.
Vstall (knots) 53.12 57.23 61.81
VPrmin (knots) 59.84 64.58 69.62
V(L/D)max (knots) 77.32 83.24 91.54
The resulting graph is shown on the following page
Velocity
Vstall = 53.12 knots
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40 50 60 70 80 90 100 110 120 130 140 15020
40
60
80
100
120
140
160
180Power Required vs Airspeed
Airspeed (knots)
Pow
er
(HP
)
0 ft5000 ft10000 ftVPrmin at sea levelVPrmin at 5000 ftVPrmin at 10000 ftVstall at sea levelVstall at 5000 ftVstall at 10000 ftTangent for sea levelTangent for 5000 ftTangent for 10000 ft
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TASK 3: POWER AVAILABLE
Discussion:
This section includes the power available is plotted as a function of velocity. The
equation used for plotting this graph is
= . ( ( ))
The various points of interest on the graph are found out using the intersection of
the power available and power required curves. The continuous lines represent
power curves at sea level, the dotted lines at 5000 ft. and dashed lines at 10000
ft. The lower end intersection of the curves represents the minimum speed and
the upper end intersection of the curves represents maximum speed. But at all
40 50 60 70 80 90 100 110 120 130 140 15020
40
60
80
100
120
140
160
180Power vs Airspeed
Airspeed (knots)
Pow
er (
HP
)
Pr at 0 ftPr at 5000 ftPr at 10000 ftPa at sea levelPa at 5000 ftPa at 10000 ft
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three altitudes the stall speed was higher than the minimum speed and so the
lowest speed at which the aircraft can fly is the stall speed.
0 = 53.12 > min 0 = 42.66 5000 = 57.23 > min 5000 = 45.92 10000 = 61.81 > min 10000 = 58.14
Minimum Speed
Speed for (L/D)max
Maximum Speed
Altitude
V (knots)
PR (hp)
PA (hp)
V (knots)
PR (hp)
PA (hp)
V (knots)
PR (hp)
PA (hp)
Sea Level
53.12 39.78 79.91 77.32 43.68 111.63 137.75 131.34 131.44
5,000 ft.
57.23 42.85 73.61 83.24 47.03 96.32 133.90 109.05 109.15
10,000 ft.
61.81 46.32 63.69 91.54 51.67 79.92 126.50 86.43 86.54
Max. Power available at sea level is 137.3378 hp.
Max. Power available at 5000 ft. is 114.4482 hp.
Max. Power available at 10000 ft. is 91.5586 hp.
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TASK 4: CLIMB PERFORMANCE
Discussion:
This section includes the graph of rate of climb vs. airspeed. In the graph below the
dependence of rate of climb in terms is airspeed is shown. The graph is plotted for three
different altitudes. Also the points corresponding to the maximum rate of climb are
shown. The equation used for plotting the graph is as follows
= ( )/
The following table shows the values of points of interest on the graph.
Properties 0 ft. 5000 ft. 10000 ft.
Max. R/C (ft./min) 853.374 615.015 364.282
Airspeed at max R/C (knots) 81.763 82.948 84.725
50 60 70 80 90 100 110 120 130 140 1500
100
200
300
400
500
600
700
800
900
1000 Climb Hodograph
Airspeed (knots)
Asc
ent
rate
(ft
/min
)
0ft
5000ft
10000ft
Max ascent rate at sea level
Max ascent rate at 5000 ft
Max ascent rate at 10000 ft
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In the above graph the blue line represents the graph at sea level, red line at 5000 ft.
and green line at 10000 ft. The graph shows that with increase in altitude the maximum
rate of climb reduces. But the velocity at max rate of climb increases. This means that
the overall graph shifts to the bottom right with increase with altitude.
= ( ) + = (2 1)/(2 1)
= (10000 0)/(364.282 853.374) = -20.44 = ( ) = 10000 ((20.44) 364.282)
The graph below shows the relationship of altitude as a function of rate of climb. This
graph is extrapolated to find out the absolute ceiling. Then the point at which the 100 ft.
