Subsonic Bomber Aircraft

SUBSONIC BOMBER AIRCRAFT

A PROJECT REPORT

Submitted by

SUDHAN.U (62205101043)

SYED MUSTHAFA.H (62205101047)

THIRUMALVALAVAN.M (62205101048)

THIRUMURUGAN.M (62205101049)

In partial fulfillment for the award of the degree

Of

BACHELOR OF ENGINEERING

In

AERONAUTICAL ENGINEERING

V.S.B. ENGINEERING COLLEGE, KARUR.


A PROJECT REPORT

Submitted by

SUDHAN.U (62205101043)




In partial fulfillment for the award of the degree

Of

BACHELOR OF ENGINEERING

In

AERONAUTICAL ENGINEERING

V.S.B. ENGINEERING COLLEGE, KARUR.

ANNA UNIVERSITY: CHENNAI 600 025

BONAFIDE CERTIFICATE

This is to certified that the project report of


is the bonafide work of

SUDHAN.U (62205101043)

who carried out the project work under my supervision.

SIGNATURE SIGNATURE

Mr. K. VELMURUGARAJAN , Mr. J.BRUCE RALPHIN ROSE,

HEAD OF THE DEPARTMENT, SUPERVISOR,

Dept. of Aeronautical Engg, Lecturer,

V. S. B. Engineering College. Dept. of Aeronautical Engg.

V. S. B. Engineering College.

Date:

Submitted this project for viva voice on……………..

External Examiner Internal Examiner








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Date:










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Date:










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Date:



ACKNOWLEDGEMENT

First and foremost, we wish to acknowledge our debt to the `HARD WORK IS THE KEY TO SUCCESS` who has given us knowledge and good health.

We would like to express to the chairman of our college,

Mr. V.S.BALUSAMY and the principal Mr. JOHN ORAL BASKAR, for providing better working environments and educational facilities.

We are much grateful Mr. K.VELMURUGARAJAN Head of the department of the Aeronautical engineering for this encouragement discussion, valuable comments and many innovative ideas in carrying out this project. Without his timely help it would have been impossible for us to complete this work.

We acknowledge in no less terms the qualified and excellent assistance rendered by Mr. J.BRUCE RALPHIN ROSE, Lecturer, Department of Aeronautical Engineering. We owe a debt of gratitude for his valuable suggestions, kind inspiration and encouragement.

We most sincerely acknowledge the staff members of Department of Aeronautical Engineering for their constant inspiration and suggestions.

We owe a debt gratitude to our parents and friends for their advice and to keep our spirits high to complete this project.

