Your Federal Quarterly Tax Payments are due April 15th

# 2 by lanyuehua

VIEWS: 9 PAGES: 6

• pg 1
```									       CALCULATION OF MAXIMUM RANGE

Range is the distance traversed by an airplane on one load of fuel. The maximum
range is given by; (Anderson 5.156)

2 C L/ 2
W               
1
2
R                               W1
 S CD
0
ct

Here, the specific fuel consumption, c t , of our engine is 0.4571/hr. S is the wing
planform area as calculated in study 3; S  19.91m 2 . The value of C L/ 2 / C D  is;
1

(Anderson 5.44)

1/ 4
 C L/ 2 
1
3 1 
          

C         4  3KC D ,0 
3
 D  m ax                

Assuming a cruise altitude of 3000m,    0.91kg / m 3 . The parasite drag
coefficient is calculated as C D ,0  0.02242 and drag due-to-lift coefficient as
K  0.0746 . Hence,

 C L/ 2 
1/ 4
1
3            1           
C  
                                    3 
 18.82
 D  max      4  3  0.0746  0.02242 

Without external fuel tanks, W fuel  1018 .3kg and recalling that W0  5337 kg , maximum
range is;

2 C L/ 2
W               
1
2
R                              W1
ct     S CD        0

R
2            2
0.457 / 3600 0.91  19.91

18.82 5337  9.81                5337  1018.3  9.81
R  2263km

With external fuel tanks with a total fuel capacity of 465.6kg (150 gallon-JP-5),
W fuel  1483 .9kg , and the improved range is;

R
2
0.457 / 3600
2
0.91  19 .91

18 .82 5337  9.81         5337  1483 .9   9.81 
R  3388km
CALCULATION OF MAXIMUM ENDURANCE

Maximum endurance comes into the scene when the main concern is staying in
the air for the longest possible time. Hence, as the word itself entails, endurance is
defined as the amount of time that an airplane can stay in the air on one load of fuel.

Endurance for jet aircraft is expressed as; (Anderson 5.167)

1 L W0
E       ln
ct D W1

To maximize this expression, L/D should be taken as L / D m ax which is
previously found to be 12.23.
Without external fuel tanks, W fuel  1018 .3kg and recalling that W0  5337 kg ,
maximum endurance is;

1 L W0    1                 5337
E       ln       12 .23 ln
ct D W1 0.457           5337  1018 .3
E  5.67hr

With external fuel tanks, W fuel  1483 .9kg and the improved endurance is;

1 L W0    1                 5337
E       ln       12 .23 ln
ct D W1 0.457           5337  1483 .9
E  8.72hr

CALCULATION OF THE STALLING SPEED

The stalling speed of an aircraft is given by; (Anderson 5.67)

2 W      1
Vstall 
  S (C L ) max

From study 3, C L m ax  2.404 with full (60) flap deflection and the wing loading
was approximated as;

W
 2600 N / m2
S
Assuming sea level conditions,    1.225 kg / m 3 . Hence the stalling speed is;
2 W     1           2  2600
Vstall                                     42.02m / s
  S (CL ) max   1.225  2.404

CALCULATION OF THE GROUND ROLL AND THE TOTAL DISTANCE FOR
TAKEOFF

The ground roll for takeoff is approximated as; (Anderson 6.95)

1.21(W / S )
sg 
g   cL max (T / W )

From study 3, thrust to weight ratio is 0.25 and CL m ax  2.173 when the flaps
are partially deflected (30 deflection). Assuming sea level conditions;

1.21(W / S )               1.21(2600)
sg                                                       482m
g   cL max (T / W ) 9.811.225 2.173 (0.25)

The distance required while airborne to clear an obstacle is expressed as; (Anderson 6.98)

sa  R sin OB

where R is the turn radius; (Anderson 6.98)

R
 2
6.96 Vstall   
g

recalculating Vstall for sea level conditions and with flaps partially deflected;

2 W      1           2  2600
Vstall                                   44.2m / s
  S (C L ) max   1.225  2.173
thus R turns out to be;

