曝光台 注意防骗
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= ~ VR (2.75)
The lift coefficient in level flight when the power required in level flight is
nunimum is as follows.
2W
CL.mp - .
= p S~2p
= ,~-
= V5Cz
(2.76)
AIRCRAFT PERFORMANCE
Similarly, we can show that
T
Emp = 0.8 6 Em (2.77)
Then, we get
8 kW2
PR.min = 3 p S /.-p
W Vmp
- (2.78)
= O 866Em
Thus, we observe that both thrust- and power-required curves assume their mini-
mum values at some velocities. In view of this, we have two intersection points
between power-available and power-required curves as shown in Fig. 2.10a for the
propeller airplane and between thrust-available and thrust-requircd curves for ajet
airplane as shown in Fig. 2.10b. These two intersection points give us two solutions
Pmin
D mln
Vmin Vm Vma^
a) Propeller aircraft
Vmin VR Vmac
b) Jet aircraft
Fig.2.10 Levelflightsolutions.
:I;g
Il;~
86 PERFORMANCE, STABILlfY, DYNAMICS, AND CONTROL
of Eqs. (2.61) and (2.62). Thus, spending the same amount of fuel and developing
the same magrutude of thmst or the power, an airplane can fly at either of these
two velocities. At the low or minimum velocity V-y."' the thrust or power available
is essentially used to overcome induced drag, whereas, at high or maximum speed
Vmvc, it is mostly used to balance the zero-lift drag.
2.4-1 Analytical Solutions of Level Flight for Propeller Airplanes
Because the power available and power required are the basic quantities for
propeller aircraft, we rewrite Eq. (2.62) as foii~:ws:
TV -DV :0
Pa - PR ~ O
For an ideal propeller airjplane, whose power developed in kilowatts
propulsive efficiency r7P are independent of fiight ve7ocity,
Pa = k'rlp P(kW)
where k' -- 1000 is the conversion constant from kW to Nm/s. Then,
k'r7pP(kW)- ~ZpSCDOV3 _ 2kW =0
pSV
(2.79)
(2.80)
P(kW) and
(2.81)
(2.82)
The above equation is of the form V4 + a V + b = O, which has no closed-form
analytical solution.ln other words, we have to solve the above level flight equation
numerically or graphically even for this simplified case where we have assumed
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