TABLE OF CONTENT

CHAPTER.NO TITLE

ABSTRACT

LIST OF SYMBOLS

INTRODUCTION

1.COMPARITIVE STUDY OF SUBSONIC BOMBER

1.1 DIMENSIONS

1.2 WEIGHT CONFIGURATION

1.3 PERFORMANCE

1.4 ENGINE CONFIGURATION

1.5 ARMAMENTS

2.SELECTION OF MAIN PARAMETERS

2.1 AIRFOIL SELECTION

2.2 CO-EFFICIENT OF LIFT VS ANGLE OF ATTACK

2.3 CO-EFFICIENT OF LIFT VS DRAG

2.4 MAX.L/D VS VELOCITY

2.5 RANGE VS VELOCITY

2.6 ALTITUDE VS VELOCITY

2.7 ASPECT RATIO VS VELOCITY

2.8 WING LOADING VS VELOCITY

2.9 SFC VS MACH NUMBER

2.10 T/W VS VELOCITY

3.WEIGHT ESTIMATION

3.1 WEIGHT CALCULATION

3.2 MISSION PROFILE

3.3 APPROXIMATE WEIGHT ESTIMATION

3.4 ACTUAL WEIGHT ESTIMATION

3.4.1 ITERATION

4.ENGINE SELECTION

4.1 ENGINE LOCATION

4.2 THRUST CALCULATION

4.2.1THRUST VS SFC

4.3 ENGINE CONFIGURATION

4.3.1 ENGINE DIMENSION

4.4 CONFIGURATION

4.4.1 ADVANTAGES OF POTTED ENGINE

4.4.2 DISADVANTAGE OF POTTED ENGINE

4.5 THRUST MATCHING

5.AIRFOIL SELECTION

5.1 CALCULATION OF CL

5.2 REYNOLD’S NUMBER

5.3 MAXIMUM CL

5.4 SKIN FRICTION DRAG FOR TURBULENT FLOW

5.5 REQUIRED CL(MAX)

5.6 NACA-63A010

5.7 THIN AIRFOIL THEORY

6.WING SELECTION

6.1 EQUIVALENT ASPECT RATIO

6.2 STRUCTURAL WEIGHT FOR VARYING

THICKNESS OF AIRFOIL

6.3 LOCATION OF CENTER OF GRAVITY

7.WETTED SURFACE AREA AND DRAG ESTIMATION

7.1 CALCULATION OF WETTED SURFACE AREA

7.1.1 FUSELAGE

7.1.2 WING AREA

7.1.3 HORIZONTAL TAIL

7.1.4 VERTICAL TAIL

7.1.5 ENGINE

7.1.6 UNDERCARRIAGE OR LANDING GEAR

7.1.7 1/4th OF FLAP

7.1.8 FULL FLAP

7.2 DRAG CALCULATION

7.2.1 DRAG POLAR

7.2.2 CRUISE

7.2.3 TAKEOFF

7.2.4 LANDING

7.3 DRAG FOR EACH ALTITUDE

7.3.1 AT SEA LEVEL

7.3.2 AT h=500m

7.3.3 AT h=1000m

7.3.4 AT h=1500m

7.3.5 AT h=2000m

8.ESTIMATION OF RATE OF CLIMB

8.1 CALCULATION OF RATE OF CLIMB

8.1.1 AT SEA LEVEL

8.1.2 AT h=500m

8.1.3 AT h=1000m

8.1.4 AT h=1500m

8.1.5AT h=2000m

8.2 TIME REQUIRED FOR REACHING THE SERVICE CEILING

9.CALCULATION OF TAKEOFF & LANDING DISTANCE

9.1 LENGTH OF TAKEOFF RUN

9.1.1 GROUND RUN

9.1.2 TRANSITION RUN

9.1.3 CLIMBING

9.2 LENGTH OF LANDING RUN

9.2.1 DESCENDING

9.2.2 TRANSITION RUN

9.2.3 GROUND RUN

10.THREE VIEW OF SUBSONIC BOMBER AIRCRAFT

10.1 FRONT VIEW

10.2 TOP VIEW

10.3 SIDE VIEW

11.BIBLIOGRAPHY

SYMBOLS USED

W-Weight of airgraft

Wo-Overall weight

Wf-weight of fuel

We-Empty weight

Lf – fuselage length

Df – diameter of fuselage

Sw - wing area

Tw - wing thickness

bw,b – wing span

Sht – horizontal tail area

tht – horizontal tail thickness

bht - horizontal tail span

AR – aspect ratio

S – Surface area

Svt – vertical tail area

tvt - vertical tail thickness

bvt – vertical tail span

Cdo – drag polar

Cd – coefficient of drag

CL - coefficient of lift

F, T – thrust

T/W-Thrust loading

W/S-Wing loading

A.R-Aspect ratio

Cr,Ct-Chord length of root, tip

Tr,Tt-Thickness of root, tip

Sp-Wetted surface area

CDp-Coefficient of drag of wetted surface area

a-Angle of attack

C.G-Center of gravity

-Dihedral angle

R-Range

E-Endurance

-Ground friction

-Free stream velocity

C-Chord

Lf-Length of fuselage

VT-Vertical tail

HT-Horizontal tail

-Density(kg/m3)

g-Gravity

s-Distance

H-Height

h-altitude

ROC – rate of climb

V, u – velocity

D – Drag

L – Lift

g – Acceleration of gravity

Wo – optimum weight

- sweep angle

Cr – root chord

ABSTRACT

First we start with design and development. Let’s take the military as an example. Suppose the military wants to design new weapons systems that will be more accurate than the system they currently have. The first thing they have to do is get a team of scientists and whatever other personnel are needed to design the system. Then, after they come up with a design they are pleased with they have to get the engineers to develop it. All these people either have to be hired or contracted. Some of the work may be sub contracted.

Then there is the acquisition of the materials needed. Some of the materials may already be on the hand while others have to be purchased domestically and others will need to be purchased from overseas. In many cases bids will be put in to various companies to see who can supply the material at the cheapest cost.

After the materials are acquired there’s the matter of storage. in many cases the material is either of so large in size that the storage facilities need to be specially built in order to adequately provide the needed space. this of course has to be figured into the equation.

Then there’s the matter of movement and distribution to may be the various military base around the country even overseas. Trucks or planes need to be acquired in order to distribute the materials if there aren’t enough read made transportation vehicles. Transportation costs alone can be astronomical.

INTRODUCTION

Airplane Design-Introduction

Three major types of airplane designs are

1. Conceptual design,2. Preliminary design,3. Detailed design.

1 .CONCEPTUAL DESIGN:

It depends on what are the major factors for the designing of aircraft.

Power plant location:The power plant location is either padded or buried type engines are

more preferred. Rear location is preferred for low drag, reduced shock and to use whole thrust.

Selection of engine:The engine to be used is selected according to the power required.

Wing selection:The selection of wing depends upon the selection of

Low wing Mid wing High wing

2. PRELIMINARY DESIGN:

Preliminary design is based on loitering. U is the mathematical method of skinning the aircraft after skinning the aircraft looks like a masked body.

Preliminary design is done with the help of FORTRAN software.

3. DETALIED DESIGN:

In the detailed design considers each and every rivets, nuts, bolts, paints, etc. In this design the connection and allocation

COMPARATIVE STUDY OF VARIOUS AIRCRAFTS

EX.NO:1 COMPARATIVE STUDY OF SUBSONIC BOMBER

OBJECTIVE:

To compare the specification and performance of various type of subsonic

bomber aircraft.

COMPARISION OF DIMENSION:

NAME LENGTH (m)

HEIGHT (m)

WING SPAN (m)

WING AREA (m2)

Kawasaki Ki-32

11.65 2.09 15.0 34

Mitsubishi B1M

9.77 3.50 14.77 59

Mitsubishi B2M

10.27 3.71 15.22 55

Yokosuka B4Y

10.15 4.36 15 50

Nakajima B5N2

10.30 3.7 15.52 37.7

Aichi D1A 9.3 3.41 11.4 34.7Aichi D3A 10.2 3.85 14.37 34.9Mitsubishi G3M2

16.45 3.68 25 75.0

Kyushu Q1W 12.09 4.12 16 38.2Nakajima G5N

31.02 -- 42.12 201.8

Mitsubishi G4M2

19.6 6 24.9 --

Radugha Kh-55

8.09 -- 3.1 --

WEIGHT CONFIGURATION:

NAME MAX.TAKEOFF

WEIGHT (Kg)

EMPTY

WEIGHT(Kg)

LOADED

WEIGHT(Kg)

WING

LOADING

(Kg/m2)

Kawasaki Ki-

32

3760 5181 -- --

Mitsubishi

B1M

2697 1442 -- 45.7

Mitsubishi

B2M

-- 2260 3600 --

Yokosuka

B4Y

3600 2000 -- 72

Nakajima

B5N2

4100 2279 3800 101

Aichi D1A 2610 1516 2610 752.161

Aichi D3A 3650 2408 8000 --

Mitsubishi

G3M2

-- 4965 -- --

Kyushu Q1W 5318 3102 4800 126

Nakajima

G5N

32000 20100 -- --

Mitsubishi

G4M2

12500 8160 -- --

Radugha Kh-

55

-- -- -- --

PERFORMANCE DETAIL:

NAME MAX.SPEED

(Km/h)

SERVICE

CEILING (m)

RANGE

(Km

)

RATE OF

CLIMB (m/s)

Kawasaki Ki-

32

423 8920 1965 --

Mitsubishi

B1M

210 4500 -- --

Mitsubishi

B2M

213 4500 1779 --

Yokosuka

B4Y

278 6000 1580 3.6

Nakajima

B5N2

367 8260 1935 6.5

Aichi D1A 309 6980 927 6.37

Aichi D3A 389 9300 1472 --

Mitsubishi

G3M2

375 9200 4400 6

Kyushu Q1W 322 4490 1342

3.81

Nakajima

G5N

420 7450 4260 --

Mitsubishi

G4M2

437 8950 4725 --

Radugha Kh-

55

571 -- 2500 --

SELECTION OF MAIN PARAMETERS

EX.NO:2 SELECTION OF MAIN PARAMETERS

CL Vs CD

The CL Vs CD was drawn in general. It is also called as drag polar. It has the major

impact on endurance of the aircraft, performance and aerodynamic properties.

