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\f min
              V
\f mai
hrust
blc
ttling
Ilg.2.13    Concept oflerrelthght at any desired speed.
                                        Example 2.3
     A propeller aircraft weighs 50,000 N and has a wing area of 30 m2. The recip-
rocating engine develops 840 kW at sea level and the adjustable pitch constant-
speed propeller has an eff/ciency of 0.85. The drag polar of the aircraft is given
by CD = 0.025 +0.05CZ, and the maximum lift coefficient is  1.75. Determine the
maximum and muumum speeds in level flight at sea Ievel.
So/ution. Wehave
       Pa = k'r7pP(kW)
PR = :;pSCDO V3 _ 2pkSW_y
We have k'  :  1000,  rIp =0.85,  P(kW) = 840,  W - 50,000 N,  S = 30 n12,  p  =
Po = 1.225 kg/m3,  CDO - 0.025, k - 0.05, and CL,    x = 1.75. Because the ana-
lytical solution is not possible, we have to use a graphical method for the given
propeller airplane.
        The power-required and power-available curves are plotted as shownin Fig. 2.14.
The stalling velocity is given by
VstrLII -
- 39.4323 m/s
AIRCRAFT PERFORMANCE
                                  Velocity rrVs
Fig. 2.14    Power curves for aircraft of Example 2.3.
91
From Fig. 2.14, we find that the minimum and maximum velocities are approxi-
mately 10 and 113.0 m/s, respectively. The fiight at Vm,n iS not possible because it
is lower than Vstall  Hence, VsiaU iS effectively the minimum speed in level flight at
sea level.
Example 2.4
      A turbojet airplane weighs 45,000 N and has a wing loading  W/S of 1450 N/m2.
The drag polar is given by CD = 0.014 + 0.038.CZ, CL,max = 1.5, and the thrust
T = 20a800 c,. Deternune the maximum and mimmum speeds in level flight at sea
level and at an altitude of 9000 m (CT :0.3813). What is the absolute ceiling of
this airplane?
So/ution.   We have
Emax - -   1
         : 2 j~o
                        1
                                  r
     == N
               :  2 /6 tiT~i~
     - 21.6777
92                PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
For level fiight at sea level (a : 1),
                                          T - 20,OOON
 TEm
z_ W
           20,000 * 21.6777
  -       45 000
-. 9.6345
VR -
Umax
llmin
. .       :[6-.t~
V  .22s l/00~4
- 62.4518 m/s
- 4.3837
    Vmax - Umax VR
              -. 273.77 m/s
- 0.2281
    Vmin : Ll:nun VR
             = 14.2459 m/s
Vstall -
- 39.7270 m/s
AIRCRAFT P,ERFORMANCE
Because  Vmin  <  Vsran, level tlight is not possible at Vnun. Therefore, the minimum
level flight speed at sea lcvelis  Vstan - 39.7270 m/s.
     At an altitude of 8000 m (<r = 0.383),
T - Toa
     - 20,000 * 0.3813 N
     - 7626.0 N
 TEm
z= w
VR -
      7626.0 * 21.6777
=        45 000
- 3.6736
-1.                  - 4[6~
　
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