/min line intersects the altitude vs. ascent rate line is the service ceiling.
Absolute ceiling = 17445.92 ft.
Service ceiling = 15403.52 ft.
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Calculation of time required to climb from sea level to 10000 ft.
= ( 17445.92)/20.4461
= /
= 20.4461 ( 17445.92)100000
0 100 200 300 400 500 600 700 800 9000
2000
4000
6000
8000
10000
12000
14000
16000
18000
Rate of Climb (ft/min)
Alti
tude
(ft)
Altitude vs Ascent rateAbsolute CeilingService Ceiling
T = 17.4051 minutes
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The Climb Hodograph for the aircraft was also plotted in this section. This was done by
plotting the rate of climb vs. horizontal speed. Then tangent was drawn to the graph
which gave us the value for the best climb angle. This can be seen in the following
graph.
In the above graph the green line represents the tangent from origin to the curve and
the blue line indicates the hodograph. The points of interest are also shown. The true
airspeeds for the points of interest are as follows.
60 80 100 120 140 160 180 200 220 2400
2
4
6
8
10
12
14
16
18
20
Horizontal Speed (ft/s)
Vert
ical S
peed (
ft/s
)
Climb Hodograph at Sea level
Vhor vs Vver
Tangent
R/Cmax
max AoA
Vmax = 67.25 knots
Vrcamax = 81.763
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The following table lists the values of the points of interest.
Rate of Climb (ft./min)
Climb Angle (degrees) Velocity (knots)
Best Rate of Climb Condition
853.374 3.5359 81.029
Best Climb Angle Condition 779.41 6.5720 66.81
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TASK 5: RANGE AND ENDURANCE
Discussion:
This section estimates the maximum range and endurance. The specifications are listed
in the handout. The only values that need to be calculated before plugging in the
Brequet range and endurance equations are the lift coefficient and the drag coefficient.
=
.
. . =
(
)
Given Constants:
=0.78;
c= 0.49 lb./hr./SHP;
e= 0.75;
Cd0= 0.03;
W0=2645 lb.
W1=2645 (50*6*0.9)
W1=2375 lb.
AR=10.53
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Calculation of Endurance:
= = = ( )/( )
= (2 2645)/(0.0017556 117.512 145.7) = 1.4978
= + ( ( )) = 0.03 + (1.49782 ( 10.53 0.75))
= 0.1204 = 0.780.49 1.49781.50.1204 2 0.0017556 145.7 (23750.5 26450.5) 3600 550
Calculation of Range:
= ( )/( ) = (2 2645)/(0.0017556 154.502 145.7)
= 0.866 = 0.03 + (0.8662 ( 10.53 0.75))
= 0.0602 = 0.780.49 0.8660.0602 (26452375) 3600 5506076
E = 10.25 hrs.
R = 803.5 nautical miles
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TASK 6: GLIDING FLIGHT
Discussion:
a.) This section includes the plot of the Glide Hodograph for the DA-40 aircraft. This
graph is plotted with the help of a relation between the glide angle and airspeed.
Also the relation between the glide angle and the airspeed is used.
=
= = =
For plotting the graph below various values of Cd were taken ranging from 0 to Clmax.
50 100 150 200 250 300 350 400 450-700
-600
-500
-400
-300
-200
-100
0
Descent
Rate
(ft
/s)
Lateral Speed (ft/s)
Glide Hodograph
Vv vs VhLine for Optimum glide angle
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The blue line represents the glide hodograph. This graph shows that initially the descent
rate increases with increase in lateral speed, but after a certain descent rate the lateral
speed decreases in the manner it increased in the upper portion of the graph. This may
be explained by the increased profile drag with increase in velocity.
b.) This section estimates the various points of interest related to the glide hodograph.