CL Vs

The general curve was drawn in CL Vs . It gives the information of

variation of lift with angle of attack and the stalling region. Reduce the curve

slope without any change of lift angle of incidence is seen in CL Vs

characteristics. Increase of stalling angle without appreciable change in

maximum lift coefficient is also seen in CL Vs graph.

CL VS CD:

The CL vs CD was drawn in general. It is also called as drag polar. It

has the major impact on endurance of the aircraft performance and aerodynamic

forces.

CL VS :

The general curve was drawn in CL vs , it gives the information of

variation of lift with angle of attack and the stalling region.

VELOCITY VS ALTITUDE:

The values from the comparative study was used to drawn graphs, the

velocity vs altitude graph was drawn and approximate or optimized value of

maximum altitude at velocity 350 KM/hr was found to be 7000m.

VELOCITY VS RANGE:

The values from the comparative study were used to plot the points in

the graph velocity vs range. The approximate value of range of the aircraft at

maximum speed of 350 km/hr was 6000m.

VELOCITY VS WING LOADING:

The graph plotted between velocity vs wing loading (w/s) was drawn

from the values obtained from the comparative study of aircraft. The main

parameters(w/s) was optimized using the graph by drawing a circle of constant

radius.

The value of wing loading is optimally 350 kg/m2.

VELOCITY VS LOADED WEIGHT:

The graph plot between velocity vs loaded weight from the

comparative study was drawn. the optimum value of loaded weight was found to

the 60kg at max speed of 350 km/hr.

VELOCITY VS THRUST-WEIGHT RATIO:

The graph plot between velocity vs thrust-weight ratio value from the

comparative study was drawn velocity vs (T/W) from the graph, the main

parameter (T/W) was optimized using the graph by drawing a circle of constant

radius.

The values of thrust-weight ratio is optimally is 0.35N/kg.

The values from the comparative study done were used to draw graphs. The

velocity Vs altitude graph was drawn and approximate or optimized values of

maximum altitude at velocity 300km/hr was found to be 2000m.

The values from the comparative study were used to plot the points in the graph,

velocity Vs range. The approximate values of range of the aircraft at max speed

300km/hr was 2000km.

The values of velocity and maximum takeoff weight (Wo) from the comparative

study were used to draw the graph velocity Vs maximum takeoff weight. From the

graph the optimum values of (WO) was found to be 5000kg at maximum speed of

300km\hr.

The plot velocity Vs wing loading (W/S) was drawn from the values

obtained from the comparative study of the aircraft. The main parameters (W/S)

was optimized sing the graph by drawing a circle of constant radius.

The value of wing loading is optimally 140Kg/m2.

The above graph was drawn between velocity and power plant from the

historical data. From the above graph for velocity 300km/hr was optimized to

680Kw.

WEIGHT ESTIMATION

EX.NO:3 WEIGHT ESTIMATION

GROSS WEIGHT:

W0 = W Crew + W payload + W fuel + W empty

Empty weight includes structure, landing gear, lift equipment and

avionic instruments.

To simply fuel weight and empty weight calculation take fraction of

them based on total weight.

W0= W Crew + W pl+ [(W f/ W0) + (We/W0)] W0

W Crew + W pl = W0 - [(W f/ W0) + (We/W0)] W0

W0 = [W Crew + W pl] / [1-(W f/ W0)+(We/W0)]

STEP 1: WPL

Weight of payload includes bombs, missiles, pods. The payload

weight that has to carried is about

WPL=11450kg

STEP 2: WTO GUESS

The arbitrary takeoff weight is calculated from the previous

experiment. (i.e) from the graph taken to be

WTOGUESS = 20000kg

STEP 3: WF

The fuel weight includes mission fuel and fuel reserved for emergency

operation.

WF= WF mission+ WF reserved

Mission fuel weight can be calculated from the fuel fraction for various mission

segments.

MISSION PROFILE

0-1 → Engine start and warm up

1-2 → Taxing

2-3 → Take off

3-4 → Climb and acceleration

4-5 → Cruise out

5-6 → Loiter

6-7 → Descent

7-8 → Combat

8-9 → Climb

9-10 → Cruise in

10-11 → Loiter

11-12 → Descent

12-13 → Landing, Taxing, and Shutoff

MISSION 0-1:

The mission 0-1 is engine start and warm up. From the historical data

it is found to be 0.99

=0.99

MISSION 1- 2:

The mission 1-2 is taxing, the fuel fraction from historical data was

0.99

= 0.99

MISSION 2-3:

The mission 2-3 is takeoff and the fuel fraction is found to be 0.995

= 0.995

MISSION 3-4:

The mission 3-4 is climb and acceleration. The fuel fraction for this

segment is found to be 0.980

= 0.980

MISSION 4-5:

The mission 4-5 is cruise. The fuel fraction for this mission was found

from the range equation (i.e) Brequet equation

R= ln

= e-[R/

Range, R is approximated from the graph of previous experiments

R = 755.06nm

cruise velocity, v = 176.29knots,88.88m/s

L/D = 14

C j=0.6

The SFC, C j is approximately to 0.6 since the armaments are placed

externally

Substitute the value in Brequet equation

= ∑

= 0.829

MISSION 5-6: [LOITER]

The fraction for this mission is found from the endurance equation

E=

Let us assume the endurance to be 20 i.e: 0.33 hr. Cj=0.5; = 15

Substitute the value in endurance equation

=0.988

MISSION 6-7: (DECENT)

The fraction of fuel for this mission was found from historical data

and to be 0.99.

MISSION 7-8: (COMBAT)

During combat, no fuel penalty or no range credit could be

encouraged if the weapons aren’t dropped. So the fraction is assumed to be

=1.0

The fuel fraction up to this stage is calculated by the consideration the

procedure up to this mission so

M ff = 0.99*0.99*0.995*0.98*0.829*0.988*.99*1

M ff =0.74

Lets calculate the weight M the bomb is dropped.

