Project Report

Airbus A-318 (Performance Analysis)

AERO VEHICLE PERFORMANCE

Submitted to : Sir IZHAR KAZMI

Submitted by : SHAHZAD MALIK

Reg # : 100101016

Department : BE-09 (AERO)

Date : 23-10-2012

Aircraft DataZero Lift Drag Coefficient= Cdo =0.0188

Wing Span = 34.1 m

Wing Area = S = 122.6 m2

Wing Sweep Back= 25o

Maximum Takeoff Weight= 68000*9.81 N

Maximum Speed= Mach 0.82 at 11000 m

Service Ceiling= 12000 m

Engines(x2) = Pratt & Whitney PW6000 series

Thrust = 200 KN (at Sea Level)

Thrust Required (Drag)

Graphical Approach1.Let the Velocity be 125 m/s

V = 125 m/s2.At a standard Altitude of 11000 m..

Density = ρ = 0.364234 kg/m3

CL = 2W/ ρV2S = 2(68000*9.81)/(0.364234*1252*122.6) = 1.913.CD = CD,O + KCL

2 = 0.0188 + (0.038*1.912) = 0.1584.TR = D = 0.5* ρ * S * CD * V2 = 55029.18 N

Similarly the same procedure is done for different Velocities at the same altitude.

Table (a)

Data at Height= 11000 m Velocity (m/s) CL CD Thrust Required (N)

115 2.259 0.21 62818.08125 1.912 0.16 55029.32135 1.639 0.12 49205.82145 1.421 0.1 44846.96155 1.244 0.08 41608.23165 1.097 0.06 39246.21175 0.976 0.05 37584.98185 0.873 0.05 36494.87195 0.786 0.04 35878.54205 0.711 0.04 35661.82215 0.646 0.03 35787.37225 0.59 0.03 36210.3235 0.541 0.03 36895.08245 0.498 0.03 37813.26255 0.459 0.03 38941.86265 0.425 0.03 40262.16275 0.395 0.02 41758.77285 0.368 0.02 43418.97295 0.343 0.02 45232.13305 0.321 0.02 47189.37315 0.301 0.02 49283.16325 0.283 0.02 51507.13335 0.266 0.02 53855.85345 0.251 0.02 56324.65355 0.237 0.02 58909.53

Thrust Required Curve for A-318 at 11000 m

Thrust Required Curve for A-318 at 11000 m, illustrating the region of velocity instavility and stability, and the direction of decreasing angle of attack with increasing velocity.

Drag Versus Velocity for a318, illustration of the variation of zero lift drag and drag due to lift at 11000 m.

Analytical ApproachTo calculate L/D(max), we will use

L/D(max) = 1/sqrt(4CDOK) = 1/sqrt(4*0.0188*0.038) = 18.71.

Variation of L/D with velocity

The value of minimum thrust required remains the same even at different altitudes, but the velocity to achieve this point is different as can be shown in the graph below.

Effect of Altitude on the point corresponding to minimum thrust required.

To Calculate minimum thrust required, we will calculate Thrust to weight Ratio minimum.

(TR/W)min = 1/(L/D)max

= 0.0535

(TR)min = 0.0535*W

= 54660.7 N

Which is analogous to graphical result.

Hence the velocity for minimum TR is

V(TR)min = sqrt((2/ρ)*(sqrt(K/CDO))*(W/S)) = 206.1 m/s

Aerodynamic RelationsCL/CD

3/2, CL/CD, CL/CD1/2

Calculations(CL/CD)(max) = 1/sqrt(4CDOK) = 1/sqrt(4*0.0188*0.038) = 18.71.