Determination of maximum range while descending from 5000 ft. to 1000 ft.
= = (5000 1000) (14.3794)
= 57517.6 .
Determination of optimum airspeed for maximum glide distance:
Optimum glide distance occurs at
=
= tan1 114.3794 = 3.9782
=
Rmax= 9.460 nautical miles
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= 2 cos 3.9782 26450.0023769 145.7 0.8600 The Cl corresponding to the was found out using the plot.
Determination of airspeed for maximum time aloft.
This is the airspeed at which the descent rate is the least.
= = 8.08 ft./s
= = 8.08sin 4.6092
Determination of Time required for descending from 5000 ft. to 1000 ft. at
= / = 4000 .8.08 ./ = 495.05
V = 78.87 knots
V = 59.57 knots
T = 8.25 minutes
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TASK 7: TURNING FLIGHT
Discussion:
This section includes the V-n diagram. This diagram is plotted with an equation which
relates the loading factor to the free stream velocity. The equation is as follows:
= ( 2 ) (2 )
The graph consists of three regions. The green demarcates the limit of the maximum
possible loading factor. The red line represents the lower limit of the loading factor and
the blue line represents the maximum possible speed at which it can turn. The region
bound by these three lines represents the region of safe cruising. In other words the
0 20 40 60 80 100 120 140 160 180-2
-1
0
1
2
3
4
Airspeed (knots)
Load F
acto
r (n
)
V-n Diagram
Positive n vs V
Maneuver point
Negative n vs V
Max speed limit
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aircraft will not stall if it operates inside the region bound by the aforementioned lines.
The point represents the maneuver point and the airspeed associated with it.
Calculation of minimum turn radius
= (2 ) ( ) = (2 2645) (32.2 0.0023769 1.90 145.7)
Calculation of maximum turn rate
= (2 ) = 32.2 3.8 1.9 0.0023769 145.7 (2 2645)
Vat maneuver point = 118.88 knots
rmin = 249.67 ft.
max = 0.7 rad/s
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TASK 8: TAKEOFF AND LANDING PERFORMANCE
Discussion:
This section includes the estimation of various values associated with landing and take-
off performance.
a.) Takeoff ground roll distance at maximum takeoff weight at standard sea level conditions.
The value of phi is used to estimate the value of Cl and Cd.
= ( ) + ( ) = (16 1.645 39.17)2 1 + (16 1.645 39.172)
= 0.3111
Now the value of phi is plugged in to find the values of Cl and Cd.
= ( ) = 1(2 0.3111) 10.53 0.75 0.02
= 0.7975 = + ( ( ))
= 0.03 + (0.79752 0.3111 ( 10.53 0.75)) = 0.038
Calculation of take-off speed
= . 27
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= 1.2 2 26450.0023769 145.7 1.90 = 107.60 / = 63.74
Calculation of thrust at 0.707 VTO
. = . 0.707 = 23707 /107.60 /
The value of . was estimated from the power curve (Task 2) graph 0.707 = 220.33
Calculation of average force
= . ( (. ) ) ( ( (. ) ))
= 220.33 (12 0.0023769 (0.707 107.6)2 145.7 0.038) 0.02 (2645 (12
0.0023769 (0.707 107.6)2 145.7 0.7975)) = 145.33 lb.
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Calculation of ground rolls distance
= = 2645 107.622 32.2 145.33
b.) Takeoff ground roll distance at maximum takeoff weight at a high altitude airport
(5000 ft.).
The only variable that changes with altitude is density and so do the values of VTO and
Favg.
Calculation of take-off speed
= . = 1.2 2 26450.0020482 145.7 1.90
= 115.91 / = 68.68
Calculation of thrust at 0.707 VTO
. = . 0.707 = 23278 /115.91 /
SG = 3272 ft.