Weight prior to bomb drop = 0.774 WTO

= 15480

Let us consider bomb is dropped weight

Immediately after bomb dropped = 15480-11250

= 4230

MISSION 8-9: (CLIMB)

The climb weight fraction is 0.980.

This ratio has to be corrected for weight change during bomb drop.

The bomb weight ratio =

=0.27

The corrected weight = [1-(1-0.98)0.27]

= 0.9945

MISSION 9-10: (CRUISE)

The fraction is found using Brequets equation =14.5, since the

bomb are dropped and profile drag would less.

C=0.6is

So is found to be 0.834

MISSION 10-11: (LOITER)

The fuel fraction for this mission is found using endurance equation.

Endurance is assumed to be 0.333hr.therefore,0.333

= (0.333) (0.6)/14.5

=0.986

MISSION 11-12: (DESCENT)

The fraction is found to be from historical data to be 0.99

=0.99

MISSION 12-13: (LANDING, TAXING & SHUTOFF)

The fuel fraction of this mission was found to be

=0.992

The fuel fraction is found from the products of every value

M ff=0.99*0.99*0.995*0.98*0.829*0.988*0.99*0.9945

*0.834*0.986*0.99*0.992

M ff= 0.622

W f = (1-0.622) WTO

W f = 7560

STEP 4:

Tentative WOE is found from following formula

WOE Tent = WTO GUESS – W f – WPL

= 2000-760-250

=990kg

STEP 5:

Tentative weight empty WE is found from the following formula,

W E tent = WOE Tent –W Reserved fuel –W crew

The weight of reserved fuel is 0.5% of gross weight

W reserved fuel = 0.005*WTO

=0.005*20000

=100

WE Tent = WE Tent –W Reserved tent- W crew

= 990-100-200

=690

CALCULATION OF GROSS WEIGHT:

From equation 4

W0=

=

W0=19489.36kg

PERCENTAGE OF ERROR:

The percentage of error between assumed gross weight and actual

gross weight is calculated

% of error = (assumed – actual)*100/assumed

= (20000 – 19489.36)*100/20000

% of error = 2.55%

ENGINE SELECTION

EX.NO:4 ENGINE SELECTION

PROCEDURE:

To calculate the maximum thrust required for takeoff

we have

(hp/w)=0.08 as reference for subsonic bomber aircraft.

The thrust required by analytical approach

(T/W)=

=

=0. 8442

(T/W)=0.8442*780.71

(T/W)=659.07 Kg

The engine must be selected such that the thrust required must equal thrust

produced by the engine.

The following engine matches the purpose.

Lycoming produced a thrust of 396 KW

Lycoming produced a thrust of 395 KW

Vedeneyer produced a thrust of 420 KW

The wet aspect ratio can be calculated using the formula

AR(WET)=

AR(WET)

=5.05/3

=1.68

The ratio SWET

/SREF

was found to be3.

THRUST MATCHING:

For a propeller driven aircraft the required takeoff (hp

/w) can be found using

=(hp/w)takeoff

(L/D)CRUISE=14

Piston engine with supercharger cruise at 75% of takeoff power.

(hp/W)Takeoff=

(hp/W)Takeoff=0.0513

Since, the reference (hp/W) and calculated (hp/W)are correct.

MODEL TYPE COMBUSTION TYPE

MAX.POWER AT SEA LEVEL

SFC AT MAX POWER

OVER ALL PRESSURE RATIO AT MAX POWER

250-B17C

AC-P Ca 420 SHP 0.66 7.2

250-B17D

AC-P Ca 420 SHP 0.66 7.2

250-B17E

AC-P Ca 420 SHP 0.66 7.2

TRS 18-1 C-J An 360 lb-t 1.26 4.7

MODEL MAX.ENVELOPE DIA (in)

MAX.ENVELOPE LENGTH (in)

DRY WEIGHT LESS TAIL PIPE (LB)

250-B17C 22.6 45 198

250-B17D 22.5 45 202

250-B17E 22.5 45 202

TRS 18-1 12 40.7 97

ADVANTAGE:

Light weight starter,

Electronic igniter,

Remote oil filter,

Air condition provision,

Optical magnets,

Fixed pitch and chard propeller.

DISADVANTAGE:

Slip stream component induces drag,

Slip stream may disturb the free air flow over the wing.

AIRFOIL SELECTION

EX.NO:5 AIRFOIL SELECTION

The airfoil can be selected from the maxLC which we can determined from the

calculation.

The maxLC for the airfoil has to be selected for choosing the specified airfoil.

Blade Element Theory for Airfoil:

In the previous section, we looked at momentum theory. This theory gave us

four useful pieces of information:

(1) Induced velocity far downstream in the rotor wake, called downwash, is twice

that at the rotor disk, called inflow.

(2) The ideal power coefficient Cp in hover equals CT3 2

2

/

(3) The induced power is minimized for a given thrust coefficient, if the induced

velocity in the far wake is uniform.

(4) The induced velocity at the rotor disk is related to the thrust coefficient in hover

by

iTv

R

C 2

Momentum theory cannot help us analyze specific rotor blades, or distinguish

between the number of blades and their other physical characteristics such as twist,

r

dr

taper, camber etc. In order to do these, we turn to a theory called blade element

theory.

Blade element theory is similar to the strip theory in fixed wing

aerodynamics. The blade is assumed to be made of several infinitesimal strips of

width ‘dr’. The lift and drag are estimated at the strip using 2-D airfoil

characteristics of the airfoil at that strip, and what we know about the local flow

magnitude, such as the angular velocity, climb speed V, and inflow v. The lift L ,

and drag D multiplied by the in-plane velocity of the rotor are integrated with

respect to r, from root to tip to obtain the thrust T and the power P consumed by a

single rotor blade. For multi-bladed rotors, this integrated expression is multiplied

by the number of blades, b.

Note: Wayne Johnson uses the symbol ‘N’ for number of blades.