V(CL/CD)max = sqrt((2/ρ)*(sqrt(K/CDO))*(W/S)) = 206.1 m/s

CL/CD3/2

(max) = (1/4)*(3/KCD,O1/3)3/4 = (1/4)(3/(0.038)(0.0188)1/3)3/4 = 17.88

V(CL3/2

/CD)max = sqrt((2/ρ)*(sqrt(K/3CDO))*(W/S)) = 156.6 m/s

CL/CD1/2

(max) = (3/4)*(1/3KCD,O1/3)3/4 = (1/4)(1/3(0.038)(0.0188)1/3)3/4 = 25.42

V(CL1/2

/CD)max = sqrt((2/ρ)*(sqrt(3K/CDO))*(W/S)) = 271.24 m/s

Hence

V(CL3/2

/CD)max : V(CL/CD)max : V(CL1/2

/CD)max = 156.6 : 206.1: 271.24 = 0.76 : 1 : 1.32

Comparison of Zero Lift Dag and Drag due to Lifta)From above we get V(CL

3/2/CD)max = 156.6 m/s

CL = 2W/ ρV2S = 2(68000*9.81)/(0.364234*156.62*122.6) = 1.22

Zero lift Drag = 0.5ρV2SCDo = 0.5*0.364234*156.62*122.6*0.0188 = 10293.95 NDrag due to lift = 0.5ρV2SKCL

2 = 0.5*0.364234*156.62*122.6*0.038*1.222 = 30969.03 N

Compairing, we get R = Zero lift Drag/ Drag due to lift = 10293.95/30969.03 = 1/3

b) Similarly, from above we get V(CL/CD)max = 206.1 m/s

CL = 2W/ ρV2S = 2(68000*9.81)/(0.364234*206.12*122.6) = 0.7Zero lift Drag = 0.5ρV2SCDo = 0.5*0.364234*206.12*122.6*0.0188 = 17830.14 NDrag due to lift = 0.5ρV2SKCL

2 = 0.5*0.364234*206.1 2*122.6*0.038*0.7 2 = 17830 N

Compairing, we get R = Zero lift Drag/ Drag due to lift = 17830/17660 = 1

c) Similarly, from above we get V(CL1/2

/CD)max = 271.24 m/s

CL = 2W/ ρV2S = 2(68000*9.81)/(0.364234*271.242*122.6) = 0.406Zero lift Drag = 0.5ρV2SCDo = 0.5*0.364234*271.242*122.6*0.0188 = 30882 NDrag due to lift = 0.5ρV2SKCL

2 = 0.5*0.364234*271.24 2*122.6*0.038*0.406 2 = 10289 N

Compairing, we get R = Zero lift Drag/ Drag due to lift = 30882/10289 = 3

Thrust Available and the Maximum Velocity of the Airplane

Sea LevelAt Sea Level, density = ρo = 1.23 kg/m3

Thrust Available = TAo = 200000 N

wing_loading = W/S = 5441.1

thrust_weight = Ta/W = 0.2998

Vmax = 374 m/s

At 11000 mAt 11000 m, density = ρ = 0.364234 kg/m3

m=1.05,

TA=55729 N

wing_loading = W/S = 5441.1

thrust_weight = 0.0835

Vmax = 343 m/s

Power Required

Graphical ApproachPower required is given by

We have TR curve from the previous section. Multiplying the values with velocities we get

Velocity (m/s) Tr (N) Pr (N.m/s)105 72854.29 7649700.48115 62322.25 7167059.02125 54531.85 6816481.43135 48743.36 6580353.61145 44468.62 6447949.42155 41378.43 6413656.28165 39248.48 6475998.93175 37926.91 6637209.46185 37314.8 6903238.43195 37354.91 7284207.55205 38026.4 7795411.88215 39344.81 8459134.76225 41368.09 9307821.08235 44211.68 10389745.11245 48080.34 11779683.81255 53336.04 13600689.69265 60655.86 16073802.64275 71460.31 19651585.73285 89407.28 25481075.4295 128846.89 38009831.33305 1829291.86 557934016.3315 7632.69 2404295.95325 7170.21 2330317.61335 6748.53 2260755.9345 6362.98 2195226.74355 6009.55 2133389.37

Analytical ApproachThe analytical formula for power required is

It can be seen from the formula that

The power required will be minimum when CL/CD3/2 is maximum.

From above, we have CL/CD3/2

(max) = 17.88

Substituting values, we get

PR = 6448799.683 Nm/s

Which is same as calculated from the graph.

Power Available & Maximum VelocityPower available is given by

Thrust available = TA = 55729 N

So

PA = 55729xV

Maximum velocity is achieved when power available is equal to the power required.

Analytically it was 343 m/s.

Now making graph of PA and PR, the point where they will cut is the max velocity.

Which is same as analytically calculated.

The velocity calculated is purely theoretical, because we have not considered transonic drag effects. This velocity will corresponds to the Mach greater than 1 which is not achievable in our aircraft.