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The value of . was estimated from the power curve (Task 2) graph 0.707 = 200.83
Calculation of average force
= . ( (. ) ) ( ( (. ) ))
= 200.83 (12 0.0020482 (0.707 115.91)2 145.7 0.038) 0.02 (2645 (12
0.0020482 (0.707 115.91)2 145.7 0.7975)) = 125.84 lb.
Calculation of ground rolls distance
= = 2645 115.9122 32.2 125.84
c.) Takeoff ground roll distance at a takeoff weight of 2400 lb. at standard sea level
conditions.
Calculation of take-off speed
= .
SG = 4385 ft.
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= 1.2 2 24000.0023769 145.7 1.90 = 102.5 / = 60.72
Calculation of thrust at 0.707 VTO
. = . 0.707 = 21481 /102.5 /
The value of . was estimated from the power curve (Task 2) graph 0.707 = 209.57
Calculation of average force
= . ( (. ) ) ( ( (. ) ))
= 209.57 (12 0.0023769 (0.707 102.5)2 145.7 0.038) 0.02 (2400 (12
0.0023769 (0.707 102.5)2 145.7 0.7975)) = 141.52 lb.
Calculation of ground roll distance
= = 2400 102.522 32.2 141.52
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d.) Landing ground roll distance at maximum weight and standard sea level
conditions.
For this calculation the only values that remain the same are the drag polar.
Calculation of touch-down speed
= . = 1.3 2 26450.0023769 145.7 1.90
= 116.57 / = 69.07
Calculation of average force
= ( (. ) ) ( ( (. ) ))
= (12 0.0023769 (0.707 116.57)2 145.7 0.038) 0.5 (2645 (12 0.0023769 (0.707 116.57)2 145.7 0.7975)) = 898.21 lb
SG = 2767 ft.
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Calculation of ground roll distance
= = 2645 116.5722 32.2 (898.21)
SL = 621.35 ft.
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CONCLUSION
From the analysis above, it can be concluded that the Diamond DA-40 is and
aerodynamically sound aircraft. Looking at the range and endurance values of the
aircraft, it can be determined that the DA-40 is thermodynamically and aerodynamically
efficient.
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REFERENCES
Anderson, John D. Introduction to Flight. New York: McGraw-Hill, 1985. Print
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APPENDIX
Task: 1 Mat lab Code
clc clear figure(1) % Values of lift coefficient CL=[0:0.001:2]; Cd0= 0.03; e=0.75; AR= 10.5305; % Calculation of Cd Cd=Cd0+ ((CL.^2)./(pi*AR*e)); % Plotting Drag Polar plot(CL,Cd) hold on plot(0,Cd0,'o') hold off title('Drag Polar') xlabel('Lift Cofficient') ylabel('Drag Coefficient') legend('Cd vs Cl','Cd0','Location','NorthWest') % Plotting Lift to Drag ratio figure(2) ldr=(CL./Cd); plot(CL,ldr) hold on xlabel('Lift Cofficient') ylabel('Lift to Drag Ratio') for i=1:length(ldr) if ldr(i)==max(ldr) plot(CL(i),ldr(i),'o') end end legend('L/D vs Cl','(L/D)max') max(ldr) Task: 2 Mat lab Code