Consider a typical element or strip shown below. The blade “sees” an in-

plane velocity UT, that is tangential to the plane of rotation. The magnitude of UT

is, of course, ∆r, where r is the radial position of the strip. This element has a pitch

angle equal to Φ. That is, the angle between the plane of rotation and the line of

zero lift is θ. If there were no climb velocity V, or induced inflow v, this would be

the section angle of attack.

These two components of velocity V and v change the flow direction by

amounts∆, as shown in the figure above. Here,

r

vVarctan

Thus, the effective angle of attack is . The airfoil lift and drag

coefficients Cl and Cd at this effective angle of attack may be looked up from a

table of airfoil characteristics. The lift and drag forces will be perpendicular to, and

along the apparent stream direction.

r

V varctan

r

V+v

Line of Zero Lift

effective =

These forces are given by

dPT

lPT

cCUUD

cCUUL

2

1

2

1

22

22

The L’ and D’ have units of force per unit span. They must be rotated in directions

normal to, and tangential to the rotor disk, respectively, and multiplied by the strip

width dr to get the thrust and drag components, as shown below.

XxT

ldPTx

dlPT

rdFdFUdP

drCCcUUdrLDdF

drCCcUUdrDLdT

sincos2

1sincos

sincos2

1sincos

22

22

Finally, the thrust and power T and P may be found by integrating dT and dP

above from root to tip (r=0 to r=R), and multiplying the results by the total number

of blades, b.

The above integration can, in general, be only numerically done since the chord

c, the sectional lift and drag coefficients may vary along the span. Finally, the

inflow velocity v depends on T. Thus, an iterative process will be needed to find

the quantity v.

Approximate expressions for thrust and power may, however, be found if we are

willing to make a number of approximations:

a) The chord c is constant,

b) The inflow velocity v and climb velocity V are small. Thus, << 1

, and << 1. We can approximate cos() by unity, and

approximate sin() by ().

c) The lift coefficient is a linear function of the effective angle of attack,

that is, . Thus,

aCl

where a is the lift curve slope. For low speeds, a may be set equal to 5.7 per

radian.

d) Cd is small. So, Cd sin(Φ) may be neglected.

e) The in-plane velocity UT is much larger than the normal component UP

over must of the rotor, except near the hub.

With these assumptions, thrust T may be expressed as

r R2 2

r 0

r R3 3

d

r 0

1 V vT cba r dr

2 r r

1 V v V vP cba C r dr

2 r r r r

To perform the integration, we need to know how the pitch angle Φ varies

with r. Many rotor blades are twisted, and it is not reasonable to assume that the

pitch angle Φ is constant. Two choices are common.

Linearly Twisted Blade:

Here, we assume that the pitch angle varies as

E Fr

where E and F are constants. Using this definition, and performing the

integration (check!), we get:

Tb

ca E FRV v

RR

bca R R

Cabc

R

a

where

bc R

V v

R

T

2

1

3

3

4 2 2 32

2 32

2 32

2 3 2 75

75 75

.

. .

/

/ /

/solidity BladeArea / DiskArea

Inflow Ratio

Notice that the thrust coefficient is linearly proportional to the pitch angle Φ at the

75% Radius. This is why the pitch angle is usually defined at the 75% R in

industry.

The expression for power may be integrated in a similar manner, if the drag

coefficient Cd is assumed to be a constant, equal to Cd0. The final expression is

(check):

C CC

P Td 0

8

The above expressions are true only for a linearly twisted rotor.

Ideally Twisted Rotor:

Here, the twist angle is inversely proportional to the radial location r. Such rotors

are hard to manufacture, but turn out to have the lowest power consumption.

t R

r

Here Φt is the pitch angle at the blade tip.

Using this in the expression for thrust given in equation (6) we get

T abcR

r

V v

rr dr abc Rt t

r

r

1

21

42 2 3

0

Or,

Ca

T t

4

The expression for the coefficient for power, for an ideally twisted rotor turns out

to be identical to that for a linearly twisted rotor.

In summary, according to the blade element theory, the following

expressions are obtained.

For a linearly twisted rotor in hover or climb,

.75TC a

2 3 2

For an ideally twisted rotor in hover or climb,

T t

aC

4

For both types of twist, the power coefficient is given by

d0P T

CC C

8

The first term in the power coefficient is identical to momentum theory, and is

called the induced power. The second term is due to power required to turn the

rotor in a viscous flow, and is called the profile power. The Figure of Merit M is

given by

T

T d0

CM

C C /8

PROCEDURE:

The airfoil can be selected from the (CL)max value which can be determined

from calculation.

The (CL)max value for the airfoil has to be selected for choosing the specified

airfoil.

DESIGN (CL)max:

The design (CL)max value for the airfoil can be calculated from the basic lift

equation

W= 21

2 SCL at steady state.

Therefore, CL=(2WS)/ 2

The value of wing loading is calculated from the experiment 2.

The density is taken from the International Standard Atmosphere(ISA) for

an altitude 4000m is to be 0.76Kg/m3 .

The velocity at cruise is 320Km/hr (88.88m/s). Hence the CL value for the

airfoil to be

CL=(2*314)/.76*(88.88)2

CL=0.104

AIRFOIL SELECTION:

The airfoil in many respects, is the heat of the airplane. The airfoil affects

the cruise speed, takeoff and landing distance, stall speed, handling qualities

(especially near the stall),overall aerodynamic efficiency during all phases of

flight.

AIRFOIL GEOMETRY:

The front of the airfoil is designed by a leading edge radius which is

tangent to the upper and lower surfaces.

An airfoil is designed to operate in supersonic flow will have a sharp

or nearly sharp leading edge to prevent a drag producing bow shock.

The chord of the airfoil is the straight line from the leading edge to the

trailing edge. It is very difficult to build a perfectly sharp trailing edge, so most

airfoils have a blunt trailing edge with some small finite thickness.

EFFECT OF THICKNESS:

The thickness distribution of the airfoil is the distance from the upper

surface to the lower surface measured perpendicular to the mean chamber line and

is a function of the distance from the leading edge.

The airfoil thickness ratio(t/c) refers to the maximum thickness of the

airfoil described by its chord.

Due to the fuselage effects, the root airfoil of a subsonic aircraft can

be as much as 20-60% thicker than tip airfoil without greatly affecting the drag.

This is very beneficial resulting in a structural weight reduction as

well as maximum volume for fuel and landing gear. this thicker root of the airfoil

should extend to not more than about 30% of the span.

