曝光台 注意防骗
网曝天猫店富美金盛家居专营店坑蒙拐骗欺诈消费者
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~
\1 .225~ 0.381317 o 0~4
=: 101.1373 m/s
N
Zlmax = ~-V
-_-
=, . -~~I
- 2.6849
中国航空网 www.aero.cn
航空翻译 www.aviation.cn
本文链接地址：
动力机械和机身手册1(54)