clc clear format compact figure(1) % Defining the flight characteristics of T-18 airfoil Cd0= 0.03; e= 0.75; S= 145.7; W= 2645; AR=10.53; Clmax= 1.9; % Densities at different altitudes rho_sl= 0.0023769; rho_1= 0.0020482;
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rho_2= 0.0017556; % Range of velocities in ft/s v=[10:0.5:400]; v_knots= v*0.592484; % Calculation of stall speed vstall= sqrt((2*W)/(rho_sl*S*Clmax))*0.592484; vstall1= sqrt((2*W)/(rho_1*S*Clmax))*0.592484; vstall2= sqrt((2*W)/(rho_2*S*Clmax))*0.592484; fprintf('The Stall speed at sea level is % 5.2f knots\n',vstall) fprintf('The Stall speed at 5000 ft is % 5.2f knots\n',vstall1) fprintf('The Stall speed at 10000 ft is % 5.2f knots\n',vstall2) % Calculation of tangent for L/D max x=0.565*v_knots; x1=0.565*v_knots; x2=0.5645*v_knots; % Calculation of power required Pr=((1/2)*rho_sl*S*Cd0*(v.^3))+((2*W^2/(rho_sl*S*pi*AR*e))./v); Pr_hp= Pr./550; Pr1=((1/2)*rho_1*S*Cd0*(v.^3))+((2*W^2/(rho_1*S*pi*AR*e))./v); Pr_hp1= Pr1./550; Pr2=((1/2)*rho_2*S*Cd0*(v.^3))+((2*W^2/(rho_2*S*pi*AR*e))./v); Pr_hp2= Pr2./550; % Extracting airspeed at minimum power required count=0; count1=0; count2=0; q=0; q1=0; q2=0; for i=1:length(Pr_hp) if Pr_hp(i)==min(Pr_hp) Vprmin= v_knots(i); end if count==0 if v_knots(i)>vstall Pstall=Pr_hp(i); count=count+1; end end if q==0 if (Pr_hp(i)-x(i))vstall1 Pstall1=Pr_hp1(j); count1=count1+1; end end
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- if q1==0 if (Pr_hp1(j)-x1(j))vstall2 Pstall2=Pr_hp2(k); count2=count2+1; end end if q2==0 if (Pr_hp2(k)-x2(k))
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Task 3 Matlab Code % Task 3 clc clear format compact figure(1) % Defining the flight characteristics of T-18 airfoil Cd0= 0.03; e= 0.75; S= 145.7; W= 2645; AR=10.53; Clmax= 1.9; % Densities at different altitudes rho_sl= 0.0023769; rho_1= 0.0020482; rho_2= 0.0017556; % Range of velocities in ft/s v=[20:0.5:400]; v_knots= v*0.592484; eta= 0.78*(1-((35./v_knots).^2)); % Calculation of stall speed vstall= sqrt((2*W)/(rho_sl*S*Clmax))*0.592484; vstall1= sqrt((2*W)/(rho_1*S*Clmax))*0.592484; vstall2= sqrt((2*W)/(rho_2*S*Clmax))*0.592484; % Calculation of tangent for L/D max x=0.565*v_knots; x1=0.565*v_knots; x2=0.5645*v_knots; % Calculation of power available SHPa= 180*eta; SHPa1= 150*eta; SHPa2= 120*eta; % Calculation of power required Pr=((1/2)*rho_sl*S*Cd0*(v.^3))+((2*W^2/(rho_sl*S*pi*AR*e))./v); Pr_hp= Pr./550; Pr1=((1/2)*rho_1*S*Cd0*(v.^3))+((2*W^2/(rho_1*S*pi*AR*e))./v); Pr_hp1= Pr1./550; Pr2=((1/2)*rho_2*S*Cd0*(v.^3))+((2*W^2/(rho_2*S*pi*AR*e))./v); Pr_hp2= Pr2./550; % Extracting Pr, Pa and Vel at stall, min speed, L/D max, max speed at 0,5000 ft, 10000 ft count=0; count1=0; count2=0; q=0; q1=0; q2=0; track=0; track1=0; track2=0; for i=1:length(Pr_hp) if count==0 if v_knots(i)>vstall Pastall=SHPa(i);