For the above condition are satisfied on the suitable airfoil is NACA

63010.

The X and Y coordinates for the selected airfoil are drawn by using

the excel software.

LOWER SURFACE UPPER SURFACE

X Y X Y

1.0 0.0 0.0 0

.95 -0.002140 0.005 0.008290

0.9 -0.006040 0.0075 0.010040

0.85 -0.010880 0.0025 0.017560

0.8 -0.016180 0.05 0.0.24400

0.75 -0.021660 0.075 0.029500

0.7 -0.027120 0.1 0.033620

0.65 -0.032340 0.15 0.039940

0.6 -0.037150 0.20 0.044450

0.55 -0.041400 0.25 0.047530

0.5 -0.044960 0.30 0.049380

0.45 -0.047660 0.35 0.05

0.4 -0.049380 0.40 0.049380

0.35 -0.05 0.45 0.047660

0.3 -0.049380 0.50 0.044960

0.25 -0.047530 0.55 0.041400

0.2 -0.044450 0.60 0.037150

0.15 -0.039940 0.65 0.032340

0.1 -0.033620 0.70 0.027120

0.075 -0.029500 0.75 0.021660

0.05 -0.024400 0.80 0.016180

0.025 -0.017560 0.85 0.010880

0.0075 -0.010040 0.90 0.006040

0.0050 -0.008290 1.0 0

NACA63A010

The above graph was drawn between the angle of attack and the co-

efficient of lift without flap. When the plain flap is deflected, the increase in lift

is due to an effective increase in camber and a virtual increase in angle of

attack.

The above graph was drawn between co-efficient of lift and co-efficient of

drag.

1. Negative values of CL pertain to negative lift, which occurs when the

angle of attack of the airplane is less than .

2. This situation is not encountered frequently in the analysis of airplane

performance, hence only that portion of the drag polar associated with

positive CL.

WING DESIGN

EX.NO:6 WING DESIGN

WING DESIGN AND CALCULATION:

To design the wing we have to find out length of the fuselage

LENGTH:

Length= aw0c

From historical data a= 0.23; c=0.5

Length =.23(19489.36).5

Length =105.3ft

SURFACE AREA:

From the historical data W/S=113.5

s=w/113.5 =19489.36/113.5=171.71m2

For our aircraft

Aspect ratio =AR=10.3=b2/s

b2=10.3*171.71

b =42.5

ROOT CHORD:

The design the aerodynamic center

C root = (2*SW)/ [b (1+ )]

= (2*171.71)/42.5(1+0.4)

C root =5.83ft

TIP CHORD:

C tip = C root =0.4*5.83

C tip =2.33ft

AERODYNAMIC CENTRE:

To design the aerodynamic centre

c = (2/3)*c root *(1+ + 2/1+ )

= (2/3)*5.833*(1+.4+.42/1+.4)

c =4.33ft

Y = (b/6) (1+2 /1+ )

=42.5/6(1+.8/1+.4)

Y =9.0107ft

The vertical and horizontal surface of the tail can be find out through the

equation

CVT=LVT.SVT/BW.SW

CHT=LHT.SHT/CW.SW

From the historical data the value of CVT & CHT for the subsonic bomber

aircraft is

CVT=.08& CHT=1.00

LENGTH OF THE VERTICAL TAIL:

The length of the vertical tail is the 50-60% of the fuselage length

LVT=.5*32.1

=16.05ft

SVT=CVT*BW.SW/LVT

= .08*42.05*171.71/16.05

SVT=35.98ft2

The length of the horizontal tail is the 25% of the fuselage length

BW=.25*32.1=8.025ft

SHT=CHTBWSW/LHT

= 1*8.025*171.71/16.05

SHT=85.855ft2

FUEL TANK:

The volume of fuel tank required

Volume of fuel =weight of fuel/density of fuel

=7560/800=9.45m3

Now we assume 50% of the fuel stored in the wing

0.5*9.45= [(t/c)* c *(.5c )*(.5*b*.72*2)]

c = 3.78

t/c = 0.0127

t =0.0127* Cr

t=0.012*5.833

t=0.069m

t/ct= 0.0127

t=0.0127*2.33

t=0.029m

ESTIMATION OF WETTED SURFACE AREA

EX.NO:7 ESTIMATION OF WETTED SURFACE AREA

WETTED SURFACE AREA:

Aircraft wetted area (Swet), the total exposed surface area, can be visualized as the

area of the external parts of the aircraft that would get wet if it were dipped into

water. The wetted area must be calculated for drag estimation, as it is the major

contributor to friction drag.

The wetted area is estimated by multiplying the true- view exposed as plan form

area (Sexposed) times a factor based upon the wing or tail thickness ratio.

If a wing or tail were papper-thin, the wetted area would be exactly twice the true

plan form area (i.e. top and bottom). The effect of finite thickness is to increase the

wetted area, as approximated. Note that the true exposed plan form area is the

projected(top-view) area divided by the cosine of the dihedral angle.

If t/c<0.05,

Swet=2.003Sexposed

If t/c>0.05,

Swet=Sexposed (1.977+0.52(t/c))

The wetted area of the fuselage can be initially estimated using just the side and

top views of the aircraft by the method. The side and top views projected areas of

the fuselage are measured from the diagram, and the values are average.

Swet=3.4(Atop+Aside)/2

A more accurate estimation of wetted area can be obtained by graphical integration

using a number of fuselage cross section. If the perimeters of the cross sections are

measured and plotted Vs longitudinal location, using the same units on the graph,

then the integrated area under the resulting curve gives the wetted area.

The cross sectional perimeters measurements should not include the portions

where components join, such as at the wing-fuselage intersection. This areas are

not “wetted”.

