Purpose Evaluate the relationships between Coefficient of Lift (CL) and Angle of Attack (AOA), Airfoil Camber, and Airfoil Thickness and the variables of the Lift Equation using graphs and equations. If you are able, review the concepts using an on-line s

Jet Aircraft Performance (Questions 1-4)

Refer to Figure 6.11 in Flight Theory and Aerodynamics (Doyle & Lewis, 1965) and assume Weight = 15,000 lb.

1. Find Climb Angle at 300 KTAS at 100 % RPM.

Angle of Climb is defined as,

sinγ = (Ta-Tr)/W

Where Ta is available thrust at 100% RPM and Tr is minimum thrust required and W is the weight of the aircraft

Climb of aircraft is visible when (Ta-Tr) is maximum, so from the graph given above we have,

Ta (max) = 4200lb and

Tr (min) = 1000lb

W = 15,000lb so, Angle of climb will be,

sinγ = (Ta-Tr)/W

= (4200-1000)/15000 =0.2133

Angle of climb is

sinγ = (Ta-Tr)/W = (4200-1000)/15000 =0.2133

γ =12.310 (Answer)

2. Find Rate of Climb at 300 KTAS at 100 % RPM.

Rate of climb is defined as,

Rate of Climb = 101.3*Vk* Sinγ

Given that,

At 300 KTAS from the graph we can determine its Ta and Tr

Thrust Available is Ta =4200lb while thrust required is Tr =1000lb

So, Rate of climb is,

Rate of Climb = 101.3*Vk* Sinγ = 101.3*300*(4200-1000)/15000

Rate of Climb =6483.2 ft/min (Answer)

3. Find Max Level Airspeed at 100% RPM.

Now the question asks about 100% RPM from the graph it is clearly visible that the maximum level air speed is,

So,

Vmax = 600 KTAS (Answer)

4. Find Max Range Airspeed (VBR) and Max Endurance Airspeed (VBE).

Now we need to calculate two different speeds maximum endurance and maximum Range speed and this can be calculated from the graph with the use of their definitions,

Max endurance airspeed V (BE) occurs when minimum thrust is required so at

Tr is minimum the speed is

V (BE) = 240 KTAS (Answer)

While for the Maximum Range speed we have,

Maximum range speed is equal to the value where the straight line drawn from the origin intersects the graph. (Look for the graph)

So,

V (BR) will be 300 KTAS (Answer)

Propeller Aircraft Performance (Questions 5-8)

Refer to Figure 8.4 in Flight Theory and Aerodynamics and assume:

· W = 20,000 lb

· S = 500 ft2

· Density Sea Level Standard Day

5. Find Max Range Airspeed (KTAS) and Max Specific Range (SRmax).

As done previously Max range airspeed occurs V(BR) occurs when the straight line drawn from the origin intersects the graph given so if we give a look to the graph we have,

V (BR) = 135 KTAS (Answer)

Now we need to calculate the maximum specific range and for maximum specific range we need to calculate,

SR = Airspeed/Fuel flow is maximum

Now from the graph we have,

V = 150 KTAS and Fuel Flow = 300 lb/hour ( From Graph 8.4)

So, Specific Range will be,

SR = 150/300n = 0.5 KTAS hour / lb (Answer)

6. Find Max Endurance Airspeed (KTAS) and Max Specific Endurance (SE = 1/FF)

Now Again from the Maximum endurance speed can be calculated as,

V (BE) = 100 KTAS (Answer)

Fuel Flow can be noted that = 240 lb/ hour

So maximum Specific Endurance will be,

Specific Endurance = 100 KTAS/ 240 lb/ Hour = 0.41 KTAS hour / lb (Answer)

7. Find Max Rate of Climb Airspeed and Max Rate of Climb.

Now for calculating maximum climb rate we need to look at graph again

From the graph we have,

Ta = 325*Pa/Vk

Pa is the full power available at sea level and Vk is the speed in KTAS

Now thrust required will be,

Tr = 325*Pr/Vk

Where Pr is power required in horse power so from the same formula we can calculate rate of climb as,

Rate of climb = 101.3*Vk* Sinγ = 101.3*Vk*(TA-Tr)/W

= 101.3*325*(Pa-Pr)/W

= 32922.5*(Pa-Pr)/W

Now from the graph we have,

Pr = 600hp

Pa = 3500hp

And W = 20,000 lb so we have

Rate of climb = 33000*(3500-600)/20000

Rate of Climb =4785 ft/min (Answer)

