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Aircraft Performance

Estimate the sea-level liftoff distance for the Beechcraft Bonanza V-tailed, single-engine light private airplane. Estimate the sea-level landing ground roll distance for the Beechcraft Bonanza V-tailed, single-engine light private airplane.

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Quach Huu Vinh
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464 views3 pages

Aircraft Performance

Estimate the sea-level liftoff distance for the Beechcraft Bonanza V-tailed, single-engine light private airplane. Estimate the sea-level landing ground roll distance for the Beechcraft Bonanza V-tailed, single-engine light private airplane.

Uploaded by

Quach Huu Vinh
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Name: Quach Huu Vinh Subject: Aircraft Performance

Student ID: 1951200028 Score:………………………….

Problem:
Consider an airplane patterned after the Beechcraft Bonanza V-tailed, single-engine light
private airplane. The characteristics of the airplane are as follows: aspect ratio = 6.2, wing
area = 181 ft2, Oswald efficiency factor = 0.91, weight = 3000 lb, and zero-lift drag
coefficient = 0.027. The airplane is powered by a single piston engine of 345 hp maximum
at sea level. Assume that the power of the engine is proportional to free-stream density.
The two-blade propeller has an efficiency of 0.83.
a. Assume a paved runway, and CL,max = 1.1 during the ground roll. When the airplane is
on the ground, the wings are 4 ft above the ground. Estimate the sea-level liftoff distance
for this airplane.
b. Assume that the airplane is landing with a weight of 2900 lb. The maximum lift
coefficient with flaps at touchdown is 1.8. After touchdown, assume zero lift. Estimate
the sea-level landing ground roll distance for this airplane.

Solution:

a)

Given data:

Aspect ratio: AR  6.2 Total weight: W  3000lb Propeller efficiency:


  0.83
Wing area: S  181 ft 2 Power of engine: Height of wings above
P  345 hp grounds: h  4 ft
Oswald efficiency factor: Zero-lift drag coefficient: Maximum lift coefficient:
e  0.91 CD0  0.027 CL,max  1.1

At sea level:   0.002377 slugs / ft 3 ; g  32.2 ft / s 2

On a paved runway: r  0.02

Wingspan: b  S  AR  181 6.2  33.5 ft


16h / b  16  4 / 33.5
2 2

Ground effect number:     0.785


1  16h / b  1  16  4 / 33.5
2 2

2W
Velocity of airplane: V  0.7VLO  0.7 1.2 Vstall  0.84 
  S  CL,max

2  3000
 0.84   94.58 ft / s
0.002377 1811.1
1 1
Dynamic pressure: q  V 2   0.002377  94.582  10.63 lb / ft 2
2 2

Lift, drag and thrust are calculated by:

L  q  S  CL,max  10.63 1811.1  2116.43 lb

 CL,max 2   1.12 
D  q  S   CD0     10.63 181  0.027  0.785    155.05 lb
  eAR     0.91 6.2 

550  PA 550   P 550  0.83  345


T    1665.18 lb
V V 94.58

So, the liftoff distance is estimated by:

1.44W 2
S LO 

g  SCL,max T   D  r  W  L   av 
1.44  30002
  569.83 ft

32.2  0.002377 1811.1 1665.18  155.05  0.02  3000  2116.43  
b)

Given data:

Maximum lift Total weight: Zero lift: L  0 lb On a paved


coefficient: W  2900lb runway, with
CL,max  1.8 braking: r  0.4

2W
Velocity of airplane: V  0.7VLO  0.7 1.3 Vstall  0.91
  S  CL,max
2  2900
 0.91  78.75 ft / s
0.002377 1811.8
1 1
Dynamic pressure: q  V 2   0.002377  78.752  7.37 lb / ft 2
2 2

Drag is calculated by: D  q  S  CD0  7.37 181 0.027  36.02 lb

So, the sea-level landing ground roll distance is estimated by:

1.69W 2
SL 
g  SCL ,max  D  r  W  L  
1.69  29002
  476.55 ft
32.2  0.002377 181 1.8  36.02  0.4   2900  0  

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