DRAG POLAR FOR CRUISE CONDITION:

Fuselage length = a 0CW

a = 0.23, c = 0.5

Fuselage length = 0.23(19490)0.5

= 32.1m

f

f

L

d = 9

df = 3.56

Fuselage Sπ = ( df2)/4

= 9.98 m2

Sw= Tw * bw

=0.0518 * 42.05

Sw = 2.180

From the historical tail area was 37.2m2

AR = b2/S

AR = 10.3

b2 = (10.3*37.2)/3

b = 11.3m

tht = 10% of wing thickness

Sht = tht*bht

= 0.00518 * 11.3

Sht =0.058594m2

Svt=tvt*bvt

= 0.00518 * 5.65

Svt=0.029267m2

Engine area = π d2/4

=3.14*(1.8)2/4

=2.5434

¼ flap Sπ=210 * *(0.2*4.33)

360

=0.0654

Under carriage = 1.1*Sπengine

= 1.1 * 2.5434

= 2.8

¾ full flap Sπ = 230 * *(0.2*4.33)

360

= 0.1962

TABULATION:

COMPONENTS Cdπ Sπ Cπ*Sπ

Fuselage 0.03 9.98 0.2994

Wing 0.008 2.18 0.01744

Horizontal Tail 0.008 0.059 0.000472

Vertical Tail 0.008 0.029 0.000232

Engine 0.01 2.54 0.0254

Under carriage 0.04 2.8 0.112

1/4flap 0.0504 0.0654 0.0033

¾ full flap 0.035 0.1962 0.00687

5

1

*otherscruise

id

doi W

C SC

S

E =0.2994 0.1744 0.000472 0.0254 0.000232

515.13

=9.70*10-4

7

1

*otherstakeoff

id

doi W

C SC

S

=0.2994 0.1744 0.000472 0.0254 0.000232 0.112 0.0033

515.13

=1.1943*10-3

6

1 1

* *otherslanding

id d

doi iW W

C S C SC

S S

=0.2994 0.1744 0.000472 0.0254 0.000232 0.112 0.00687

515.13

= 1.2*10-3

Cd0 = Cd0 cruise+ C do takeoff+ C do landing

=9.7*10-4+1.1943*10-3+1.20*10-3

=3.36*10-3

CD0=Cfe * wet

ref

S

S

= 0.0030*3

= 0.009

K=1

AR = 1/(3.14*0.79*10.3)

=0.039

Where =0.79 for subsonic

AT h=0, a=340 m/s

V Cl M Cdto Cdw Cd0 Cdt D(kN)

20 14 0.05 0.0012 0.024 0.025 8.04 109.8

40 3.5 0.117 0.0012 0.006 0.007 0.511 27.9

60 1.56 0176 0.0012 0.0026 0.003 0.12 14.715

83.3 0.8 0.2345 0.0012 0.0014 0.002 0.04 8.83

AT h = 500m, a=338m/s

V Cl M Cdto Cdw Cdo Cdt D(KN)

20 14.7 0.06 0.0012 0.0025 0.026 8.85 11.7

40 3.68 0.118 0.0012 0.006 0.007 0.56 2.96

60 1.63 0.177 0.0012 0.0027 0.003 0.118 1.41

83.3 0.85 0.246 0.0012 0.0014 0.002 0.04 0.9

AT h=1000, a=336m/s

V Cl M Cdto Cdw Cd0 Cdt D(KN)

20 15.4 0.06 0.0012 0.0026 0.027 9.72 12.3

40 3.86 0.119 0.0012 0.007 0.008 0.62 3.13

60 1.71 0.178 0.0012 0.0029 0.004 0.13 1.48

83.3 0.89 0.247 0.0012 0.0015 0.003 0.04 0.875

AT h=1500, a=334m/s

V Cl M Cdto Cdw Cd0 Cdt D(KN)

20 16.22 0.06 0.0012 0.027 0.028 10.78 12.9

40 4.06 0.119 0.0012 0.007 0.008 0.68 3.26

60 1.8 0.179 0.0012 0.003 0.004 0.14 1.515

83.3 0.93 0.249 0.0012 0.0015 0.003 0.05 1.047

AT h =2000, a=332m/s

V Cl M C dto C dw Cd0 C dt D(KN)

20 17.05 0.06 0.0012 0.028 0.029 11.19 12.8

40 4.26 0.12 0.0012 0.007 0.008 0.75 3.43

60 1.89 0.18 0.0012 0.003 0.004 0.16 1.65

83.3 0.983 0.25 0.0012 0.0016 0.003 0.05 .991

Cl Cdt cruise Cdt take off Cdt landing

0 1.0185*10-3 1.19*10-3 1.2*10-3

1 0.0419 0.04214 0.04215

2 0.1648 0.16499 0.165

3 0.369 0.369 0.369

3.9 0.624 0.624 0.624

CL vs Cdt

Cdt

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

CL

0

1

2

3

4

5

Cdt cruise vs CL

Cdt takeoff vs CL

Cdt landing vs CL

The above graph is drawn in between CL Vs Cdt. From the above graph we

should understand that the drag is low for takeoff. The drag value is maximum

for the landing and drag value for landing. The drag value for the cruise is in

between the drag value of take-off and landing.

DRAG VS VELOCITY

VELOCITY

0 20 40 60 80 100

DR

AG

0

2

4

6

8

10

12

14

h=0(sea level)h=500mh=1000mh=1500mh=2000m

The above graph is drawn between velocity and drag. From the above graph we

should understand that as velocity increases the also increases relatively up to

certain velocity and then the drag value decreases.

ESTIMATION OF RATE OF CLIMB

EX.NO:8 ESTIMATION OF RATE OF CLIMB

RATE OF CLIMB:

At steady state condition of flight for increase in power of engine there will have a

corresponding increase of lift, if the elevators are operated suitably this increase in

power can be used for the climbing flight with the nose up.

For such a flight L=W sinθ. The vertical velocity component Vsinθ is called rate of

climb. In a climbing flight power is required for both of the following

1. To pull the aircraft weight up

2. To overcome the drag force,

Rate of climb = T - W / W V

=Excess power/W

This relation is true for small angles of climb.(say θ<200)

Therefore, From the engine selection.