R
 2
6.96 Vstall


6.96 (44 .2)
 31 .36 m
g            9.81

and assuming an obstacle height of 10 m;

        hOB       1   10 
OB  cos1 1                cos 1          47 .07
         R            31 .36 
Hence, sa  R sin OB  31 .36 sin 47 .07  23 m

The total takeoff distance is the sum of the ground roll and the distance required
while airborne to clear an obstacle, s g  s a .

Total takeoff distance  s g  sa  482  23  505 m

CALCULATION OF THE GROUND ROLL AND THE TOTAL DISTANCE FOR
LANDING

The landing ground roll is given by; (Anderson 8.28)

2 W       1          j 2  (W / S )
S g  jN                         
  S


 Cl max g    Cl max   r

Assuming sea level conditions,   =1.225 kg/m3. j n be taken as 1.1 for military jets. For
flaps in fully deflected position,             CL max  2.404 . Taking N=2 s and    r =0.4 then;

2                  1         1.12  (2600)
S g  1.1  2              2600                                   364.7 m
1.225              2.404 9.81 1.225  2.404  0.4

Assuming an approach angle of  a  3 , and an average velocity during flare
as; V f  1.23Vstall  1.23  42 .02  51 .68 m / s . The resulting flight path radius during flare
is; (Anderson 6.107)

V f2          51.82
R                            1367.6m
0.2 g       0.2  9.81

Using Eq. 6.106 from Anderson, the flare height can be calculated as;

h f  R 1  cos  a   1367 .61  cos 3  1.9m

Then, the approach distance to clear a 15m can be found from Eq. 6.108;

15  h f         15  1.9
sa                              250m
Tan a            Tan3

And the flare distance can be obtained by multiplying the flight path radius, R, by
sin  a ;
s f  R sin  a  1367 .6 sin 3  71 .6m

Hence, the total landing distance is found by summing the ground roll, and the
approach and flare distances;

Total Landing Distance  s g  s f  sa  364 .7  250  71 .6  686 .3m

VELOCITY MODIFICATION

The velocity used for maximum range calculation is given by; (Anderson 5.45)

1/ 2
 2               3K W 
V(C1 / 2 / C )                            
D max                C D,0 S 
                         
L

Substituting    0.91kg / m 3 , C D ,0  0.02242 , K  0.0746 , and
W / S  2600 N / m2 ;

1/ 2
 2    3  0.0746      
V( C 1 / 2 / C              
 0.91 0.02242    2600 
                     134 .4m / s
D ) m ax
                      
L

Recall that Vstall  42 .02 m / s . Thus, V(C1 / 2 / C                                Vstall .
L      D ) max

The velocity used for the maximum endurance calculation corresponds to the
velocity at L / D m ax and can be calculated as; (Anderson 5.34)

1/ 2
 2         K W
V( L / D ) max                         
        C D,0 S 
                  

Sea level conditions apply for maximum endurance calculation, hence   should
be taken as 1.225 kg / m 3 ;

1/ 2
 2          0.0746       
V( L / D ) m ax      
 1.225              2600 
                  88 m / s
            0.02242      

Thus V( L / D) max  Vstall .

Coming to the velocity used at rate of climb calculations, for jet propelled
airplane, Anderson gives V R / C  m ax as (5.114);
1/ 2
 (T / W )(W / S ) Z 
V( R / C ) m ax                      
      3  C D , 0  

3
where Z  1  1                                        . Recall that T / W  0.25 and
L / D  T / W 2
2
max

L / D m ax  12 .23 . Hence,

3
Z 1 1                        2.15
12.23 0.25
2    2

1/ 2
 (0.25)(2600)2.15 
V( R / C ) max                                      130.2m / s
 3  1.225  0.02242


Thus, V( R / C ) max  Vstall as expected.

As illustrated, the results favored the design and there is no need to make any
modifications.

```
To top