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Prstall=Pr_hp(i); count=count+1; end end if q==0 if (Pr_hp(i)-x(i)) Pr_hp(i) Vmin= v_knots(i); Pamin= SHPa(i); Prmin= Pr_hp(i); track=track+1; end end if track==1 if SHPa(i)< Pr_hp(i) Vmax= v_knots(i); Pamax= SHPa(i); Prmax= Pr_hp(i); track=track+1; end end end for j=1:length(Pr_hp1) if count1==0 if v_knots(j)>vstall1 Pastall1=SHPa1(j); Prstall1=Pr_hp1(j); count1=count1+1; end end if q1==0 if (Pr_hp1(j)-x1(j)) Pr_hp1(j) Vmin1= v_knots(j); Pamin1= SHPa1(j); Prmin1= Pr_hp1(j); track1=track1+1; end end if track1==1 if SHPa1(j)< Pr_hp1(j) Vmax1= v_knots(j); Pamax1= SHPa1(j);
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Prmax1= Pr_hp1(j); track1=track1+1; end end end for k=1:length(Pr_hp2) if count2==0 if v_knots(k)>vstall2 Pastall2=SHPa2(k); Prstall2=Pr_hp2(k); count2=count2+1; end end if q2==0 if (Pr_hp2(k)-x2(k)) Pr_hp2(k) Vmin2= v_knots(k); Pamin2= SHPa2(k); Prmin2= Pr_hp2(k); track2=track2+1; end end if track2==1 if SHPa2(k)< Pr_hp2(k) Vmax2= v_knots(k); Pamax2= SHPa2(k); Prmax2= Pr_hp(k); track2=track2+1; end end end fprintf('The Stall speed at sea level is % 5.2f knots\n',vstall) fprintf('The Power available at stall speed at sea level is % 5.2f hp\n',Pastall) fprintf('The Power required at stall speed at sea level is % 5.2f hp\n',Prstall) fprintf('\nThe Stall speed at 5000 ft is % 5.2f knots\n',vstall1) fprintf('The Power available at stall speed at 5000 ft is % 5.2f hp\n',Pastall1) fprintf('The Power required at stall speed at 5000 ft is % 5.2f hp\n',Prstall1) fprintf('\nThe Stall speed at 10000 ft is % 5.2f knots\n',vstall2) fprintf('The Power available at stall speed at 10000 ft is % 5.2f hp\n',Pastall2)
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fprintf('The Power required at stall speed at 10000 ft is % 5.2f hp\n',Prstall2) fprintf('\nThe airspeed for maximum lift to drag ratio at sea level is % 5.2f knots\n',Vldmax) fprintf('The Power available for maximum lift to drag ratio at sea level is % 5.2f hp\n',Paldmax) fprintf('The Power required for maximum lift to drag ratio at sea level is % 5.2f hp\n',Prldmax) fprintf(' \nThe airspeed for maximum lift to drag ratio at 5000 ft is % 5.2f knots\n',Vldmax1) fprintf('The Power available for maximum lift to drag ratio at 5000 ft is % 5.2f hp\n',Paldmax1) fprintf('The Power required for maximum lift to drag ratio at 5000 ft is % 5.2f hp\n',Prldmax1) fprintf(' \nThe airspeed for maximum lift to drag ratio at 10000 ft is % 5.2f knots\n',Vldmax2) fprintf('The Power available for maximum lift to drag ratio at 10000 ft is % 5.2f hp\n',Paldmax2) fprintf('The Power required for maximum lift to drag ratio at 10000 ft is % 5.2f hp\n',Prldmax2) fprintf('\nThe min speed at sea level is % 5.2f knots\n',Vmin) fprintf('The Power available at min speed at sea level is % 5.2f hp\n',Pamin) fprintf('The Power required at min speed at sea level is % 5.2f hp\n',Prmin) fprintf('\nThe min speed at 5000 ft is % 5.2f