Thrust required at altitude = F*σ^1.15

Where

σ= [(20-h)/(20+h)]

At sea level h=0,

F=3*[(20-0)/ (20+0)] 1.15

F=300KN

At h=500m

F=3*[(20-0.5)/ (20+0.5)] 1.15

F=283KN

At h=1000m

F=3*[(20-1)/ (20+1)] 1.15

F=267KN

At h=1500m

F=3*[(20-1.5)/ (20+1.5)] 1.15

F=252KN

At h=2000m

F=3*[(20-2)/ (20+2)]1.15

F=238KN

At h=0,

V D T R/C=[(T-D)*60]/ (W0*9.81)

20 11.19 3 -2.5640 2.845 3 0.0460 1.5 3 0.47083.3 0.9 3 0.66

At h=500m

V D T R/C20 11.7 2.83 -2.7840 2.96 2.83 -0.0460 1.41 2.83 0.4483.3 0.9 2.83 0.6

At h=1000m

V D T R/C20 12.3 2.67 -3.0240 3.13 2.67 -0.1460 1.48 2.67 0.3783.3 0.875 2.67 0.56At h=1500m

V D T R/C20 12.9 2.52 -3.2540 3.26 2.52 -0.2560 1.515 2.52 0.3183.3 1.047 2.52 0.46

At h=2000m

V D T R/C

20 12.8 2.38 -3.27

40 3.43 2.38 -0.33

60 1.65 2.38 0.23

83.3 0.991 2.38 0.435

ALTITUDE VS R/C

R/C

-0.8 -0.7 -0.6 -0.5 -0.4 -0.3

AL

TIT

UD

E

0

1

2

3

ALTITUDE VS R/C

1. The service ceiling can be defined as the altitude at which the rate of

climb is 100 units/units time.

2. The maximum rate of climb will not occur at maximum angle of climb.

The above graph is drawn in between altitude and rate of climb from this

above graph we should understand that rate of climb decreases with increase in

altitude.

3. Absolute ceiling is termed as generally used to indicate the maximum

height to which airplane can reach. At that altitude the power available

and power required will be equal

VELOCITY VS R/C

VELOCITY

10 20 30 40 50 60 70 80 90

R/C

-4

-3

-2

-1

0

1

at h=0(sea level)at h=500mat h=1000mat h=1500mat h=2000m

The above graph is drawn in between velocity and rate of climb. From the above

graph we understand that

1. For steady climbing flight, lift and coefficient of lift are smaller than that

in a steady level flight. Hence induced drag also smaller.

2. Consequently total drag for climbing flight becomes smaller than total

drag for level flight at same velocity.

Hence a plot of drag power (i.e., drag x velocity) against velocity of flight for

climbing flight will be totally different from same plot for level flight, as the drag

is smaller during climbing than level flight at same flight velocity.

ALTITUDE VS 1/(R/C)

ALTITUDE

0 500 1000 1500 2000 2500 3000

1/(R

/C)

-4

-3

-2

-1

ALTITUDE VS 1/(R/C)

The above graph is drawn in between the 1/ROC and altitude. From this above

graph we should understand that as altitude increases the 1/ROC value increases up

to 4000m and after this altitude the value of 1/ROC value starts to decreases.

TAKE OFF AND LANDING DISTANCE

EX.NO:9 TAKE OFF AND LANDING DISTANCE

LENGTH OF TAKEOFF RUNWAY:

(a)Ground run(S1):

( ) ( / )( / )T D W L W g dv dt

( ) ( ) ( / )( / )T D W L W g vdv ds

1

1

0

s

S ds

1

0

( / )

( ) ( )

vW g

vdvT D W L

2

1( / )

( ) ( )

W V g

T D W L

V1=1.2*VStall

S1 19489.36(83.88 / 2)

(3 8.83) 0.4(191.19 307.93)

S1 = 1.657 km

(b)TRANSITION RUN (S2):

µ(W-L)=0,(W/L)is assumption

(T-D) = (W/g)(v dv/ds)

ds = (W/g)(v dv/(T-W))

2 2

1

2

0 ( )

S V

V

WS ds vdv

g T D

2 2

2 1( )

2 ( )

W V V

g T D

Where

V1=20 m/s

V2=40 m/s

D = 3.43KN2 2

2

191.19 (40 20 )*

(3 3.43) 2S

2 292.39S m

(c)CLIMBING(S3):

3

3

( ) ( ) 0

sin ( ) /

T D WSin

T D W

HTan

S

S HCot

CosCot

Sin

21 Sin

Sin

2

( )1

(( ) / )

T DW

T D W

2 2

3

( )

( )

W T DS H

T D

First segment:

2 2

3

(191.19) (3 12.9)2000

(3 12.9)S

3 1.732S km

Finally

1 2 3

3.681

S S S S

S km

TAKEOFF PROFILE

FOR LANDING:

(a)Descend(S1):

2 2

1

( )

( )

W T DS H

T D

2 2

1

(191.19) (395.76 8.83)2000

(395.76 8.83)S

1 1.738S km

(b)Transition(S2):

2 22 1

2

2 2

2

2

( )

2 ( )

19489.36(40 20 )

2(395.76 11.7)

30.477

W V VS

g T D

S

S m

(c): Ground run (S3):

21

3

2

3

3

( )2

( ) ( )

19489.36(83.33 / 2)

(395.76 109.8) 0.4(191.19 307.935)

203.41

VW

gS

T D W L

S

S m

Finally,

1 2 3

2.245

S S S S

S km

LANDING PROFILE

THREE VIEW OF SUBSONIC BOMBER AIRCRAFT

FRONT VIEW

TOP VIEW

SIDE VIEW

BIBLIOGRAPHY:

1. Daniel p. Ramer, “Aircraft conceptual design”, seventh edition.

2. L.M. Milne-Thomson, “Theoretical Aerodynamics”, second edition.

3. J.D. Anderson, Aircraft Performance”.

4. Thomas cork, “Preliminary Aircraft Design”.

5. John F. Fielding, “Airplane Design”.

6. Jan Roshkam, “Airplane Design”, All seven edition.

7. Ira. H. Abbot, “Theory of wing selections”.

8. Taylor J. Janes, “All the World Aircraft”, Janes’s, London, England, UK,

1976.

9. Courtland D. Perkins & Robert E. Hage, “Airplane Performance and

Stability control”.

10.Aviation weeks- January 2008 edition.

Few websites:

www.NASA.org

www.ZAP16.com

www.Propulsion.org

www.BOMBERAIRCRAFT.com

www.ADL.GETCH.edu

http://www.NASA.org/

http://www.ADL.GETCH.edu/

http://www.BOMBERAIRCRAFT.com/

http://www.Propulsion.org/

http://www.ZAP16.com/

Documents

Subsonic Bomber Aircraft