Now maximum rate of Climb air speed will be,

Vk = 150 KTAS (Answer)

8. Find new Max Range Airspeed after fuel has burned down and W = 15,000 lb.

For decreased weight of the aircraft less Maximum range Airspeed is required so it will be less than, 150 KTAS and can be calculated as,

W1/W2 = V1/V2

20,000/15,000 = 150/V2

V2 = 150*15/20

V2 = 112.5 KTAS (Answer)

Landing Performance (Questions 9-14)

W = 15,000 lb CLMax = 1.5

S = 230 ft2 Thrust @ Idle = 500 lb

75% of Weight on Main Tires 25 % of Weight on Nose Tire

Average Drag = 500 lb Sea level standard day

9. Find Approach Speed if V approach = 1.2 V stall (KTAS) (For landing and takeoff performance use KTAS. At sea level standard day KEAS=KTAS).

Given that

CLmax = 1.5

From the formula we have,

CLmax = W/(q*S)

Where

q =σ*(Vk )2/295

But given that,

σ =1

And

w =15000 lb

S =230 ft2

So from the graph we have,

Vstall = 76.81 KTAS

Velocity of approach will be,

Vapp = 1.2*VStall

Vapp = 92.17 KTAS (Answer)

10. Find Average Rolling Friction (lb) on Nose Tire during Landing Rollout if Rolling Coefficient of Friction = 0.02.

Now from the graph 8.9 we need to calculate Average rolling friction on nose tire

Ff = µ*0.25*W =0.02*0.25*15000

Ff = 75 lb (Answer)

11. Find Average Braking Friction (lb) on Main Tires during Landing Rollout on Dry Concrete. Use Figure 13.9 assume 10% Slip

Maximum friction coefficient for then aircraft on landing is given as

µ=0.7 at 10% slip (Graph 8.9)

Minimum friction coefficient for then aircraft on landing is given as

µ=0.5 at 100% slip

Average breaking friction coefficient for the aircraft on landing is given as

µ=0.6 and

Average braking friction on main tires of the aircraft is given as,

µ = 0.75

So, we have

Ff = µ*0.75*W =0.6*0.75*15000

Ff = 6750 lb (Answer)

12. Find Average Deceleration (ft/s2) during landing rollout. Assume Lift on wing is zero during landing rollout. Account for residual Thrust, Rolling Friction, Drag, and Braking Friction.

Total Force applicable on the aircraft can be calculated as = W*a

Total force = Residual thrust on the Aircraft - Rolling Friction on nose tire – Drag on the aircraft - Breaking Friction on main tires so,

=500 - 750 - 500 - 6750 = -7500 lb

In Newtons we have,

=-7500*0.4536

= -3402 kg

=-3402*9.81

= -33373.62 N

W =15000 lb

In Newtons we have,

=15000*0.4536

= 6804 kg

a = F/W

=-33373.62/6804

= -4.905 m/s2

= -4.905*3.28

Retardation = -16.09 ft/s2 (Answer)

13. Find Landing Distance Ground Roll-out (ft)

Now for calculating Landing Distance for the aircraft we have from from Newton’s third equation of motion we have,

V2 = -2as

So,

Vapp =92.17 KTAS

In ft/s we have,

= 92.17*1.688

= 155.58 ft/s

a = -16.09 ft/s2

So, Landing Distance will be

s = -V2 /2a

s = 752ft (Answer)

14. Find Landing Distance (ft) if Runway is at 5,000 ft Density Altitude. Assume Residual Average Thrust and Average Drag remains the same.

At 5000 ft density aircraft track is different from on the ground, so for this kind of track we have

Total force = Residual thrust on the Aircraft - Rolling Friction on nose tire – Drag on the aircraft - Breaking Friction on main tires so,

=500-288-750-6750

= -7288 lb

=-7288*0.4536

= -3305.84 kg

= -3305.84*9.81

= -32430.26 N

W =6804 kg (calculated before)

Now acceleration of the aircraft can be calculated as,

Acceleration from Newton’s second law

a = F/W

=-32430.3/6804 kg

=- 4.76 m/s2

= -4.76*3.28 ft/s2

=-15.6 ft/s2

Initial speed of the aircraft will be,

Vini = 92.17 KTAS =155.58 ft/s

So total distance travelled will be,

s = -V2 /2a

= 155.58/ (2*15.634)

s = 774 ft (Answer)

This document was developed for online learning in ASCI 309.

File name: Ex_5_Acft_Performance

Updated: 05/31/2014