knots\n',Vmin1) fprintf('The Power available at min speed at 5000 ft is % 5.2f hp\n',Pamin1) fprintf('The Power required at min speed at 5000 ft is % 5.2f hp\n',Prmin1) fprintf('\nThe min speed at 10000 ft is % 5.2f knots\n',Vmin2) fprintf('The Power available at min speed at 10000 ft is % 5.2f hp\n',Pamin2) fprintf('The Power required at min speed at 10000 ft is % 5.2f hp\n',Prmin2) fprintf('\nThe max speed at sea level is % 5.2f knots\n',Vmax) fprintf('The Power available at max speed at sea level is % 5.2f hp\n',Pamax) fprintf('The Power required at max speed at sea level is % 5.2f hp\n',Prmax) fprintf('\nThe max speed at 5000 ft is % 5.2f knots\n',Vmax1) fprintf('The Power available at max speed at 5000 ft is % 5.2f hp\n',Pamax1) fprintf('The Power required at max speed at 5000 ft is % 5.2f hp\n',Prmax1) fprintf('\nThe max speed at 10000 ft is % 5.2f knots\n',Vmax2) fprintf('The Power available at max speed at 10000 ft is % 5.2f hp\n',Pamax2) fprintf('The Power required at max speed at 10000 ft is % 5.2f hp\n',Prmax2) plot(v_knots,Pr_hp,v_knots,Pr_hp1,':r',v_knots,Pr_hp2,'--g') hold on
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plot(v_knots,SHPa,'c',v_knots,SHPa1,':m',v_knots,SHPa2,'--k') title('Power Required vs Airspeed') xlabel('Airspeed (knots)') ylabel('Power (HP)') legend('Pr at 0 ft','Pr at 5000 ft','Pr at 10000 ft','Pa at sea level','Pa at 5000 ft','Pa at 10000 ft','Tangent for sea level','Tangent for 5000 ft','Tangent for 10000 ft','Pa at 0 ft','Pa at 5000 ft','Pa at 10000 ft','Location','NorthWest') Task 4 Matlab Code clc clear format compact figure(1) % Defining the flight characteristics of T-18 airfoil at all the three altitudes % power in hp SHPa= 180; SHPa1= 150; SHPa2= 120; Cd0= 0.03; e= 0.75; S= 145.7; W= 2645; AR=10.53; % Values of density of air at specified heights rho_sl= 0.0023769; rho1= 0.0020482; rho2= 0.0017556; % velocity in ft/s v=[70:0.5:300]; % vel in knots v_knots= v*0.592484; eta= 0.78*(1-((35./v_knots).^2)); % Power available in hp Pa_hp=eta*SHPa; Pa_hp1=eta*SHPa1; Pa_hp2=eta*SHPa2; %Power available in ft-lb/s Pa= eta*SHPa*550; Pa1= eta*SHPa1*550; Pa2= eta*SHPa2*550; % Calculation of required power in lb-ft/s Pr=((1/2)*rho_sl*S*Cd0*(v.^3))+((2*W^2/(rho_sl*S*pi*AR*e))./v); Pr1=((1/2)*rho1*S*Cd0*(v.^3))+((2*W^2/(rho1*S*pi*AR*e))./v); Pr2=((1/2)*rho2*S*Cd0*(v.^3))+((2*W^2/(rho2*S*pi*AR*e))./v); % Conversion to power in hp Pr_hp= Pr./550; Pr_hp1= Pr1./550; Pr_hp2= Pr2./550;
