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VR -.
(2.189)
Note that the flight velocity is held constant at this value throughout the flight.
    The velocities for maximum range and endurance are schematically shown for
propeller and jet aircraft in Figs. 2.24 and 2.25.
                                      Example 2.7
    A turbojet airplane weighs 80,343.9 N, has a wing area of 37.2 m2, and has a
specific fuelconsumption equal to 1.3 N/Nh at an altitude oflO,OOOm (a  : 0.374).
The airplane drag polaris given by CD = 0.02+0.06CZ. Determine the fuelloads
for  1) range of 2400 km and 2) endurance of 5 h. With fuel load as obtained in
AIRCRAFT PERFORMANCE
117
               VR
                      t x,Rmax(Brcguet)  V
Fig. 2.25    Velocities for maximum range and maximum endurance ofjet aircraft
1), find the change in the range if the airplane encounters a steady headwind of
10 m/s throughout the flight.
    So/ution.   We will solve this range problem using both constant-altitude and
constant-velocit)r (Breguet) methods.
    The maximum Breguet range is given by Eq. (2.150).
Rmax = ( V, )E,  e.~v
We have  Wo = 80,343.9 N, S  - 37.2 m2, c = 1.3 N/Nh, CDO = 0.02,  k - 0.06,
ht -. 10 km, Gwuc -. 5 h, Rmax - 2400 km, and a, - 0.374.
     The specific fuel consumption has the units of N/Nh, and we have to express it
in terms ofN/Ns as follows.
We have
c = 1.3 N
  Nh
    1.3 N
  = 3600 Ns
- 14.433
              1
Em - 2~
                  1
N~
 2^/6 fiB_
118            PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
VR -
                                                     -.  127.8108 m/s
For Rm = 2400 * l03 m,
                                                           127.8108 * 36(
               2400*10? =  .           ;600),14.3, e.vX
                  1.3
Solving, Wi - 50,227.5 N, and W f = Wo - Wi - 30,116.41 N.
     rfhe approximate height gained in cruise climb is obtained as follows.
                Wl
                         o-/ - Woai
    50,227.5
                                                                   =  80 343 9 * 0.374
                                     - 0.23378
Then,
                                    hf = 44.3(1 - ul.235)
                                                             = 44.3(1 _ 0.233780.23S)
　
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