da-40 analysis

PERFORMANCE ANALYSIS OF THE DIAMOND DA-40

Indu Shekhar Kumar

AEROENG 2200

12/02/2013

For

Dr. James W. Gregory

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)

2

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.

3

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)

4

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)

5

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;

6

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.

7

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

8

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

9

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

10

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

11

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

12

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

13

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.

14

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

15

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.

16

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

17

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

18

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

19

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

20

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

21

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

22

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

23

= 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

24

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

25

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

26

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

= 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.

28

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.


= . = 1.2 2 26450.0020482 145.7 1.90

= 115.91 / = 68.68


. = . 0.707 = 23278 /115.91 /

SG = 3272 ft.

29



= . ( (. ) ) ( ( (. ) ))

= 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.


= .

SG = 4385 ft.

30

= 1.2 2 24000.0023769 145.7 1.90 = 102.5 / = 60.72


. = . 0.707 = 21481 /102.5 /



= . ( (. ) ) ( ( (. ) ))

= 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

31

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


= ( (. ) ) ( ( (. ) ))

= (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.

32

Calculation of ground roll distance

= = 2645 116.5722 32.2 (898.21)

SL = 621.35 ft.

33

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.

34

REFERENCES

Anderson, John D. Introduction to Flight. New York: McGraw-Hill, 1985. Print

35

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;

36

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

37

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))

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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fprintf(' The airspeed at maximum ascent rate at 5000 ft is %5.3f knots\n',Vrcmax2) fprintf(' The airspeed at maximum ascent rate at 10000 ft is %5.3f knots\n',Vrcmax3) % Absolute and Service ceilings figure(2) alti=[0,5000,10000,17448.12]; Hor=linspace(0,2500,250); v_ftmin= [max(rc1),max(rc2),max(rc3),0]*60; slope= (10000/((max(rc3)-max(rc1))*60)); plot(v_ftmin,alti,0,17448.12,'o',100,15403.52,'s') legend('Altitude vs Ascent rate','Absolute Ceiling','Service Ceiling') xlabel('Rate of Climb (ft/min)') ylabel('Altitude (ft)') % Absolute ceiling AbsC= alti(3)-(slope*v_ftmin(3)); fprintf('\n The absolute ceiling of the aircraft is %7.2f ft\n',AbsC) Ver=slope*Hor+AbsC; % Service ceiling SerC= slope*100+AbsC; fprintf('The Service ceiling of the aircraft is %7.2f ft\n',SerC) % Climb hodo graph figure(3) Vhor= sqrt(abs((v.^2)-(rc1.^2))); % Slope creation for theta max tangent= 0.1152*Vhor; plot(Vhor,rc1,Vhor,tangent) xlabel('Horizontal Speed (ft/s)') ylabel('Vertical Speed (ft/s)') title(' Climb Hodograph') for j= 1:length(tangent) if (tangent(j)-rc1(j))

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)

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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da-40 analysis

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