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% Calculation of ascent rate rc1= (Pa-Pr)./W; rc1_ftmin= rc1*60; rc2= (Pa1-Pr1)./W; rc2_ftmin= rc2*60; rc3= (Pa2-Pr2)./W; rc3_ftmin= rc3*60; % plotting the graph plot(v_knots,rc1_ftmin,'b',v_knots,rc2_ftmin,'r',v_knots,rc3_ftmin,'g') hold on title(' Climb Hodograph') xlabel('Airspeed (knots)') ylabel('Ascent rate (ft/min)') % Max R/c max values fprintf(' The maximum ascent rate at sea level is %5.3f ft/min\n',max(rc1_ftmin)) fprintf(' The maximum ascent rate at 5000 ft is %5.3f ft/min\n',max(rc2_ftmin)) fprintf(' The maximum ascent rate at 10000 ft is %5.3f ft/min\n',max(rc3_ftmin)) % Extracting airspeed at max R/c value track=0; track1=0; track2=0; for i=2:length(rc1) if track==0 if rc1(i-1)>rc1(i) rcmax1= rc1_ftmin(i); Vrcmax1=v_knots(i); track=track+1; end end if track1==0 if rc2(i-1)>rc2(i) rcmax2= rc2_ftmin(i); Vrcmax2=v_knots(i); track1=track1+1; end end if track2==0 if rc3(i-1)>rc3(i) rcmax3= rc3_ftmin(i); Vrcmax3=v_knots(i); track2=track2+1; end end end plot(Vrcmax1,rcmax1,'m*',Vrcmax2,rcmax2,'sk',Vrcmax3,rcmax3,'dc') legend('0ft','5000ft','10000ft','Max ascent rate at sea level','Max ascent rate at 5000 ft','Max ascent rate at 10000 ft') hold off fprintf(' \nThe airspeed at maximum ascent rate at sea level is %5.3f knots\n',Vrcmax1)
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Task 5 Matlab Code clc clear eta=0.78; SHP= 120; % Units lbs/he/SHP e=0.75; S= 145.7; AR=10.53; c= 0.49; rho= 0.0017556; W0= 2645; Vtrmin= 91.54/0.592484; Vprmin= 69.62/0.592484; Cl1= 2*W0/(rho*Vprmin*Vprmin*S) Cd1= 0.03+ (Cl1^2/(pi*AR*e)) E= eta*(Cl1^1.5)*sqrt(2*rho*S)*((30.048^(-0.5))-300.48^(-0.5))/(c*Cd1) % in hours Cl2= 2*W0/(rho*Vtrmin*Vtrmin*S) Cd2= 0.03+ (Cl2^2/(pi*AR*e)) R= (eta*Cl2*log(300/30)*(3600/6076.12))/(Cd2*c) Task 7 Matlab code %Task 7 clc clear rho=0.0023769; S=145.7; W=2645; V=[0:300]; Clmax= 1.90 Vstar= sqrt((2*3.8*W)/(rho*Clmax*S)); line=linspace(-1.52,3.8,300); for i=1:length(V) p(i)= rho*(V(i)^2)*S*Clmax./(2*W); if p(i)(-1.52) N(j)=n(j); else N(j)= -1.52; end end
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plot(V*0.592484,N,'r') plot(178,line,'b-') hold off xlabel('Airspeed (knots)') ylabel('Load Factor (n)') title('V-n Diagram') legend('Positive n vs V','Maneuver point','Negative n vs V','Max speed limit','Location','NorthWest') Task 6 Matlab code
% Task 6 clc clear Cl=linspace(0,1.90,300); Cd=0.03+((Cl.^2)./(pi*10.53*0.75)); theta=atand(1./(Cl./Cd)); theta1= atand(1/14.3794) slope=tand(theta1) v=sqrt((2.*2645.*cosd(theta))./(0.0023769.*Cl.*145.7)); vh= v.*cosd(theta); vv= (-1).*v.*sind(theta); x= (-1)*slope*vh; plot(vh,vv,vh,x) ylabel('Descent Rate (ft/s)') xlabel('Lateral Speed (ft/s)') title('Glide Hodograph') legend('Vv vs Vh','Line for Optimum glide angle') for i=1:length(Cl) if abs(theta(i)-theta1)
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for i=1:length(V) p(i)= rho*(V(i)^2)*S*Clmax./(2*W); if p(i)(-1.52) N(j)=n(j); else N(j)= -1.52; end end plot(V*0.592484,N,'r') plot(178,line,'b-') hold off xlabel('Airspeed (knots)') ylabel('Load Factor (n)') title('V-n Diagram') legend('Positive n vs V','Maneuver point','Negative n vs V','Max speed limit','Location','NorthWest')
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